Answer:
The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.
Explanation:
Hi there!
The position of the patrol car at a time "t" can be calculated using this equation:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the patrol car at a time "t"
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:
x = x0 + v · t
Where "v" is the constant velocity.
First, let´s convert the velocity units into m/s:
140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s
105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s
We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.
When the patrol car catches the speeder, the position of both cars is the same:
position of the patrol car = position of the speeder
x0 + v0 · t + 1/2 · a · t² = x0 + v · t
if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:
x = v · t
x = 38.9 m/s · 1 s = 38.9 m
When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:
x0 + v0 · t + 1/2 · a · t² = x0 + v · t
0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t
Let´s agrupate terms and equalize to zero:
-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0
-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0
Solving the quadratic equation for t using the quadratic formula:
t = 8.24 s (the other solution is discarded because it is negative)
The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.
The incoming radiation from the sun interacts with Earth's surface in different ways. Describe a surface that would have a high albedo and a surface that would have a low albedo.
Answer:
Albedo Alto: Snow
Albedo bass: ocean
Explanation:
Albedo is the fraction of solar radiation reflected by a surface. The term has its origin from the Latin word albus, which means "white." It is quantified as the proportion or percentage of solar radiation of all wavelengths reflected by a body or surface to the amount incident on it.
The color of the soil certainly affects the reflectivity, the lighter colors that have greater albedo than the dark colors and, therefore, present greater albedo. Soil texture is also a factor that affects albedo. Some studies have shown that snow-covered soils, being a light color, reflect most of the light and therefore do not absorb it and present the greatest amount of terrestrial surfaces. While with the ocean and seas most of the radiation is absorbed it has a dark color increasing with the depth of the place and therefore has the lowest surface albedo. In the end of the year the snow has 86% of reflected light while the oceans 8% of reflected light.
A major leaguer hits a baseball so that it leaves the bat at a speed of 31.3 m/s and at an angle of 36.7 ∘ above the horizontal. You can ignore air resistance.
A) At what two times is the baseball at a height of 9.00 m above the point at which it left the bat?
t1,2 = _____ s
B) Calculate the horizontal component of the baseball's velocity at each of the two times you found in part (a).
vh1,2 = _____ m/s
C) Calculate the vertical component of the baseball's velocity at each of the two times you found in part (a).
v v1,2 = _____ m/s
D) What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?
v = ____ m/s
E) What is the direction of the baseball's velocity when it returns to the level at which it left the bat?
= ______ degrees below the horizontal
Answer:
A) t₁ = 0.56 s t₂ =3.26 s
B) vh₁=vh₂ = 25.1 m/s
C) v₁ = 13.2 m/s v₂ = -13.2 m/s
D) v = 31. 3 m/s
E) 36.7º below horizontal.
Explanation:
A) As the only acceleration of the baseball is due to gravity, as it is constant, we can apply the kinematic equations in order to get times.
First, we can get the horizontal and vertical components of the velocity, as the movements along these directions are independent each other.
v₀ₓ = v* cos 36.7º = 31.3 m/s * cos 36.7º = 25.1 m/s
v₀y = v* sin 36.7º = 31.3 m/s * sin 36.7º = 18.7 m/s
As in the horizontal direction, movement is at constant speed, the time, at any point of the trajectory, is defined by the vertical direction.
We can apply to this direction the kinematic equation that relates the displacement, the initial velocity and time, as follows:
Δy = v₀y*t -1/2*g*t²
We can replace Δy, v₀y and g for the values given, solving a quadratic equation for t, as follows:
4.9*t²-18.7t + 9 = 0
The two solutions for t, are just the two times at which the baseball is at a height of 9.00 m above the point at which it left the bat:
t = 1.91 sec +/- 1.35 sec.
⇒ t₁ = 0.56 sec t₂= 3.26 sec.
B) As we have already told, in the horizontal direction (as gravity is always downward) the movement is along a straight path, at a constant speed, equal to the x component of the initial velocity.
⇒ vₓ = v₀ₓ = 25.1 m/s
C) In order to get the value of the vertical components at the two times that we have just found, we can apply the definition of acceleration (g in this case), solving for vfy, as follows:
vf1 = v₀y - g*t₁ = 18.7 m/s - (9.8m/s²*0.56 sec) = 13.2 m/s
vf₂ = v₀y -g*t₂ = 18.7 m/s - (9.8 m/s²*3.26 sec) = -13.2 m/s
D) In order to get the magnitude of the baseball's velocity when it returns to the level at which it left the bat, we need to know the value of the vertical component at this time.
We could do in different ways, but the easiest way is using the following kinematic equation:
vfy² - v₀y² = 2*g*Δh
If we take the upward path, we know that at the highest point, the baseball will come momentarily to an stop, so at this point, vfy = 0
We can solve for Δh, as follows:
Δh = v₀y² / (2*g) = (18.7m/s)² / 2*9.8 m/s² = 17.8 m
Now, we can use the same equation, for the downward part, knowing that after reaching to the highest point, the baseball will start to fall, starting from rest:
vfy² = 2*g*(-Δh) ⇒ vfy = -√2*g*Δh = -√348.9 = -18. 7 m/s
The horizontal component is the same horizontal component of the initial velocity:
vx = 25.1 m/s
We can get the magnitude of the baseball's velocity when it returns to the level at which it left the bat, just applying Pythagorean Theorem, as follows:
v = √(vx)² +(vfy)² = 31.3 m/s
E) The direction below horizontal of the velocity vector, is given by the tangent of the angle with the horizontal, that can be obtained as follows:
tg Ф = vfy/ vx = -18.7 / 25. 1 =- 0.745
⇒ Ф = tg⁻¹ (-0.745) = -36.7º
The minus sign tell us that the velocity vector is at a 36.7º angle below the horizontal.
A major leaguer hits a baseball and the question asks for the time, velocity components, velocity magnitude, and direction at different points of the ball's trajectory when it reaches a certain height.
Explanation:A) Let's consider the vertical motion of the baseball. The equation to determine the time it takes for an object to reach a certain height is:
h = v0t + 0.5gt2
Where:
h is the height = 9.00 mv0 is the initial vertical velocity = 31.3 m/s * sin(36.7º)g is the acceleration due to gravity = 9.8 m/s2By rearranging the equation, we can solve for t:
t2 + (2v0/g)t - (2h/g) = 0
Using the quadratic formula, we can plug in the values and solve for t. The two possible values of t will correspond to the two times when the ball is at a height of 9.00 m above the point at which it left the bat.
B) The horizontal component of velocity remains constant throughout the motion. Therefore, the horizontal component of velocity at each of the two times found in part (a) will be the same as the initial horizontal velocity, which is 31.3 m/s * cos(36.7º).
C) The vertical component of velocity changes due to the acceleration of gravity. At each of the two times found in part (a), the vertical component of velocity can be found using the equation:
vv1,2 = v0 + gt
Where:
v0 is the initial vertical velocity = 31.3 m/s * sin(36.7º)g is the acceleration due to gravity = 9.8 m/s2t is the time at each height, which were found in part (a)D) When the baseball returns to the level at which it left the bat, its vertical velocity will be -v0, which means it will be moving downward with the same magnitude as the initial velocity but in the opposite direction. The horizontal component of velocity remains unchanged. Therefore, the magnitude of the baseball's velocity when it returns to the level at which it left the bat will be the square root of the sum of the squares of the horizontal and vertical components, which can be calculated using the formula:
v = √(vh2 + vv2)
Where:
vh is the horizontal component of velocity, which was found in part (b)vv is the vertical component of velocity when the ball returns to the level of the bat, which is -v0E) The direction of the baseball's velocity when it returns to the level at which it left the bat can be determined using trigonometry. The angle below the horizontal can be found as:
θ = tan-1(vv/vh)
Where:
vv is the vertical component of velocity when the ball returns to the level of the bat, which is -v0vh is the horizontal component of velocity, which was found in part (b)A 6-inch diameter smooth plastic pipe 3100 ft in length delivers 2.2 cfs of water between two reservoirs. Water viscosity is 11.11 x 10-6 ft2/s. What is the elevation difference (ft) in water surface between the reservoirs
Answer:
h = 6.39 m
Explanation:
radius = 6/2 =3in= 3/12 in/ft = 0.25 ft
apply continuity equation to evaluate the velocity
flow rate = area * velocity
2.2 = π(0.25^2) * V
⇒ V = 11.2 ft/s
*assuming there is no applied pressure difference along the pipe
applying Energy conservation law
initial potential energy = final kinetic energy
pgh = (1/2)*pv^2
where p = density
v = velocity
plugging in the values:
p cancels out on both the sides
⇒ 9.81 * h = (11.2^2)/2
h = 6.39 m
A particle's position along the x-axis is described by: x(t)=A t + B t2
where t is in seconds, x is in meters, and the constants A and B are given below.
Randomized Variables
A = -4.3 m/s
B = 4.9 m/s2
(a) What is the velocity, in meters per second, of the particle at t1 = 3.0 s?
(b) What is the velocity, in meters per second, of the particle when it is at the origin (x=0) at t0 > 0?
Answer:
a)V = 25.1 m/s
b)V = 4.226 m/s
Explanation:
Given that
x(t)=A t + B t²
A = -4.3 m/s
B = 4.9 m/s²
x(t)= - 4.3 t +4.9 t²
The velocity of the particle is given as
[tex]V=\dfrac{dx}{dt}[/tex]
V=-4.3 + 4.9 x 2 t
V= - 4.3 + 9.8 t m/s
Velocity at point t= 3 s
V= - 4. 3 + 9.8 x 3 m/s
V= - 4.3 + 29 .4 m/s
V = 25.1 m/s
At origin :
x= 0 m
0 = - 4.3 t +4.9 t²
0 = - 4.3 + 4.9 t
[tex]t=\dfrac{4.3}{4.9}\ s[/tex]
t=0.87 s
The velocity at t= 0.87 s
V= - 4.3 + 9.8 t m/s
V= - 4. 3 + 9.8 x 0.87 m/s
V= - 4.3 + 8.526 m/s
V = 4.226 m/s
a)V = 25.1 m/s
b)V = 4.226 m/s
The velocity of the particle at t = 3.0s is 25.1 m/s.
The velocity of the particle when it is at the origin is 4.324 m/s.
The given parameters:
Position of the particle, x(t) = At + Bt²A = -4.3 m/s and B = 4.9 m/s²The velocity of the particle at t = 3.0s is calculated as follows;
[tex]v = \frac{dx}{dt} = A + 2Bt\\\\v = -4.3 + 2(4.9\times 3)\\\\v = 25.1 \ m/s[/tex]
The velocity of the particle when it is at the origin (x = 0)
[tex]0 = -4.3t + 4.9t^2\\\\0 = t(-4.3 + 4.9t)\\\\t = 0 \ \ \ or \ \ -4.3 + 4.9t = 0\\\\4.9t = 4.3\\\\t = \frac{4.3}{4.9} \\\\t = 0.88 \ s\\\\v = A + 2Bt\\\\ v = -4.3 + (2\times 4.9 \times 0.88)\\\\v = 4.324 \ m/s[/tex]
Learn more about velocity of particle here: https://brainly.com/question/80295?source=archive
Water, with a density of 1000 kg/m3, flows out of a spigot, through a hose, and out a nozzle into the air. The hose has an inner diameter of 2.25 cm. The opening in the nozzle that the water comes out of has a diameter of 2.00 mm. The water coming out of the nozzle, which is held at a height of 7.25 meters above the height of the spigot, has a velocity of 11.2 m/s. Neglecting viscosity and assuming that the water flow is laminar (not necessarily good assumptions, but let's not make this any harder than it already is), what is the pressure of the water in the hose right after it comes out of the spigot where the water enters the hose (to three significant digits)? Assume that ????=9.80 m/s2 and that the surrounding air is at a pressure of 1.013×105 N/m2
Final answer:
To calculate the pressure of the water in the hose at the spigot, you can use Bernoulli's equation, which relates pressure, density, velocity, and height of a fluid. By assuming the height at the spigot is the same as the nozzle and plugging in the given values, the pressure can be calculated.
Explanation:
The pressure of the water in the hose right after it comes out of the spigot can be calculated using Bernoulli's equation. Bernoulli's equation relates the pressure, density, velocity, and height of a fluid. In this case, the pressure can be found using the equation:
P1 + 1/2 ρv1² + ρgh1 = P2 + 1/2 ρv2²+ ρgh2
Where P1 is the pressure at the spigot, P2 is the pressure at the nozzle, ρ is the density of water, v1 is the initial velocity of water at the spigot, v2 is the velocity at the nozzle, h1 is the height of the spigot, and h2 is the height of the nozzle.
Since the height of the spigot is not given, we can assume it is at the same level as the nozzle, which means h1 = h2 = 7.25 m. The density of water, ρ, is 1000 kg/m³. The velocity at the nozzle, v2, is given as 11.2 m/s. Given these values, we can solve for the pressure at the spigot, P1.
Final answer:
The pressure of the water in the hose right after it comes out of the spigot can be calculated using Bernoulli's equation, which relates the pressure, velocity, and height of a fluid. Assuming the flow is laminar and neglecting viscosity, the pressure is found to be 229000 N/m².
Explanation:
The pressure of the water in the hose right after it comes out of the spigot can be calculated using Bernoulli's equation, which relates the pressure, velocity, and height of a fluid. Assuming the flow is laminar and neglecting viscosity, we can use the equation:
P + 1/2ρv² + ρgh = constant
Where P is the pressure, ρ is the density of water, v is the velocity of the water, g is the acceleration due to gravity, and h is the height difference between two points. Since the hose and nozzle are connected, we can assume that the pressures at these two points are the same. Also, the velocity of the water inside the hose can be considered negligible compared to the velocity at the nozzle. Therefore, we can simplify the equation to:
1/2ρv²+ 0 + ρgh = constant
The pressure at the spigot is atmospheric pressure, which is given as 1.013x10⁵ N/m². Rearranging the equation and solving for P, we have:
P = 1.013x10⁵ N/m² + 1/2ρv² + ρgh
Using the given values, we can substitute them into the equation and calculate the pressure
P = 1.013x10⁵ N/m² + 0.5x1000 kg/m³x(11.2 m/s)² + 1000 kg/m³x9.8 m/s²x7.25 m
P = 1.013x10⁵ N/m²+ 62656.96 N/m² + 68300 N/m²
P = 228957.96 N/m²
Rounding to three significant digits, the pressure of the water in the hose right after it comes out of the spigot is 229000 N/m².
A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 44 cm/s.
What is the amplitude of the subsequent oscillations? Answer should be in cm.
What is the block's speed at the point where x= 0.70 A? Answer should be in cm/s.
Answer:
8.8 cm
31.422 cm/s
Explanation:
m = Mass of block = 0.6 kg
k = Spring constant = 15 N/m
x = Compression of spring
v = Velocity of block
A = Amplitude
As the energy of the system is conserved we have
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm[/tex]
Amplitude of the oscillations is 8.8 cm
At x = 0.7 A
Again, as the energy of the system is conserved we have
[tex]\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s[/tex]
The block's speed is 31.422 cm/s
Final answer:
This physics problem involves calculating the amplitude of oscillations and the speed of a block at a specific displacement by applying conservation of mechanical energy in harmonic motion. The amplitude is found to be approximately 6.37 cm, and the block's speed at 0.70 A (amplitude) is approximately 28.4 cm/s.
Explanation:
The problem involves a 0.600 kg block attached to a spring with a spring constant of 15 N/m that is initially at rest and is then given a speed of 44 cm/s. To find the amplitude of the subsequent oscillations and the block's speed when x = 0.70 A, where A is the amplitude, we first use the principle of conservation of mechanical energy in harmonic motion. The initial kinetic energy given to the block will be equal to the potential energy of the spring at maximum displacement, which allows us to calculate the amplitude. Next, we determine the speed of the block at a displacement of 0.70 A using the relationship between kinetic and potential energy at any point during the oscillation.
The amplitude can be found using: KE = 1/2 k A^2, where KE is the kinetic energy of the block. Converting 44 cm/s to m/s gives 0.44 m/s. The kinetic energy of the block is KE = 1/2 mv^2 = 1/2 (0.600 kg)(0.44 m/s)^2, and solving for A gives the amplitude in meters, which can then be converted back to centimeters. To find the block's speed at x = 0.70 A, we use the conversion of potential energy at this displacement back to kinetic energy, considering the total mechanical energy of the system remains constant.
Using these principles, calculations show the amplitude of the oscillations to be approximately 6.37 cm, and the block's speed at x = 0.70 A is about 28.4 cm/s.
Two 6-in wide wooden boards are to be joined by splice plates that will be fully glued on the contact surfaces.The glue to be used can safely provide a shear strength of 120 psi. Determine the smallest allowable length L that can be used for the splice plates for an applied load of P = 15,000 lb. Note that a gap of 0.5 in is required between boards 1 and 2.which one of these is the right answer and show work to prove why.a) 19.7 inb) 15.7 inc) 24.3 ind) 21.3 ine) 11.6 in
Answer:
D
Explanation:
The smallest possible allowable length that can be used:
Calculate area of contact between plates A
A = 2 * 6 * a
A = 12a in^2
Where a is an arbitrary constant length along the direction L.
Shear stress = Shear force / Area
Area = 15000 lb / 120 psi = 12*a
Evaluate a
a = 10.411 in
Total Length = 2*a + gap = 2*10.411 + 0.5 = 21.33 in
Answer: L = 21.33
To find the smallest allowable length for the splice plates, the required shear area was calculated using the provided load and the shear strength of the glue, then divided by the effective board width. The resulting length per plate was approximately 11.35 inches, making the smallest length that can safely be used 11.6 inches.
To determine the smallest allowable length L that can be used for the splice plates, we must first calculate the shear area required to resist the applied load P of 15,000 lb with a glue shear strength of 120 psi.
The total shear force that the splice plates must resist is equal to the applied load P. The shear strength per unit area provided by the glue is 120 psi. Therefore, the required shear area A can be found using the formula:
A = P / Shear Strength
A = 15,000 lb / 120 psi = 125 square inches
Each board has a width of 6 inches, but we account for a 0.5-inch gap between the boards, so the effective width w for the splice is (6 - 0.5) inches = 5.5 inches. Then, we can calculate the length L needed for each splice plate as follows:
L = A / w
L = 125 square inches / 5.5 inches ≈ 22.7 inches
But since the glue is applied to two plates, we divide this length by 2 to get the length required for one splice plate:
L = 22.7 inches / 2 ≈ 11.35 inches
To ensure safety, the smallest allowable length should be the next highest option available, which is 11.6 inches (Option E).
How much heat (in J) must be added to raise the temperature of 2.70 mol of air from 22.0°C to 32.0°C at constant volume? Assume air is completely diatomic.
Answer:
Heat required to raise the temperature will be 563.625 J
Explanation:
We have given number of moles n = 2.70 mole
Temperature is raises from 22°C to 32°C
So increase in temperature [tex]\Delta T=32-22=10^{\circ}C[/tex]
It is given that air is diatomic so [tex]c_v=\frac{5}{2}R=2.5\times 8.314=20.875[/tex]
We know that heat is given by [tex]Q=nc_v\Delta T[/tex]
So heat will be equal to [tex]Q=2.70\times 20.875\times 10=563.625J[/tex]
So heat required to raise the temperature will be 563.625 J
A 0.25 kg ideal harmonic oscillator has a total mechanical energy of 9.8 J. If the oscillation amplitude is 20.0 cm, what is the oscillation frequency?a. 4.6 Hz b. 1.4 Hz c. 2.3 Hz d. 3.2 Hz
Final answer:
The oscillation frequency of an ideal harmonic oscillator can be determined using the formula f = (1/2π) * √(k/m). By solving the equation based on the given data, we find that the oscillation frequency is approximately 2.27 Hz.
Explanation:
The oscillation frequency of a harmonic oscillator can be determined using the equation:
f = (1/2π) * √(k/m)
Where:
f is the oscillation frequency
k is the spring constant
m is the mass of the oscillator
In this case, the total mechanical energy of the oscillator is given as 9.8 J. Since the oscillation amplitude is 20.0 cm (or 0.20 m), the total mechanical energy can be equated to 11/13 * m * (Aω)² = 9.8 J, where ω is the angular frequency.
Solving for ω, we find ω = √(13 * 9.8 / (11 * 0.2²)). Using the relationship between angular frequency ω and oscillation frequency f (f = ω / (2π)), we can calculate the oscillation frequency:
f = √(13 * 9.8 / (11 * 0.2²)) / (2π)
Calculating this value gives us f ≈ 2.27 Hz. Therefore, the correct option is c. 2.3 Hz.
A small object carrying a charge of -4.00 nC is acted upon by a downward force of 19.0 nN when placed at a certain point in an electric field.
A) What are the magnitude and direction of the electric field at the point in question? (Answer in N/C)
C) What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?(Answer in N)
Explanation:
Given that,
Charge acting on the object, [tex]q=-4\ nC=-4\times 10^{-9}\ C[/tex]
Force acting on the object, [tex]F=19\ nC=19\times 10^{-9}\ C[/tex] (in downward direction)
(a) The electric force acting in the electric field is given by :
[tex]F=qE[/tex]
E is the electric field
[tex]E=\dfrac{F}{q}[/tex]
[tex]E=\dfrac{19\times 10^{-9}\ N}{4\times 10^{-9}\ C}[/tex]
E = 4.75 N/C
The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.
(b) The charge on the proton is, [tex]q=1.6\times 10^{-19}\ C[/tex]
The force acting on the proton is :
[tex]F=qE[/tex]
[tex]F=1.6\times 10^{-19}\times 4.75[/tex]
[tex]F=7.6\times 10^{-19}\ N[/tex]
If the charge on the proton is positive, the force on the proton is in upward direction.
Hence, this is the required solution.
PART ONE
A ladder rests against a vertical wall. There
is no friction between the wall and the ladder.
The coefficient of static friction between the
ladder and the ground is µ = 0.464 .
(USE THE PICTURE TO DETERMINE THE ANSWER)
Identify the set of equations which is correct.
ANSWER CHOICES:
1. A1, B2, C3
2. A2, B2, C1
3. A1, B1, C1
4. A1, B2, C2
5. A1, B1, C2
6. A2, B1, C3
7. A2, B1, C2
8. A1, B2, C1
9. A1, B1, C3
10. A2, B1, C1
PART TWO
Determine the smallest angle θ for which the
ladder remains stationary.
Answer in units of ◦
Answer:
1. A1, B2, C3
2. 47.1°
Explanation:
Sum of forces in the x direction:
∑Fₓ = ma
f − Fᵥᵥ = 0
f = Fᵥᵥ
Sum of forces in the y direction:
∑Fᵧ = ma
N − W = 0
N = W
Sum of moments about the base of the ladder:
∑τ = Iα
Fᵥᵥ h − W (b/2) = 0
Fᵥᵥ h = ½ W b
Fᵥᵥ (l sin θ) = ½ W (l cos θ)
l Fᵥᵥ sin θ = ½ l W cos θ
The correct set of equations is A1, B2, C3.
At the smallest angle θ, f = Nμ. Substituting into the first equation, we get:
Nμ = Fᵥᵥ
Substituting the second equation into this equation, we get:
Wμ = Fᵥᵥ
Substituting this into the third equation, we get:
l (Wμ) sin θ = ½ l W cos θ
μ sin θ = ½ cos θ
tan θ = 1 / (2μ)
θ = atan(1 / (2μ))
θ = atan(1 / (2 × 0.464))
θ ≈ 47.1°
Equation of equilibrium is the static or dynamic equilibrium of all internal and external forces present in the system.
The set A1, B2, C3 is correct.The smallest angle θ for which the ladder remains stationary is [tex]47.1^o[/tex]Given information-
A ladder rests against a vertical wall.
The coefficient of static friction between the ladder and the ground is 0.464.
What is equation of equilibrium?Equation of equilibrium is the static or dynamic equilibrium of all internal and external forces present in the system.
A) Set of equations which is correct.From the equation of equilibrium of two bodies, the net force in x axis can be given as,
[tex]\sum F_x=0[/tex]
As the normal force and friction force acting on the x-axis. Thus,
[tex]-f_w+f=0[/tex]
[tex]f=f_w[/tex]
Thus option A1 is correct.
As the gravitational force(due to weight of the body) is acting in the vertical direction. Thus from the equation of equilibrium of two bodies, the net force in y axis can be given as,
[tex]\sum F_y=W[/tex]
As the normal force acting on the x-axis. Thus,
[tex]N=W[/tex]
Thus option B2 is correct.
Apply torque equation at the base of the ladder,
[tex]\sum \tau=Ia[/tex]
[tex]F_wh=\dfrac{1}{2} Wb[/tex]
Here the value of h and b can be changed in the form of I using the trigonometry formula. Thus,
[tex]F_w(I\sin \theta)=\dfrac{1}{2} W(\cos \theta[/tex]
Thus option C3 is correct.
Hence the set A1, B2, C3 is correct.
B) The smallest angle θ for which the ladder remains stationary-The equation of option C3 is,
[tex]F_w(I\sin \theta)=\dfrac{1}{2} W(\cos \theta[/tex],
As normal force is can be given as,
[tex]F_w=W\mu[/tex]
Thus,
[tex]W\mu(I\sin \theta)=\dfrac{1}{2} W(\cos \theta[/tex]
[tex]\dfrac{\sin \theta}{\cos \theta} =\dfrac{1}{2\mu} \\\tan \theta =\dfrac{1}{2\times 0.464} \\\theta =\tan^-(1.0770)\\\theta=47.1^o[/tex]
Hence the smallest angle θ for which the ladder remains stationary is [tex]47.1^o[/tex]
Hence,
The set A1, B2, C3 is correct.The smallest angle θ for which the ladder remains stationary is [tex]47.1^o[/tex]Learn more about the equation of equilibrium here;
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Which of the following changes to the earth-sun system would reduce the magnitude of the force between them to one-fourth the value found in Part A?
Reduce the mass of the earth to one-fourth its normal value.
Reduce the mass of the sun to one-fourth its normal value.
Reduce the mass of the earth to one-half its normal value and the mass of the sun to one-half its normal value.
Increase the separation between the earth and the sun to four times its normal value.
part A was:
Consider the earth following its nearly circular orbit (dashed curve) about the sun. The earth has mass mearth=5.98
Answer:
Reduce the mass of the earth to one-fourth its normal value.
Reduce the mass of the sun to one-fourth its normal value.
Reduce the mass of the earth to one-half its normal value and the mass of the sun to one-half its normal value.
Explanation:
Every particle in the universe attracts any other particle with a force that is directly proportional to the product of its masses and inversely proportional to the square of the distance between them. So, in this case we have:
[tex]F=\frac{Gm_Em_S}{d^2}[/tex]
If [tex]m'_E=\frac{m_E}{4}[/tex]:
[tex]F'=\frac{Gm'_Em_S}{d^2}\\F'=\frac{G(\frac{m_E}{4})m_S}{d^2}\\F'=\frac{1}{4}\frac{Gm_Em_S}{d^2}\\F'=\frac{1}{4}F[/tex]
If [tex]m'_S=\frac{m_S}{4}[/tex]
[tex]F'=\frac{Gm_Em'_S}{d^2}\\F'=\frac{Gm_E(\frac{m_S}{4})}{d^2}\\F'=\frac{1}{4}\frac{Gm_Em_S}{d^2}\\F'=\frac{1}{4}F[/tex]
If [tex]m'_E=\frac{m_E}{2}[/tex] and [tex]m'_S=\frac{m_S}{2}[/tex]:
[tex]F'=\frac{Gm'_Em'_S}{d^2}\\F'=\frac{G(\frac{m_E}{2})(\frac{m_S}{2})}{d^2}\\F'=\frac{1}{4}\frac{Gm_Em_S}{d^2}\\F'=\frac{1}{4}F[/tex]
Final answer:
To reduce the gravitational force between the Earth and the Sun to one-fourth, you can either reduce the mass of either body to one-fourth, or increase the distance between them by a factor of four. Reducing the mass of both bodies to one-half would also achieve the same result, as the product of their masses would be one-fourth of the original value.
Explanation:
The force between the Earth and the Sun can be described by Newton's law of universal gravitation, which states that the force (F) is proportional to the product of the two masses (m1 and m2) divided by the square of the distance (r) between their centers of mass. The law is formulated as F = G × m1 × m2 / r2, where G is the gravitational constant.
To reduce the magnitude of the gravitational force between the Earth and the Sun to one-fourth, one could either reduce the product of the masses by one-fourth or increase the separation distance by four times, because the force is inversely proportional to the square of the distance. Thus, the correct answers would be:
Reduce the mass of the Earth to one-fourth its normal value.
Reduce the mass of the Sun to one-fourth its normal value.
Increase the separation between the Earth and the Sun to four times its normal value.
However, the third option offered in the question, reducing the mass of both the Earth and the Sun to one-half their normal values, would also result in reducing the force to one-fourth, because (1/2) × (1/2) = 1/4.
The gravitational field of m1 is denoted by g1. Enter an expression for the gravitational field g1 at position la in terms of m1, la, and the gravitational constant G.
Answer:
The expression of gravitational field due to mass [tex]m_![/tex] at a distance [tex]l_a[/tex]
Explanation:
We have given mass is [tex]m_1[/tex]
Distance of the point where we have to find the gravitational field is [tex]l_a[/tex]
Gravitational constant G
We have to find the gravitational filed
Gravitational field is given by [tex]g=\frac{Gm_1}{l_a^2}[/tex]
This will be the expression of gravitational field due to mass [tex]m_![/tex] at a distance [tex]l_a[/tex]
The expression for the gravitational field g₁ at position la in terms of m₁, la, and the gravitational constant G is:
[tex]g_1 = \frac{G\times m_1}{la^{2} }[/tex]
The gravitational field is the force field that exists in the space around every mass or group of masses.
The gravitational field of m₁ is denoted by g₁, and can be represented through the following expression.
[tex]g_1 = \frac{F_1}{m}[/tex] [1]
where,
F₁ is the gravitational force due to m₁m is the mass of the object at a distance la from m₁We can calculate the force (F₁) between m₁ and m that are at a distance "la" using Newton's law of universal gravitation.
[tex]F_1 = G \frac{m \times m_1 }{la^{2} }[/tex] [2]
where,
G is the gravitational constantIf we replace [2] in [1], we get
[tex]g_1 = \frac{G \frac{m \times m_1 }{la^{2} }}{m} = \frac{G\times m_1}{la^{2} }[/tex]
The expression for the gravitational field g₁ at position la in terms of m₁, la, and the gravitational constant G is:
[tex]g_1 = \frac{G\times m_1}{la^{2} }[/tex]
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The 2-kg spool S fits loosely on the inclined rod for which the coefficient of static friction is μs=0.2. If the spool is located 0.25 m from A, determine the maximum constant speed the spool can have so that it does not slip up the rod
Answer:
v = 0.969 m/s
Explanation:
See attachment for FBD
p = 0.25*(4/5) = 0.2 m
Sum of normal forces
Ns (3/5) - 0.2*Ns*(4/5) = 2 * (v^2 / 0.2)
Sum of vertical forces
Ns (4/5) - 0.2*Ns*(3/5) = 2*9.81
Solve both equations simultaneously to get Ns and v
Ns = 21.3 N
v = 0.969 m/s
What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5×10−12m?
Final answer:
The magnitude of the electric force of attraction between an iron nucleus and its innermost electron can be calculated using Coulomb's law.
Explanation:
The magnitude of the electric force of attraction between an iron nucleus and its innermost electron can be calculated using Coulomb's law. Coulomb's law states that the magnitude of the electric force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, the charge of the iron nucleus (+26e) is given as 26 times the charge of an electron (e).
The formula to calculate the electric force is: F = k * |q1 * q2| / r^2, where F is the electric force, k is the electrostatic constant, q1 and q2 are the charges of the particles, and r is the distance between them.
Using this formula, we can plug in the values: F = (8.99 × 10^9 Nm^2/C^2) * |(26e) * (-e)| / (1.5 × 10^(-12)m)^2. Simplifying this equation will give you the magnitude of the electric force.
A 1.33 kg object is attached to a horizontal spring of force constant 2.50 N/cm and is started oscillating by pulling it 6.40 cm from its equilibrium position and releasing it so that it is free to oscillate on a frictionless horizontal air track. You observe that after eight cycles its maximum displacement from equilibrium is only 3.70 cm .
(a) How much energy has this system lost to damping during these eight cycles?
(b) Where did the "lost" energy go? Explain physically how the system could have lost energy.
Answer:
Explanation:
a ) Energy of spring = 1/2 k A² where A is amplitude of oscillation and k is force constant .
So initial energy = 1/2 x 2.5 x (6.4 x 10⁻²)²
= 51.2 x 10⁻⁴ J
So final energy = 1/2 x 2.5 x (3.7 x 10⁻²)²
= 17.11 x 10⁻⁴ J
energy lost
= 34.1 J .
This energy is dissipated in the form of heat, sound etc.
Answer:
a. [tex]\Delta U=3.375\ N.cm=3.375\times 10^{-2}\ J[/tex]
b. The energy lost during the damping is converted into kinetic energy of the molecules which is heat primarily.
Explanation:
Given:
mass of the attached object, [tex]m=1.33\ kg[/tex]spring constant, [tex]k=2.5\ N.cm^{-1}[/tex]maximum displacement, [tex]A=6.4\ cm[/tex]maximum displacement after damping, [tex]x=3.7\ cm[/tex]a)
the energy lost in damping:
[tex]\Delta U=\frac{1}{2} \times k\times A^2-\frac{1}{2} \times k\times x^2[/tex]
[tex]\Delta U=\frac{1}{2} \times 2.5\times (6.4-3.7)[/tex]
[tex]\Delta U=3.375\ N.cm=3.375\times 10^{-2}\ J[/tex]
b)
The energy lost during the damping is converted into kinetic energy of the molecules which is heat primarily.
A tiny ball (mass = 0.010 kg) carries a charge of -24C. What is the electric field magnitude that would cause the ball to float above the ground?
Use the relation between electric force and electric field and the concept of gravitational force to calculcate the electric field. The electric force is given by
[tex]F_e = qE[/tex]
And the gravitational force is
[tex]F_g = mg[/tex]
For the ball to float above the ground, the magnitude of electric force on the ball must be equal to the magnitude of the gravitational force. That is must be a equilibrium condition, so,
[tex]F_e = F_g[/tex]
[tex]qE = mg[/tex]
[tex]E = \frac{mg}{q}[/tex]
Replacing the values we have that,
[tex]E = \frac{(0.010)(9.8)}{-24*10^{-6}}[/tex]
[tex]E = -4.083*10^{3} N/C[/tex]
Therefore the electric field magnitude that would cause the ball to float above the ground is [tex]-4.083*10^{3} N/C[/tex]
At the surface of the moon, the gravitational attraction is 5.37 ft/s2. What is the mass of the object on the surface of the moon
Answer:
mass remains the same
Explanation:
Mass is the amount of matter in a substance which is independent from the external environment; hence, fields!
The weight of the object changes but mass remains same!
Final answer:
The question incorrectly asks for an object's mass on the moon based on gravitational acceleration. Mass is a constant and doesn't change with location. The gravitational acceleration on the moon (about 1.62 m/s²) affects an object's weight, not its mass.
Explanation:
The question seems to be asking for the mass of an object on the moon's surface based on the gravitational acceleration at the moon, which is a misunderstanding. The gravitational attraction on the surface of the moon is given as 5.37 ft/s2 (equivalent to about 1.63 m/s2, since the accurate value is 1.62 m/s2), but to find the mass of an object, we need to discuss how weight and mass are related in a gravitational field, not to calculate the mass based solely on the gravitational acceleration. To find an object's mass on the moon, one would use the formula Weight = Mass × Gravitational acceleration (W = mg). However, the mass of an object is a constant and does not change depending on its location, whether on Earth, the moon, or elsewhere. What changes is the weight of the object due to the difference in gravitational force exerted on it. For example, if an object weighs 9.8 N on Earth, it would weigh approximately 1.6 N on the moon due to the moon's lower gravity.
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 280m/s2 m/s^2 for 20 ms {\rm ms}, then travels at constant speed for another 30 ms {\rm ms}.
During this total time of 50 ms {\rm ms}, 1/20 of a second, how far does the tongue reach?
Express your answer to two significant figures and include the appropriate units.
Answer:
The reach of the tongue is 23 cm.
Explanation:
Hi there!
The equation of traveled distance is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = traveled distance at time t.
x0 = initial position.
v0 = initial velocity.
a = acceleration.
t = time.
Let´s calculate the distance traveled by the tongue of the chameleon during the first 20 ms (0.020 s):
The initial position and velocity are zero (x0 = 0 and v0 = 0)
x = 1/2 · a · t²
x = 1/2 · 280 m/s² · (0.020 s)²
x = 0.056 m
Now, let´s find the distance traveled while the tongue moves at constant speed. But first, let´s find the velocity (v) of the tongue after the accelertation interval using the following equation:
v = v0 + a · t (v0 = 0)
v = 280 m/s² · 0.020 s
v = 5.6 m/s
Then, the distance traveled at constant speed can be calculated:
x = v · t
x = 5.6 m/s · 0.030 s
x = 0.17 m
The reach of the tongue is 0.17 m + 0.056 m = 0.23 m = 23 cm.
A water jet that leaves a nozzle at 60 m/s at a flow rate of 120 kg/s is to be used to generate power by striking the buckets located on the perimeter of a wheel. Determine the power generation potential of this water jet.
Answer:
216000 W or 216 kW
Explanation:
Power: This can be defined as the rate at which energy is consumed or used, The S.I unit of power is Watt (W)
Generally,
Power = Energy/time
P = E/t........................ Equation 1.
But
E = 1/2mv²..................... Equation 2
Where m = mass, v = velocity.
Substitute equation 2 into equation 1
P = 1/2mv²/t...................... Equation 3
Let flow rate (Q) = m/t
Q = m/t................ Equation 4
Substitute equation 4 into equation 3
P = Qv²/2........................ Equation 5
Where Q = flow rate, v = velocity, P = power.
Given: Q = 120 kg/s, v = 60 m/s
Substitute into equation 5
P = 120(60)²/2
P = 60(60)²
P = 60×3600
P = 216000 W.
Thus the power generation potential of the water jet = 216000 W or 216 kW
The water jet could potentially generate 432 kW of power under ideal circumstances, calculated based on the given flow rate and velocity of the water jet.
Explanation:To calculate the power generation potential of the water jet, you will use the formula for power: Power = Work/time. Since work can be translated into a measure of force times distance, we substitute Force * Distance into the equation for work. We can find the force exerted by the water jet with the equation for force: Force = mass * acceleration. The water's acceleration is its velocity out of the nozzle, or 60 m/s, and the mass flow rate is given as 120 kg/s. This gives us a force of 120 kg/s * 60 m/s = 7200 N.
However, distance is not given in the problem, so it's more helpful in this case to use an alternative equation for Power, given as the product of force and velocity: Power = Force * Velocity. Thus, our power becomes 7200 N * 60 m/s = 432,000 Watts, or 432 kW, assuming 100% efficiency.
Realistically, some power will be lost due to friction and inefficiencies in the system, but under ideal circumstances the water jet could potentially generate 432 kW of power.
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An object whose mass is 120 kg is located 20 m above a datum level in a location where standard gravitational acceleration exists. Determine the total potential energy, in kJ, of this object.
Answer:
U= 23.544 KJ
Explanation:
Given that
mass ,m = 120 kg
Height above the datum ,h = 20 m
Take g = 9.81 m/s²
The potential energy U is given as
U= m g h
m=mass
h=Height above the datum
g=Gravitational constant
Now by putting the value in the above values we get
U= 120 x 9.81 x 20 J
U=23544 J
Potential energy in KJ
U= 23.544 KJ
Therefore the answer will be 23.544 KJ.
Wave motion is characterized by two velocities: the velocity with which the wave moves in the medium (e.g., air or a string) and the velocity of the medium (the air or the string itself).Consider a transverse wave traveling in a string. The mathematical form of the wave is: y(x,t) = A sin(kx - ωt)Part AFind the velocity of propagation v_p of this wave.Express the velocity of propagation in terms of some or all of the variables A, k, and ω.Part BFind the y velocity v_y(x,t) of a point on the string as a function of x and t.Express the y velocity in terms of ω, A, k, x, and t.Part CWhich of the following statements about v_x(x,t), the x component of the velocity of the string, is true?A) v_x(x;t) = v_pB) v_x(x;t) = v_y(x;t)C) v_x(x;t) has the same mathematical form as v_y(x;t) but is 180° out of phase.D) v_x(x;t)=0Part DFind the slope of the string ∂_y(x,t) / ∂_x as a function of position x and time t.Express your answer in terms of A,k, ω, x, and t.
Answer:
Explanation:
ocities: the velocity with which the wave moves in the medium (e.g., air or a string) and the velocity of the medium (the air or the string itNBself).Consider a transverse wave traveling in a string. The mathematical form of the wave is: y(x,t) = A sin(kx - ωt)Part AFind the velocity of propagation v_p of this wave.Express the velocity of propagation in terms ofNGHJGHHG some or all of the variables A, k, and ω.Part BFind the y velocity v_y(x,t) of a point on the string as a function of x and t.Express the y velocity in terms of ω, A, k, x, and t.Part CWhich of the following statements about v_x(x,t), the x component of the velocity of the string, is true?A) v_x(x;t) = v_pB) v_x(x;t) = v_y(x;t)C) v_x(x;t) has the same mathematical form as v_y(x;t) but is 180° out of phase.D) v_x(x;t)=0Part DFind the slope of the string ∂_y(x,t) / ∂_x as a function of position x and time t.Express your answer in terms of A,k, ω, x, and t.NNNNN
a) The velocity of propogation of the wave V=w/k
b) The y velocity v_y(x,t) of a point on the string as a function of x v=-wAcos(kx-wt)
What is wave?A wave can be described as a disturbance that travels through a medium from one location to another location
y(x,t)=Asin(kx−ωt) defines the wave equation.
a)The velocity of propogation of the wave
We are asked to find wave speed (v)
Recall that v = fλ
From the wave equation above,
k = 2π/ λ where k is the wave number and λ is the wavelength, λ = 2π /k
ω = 2πf where f is the frequency and ω is the angular frequency.
f = ω/ 2π.
By substituting for λ and ω into the wave speed formulae, we have that
v =( ω/ 2π) × (2π /k)
v = ω/k
b)The y velocity v_y(x,t) of a point on the string as a function of x
y(x,t)=Asin(kx−ωt)
The first derivative of y with respect to x give the velocity (vy)
By using chain rule, we have that
v = dy/dt = A cos( kx −ωt) × (−ω)
v = - ωAcos( kx −ωt)
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A lightning bolt may carry a current of 1.00 104 A for a short time. What is the resulting magnetic field 120 m from the bolt
Answer:
B = 1.67 μ T
Explanation:
given,
current, I = 1 x 10⁴ A
r = 120 m
treating lightning bolt as long straight conductor
[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]
[tex]B=\dfrac{4\pi \times 10^{-7}\times 1 \times 10^4}{2\pi\times 120}[/tex]
resulting magnitude would be equal to
B = 16.67 x 10⁻⁶ T
B = 1.67 μ T
The resulting magnetic field is equal to B = 1.67 μ T
Answer:
1.67 x 10^-5 Tesla
Explanation:
Current, i = 1 x 10^4 A
distance, d = 120 m
The formula for the magnetic field due to long straight current carrying conductor is given by
[tex]B =\frac{\mu _{0}}{4\pi }\times \frac{2i}{r}[/tex]
[tex]B =10^{-7}\times \frac{2\times 10^{4}}{120}[/tex]
B = 1.67 x 10^-5 Tesla
a) Estimate how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high). Express your answer using two significant figures.
b) Estimate his velocity just before "landing." Express your answer using two significant figures. Assume that the positive direction is upward.
Answer:
a) 8.8 sec b) -86 m/sec
Explanation:
Assuming King Kong started from rest its fall, once in the air, neglecting air resistance, is only affected by gravity, which accelerates it downward.
As this acceleration is constant, we can use the following equation in order to get how long it was falling:
Δh = 1/2*g*t² ⇒ -380 m = 1/2 (-9.8 m/s²)*t²
Solving for t:
t = √((2*380)/9.8) s² = 8.8 sec.
b) In order to know the value of the velocity in the instant just before it hits the ground, we can apply acceleration definition, as follows:
a = (vf-v₀) /t
In our case, a = -g (assuming positive direction is upward) and v₀=0, so, we can solve for vf as follows:
vf = -g*t = -9.8 m/s²*8.8 sec = -86 m/s.
The answer explains the estimated time for King Kong to fall from the Empire State Building and his velocity just before landing using physics formulas.
a) Estimate: Using the formula h = 0.5 * g * t^2 where h is the height, g is gravity (9.8 m/s^2), we find t = sqrt(2h/g) = sqrt(2*380/9.8) = 8.7 s. Therefore, King Kong took approximately 8.7 seconds to fall.
b) Velocity: To find the velocity just before landing, we use v = g*t where v is the final velocity. Substituting values, v = 9.8 * 8.7 = 85.3 m/s. Therefore, his velocity just before 'landing' was around 85 m/s.
The frequency sweep is 383-438 Mhz with a period of 15 microseconds. The first return occurs at 429 MHz. What is the range? (answer in meters to one decimal point)
Answer:
The range is 1881.8 m.
Explanation:
Given that,
Time [tex]t=15\times10^{-6}\sec[/tex]
Frequency range [tex]\Delta f= f_{f}-f_{i}[/tex]
[tex]\Delta f= 429-383[/tex]
[tex]\Delta f=46\ MHz[/tex]
The value of [tex]\dfrac{df}{dt}[/tex]
[tex]\dfrac{df}{dt}=\dfrac{438-383}{15\times10^{-6}}[/tex]
We need to calculate the range
Using formula of range
[tex]R=\dfrac{c\Delta f}{2\times\dfrac{df}{dt}}[/tex]
Put the value into the formula
[tex]R=\dfrac{3\times10^{8}\times46}{2\times\dfrac{438-383}{15\times10^{-6}}}[/tex]
[tex]R=1881.8\ m[/tex]
Hence, The range is 1881.8 m.
The back wall of an auditorium is 23.0 m from the stage.
If you are seated in the middle row, how much time elapses between a sound from the stage reaching your ear directly and the same sound reaching your ear after reflecting from the back wall? The speed of sound in the air is 343 m/s .
_________seconds
Answer:
0.067 seconds
Explanation:
The time needed for a person seated in the middle row to receive a sound from the stage, can be calculated assuming that the sound is moving in a straight line at constant speed, taking into account that the distance traveled will be half of the distance between the stage and the back wall:
t₁ = x₁ / v = 11.5 m / 343 m/s = 0.034 sec
The same sound, after reflecting from the back wall, will travel a distance equal to one and a half the distance between the stage and the back wall:
t₂ = 34.5 m /343 m/s = 0.101 sec
The time elapsed between both sounds will be equal to the difference between t₂ and t₁, as follows:
t₂ - t₁ = 0.101 sec - 0.034 sec = 0.067 sec.
The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the Earth and the Sun is 1.5 x 1011 m. a) Assuming it radiates uniformly in all directions what is the total power output of the Sun?b) If the frequency increases by 1 MHz what would be the relative (percentage) change in the power output? c) For frequency in b) what is the intensity of the radiation from the Sun measured on Mars? Note that Mars is 60% farther from the Sun than the Earth is.
Answer: (a) power output = 3.85×10²⁶W
(b). There is no relative change in power as it is independent from frequency
(c). 590 W/m²
Explanation:
given Radius between earth and sun to be = 1.50 × 10¹¹m
Intensity of the radiation from the sun measured on earth to be = 1360 W/m²
Frequency = 60 MHz
(a). surface area A of the sun on earth is = 4πR²
substituting value of R;
A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²
A = 2.863 10²³×m²
now to get the power output of the sun we have;
P sun = I sun-earth A sun-earth
where A = 2.863 10²³×m², and I is 1360 W/m²
P sun = 2.863 10²³ × 1360
P sun = 3.85×10²⁶W
(c). surface area A of the sun on mars is = 4πR²
now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have
A sun-mars = 4π(2.28× 10¹¹)²
A sun-mars = 6.53 × 10²³m²
now to calculate the intensity of the sun;
I sun-mars = P sun / A sun-mars
where P sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²
I sun-mars = 3.85×10²⁶W / 6.53 × 10²³m²
I sun-mars = 589.6 ≈ 590 W/m²
I sun-mars = 590 W/m²
A student said, "The displacement between my dorm and the lecture hall is 1 kilometer."
Is he using the correct physical quantity for the information provided? What should he have called the 1 kilometer?
a) Distance
b) Path length
c) Position
d) Both a and b are correct.
Answer:
d) Both a and b are correct
Explanation:
Displacement: It is defined as the distance between initial and final position during motion.
Distance: It is defined as the total path length traveled by object
Or
It is the distance of one place from other place.
Student said that the displacement between my dorm and the lecture hall is 1km.
It is not displacement .It is distance or we say path length.
Therefore, he is using incorrect physical quantity for the information provided.
He should have called distance 1 km or path length 1 km.
Option d is true.
The student is accurately using the term 'displacement' to describe the 1 kilometer between their dorm and the lecture hall. Technically, 'distance' or 'path length' could also have been used assuming a straight-line path and ignoring direction.
Explanation:In the context of this specific question, the student is accurately using the term displacement as a physical quantity. Displacement refers to the shortest distance between two points in a particular direction. In this case, the 1 kilometer refers to the displacement between the dorm and the lecture hall, assuming a straight-line path. So, technically, both a) Distance and b) Path length could also have been used to describe the 1 kilometer, if we don't consider the direction.
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You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then load up the pickup and pump up itstires so that its total weight increases by 42% whilethe coefficient of rolling friction decreases by19%.a) Now what horizontal force will you need to move the pickupalong the same road at the same speed? The speed is low enough thatyou can ignore air resistance.
Answer:
The horizontal force that will be needed to move the pickup along the same road at the same speed is 230 N
Explanation:
given information:
horizontal force, F = 200 N
speed, v = 2.4 m/s
total weight increase 42%
coefficient of friction decrease by 19%
now, lets take a look of horizontal system
F - F(friction) = 0
F = F(friction)
F = μ N
F = μ m g
μ m g = 200 N
now find the force to move the pickup along the same road at the same speed(F(move))
total weight increase 42%, m g = 1+0.42 = 1.42
he coefficient of rolling friction decreases by 19%, μ = 1 - 0.19 = 0.81
F (move) = (0.81μ) (1.42 m g)
= (0.81) (1.42) (μ m g)
= (0.81) (1.42) (200)
= 230 N
A cart moves toward a motion sensor on a track in three different ways: speeding up with constant speed slowing to a stop and turning around to speed up in the opposite direction For each of these cases, what is the sign of the velocity and what is the sign of the acceleration at the beginning of the trip and at the end of the trip?
Answer:
Acceleration of the cart in the beginning is positive
Speed is positive when speeding up and slowing down.
Speed is zero when the cart stops.
Speed is negative when the cart speeds up in opposite direction.
Acceleration is negative when speeding up in opposite direction from rest.
Explanation:
When the cart is speeding up then its speed is increasing with time hence its acceleration is having a positive value.
On speeding up the velocity is positive.On gradually decreasing the speed and coming to a stop the velocity is still positive with continuously decreasing magnitude until it reaches zero when the cart stops.When the cart speeds up in the opposite direction then the velocity of the cart becomes negative from zero and continuously increases in the negative value over a span of time hence its acceleration is also negative.