Answer:
The answer is C "think about the problem first, systematically consider all factors, and form a hypothesis"
Explanation:
In physics there is some basic fomula that sir Isacc Newton proposed under the topic of motion. The three formulas are below;
1) v=u+at
2)v^2=u^2+2as
3)s=ut+(1/2)(at^2)
the variables are explained below;
u= initial velocity of the body
a=acceleration/Speed of the body
t= time taken by the body while travelling
s= displacement of the body.
Therefore to solve keatons problem, the factors(variables) in the formulas above need to be systematically considered. Since the ball was dropped from the top of the building, the initial velocity is 0 because the body was at rest. Also the acceleration will be acceleration due to gravity (9.8m/s^2)
An average freight train traveling at __________ mph needs a stopping distance of more than 1/2 mile.
Answer:
30 mph
Explanation:
An average freight train traveling at 30 mph needs a stopping distance of more than 1/2 mile.
Ian infinite nonconducting sheet has surface charge density on one side. How far apart are equi-potential surfacews whose potentials sdiffer by 50 V?
Answer:
8.85 × 10 ⁻³ m
Explanation:
The electric field produced by infinite sheet of charge relates by
E = σ / 2ε₀
where σ = surface charge density = 0.1 μC/m² = 0.1 × 10⁻⁶C/m², ε₀ = 8.85 × 10⁻¹² C⁻²/Nm²
also
V = E Δs where Δs is the distance apart in meters, V is the voltage of the
make E subject of the formula
V/Δs = E
equate the two equation
V / Δs = σ / 2ε₀
make Δs subject of the formula
V2ε₀ / σ = Δs
substitute the values into the expression
Δs = (50 × 2 × 8.85 × 10⁻¹²) / (0.1 × 10⁻⁶) = 8.85 × 10 ⁻³ m
The images output from your new color laser printer seem to be a little too blue. What can you do to fix this?
Answer:
The images output from your new color laser printer seem to be a little too blue. to fix this problem we need to calibrate the printer.
Explanation:
This can be done by opening the toolbox, clicking in the device setting folder their you get print quality page click on it. Under the print quality option click on the calibrate next to calibrate now. Then click OK unless when the 'your request has been sent to the device' appears on the screen. When the calibration ends again try printing. calibrating is useful for managing the proper alignment of the inkjet cartridge nozzle to the paper and each other, without proper calibration the print quality deteriorates.
Which of the following terms describes one’s ability to understand how another person is feeling?
a) empathy
b) appreciation
c) cooperation
d) commitment
Answer:
Empathy
Explanation:
Empathy: to understand the thoughts, feelings, or emotional state of another person
Appreciation: to show gratitude
Cooperation: is the association for mutual benefit
Commitment: is the agreement to do something.
From the definitions above, empathy is the term that describes one’s ability to understand how another person is feeling.If 1.00 kilograms of coal is burned, approximately 3.00 x 107 joules (J) of energy is released. How many tons of coal, when burned, would produce approximately 9.00 x 1016 J?
Answer:
[tex]3.00\times10^{6}\,tons [/tex]
Explanation:
We can resolve this problem using a proportion because the ratio tons burned and energy produced is constant, using x for tons burned of coal to produce [tex] 9.00\times10^{16}J [/tex]:
[tex]\frac{x}{9.00\times10^{16}J}=\frac{1.00\,kg}{3.00\times10^{7}J} [/tex]
solving the proportion:
[tex]x*3.00\times10^{7}J=1.00\,kg*9.00\times10^{16}J [/tex]
[tex] \frac{x*\cancel{3.00\times10^{7}J}}{\cancel{3.00\times10^{7}J}}=\frac{1.00\,kg*9.00\times10^{16}J}{3.00\times10^{7}J}[/tex]
[tex]x=\frac{9.00\times10^{16}kg*\cancel{J}}{3.00\times10^{7}\cancel{J}}=3.00\times10^{9}\,kg [/tex]
Now we have the answer but in kilograms we should convert this knowing that 1 ton = 1000 kg:
[tex]3.00\times10^{9}\,kg=3.00\times10^{9}\,\cancel{kg}*\frac{1\,ton}{1000\,\cancel{kg}} [/tex]
[tex]3.00\times10^{9}\,kg=3.00\times10^{6}\,tons [/tex]
Answer:
3.3 x 106 tons
Explanation:
Since you know that E = mc2, and both equations possess the constant speed of light, set the two equations as equivalencies. For instance, E1/m1 = E2/m2. Solve for m2.
Took the test and got it correct
A 1.0 kg block slides along a frictionless horizontal surface with a speed of 7.0 m/s. After sliding a distance of 2.0 m, the block makes a smooth transition to a frictionless ramp inclined at an angle of 40° to the horizontal.
How far up the ramp does the block slide before coming momentarily to rest?
Answer:
3.9 m
Explanation:
The principle of work and energy
ΔE = W Formula (1)
where:
ΔE: mechanical energy change (J)
W : work of the non-conservative forces (J)
ΔE = Ef - E₀
Ef : final mechanical energy
E₀ : initial mechanical energy
Ef = K f+ Uf
E₀ = K₀ + U₀
K =(1/2 )mv² : Kinetic energy (J)
U = mgh :Potential energy (J)
m: mass (kg)
v : speed (m/s)
h: hight (m)
Known data
m = 1 kg : mass of the block
v₀ = mg(h).0 m/s Initial speed of the block
vf = 0 = Final speed of the block
θ =40° :angle θ of the ramp with respect to the horizontal direction
μk=0 : coefficient of kinetic friction
g = 9.8 m/s² : acceleration due to gravity
Problem development
W = 0 , Because the friction force (non-conservative force ) is zero
Principle of work and energy to the Block:
ΔE = W
Ef - E₀ = 0 Equation (1)
Ef = K f+ Uf =(1/2 )m(0)² + mg(h)= mg(h) (Joules)
E₀ = K₀ + U₀ = (1/2 )m(7)² + mg(0) = 24.5m (Joules)
In the equation (1) :
Ef = E₀
mg(h) = 24.5m
We divided by m both sides of the equation
g(h) = 24.5
h = 24.5 / g
h = 24.5 / 9.8
h = 2.5 m
We apply trigonometry at the ramp to calculate how far up the ramp (d) does the block slide before coming momentarily to rest :
sinθ = h/d
d = h / sinθ
d = 2.5 m / sin40°
d = 3.9 m
The distance the block slides up the ramp can be determined using the principle of energy conservation: initial kinetic energy is converted into gravitational potential energy. The distance along the ramp is the height (calculated using the energy conservation equation) divided by the sine of the ramp's angle of inclination.
Explanation:This is a problem of energy conservation. Initially, the block has some kinetic energy due to its speed and no potential energy. When it starts climbing the ramp, kinetic energy is transformed into gravitational potential energy. The block will come to rest when all the kinetic energy has been converted into potential energy.
To determine how far up the ramp the block will slide, we use the equation of energy conservation: 1/2 * mass * speed^2 = mass * gravity * height. Given that we know the mass of the block (1.0 kg), its initial speed (7.0 m/s), and the acceleration due to gravity (approximately 9.8 m/s^2), we can solve this equation to get the height as: height=1/2 * mass * speed^2 / (mass * gravity), which simplifies to height= 1/2 * speed^2 / gravity.
Once we get the height, we need to determine the distance along the ramp, which is the height divided by the sine of the ramp’s angle of inclination (40°). Thus, the distance = height / sin(40°).
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How much elastic potential energy is stored in a bungee cord with a spring constant of 10.0 N/m when the cord is stretched 2.00 m?
Answer:
The elastic potential energy stored in the bungee cord = 20 J
Explanation:
potential energy: This is the energy possessed by a body due to its position. The S.I unit of energy is Joules. The mathematical expression for elastic potential energy is given below
E = 1/2ke²................ Equation 1
Where E = elastic potential energy of the spring, k = force constant of the spring, e = extension
Given: K = 10 N/m, e = 2.00 m
Substituting these values into Equation 1
E = 1/2(10)(2)²
E = 5×4
E = 20 Joules.
Therefore the elastic potential energy stored in the bungee cord = 20 J
Final answer:
The elastic potential energy stored in a bungee cord with a spring constant of 10.0 N/m and stretched 2.00 m is 20.0 joules.
Explanation:
The amount of elastic potential energy stored in a bungee cord can be calculated using the formula for the potential energy in a spring: E = 1/2 kx², where E is the elastic potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position. In this case, with a spring constant of 10.0 N/m and a displacement of 2.00 m . Putting all these values into the equation we get the energy stored is E = 1/2 (10.0 N/m)(2.00 m)² = 20.0 J.
As the wavelength of an electromagnetic wave ___ the frequency of the wave ________.
Answer:
As the wavelength of an electromagnetic wave _decrease__ the frequency of the wave _increase_______.
Explanation:
What is the relationship between frequency and wavelength?
Wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength.
That number, also known as the frequency, will be larger for a short-wavelength wave than for a long-wavelength wave. The equation that relates wavelength and frequency is:
V= fλ
where v= velocity
f= frequency
λ = wavelength
⇒ f = v/λ
also f ∝ 1/λ
For electromagnetic radiation, the speed is equal to the speed of light, c, and the equation becomes:
C= fλ
where c= Speed of light
f= frequency
λ = wavelength
⇒ f = v/λ
also f ∝ 1/λ
The heat of combustion of bituminous coal is 2.50 x 10⁴ J/g. What quantity of the coal is required to produce the energy to convert 106.9 pounds of ice at 0.00°C to steam at 100.°C?specific heat (ice) = 2.10 J/g°Cspecific heat (water) = 4.18 J/g°Cheat of fusion = 333 J/gheat of vaporization = 2258 J/g
Answer:
mc=5.84kg
Explanation:
Given
m i = 106.9 l b s
m i = 48480.7 g
T 1 = T i = 0 °C
T 2 = T s = 100 °C
C p i = 2.10 J / g − C
C p w = 4.18 J / g − C
H f = 333 J / g
H v = 2258 J / g
H c = 25000.00 J / g
According to the first law of thermodynamics, the heat supplied by the coal must be equal to the heat required to convert the ice into water and then the water into steam:
Q c = Q i + Q w + Q s . . . . . . . . . . . . . . . . . . .(eq:1)
Since the water will undergo a change in temperature (0 C to 100 C )
Q i = m i ∗ H f..........(eq:2)
Q s= m s ∗ H v..........(eq:3)
heat allows for the change in temperature, calculated as:
Q w = m w ∗ C p w ∗ ( T 2 − T 1 )...........(eq:4)
Integrating Equations 2, 3 and 4 into Equation 1, we get:
Q c = m i ∗ H f + m w ∗ C p w ∗ ( T 2 − T 1 ) + m s ∗ H v ...........(eq:5)
m i = m w = m s = m
Q c = m ∗ ( H f + C p w ∗ ( T 2 − T 1 ) + H v )
Q c = 48480.7 ∗ ( 333 + 4.18 ∗ ( 100 − 0 ) + 2258 ) J
Q c = 145878426.3 J
The heat of the coal is the product of the mass of the coal and its heat of combustion.
Thus:
m c = Q c / H c
m c = 145878426.3/ 2.5 x 10 4 g
m c = 5835.137052 g
mc=5.84kg
In a popular classroom demonstration, a cotton ball is placed in the bottom of a strong test tube. A plunger fits inside the tube and it makes an air-tight seal. It is then pushed down very rapidly, and the cotton flashes and burns. This happens because ______.
Answer:
An increase in air temperature because of its compression.
Explanation:
The Gay-Lussac's Law states that a gas pressure is directly proportional to its temperature in an enclosed system to constant volume.
[tex] P = kT [/tex]
where P: is the gas pressure, T: is the gas temperature and k: is a constant.
Therefore, due to Gay-Lussac's Law, when the plunger is pushed down very rapidly, the pressure of the air increase, which leads to its temperature increase. That is why cotton flashes and burns.
I hope it helps you!
Early LCD backlights use _______________ technology, popular for its low power use, even brightness, and long life.
Answer:
cold cathode fluorescent lamp
Explanation:
A cold cathode fluorescent lamp (CCFL) is a lighting system that uses two phenomena: electron discharge and fluorescence.
CCFLs are mainly used as light sources for backlights, since they are compact and durable than ordinary fluorescent lamps. They exhibit a wide range of brightness and color (color temperature and chromaticity) that can be achieved by varying the pressure and type of the material injected into the glass tube. The thickness and type of the phosphor used to coat the inner wall of the tube also plays a role in altering the color and brightness.
Final answer:
Early LCD backlights used CCFL technology for its efficiency and longevity. LCDs rely on a backlight to illuminate the liquid crystals in pixels. Each pixel can control red, blue, or green filters to create the full color spectrum on the screen.
Explanation:
Early LCD backlights use CCFL (Cold Cathode Fluorescent Lamp) technology, popular for its low power use, even brightness, and long life. LCD, which stands for liquid crystal display, relies on the light modulating properties of liquid crystals. These crystals themselves do not emit light. In flat screen LCD televisions, the backlight illuminates the screen through millions of tiny units called pixels. Each pixel has three cells with red, blue, or green filters, each controlled independently to vary the picture contrast and color representation by adjusting the voltage applied to the liquid crystal.
BOTH INTERNATIONAL & INLAND A vessel is "engaged in fishing" when ________.
Select one:
a. she is using any type of gear, other than lines
b. she is using fishing apparatus which restricts her maneuverability
c. her gear extends more than 100 meters from the vessel
d. she has any fishing gear on board
Answer:
(b) she is using fishing apparatus which restricts her manoeuvrability
Explanation:
The explanation is spelled out in Rule 3(d) of the inland and water ways laws that state that: A vessel engage in fishing is one whose fishing apparatus restricts manoeuvrability. I f the fishing gear does not hamper her movements, then she is engage in fishing. The rules state categorically that a troller with hook-spangled lines following in her wake is not restricted by her gear. Rule 3(d) might appear to be splitting hairs, but it has a point to make. Although a vessel might not be hampered as to be unable to keep clear of others, her fishing gear could still make following a script difficulty
An object has a position given by r = [2.0 m + (5.00 m/s)t] i^ + [3.0m−(2.00 m/s2)t2] j^, where all quantities are in SI units. What is the magnitude of the acceleration of the object at time t = 2.00 s?
Final answer:
The magnitude of the acceleration of the object at time t = 2.00 s is 6.40 m/s².
Explanation:
The magnitude of acceleration can be found by taking the second derivative of the position function with respect to time.
Given the position function r = [2.0 m + (5.00 m/s)t] i^ + [3.0m−(2.00 m/s²)t²] j^, we can differentiate it twice to get the acceleration function a(t) = 5.00 m/s i^ - 4.00 m/s² j^.
Plugging in t = 2.00 s gives us a(2.00 s) = 5.00 m/s i^ - 4.00 m/s² j^, and calculating the magnitude of this vector gives us 6.40 m/s² as the magnitude of acceleration at t = 2.00 s.
The magnitude of the acceleration of the object at [tex]\( t = 2.00 \, \text{s} \) is \( 4.00 \, \text{m/s}^2 \).[/tex]
To find the magnitude of the acceleration of the object at time [tex]\( t = 2.00 \, \text{s} \)[/tex], we first need to determine the acceleration vector from the given position vector. The position vector[tex]\( \mathbf{r}(t) \)[/tex]is given by:
[tex]\[\mathbf{r}(t) = [2.0 \, \text{m} + (5.00 \, \text{m/s})t] \hat{i} + [3.0 \, \text{m} - (2.00 \, \text{m/s}^2)t^2] \hat{j}\][/tex]
Find the Velocity Vector
The velocity vector [tex]\( \mathbf{v}(t) \)[/tex] is the time derivative of the position vector [tex]\( \mathbf{r}(t) \)[/tex]:
[tex]\[\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}\][/tex]
Differentiate each component with respect to time t :
For the [tex]\( \hat{i} \)[/tex] component:
[tex]\[v_x(t) = \frac{d}{dt}[2.0 + (5.00 \, \text{m/s})t] = 5.00 \, \text{m/s}\][/tex]
For the[tex]\( \hat{j} \)[/tex] component:
[tex]\[v_y(t) = \frac{d}{dt}[3.0 - (2.00 \, \text{m/s}^2)t^2] = -2.00 \times 2t = -4.00 \, t \, \text{m/s}\][/tex]
So the velocity vector is:
[tex]\[\mathbf{v}(t) = (5.00 \, \text{m/s}) \hat{i} + (-4.00 \, t) \hat{j}\][/tex]
Find the Acceleration Vector
The acceleration vector[tex]\( \mathbf{a}(t) \)[/tex] is the time derivative of the velocity vector [tex]\( \mathbf{v}(t) \)[/tex]:
[tex]\[\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt}\][/tex]
Differentiate each component with respect to time t:
For the [tex]\( \hat{i} \)[/tex]component:
[tex]\[a_x(t) = \frac{d}{dt}[5.00] = 0 \, \text{m/s}^2\][/tex]
For the [tex]\( \hat{j} \)[/tex] component:
[tex]\[a_y(t) = \frac{d}{dt}[-4.00 \, t] = -4.00 \, \text{m/s}^2\][/tex]
So the acceleration vector is:
[tex]\[\mathbf{a}(t) = 0 \hat{i} - 4.00 \hat{j}\][/tex]
Calculate the Magnitude of the Acceleration
The magnitude of the acceleration [tex]\( |\mathbf{a}(t)| \)[/tex] is:
[tex]\[|\mathbf{a}(t)| = \sqrt{a_x^2 + a_y^2}\][/tex]
Substitute[tex]\( a_x = 0 \) and \( a_y = -4.00 \)[/tex]:
[tex]\[|\mathbf{a}(t)| = \sqrt{0^2 + (-4.00)^2} = \sqrt{16.00} = 4.00 \, \text{m/s}^2\][/tex]
The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?
Answer:20 cal/s
Explanation:
Given
Heat transfer rate is [tex]\dot{Q}=10 cal/s[/tex]
Also heat rate is given by
[tex]\dot{Q}=kA\frac{dT}{dx}[/tex]
where [tex]k=thermal\ conductivity[/tex]
[tex]A=area\ of\ cross-section[/tex]
[tex]dT=change\ in\ temperature[/tex]
[tex]dx=change\ in\ length[/tex]
[tex]10=k\frac{\pi }{4}d^2\frac{dT}{L}----1[/tex]
For [tex]d'=2d, Length L'=2L[/tex]
[tex]\dot{Q}=k\frac{\pi }{4}(2d)^2\frac{dT}{2L}---2[/tex]
dividing 1 and 2 we get
[tex]\frac{10}{\dot{Q}}=\frac{2d^2}{4d^2}[/tex]
[tex]\dot{Q}=20 cal/s[/tex]
The power rate of the steel is mathematically given as
P=20cal/s
What rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?Question Parameter(s):
The rod conducts heat from one end to the other at a rate of 10 cal/s
Generally, the equation for the Power is mathematically given as
Power = (kA∆T)/l
Therefore
Power = [k × 4(πd^2)/4 × ∆T]/2l
Therefore
Power = 2(kA∆T)/l
Hence, Initial Power
iP = (kA∆T)/l
iP= 10cal/s
Where
dT is the same
2[(kA∆T)/l] = 2 × 10
P=20cal/s
In conclusion, The power rate is
P=20cal/s
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A wire connected to a 9 V battery records a current of 3.6 A in an ammeter. What is the resistance of the wire?
Answer:
R=2.5 ohm
Explanation:
Given that :
Voltage difference = 9 V
Current ,I= 3.6 A
As we know that from Ohm's law
V= I R
V =Voltage difference
I=Current
R=Resistance of the wire
Now by putting the values in the above equation we get
9 = 3.6 R
[tex]R=\dfrac{9}{3.6}[/tex]
R=2.5 ohm
Therefore the resistance of the wire will be 2.5 ohm.
Classify each of the following as a physical change or a chemical change. a) grinding glass: b) rusting of iron: c) burning a candle: d) sanding wood:
Answer:
Physical change: (a) grinding of glass, (c) burning of candle, (d) sanding wood
Chemical change: (b) rusting of iron.
Explanation:
Physical Change: This is the type of change a matter undergoes, it is easily reversible and no new substance is formed.
Chemical Change: Is one which is not easily reversible and a new substance is formed.
From the question above, the following changes can be classified into physical or chemical change
Physical change: (a) grinding of glass, (c) burning of candle, (d) sanding wood
Chemical change: (b) rusting of iron
Classification is as follows Physical change includes grinding of glass, burning of candles, and sanding wood. The chemical change includes rusting of iron.
Physical Change: This is the type of change a matter undergoes, it is easily reversible and no new substance is formed.
Chemical Change: This is one which is not easily reversible and a new substance is formed.
Therefore, the Classification is as follows Physical change includes grinding of glass, burning of candles, and sanding wood. The chemical change includes rusting of iron.
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When a car turns a corner on a level road, which force provides the necessary centripetal acceleration?
Answer:
The frictional force between the tire made with the road
Explanation:
This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.
The force providing the necessary centripetal acceleration when a car turns a corner on a level road is the force of friction between the tires and the road. This force points towards the center of the curve, counteracting the centrifugal force and facilitating the turn.
Explanation:When a car turns a corner on a level road, the force that provides the necessary centripetal acceleration is the force of friction between the car's tires and the road. This frictional force, which points towards the center of the curved path, prevents the car from moving in a larger-radius curve and leaving the road. The sharper the curve and the greater the car's speed, the more substantial this centripetal force becomes. For instance, your car pushes you toward the center of the turn when turning at high speeds or in sharp curves. This friction also counteracts the centrifugal force experienced by you which appears as if you are being pushed away from the center of the turn.
The force of friction is calculated using the equation f = Fc = μ₂N, where μ₂ is the coefficient of friction and N is the normal force. The normal force equals the car's weight on level ground (N = mg), and therefore, in such a scenario, the centripetal force is Fc = f = μ₂mg.
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Suppose an object is moving in a straight line at 50 mi/hr. According to Newton's first law of motion, the object will:
Answer:
The object will continue to move in straight line at 50 mi/hr except it is acted upon by an external force
Explanation:
Newton First Law of motion: A body will continue in its present state of rest or, if it is in motion will continue to move with uniform speed in a straight line unless it is acted upon by a force. Newton first law of motion is also called the law of inertia.
Inertia: This is the tendency of a body to remain in its states of rest or uniform motion.
From the question, The object moving in a straight line at 50 mi/hr will continue to move in a straight line at 50 mi/hr, except it is acted upon by external force that will change the speed of the object.
Final answer:
An object moving in a straight line at a constant speed will continue this state of motion until an external force is applied, according to Newton's first law.
Explanation:
According to Newton's first law of motion, an object moving at a constant speed in a straight line will continue to do so until a net external force acts upon it. In the case of an object moving at 50 mi/hr in a straight line, it will maintain this motion and speed as long as no forces such as friction, air resistance, or another applied force interfere with its movement. This principle indicates that without external forces, there is no change in velocity (speed and direction).
If the moon did not rotate at the same rate that it revolved, what would happen to the gravity of the earth?
Answer:
The gravity, which is an acceleration to the center of the earth, will be the same.
Explanation:
The gravity on earth depends only on the masses and distance, between two objects. We can see it in the gravitational force equation.
[tex]F=G\frac{m\cdot M}{r^{2}}[/tex]
Now if we put a man, with mass m, on the surface of the earth, with mass M, the distance from the center of mass and the man will be R (earth radius). Knowing that F = m*a, we can find the accelerations due to this mass M and this value will be 9.81 m/s².
On the other hand, the moon has a gravity value and is less than the earth, because its mass, and affects the water sea due to the gravitational force between earth and moon. If the moon changes the rate of its rotate it changes probably the distance between them, let's recall they must conserve angular momentum, but the gravity won't be affected.
Therefore, the gravity, which is an acceleration to the center of the earth, will be the same.
I hope it helps you!
What is the volume of water in 150ml of the 35% w/w of sucrose solution with a specific gravity of 1.115?
Answer:108.71 mL
Explanation:
Given
Volume of sample V=150 mL
concentration of sucrose solution 35 % w/w i.e. In 100 gm of sample 35 gm is sucrose
specific gravity =1.115
Density of solution [tex]\rho _s=1.115\times density\ of\ water[/tex]
Thus
[tex]\rho _s=1.15\times 1 gm/mL=1.115\ gm/mL[/tex]
mass of sample [tex]M=1.115\times 150=167.25\ gm[/tex]
mass of sucrose [tex]m_s=0.35\times 167.25=58.53\ gm[/tex]
mass of Water [tex]m_w=108.71 gm[/tex]
Volume of water [tex]=108.71\times 1=108.781 mL[/tex]
What force must be provided to accelerate a 64-lb object upward at a rate of 2 ft/s2? (Use g = 32 ft/s2.)
Answer:
The force that must be provided to accelerate the object = 302.196 N
Explanation:
Force: Force can be defined as the product of mass and acceleration. The S.I unit of force is Newton (N).
Ft - W = ma............ Equation 1
Ft = W + ma .......... Equation 2
Where Ft = force provided, W = weight of the object, m= mass of the object, a = acceleration of the object.
Given: mass = 64 lb, g = 32 ft/s² a = 2 ft/s²
Conversion: (i) from 64 pounds to Kg = 64/2.2046
= 29.03 kg
(ii) from 2 ft/s² to m/s² = 0.3048 × 2 = 0.6098 m/s²
(iii) from 32 ft/s² to m/s² = 0.3048×32 = 9.8 m/s².
and W = mg = 29.03 × 9.8 = 284.494 N,
Substituting these values into equation 2,
Ft = 284.494 + 29.03×0.6098
Ft = 284.494 + 17.7
Ft = 302.196 N
Therefore, the force that must be provided to accelerate the object = 302.196 N
To accelerate a 64-lb object upward at a rate of 2 ft/s², a force of approximately 3.974 lbf is required.
Explanation:To find the force required to accelerate a 64-lb object upward at a rate of 2 ft/s², we need to use Newton's second law of motion. The formula for force is F = m * a, where F is the force, m is the mass, and a is the acceleration.
First, convert the mass of the object from pounds to slugs. Since 1 slug is equal to the mass that accelerates at 1 ft/s² when acted upon by 1 lb of force, 1 slug is equal to 32.2 lbs. Therefore, the mass of the object is 64 lbs divided by 32.2 lbs/slugs, which is approximately 1.987 slugs.
Now, substitute the mass (m = 1.987 slugs) and the acceleration (a = 2 ft/s²) into the formula F = m * a. The force required to accelerate the object upward is approximately 3.974 lb-ft/s² or 3.974 lbf.
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If the distance between two asteroids is halved, the gravitational force they exert on each other will
a) Also be doubled
b) Be half as great
c) Be one fourth as great
d) Will be 1/16 as great
e) Be four times greater
Answer:
e) Be four times greater
Explanation:
Here we have to use Newton's gravitational law that relates the gravitational force between two objects with their masses ([tex]m_{1} [/tex] & [tex]m_{2} [/tex]) and the distance between them ([tex]r[/tex]) in the next way:
[tex] F=G\frac{m_{1}m_{2}}{r^{2}}[/tex] (2)
Now if distance between asteroids is halved:
[tex]F_{2}=G\frac{m_{1}m_{2}}{(\frac{r}{2})^{2}} [/tex]
[tex]F_{2}=G\frac{m_{1}m_{2}}{\frac{r^{2}}{4}} [/tex]
[tex]F_{2}=G\frac{4m_{1}m_{2}}{r^{2}}=4G\frac{m_{1}m_{2}}{r^{2}} [/tex]
Note that [tex] G\frac{m_{1}m_{2}}{r^{2}} [/tex] because (1) is F so:
[tex]F_{2}=4F [/tex]
It's four times greater!
Final answer:
The gravitational force between two asteroids will be four times greater if the distance between them is halved, due to the inverse square law of gravitation.
Explanation:
When considering gravitational force between two objects, such as asteroids, we refer to Newton's law of universal gravitation, which states that the force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between their centers.
When the distance between two asteroids is halved, the gravitational force they exert on each other is not simply doubled or halved; it changes more drastically because the force is inversely proportional to the square of the distance between them. In this case, the force becomes stronger by a factor of the square of the inverse of the change in distance. Halving the distance increases the force by a factor of four, since (1/2)^2 = 1/4, and inversely, the force is four times greater.
The correct answer to the given question is that if the distance between two asteroids is halved, the gravitational force they exert on each other will be four times greater. Therefore, the correct option is:
e) Be four times greategreaterA spring that is stretched 23 cm from its equilibrium point experiences a force of 103 N .Now much force is stored in the spring?
Answer
given,
spring is stretched = 23 cm
x = 0.23 m
Force experiences = 103 N
we know,
F = k x
where k is the spring constant
[tex]k =\dfrac{F}{x}[/tex]
[tex]k =\dfrac{103}{0.23}[/tex]
k = 447.83 N/m
Assuming we have to calculate the energy stored in the spring
energy stored in the spring
[tex]U =\dfrac{1}{2}kx^2[/tex]
[tex]U =\dfrac{1}{2}\times 447.83 \times 0.23^2[/tex]
U = 11.85 J
hence, energy stored in the spring is equal to U = 11.85 J
Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 27.0°below the horizontal. If it strikes the ground 59.0 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)(a) the time of flight.(b) the inital.(c) the speed and angle of the velocity vector with respect to the horizontal at impact.
a) The time of flight is 3.78 s
b) The initial speed is 17.0 m/s
c) The speed at impact is 46.4 m/s at [tex]70.3^{\circ}[/tex] below the horizontal
Explanation:
The picture of the previous problem (and some data) is missing: find it in attachment.
a)
The motion of the ball is a projectile motion, which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction
We start by considering the vertical motion, to find the time of flight of the ball. We do it by using the following suvat equation: for the y-displacement:
[tex]y=u_y t+\frac{1}{2}at^2[/tex]
where we have:
y = -45.0 m is the vertical displacement of the ball (the height of the building)
[tex]u_y=u sin \theta[/tex] is the initial vertical velocity, with u being the initial velocity (unknown) and [tex]\theta=-27.0^{\circ}[/tex] the angle of projection
t is the time of the fall
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity
Along the x-direction, the equation of motion is instead
[tex]x=(u cos \theta)t[/tex]
where [tex]ucos \theta[/tex] is the horizontal component of the velocity. Rewriting this equation as
[tex]t=\frac{x}{ucos \theta}[/tex]
And substituting into the previous equation, we get
[tex]y=xtan \theta + \frac{1}{2}gt^2[/tex]
And using the fact that the horizontal range is
x = 59.0 m
And solving for t, we find the time of flight:
[tex]t=\sqrt{\frac{y-x tan \theta}{g}}=\sqrt{\frac{-45-(59.0)(tan(-27^{\circ}))}{-9.8}}=3.78 s[/tex]
b)
We can now find the initial speed, u, by using the equation of motion along the x-direction
[tex]x=u cos \theta t[/tex]
where we know that:
x = 59.0 m is the horizontal range
[tex]\theta=-23^{\circ}[/tex] is the angle of projection
[tex]t=3.78 s[/tex] is the time of flight
Solving for u, we find the initial speed:
[tex]u=\frac{x}{cos \theta t}=\frac{59.0}{(cos (-23^{\circ}))(3.78)}=17.0 m/s[/tex]
c)
First of all, we notice that the horizontal component of the velocity remains constant during the motion, and it is equal to
[tex]v_x = u cos \theta = (17.0)(cos (-23^{\circ})=15.6 m/s[/tex]
The vertical velocity instead changes according to the equation
[tex]v_y = u sin \theta + gt[/tex]
Substituting all the values and t = 3.78 s, the time of flight, we find the vertical velocity at the time of impact:
[tex]v_y = (17.0)(sin (-23^{\circ}))+(-9.8)(3.78)=-43.7 m/s[/tex]
Where the negative sign means it is downward.
Therefore, the speed at impact is
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(15.6)^2+(-43.7)^2}=46.4 m/s[/tex]
while the direction is given by
[tex]\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-43.7}{15.6})=-70.3^{\circ}[/tex]
So, [tex]70.3^{\circ}[/tex] below the horizontal.
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How can a surface be smooth/polished for some waves and not for others? (This is an important question, refer to the homework on WebAssign and question regarding whether a surface is smooth/polished for some electromagnetic waves verses others)
Answer:
Depending upon the size of the irregularities on the surface and the wavelength of the surface a surface can be smooth for the waves of smaller wavelengths than the size of the irregularities.
Explanation:
Depending upon the size of the irregularities on the surface and the wavelength of the surface a surface can be smooth for some of the waves but not for the others.
More specifically, when a surface has irregularities of the size greater than or equal to the wavelength of a wave then the surface is rough for that particular wave and causes irregular reflection in the incident light.Contrarily, when the surface has the irregularities less than the size of the wavelength then it reflects most of the light in a uniform manner.
At what part of the cardiac conduction system does the electrical impulse travel most rapidly?
Answer:
Purkinje Fiber
Explanation:
In normal circumstances, the SA(sinoatrial) node the heart natural pacemaker, produces electrical activity automatically. This electrical impulse is transmitted across the whole right atrium and into the bundle of Bachmann to the left atrium, arousing the atria's myocardium to contract. And the part through which the electrical impulse travel most rapidly is called Purkrnje fibers, which is a specialized conducting fiber consisting of electrically excitable cells and these conduct cardiac action potentials more quickly and efficiently than any other cells in the heart.
What will the temperature of the cosmic microwave background be when the average distances between galaxies are 1.7 times as large as they are today? (Hint: The peak wavelength of photons in the background will then also be 1.7 times as large as it is today.)
Answer:
[tex]T'=1.58K[/tex]
Explanation:
As the Universe expands, the photons of the cosmic microwave background increase its wavelength, making its temperature inversely proportional to the scale factor of the Universe. That is, as the average distances between galaxies increase, the temperature of the cosmic microwave background decreases by the same factor. Therefore, the temperature when the distances between galaxies are 1.7 times as large as they are today will be:
[tex]T'=\frac{T_{now}}{1.73}\\T'=\frac{2.725K}{1.73}\\T'=1.58K[/tex]
When the average distances between galaxies are 1.7 times their present distances, the temperature of the cosmic microwave background (CMB), as a result of Wien's Law, will be about 1.61 Kelvin, which is 1.7 times lower than its current close to 2.73 Kelvin figure.
Explanation:The cosmic microwave background radiation (CMBR), an aftermath of the Big Bang event, exhibits a relationship between wavelength and temperature called Wien's Law. At present, the temperature of the CMBR is approximately 2.73 Kelvin (K), a finding made possible by scientists Penzias and Wilson who associated radiation intensity at different wavelengths to a temperature of roughly 3.5 K.
As the galaxies continue to recede from one another, their emitted radiation wavelengths become redshifted, effectively stretching them out. This movement is also reflected in the increase in the peak wavelength of the CMBR. If the average distances between galaxies were to magnify by 1.7 times, the peak wavelength of the CMBR would equally elongate by the same factor.
Applying Wien's Law, this increase will consequently result in a decrease in the CMB's temperature. Thus, when galaxies have expanded such that their average distances are 1.7 times larger than their current distances, the temperature of the cosmic microwave background will be 1.7 times lower than its current temperature of approximately 2.73 K. Hence, the new temperature would approximately be 1.61 K.
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A 1.6-cm-thick wood floor covers a 4.0m×5.5m room. Part A The subfloor on which the flooring sits is at a temperature of 16.1 ∘C, while the air in the room is at 19.5 ∘C. What is the rate of heat conduction through the floor? Use 0.200 W/m⋅K for the thermal conductivity of wood.
Answer:
Rate of heat conduction will be 935 Watt
Explanation:
We have given thickness of the floor d = 1.6 cm = 0.016 m
Area [tex]A=4\times 5.5=22m^2[/tex]
Temperature difference [tex]\Delta T=19.5-16.1=3.4^{\circ}C[/tex]
Thermal conductivity of the wood k = 0.200 w/M-K
We have to find heat conduction
Heat conduction is given by [tex]\frac{Q}{t}=\frac{kA\Delta T}{d}=\frac{0.200\times 22\times 3.4}{0.016}=935Watt[/tex]
Rate of heat conduction will be 935 Watt
A person is in a closed room (a racquetball court) with ????=453m3 hitting a ball (m=42.0g) around at random without any pauses. The average kinetic energy of the ball is 2.30 J. (a) What is the average value of ????2x? Does it matter which direction you take to be x? (b) Applying the methods of this chapter, find the average pressure on the walls? (c) Aside from the presence of only one "molecule" in this problem, what is the main assumption in Pressure, Temperature, and RMS Speed that does not apply here?
Answer:
(a) [tex]v^{2} _{x} = 36.508 m^2/s^2[/tex]
(b) 0.1235 Pa
(c) The volume of the ball is significant when compared with the volume of the closed room. Thus, the constant 'b' of the equation of Van der Waal is greater than zero.
Explanation:
In the question, we are given the following variables:
V = 453 m^3
[tex]m_{ball} = m = 42.0 g[/tex]
Kinetic energy (K.E) = 2.30 J
(a) The equation for kinetic energy is:
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
Since the average velocity components of the ball in the three dimensions are equal. Thus, we have:
[tex]v_{x} ^{2} = v^{2} _{y} =v^{2} _{z}[/tex]
In addition:
[tex]v^{2} =[/tex][tex]v_{x} ^{2} + v^{2} _{y} +v^{2} _{z}[/tex] = [tex]3v^{2} _{x}[/tex]
Therefore:
[tex]K.E = \frac{1}{2}m[3v^{2} _{x}][/tex] = [tex]\frac{3}{2}mv^{2} _{x}[/tex]
Thus:
[tex]v^{2} _{x} = \frac{KE}{1.5*m} = \frac{2.3}{1.5*0.042} = 36.508 m^2/s^2[/tex]
It does not matter which direction we take as x-direction because the average velocity component in every direction is the same.
(b) The average pressure on the walls can be calculated using the equation below:
[tex]P =\frac{mv^{2} _{x} }{V}[/tex]
where V = 453 m^3
Thus: P = (0.042*36.508^2)/453 = 0.1235 Pa
(c) The volume of the ball is significant when compared with the volume of the closed room. Thus, the constant 'b' of the equation of van der Waal is greater than zero.
The average value of u² is 54.8 m²/s². It does not matter which direction is taken as x. The average pressure on the walls can be found using the force and area equations.
Explanation:(a) The average value of u² can be found by dividing the average kinetic energy by the mass of the ball: u² = KE / m. In this case, u² = 2.30 J / 0.042 kg = 54.8 m²/s².
It does not matter which direction you take to be x. The average value of u² will be the same regardless of the direction.
(b) The average pressure on the walls can be found using the equation P = F / A, where P is the pressure, F is the force, and A is the area of the walls. The force exerted on the walls can be found using the equation F = ∆p / ∆t, where ∆p is the change in momentum and ∆t is the change in time. Since there are no pauses in the ball's motion, ∆p = m * ∆v, where m is the mass of the ball and ∆v is the change in velocity. The average pressure will be the force divided by the area of the walls.
(c) The main assumption in Pressure, Temperature, and RMS Speed that does not apply here is that there are a large number of gas molecules moving in random directions. In this problem, there is only one molecule, so the assumption of a large number of molecules does not hold.
If Air Traffic Control advises that radar service is terminated when the pilot is departing Class C airspace, the transponder should be set to code?
Explanation:
1200 feet.
In the given scenario we would be operation under VFR.
VFR is visual flight rules are set of regulations under which the pilot operates aircraft in a weather which generally clear enough that facilitates pilot to see the flight where it is going.