Answer:
The depth of the crater is 20[m]
Explanation:
We can solve this problem using the kinematics equations, it should be noted that the negative sign of acceleration means that the drop of the object is down.
Data:
a = 1.6[m/s^2]
t = 5 [s]
[tex]y=y_{0} +v_{0} *t-\frac{1}{2} *g*(t)^{2}[/tex]
v0= 0, because there is not initial veocity
y = final distance = 0 the bottom of the crater
g = a = acceleration of the crater [m/s^2]
y0= Point where the ball was dropped. = Distance from the original point to the bottom of the crater.
[tex]y_{0} =\frac{1}{2}*1.6*(5)^{2}\\ y_{0} = 20 [m][/tex]
A machine part is undergoing SHM with a frequency of 3.20 Hz and amplitude 1.80 cm. How long does it take the part to gofrom x=0 to x= -1.80 cm?
The time it takes for the machine part to move from x = 0 to x = -1.80 cm is approximately 0.156 seconds.
To find out how long it takes for a machine part undergoing simple harmonic motion (SHM) to go from position x = 0 to x = -1.80 cm, we need to follow these steps:
The frequency (f) of the oscillation is 3.20 Hz. The amplitude (A) is given as 1.80 cm (or 0.018 m).Calculate the Period (T):
The period of oscillation (T) is the time taken to complete one full cycle. The period is calculated using the formula:
[tex]T = \frac{1}{f}[/tex]
Substituting the frequency:
[tex]T = \frac{1}{3.20} \approx 0.3125 \text{ seconds}[/tex]
Find the Phase for SHM:
For simple harmonic motion, the position as a function of time (x(t)) can be expressed as:
[tex]x(t) = A \cos(\omega t)[/tex]
where [tex]\omega[/tex] (angular frequency) is given by:
[tex]\omega = 2\pi f[/tex]
Therefore:
[tex]\omega = 2\pi \times 3.20 \approx 20.1 \text{ radians/second}[/tex]
Calculate the Time to Reach -1.80 cm:
Since the part starts at x = 0, we want to find the time it takes to reach x = -1.80 cm (or -0.018 m). Using the cosine function, we can write:
[tex]-0.018 = 0.018 \cos(20.1 t)[/tex]
This simplifies to:
[tex]\cos(20.1 t) = -1[/tex]
Determine the Time from Cosine Value:
The cosine function equals -1 at odd multiples of [tex]\pi[/tex]:
[tex]20.1 t = (2n + 1)\pi[/tex]
The first instance (n=0) is:
[tex]20.1 t = \pi \implies t = \frac{\pi}{20.1} \approx 0.156 \text{ seconds}[/tex]
A student attaches a rope to a 20.0 kg box of books. He pulls with a force of 90.0 N at an angle of 30 degrees with the horizontal. To make his job easier, he oiled the bottom of the box, reducing the coefficient of friction to 0.250. What is the magnitude of the acceleration of the box?
Answer:
The magnitude of the acceleration of the box is 2.01 m/s².
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem.
We have the following horizontal forces:
Fr = friction force.
Fx = Horizontal component of the applied force, F.
And we have the following vertical forces:
Fy = vertical component of the applied force.
N = normal force exerted on the box.
W = weight of the box.
According to Newton´s second law:
∑F = m · a
Then, in the horizontal direction:
Fx - Fr = m · a
Where "m" is the mass of the box and "a" its acceleration.
Fx can be obtained by trigonometry (see figure):
Fx = F · cos 30°
Fx = 90.0 N · cos 30°
Fr is calculated as follows:
Fr = μ · N
Where μ is the coefficient of friction and N the normal force.
So, we have to find the magnitude of the normal force.
Using Newton´s second law in the vertical direction:
∑F = N + Fy - W = m · a
Notice that the box has no vertical acceleration, then:
N + Fy - W = 0
Solving for N:
N = W - Fy
The weight is calculated as follows:
W = m · g
Where g is the acceleration due to gravity:
W = 20.0 kg · 9.8 m/s² = 196 N
And the vertical component of the applied force can be obtained by trigonometry:
Fy = F · sin 30°
Fy = 90.0 N · sin 30°
The normal force will be:
N = W - Fy = 196 N - 90.0 N · sin 30°
N = 151 N
Now, we can calculate the friction force:
Fr = μ · N
Fr = 0.250 · 151 N
Fr = 37.8 N
And now, we can obtain the acceleration of the box:
Fx - Fr = m · a
(Fx - Fr) / m = a
(90.0 N · cos 30° - 37.8 N ) / 20.0 kg = a
a = 2.01 m/s²
The magnitude of the acceleration of the box is 2.01 m/s².
The resistance (R) of a copper wire varies directly as its length (L). Write this relation as a formula using k as the constant of variation.
Answer:
Explanation:
According to ohm's law current flowing in a conductor is directly proportional to the voltage applied across two end of conductor.
i.e. [tex]V\propto R[/tex]
[tex]V=R I[/tex]
where R=resistance
[tex]R\propto L[/tex]
[tex]R\propto \frac{1}{d^2}[/tex]
whee L and d are length and Diameter
thus [tex]R=k \frac{L}{d^2}[/tex]
where k=constant of Variation
Please help, it’s much needed
Answer:
Displacement = 72m
Acceleration = 4 [tex]m/s^{2}[/tex]
Explanation:
The jet ski increases its speed from zero to 24 m/s in 6 seconds.
We are supposed to find the displacement and its acceleration rate.
Displacement is the distance it covers in this time interval.
Acceleration is the rate of increase of speed.
Acceleration = [tex]\frac{change in velocity}{time}[/tex]
Acceleration = [tex]\frac{24 - 0}{6}[/tex]
Acceleration = 4 [tex]m/s^{2}[/tex]
According to equation of motion ,
[tex]v^{2}-u^{2} = 2\times a\times s[/tex]
[tex]24^{2} - 0 = 2\times 4\times s[/tex]
s = [tex]\frac{576}{8}[/tex]
Displacement , s = 72 m
The two blocks in oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slip when the amplitude is increased to 40 cm. What is the coefficient of static friction between the two blocks?
Answer:
0.72
Explanation:
[tex]T[/tex] = Time period of oscillation = 1.5 s
Angular frequency is given as
[tex]w = \frac{2\pi }{T}\\w = \frac{2(3.14) }{1.5}\\w = 4.2 rad/s[/tex]
[tex]A[/tex] = Amplitude of oscillation = 40 cm = 0.40 m
[tex]\mu[/tex] = Coefficient of static friction = ?
[tex]a[/tex] = acceleration of the block
[tex]m[/tex] = mass of the block
Maximum acceleration of the block is given as
[tex]a = Aw^{2}[/tex]
frictional force is given as
[tex]f = \mu mg[/tex]
As per newton's second law
[tex]f = ma \\\\\mu mg = ma \\\mu g = a\\\mu g = Aw^{2}\\\mu (9.8) = (0.40)(4.2)^{2}\\\mu = 0.72[/tex]
The coefficient of static friction between the two blocks is 0.72.
How to calculate the coefficient?From the information given, the angular frequency will be:
= 2π/T
= 2(3.14)/1.5
= 4.2rad/s
The maximum acceleration of the block is Aw² and the frictional force is umg. In this case,
umg = ma
u(9.8) = 0.40 × 4.2².
u = (0.40 × 4.2²) / 9.8
u = 0.72
Therefore, the coefficient of static friction between the two blocks is 0.72.
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The three stages of a train route took 1 hour ,2 hours ,and 4 hours . The first two stages were 80km and 200km of the train average speed over the course of the trip was 100km/h how many kilometers long was the third stage
Answer:
the third stage was 480 km long
Explanation:
Stage 1:
Time = 1 hours
Speed = 80km
Stage 2:
Time = 2 hours
Speed = 200km
Stage 3:
Time = 4 hours
Let the Distance at the stage 3 be x
Average speed of the train route = 100 km/h
So
[tex]\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 0[/tex]
[tex]\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 100[/tex]
Lets find the speed at stage 1
Speed = [tex]\frac{Distance }{Time}[/tex]
Speed = [tex]\frac{80}{1}[/tex]
Speed 1= 80 km/hr
The speed at stage 2
Speed = [tex]\frac{Distance }{Time}[/tex]
Speed = [tex]\frac{200}{2}[/tex]
Speed 2 = 100 km/hr
The speed at stage 3
Speed = [tex]\frac{Distance }{Time}[/tex]
Speed = [tex]\frac{x}{4}[/tex]
Speed 3 = [tex]\frac{x}{4}[/tex]
we kow that average is ,
[tex]\frac{ \text{speed 1} + \text{speed 2} + \text{speed 3}}{3} = 100[/tex]
[tex]\frac{ 80 + 100+ \frac{x}{4} }{3} = 100[/tex]
[tex]\frac{ 180 + \frac{x}{4} }{3} = 100[/tex]
[tex]\frac{ \frac{720 +x}{4} }{3} = 100[/tex]
[tex]\frac{720 +x}{4} \times \frac{1}{3} = 100[/tex]
[tex]\frac{720 +x}{12} = 100[/tex]
[tex]720 +x = 100 \times 12[/tex]
[tex]720 +x = 1200[/tex]
[tex]x = 1200- 720[/tex]
x = 480
To calculate the length of the third stage of the train's journey, we use the average speed over the total trip. Summing the known distances of the first two stages and subtracting this from the total distance traveled, which is 700 km, gives us the length of the third stage as 420 km.
Explanation:To find the length of the third stage in a train's journey when the average speed over the entire trip is known, we can use the formula for average speed, which is the total distance traveled divided by the total time taken. According to the given problem, the first two stages took 1 and 2 hours and were 80 km and 200 km long, respectively. Adding the times gives us 3 hours for the first two stages.
The third stage took 4 hours, making the total time for the full journey 7 hours (3 hours + 4 hours). Since the average speed of the train over the course of the trip was 100 km/h, we multiply this by the total time to find the total distance. Thus, the total distance is 700 km (100 km/h * 7 h).
To find the distance covered in the third stage, we subtract the sum of the first two stages (80 km + 200 km) from the total distance. The length of the third stage is 700 km - 280 km = 420 km.
Do molecules crossing a membrane using active transport with or against the concentration gradient?
Answer:
Against the concentration gradient.
Explanation:
Active transport implies the movement of molecules across a membrane from a lower concentration to a higher concentration, that is to say, against the concentration gradient. The molecules requires energy to go against the concentration gradient, which is usually coming from ATP or an electrochemical gradient.
In passive transport, the molecules move across a membrane down a concentration gradient without the need for energy.
Have a nice day!
Answer:
against
Explanation:
because it cannot go with the membrane. hope this helps you.
lood flows through a section of a horizontal artery that is partially blocked by a deposit along the artery wall. As a hemoglobin molecule moves from the narrow region into the wider region, its speed changes from v2 = 0.800 m/s to v1 = 0.475 m/s. What is the change in pressure, P1 - P2, that it experiences? The density of blood is 1060 kg/m3.
Answer:
[tex]P_1 - P_2 =219.62\ Pa[/tex]
Explanation:
given,
density of blood = 1060 kg/m³
v₂ = 0.800 m/s
v₁ = 0.475 m/s
change in pressure calculation
using Bernoulli's equation
[tex]P_1 +\dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2[/tex]
both are at same level
[tex]P_1 - P_2 =\dfrac{1}{2}\rho (v_2^2- v_1^2)[/tex]
[tex]P_1 - P_2 =\dfrac{1}{2}1060\times (0.8^2- 0.475^2)[/tex]
[tex]P_1 - P_2 =219.62\ Pa[/tex]
the change in pressure is equal to 219.62 Pa
Please help me I need the answer before tomorrow and Please explain how you got your answer!!! While playing basketball in PE class, 67 kg Johan lost his balance after making a lay-up at 5.4 m/s and collided with the padded wall behind the basket and came to rest in 0.24 seconds. What force acted on his body?
A: -1300
B: -1500
C: -1700
D: -1900
Answer:
Option B
-1500 N
Explanation:
We know that force, F=ma where m is the mass of Johan and a is the acceleration
Since acceleration, [tex]a=\frac {v-u}{t}[/tex] where v and u are final and initial velocities, t is time taken. Considering that he comes to rest, v=0 and u is given as 5.4 m/s, t=0.24 s.
Therefore
[tex]F=ma=m\frac {v-u}{t}=67\times \frac {0-5.4}{0.24}=-1507.5\approx -1500 N[/tex]
Voltage is shared by components that are __________. A. Together B. In parallel C. In series D. In perpendicular
Voltage is shared by components that are in series, therefore the correct answer is option C
What is resistance?Resistance is the obstruction of electrons in an electrically conducting material.
The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.
V=IR
In a series circuit, the components share the supply voltage.
The voltage of the supply is equal to the sum of the voltages across components connected in series.
Each component in a series has a voltage across it that is proportional to the resistance of that component. This implies that the supply voltage is divided equally across two identical components connected in series.
For example, if a 12-volt battery is supplying current in an electrical circuit for two series resistors that have equal resistance of 3 Ω then the voltage drop across them would be 6 volts each.
Thus, The voltage is shared by components that are in series, therefore the correct answer is option C
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5. Combustion gases enter a gas turbine at 627 C and 1.2 MPa at a rate of 2.5 kg/s and leave at 527 C and 500 kPa. It is estimated that heat is lost from the turbine at a rate of 20 kW. Using air properties for the combustion gases and assuming the surroundings to be at 25 C and 100 kPa, determine
This Question is incomplete.The complete question is
Combustion gases enter a gas turbine at 627∘C and 1.2 MPa at a rate of 2.5 kg/s and leave at 527∘C and 500 kPa. It is estimated that heat is lost from the turbine at a rate of 20 kW. Using air properties for the combustion gases and assuming the surroundings to be at 25∘C and 100 kPa, determine (a) the actual and reversible power outputs of the turbine. (b) the exergy destroyed within the turbine, and (c) the second-law efficiency of the turbine.
Answer:
(a) W=257.5kW Wrev=367.3kW
(b) X=109.8kW
(c) n=0.7
Explanation:
For Part(a)
Actual power output is determined from the energy balance. The final and initial enthalpy are taken from A-17 for the given temperature
mh₁=W+Q+mh₂
W=m(h₁-h₂)-Q
W=2.5(939.93-821.95)-20
W=257.5kW
The reversible power output is determined from the rate of energy destruction
Wrev=W+X
Wrev=W+T₀(m(s₂-s₁-Rln(p₂/p₁)+Q/T₀)
Wrev=257.5+298( 2.5 (2.71787-2.84856-0.287*ln(500/1200)+20/298)
Wrev=367.3kW
For part(b)
X=Wrev-W
X=367.3-257.5
X=109.8kW
For part (c)
Second-law efficiency is determined from the ratio of the actual and reversible power output:
n=W/Wrev
n=257.5/367.3
n=0.7
A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)Assume that the tension of the string is constant and equal to W. (a) How much time does it take a pulse to travel the fulllength of the string? (b) What is the weight W? (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling down the string?
Answer:
Explanation:
mass of string = .0125 / 9.8
= 1.275 x 10⁻³ kg
Length of string l = 1.5 m .
m = mass per unit length
= ( .1.275 / 1.5) x 10⁻³ kg/m
m = .85 x 10⁻³ kg/m
wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)
compare with equation of wave
y(x,t) = Acos(K x − ω t)
ω ( angular velocity ) = 4830 rad/s
k = 172 rad/m
Velocity = ω / k
= 4830/172 m /s
= 28.08 m /s
velocity of wave = [tex]\sqrt{\frac{W}{m } }[/tex]
28.08 = [tex]\sqrt{\frac{W}{.85\times10^{-3} } }[/tex]
788.48 = W / .85 X 10⁻³
W = 670 x 10⁻³ N .
c ) wave length
wave length =2π / k
= 2 x 3.14 / 172
= .0365 m
no of wave lengths over whole length of string
= 1.5 / .0365
= 41
d )
equation for waves traveling down the string
= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)
For the string, which tied to the ceiling at its upper end support a weight has,
(a) Time does it take a pulse to travel the full length of the string is 0.0534 seconds.(b) The weight W is 0.67 N.(c) Number of wavelengths are on the string at any instant of time is 41.(d) The equation for waves traveling down the string is,[tex]\rm y(x,t) = (8.50 mm)cos(172x rad/m - 4830t rad/s \;)[/tex]
What is equation of motion of standing wave?The equation of the motion of standing wave can be given as,
[tex]\rm y(x,t) = Acos(kx-\omega t)[/tex]
Here, (k) is the wave number and ([tex]\omega[/tex]) is the angular velocity.
A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W.
(a) Time does it take a pulse to travel the full length of the string-When you pluck the string slightly, the waves traveling up the string obey the equation:
[tex]y(x,t) = (8.50 )\cos(172x -4830)[/tex]
Compare this equation with the general equation of the motion of standing wave, we get,
Angular velocity ([tex]\omega[/tex]) is 4830 rad/s and the wave number (k) is 172 rad/m.
Now the velocity is the ratio of angular velocity to the wave number. Thus, velocity is,
[tex]v=\dfrac{\omega}{k}\\v=\dfrac{4830}{172}\\v=28.08\rm m/s[/tex]
The length of the string is 1.50 m. Thus, time does it take a pulse to travel the full length of the string is,
[tex]t=\dfrac{1.05}{28.08}\\t=0.0534\rm s[/tex]
(b) The weight W-For the string, force can be given as,
[tex]F=\dfrac{v^2}{\mu}\\F=\dfrac{28.08^2}{0.85\times10^{-3}}\\F=0.67\rm N[/tex]
(c) Wavelengths are on the string at any instant of time-The wavelength of the wave is twice the ratio of pi to the wave number.Thus, the wavelength is,
[tex]\lambda=\dfrac{2\times \pi}{k}\\\lambda=\dfrac{2\times \pi}{172}\\\lambda=0.0365\rm m[/tex]
As the length of the string is 1.50 m. Thus, number of wavelengths are on the string at any instant of time is,
[tex]\lambda=\dfrac{1.5}{0.0365}\\\lambda=41[/tex]
(d) The equation for waves traveling down the string-The equation can be given as,
[tex]\rm y(x,t) = (8.50 mm)cos(172x rad/m - 4830t rad/s \;)[/tex]
Thus, for the string, which tied to the ceiling at its upper end support a weight has,
(a) Time does it take a pulse to travel the full length of the string is 0.0534 seconds.(b) The weight W is 0.67 N.(c) Number of wavelengths are on the string at any instant of time is 41.(d) The equation for waves traveling down the string is,[tex]\rm y(x,t) = (8.50 mm)cos(172x rad/m - 4830t rad/s \;)[/tex]
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why is 21-cm radiation so important to the study of interstellar matter and the galaxy
Answer:
Because its the most suitable radiation for that use
Explanation:
Astronomers use the 21 cm radiation for interstellar and cosmological research as this type of radiation is best suitable for the purpose its intended for. It can easily go through gas and dust molecules and other interstellar bodies that causes a blur vision of the content of space.
Which of the following is MOST useful to scientists in measuring the size of asteroids?
A) The speed at which the asteroid is traveling.
B) The location of the asteroid in the Solar System.
C) The gravitational effect when spacecraft flies close to the asteroid.
D) The composition of the asteroid.
Answer:c-The gravitational effect when spacecraft flies close to the asteriod
Explanation:
Gravitational effect on the spacecraft gives an estimate that how big is the asteroid by experiencing its gravitational pull.
The amount of extra thrust required to maintain the trajectory of the spacecraft during its motion hints at the scientist about the size of the asteroid.
Gravitational pull is directly proportional to the mass of object so greater the mass, greater will be the pull.
Answer:
C
Explanation:
The gravity of a planet, or in this case an asteroid, is determined by two factors: it's mass and its size. The bigger the mass, the stronger the gravity is. But the bigger the size of the mass is, the smaller the gravity is, because of the distance from the center of the mass. As you can see by measuring the gravitational effect when a spacecraft is close to an asteroid can help determine its size.
Help please!
A cat is chasing a mouse across a 1.3 m tall dining table. The mouse darts to the side and the cat accidentally slides off, landing clumsily 0.75 m away from the edge of the table. How long did it take the cat to hit the ground?
0.78 seconds
0.51 seconds
0.27 seconds
0.95 seconds
Using the information you solved for in problem 9, how fast was the cat moving when it fell off the table?
2.97 m/s
1.45 m/s
3.81 m/s
6.93 m/s
Answer:
The will hit the ground in t = 0.51[sg] and he was moving with a velocity of 1.45 [m/s]
Explanation:
In the attached image, we can see a sketch of the conditions of the problem, we are interested to know how long the cat takes to fall from the table to the ground and then using that time we can find the initial velocity of the cat when it arrives at the edge of the table.
The initial condition of this problem is that when the cat reaches the edge of the table, it will only have initial velocity on the X-axis component, its velocity on the y-axis component will be zero.
Therefore:
[tex]y=(v_{y})_{0}[/tex]
In this way, we can use the following equation of kinematics and find the time t.
[tex]y=(v_{y})_{0}*t - \frac{1}{2} * g*t\\where:\\g =9.81[m/s^{2} ] gravity\\y = 1.3 [m] table height[/tex] =0[/tex]
We note that gravity is taken as a negative quantity since the movement is downwards, likewise the value and the equation will have negative value as the reference point will be taken as the edge of the table, therefore, y = -1.3 [m]
[tex]-1.3 = (0)*t - \frac{1}{2} * (9.81)*t^{2} \\t = \sqrt{\frac{2*1.3}{9.81} }\\ t= 0.51 [m/s]\\[/tex]
Now using this time and replacing in the next kinematics equation we will have:
[tex]x=(v_{x} )_{0} *t\\where\\x=0.75 [m][/tex]
We will have:
[tex]x=(v_{x} )_{0} *t\\\\(v_{x} )_{0}=\frac{x}{t} \\(v_{x} )_{0}=\frac{0.75}{0.51}\\ (v_{x} )_{0}=1.45 [m/s][/tex]
earth's troposphere hydrosphere and lithosphere contain large amounts of which element?
Answer: the element is Oxygen.
Explanation: Earth's troposphere hydrosphere and lithosphere contain large amounts of Oxygen. it is the most abundant element in the Earth's crust. Oxygen makes up 467,100 ppm (parts per million) of the Earth's crust, or 46.6%.
Oxygen is the primary element found in Earth's troposphere, hydrosphere, and lithosphere, as part of the air, water (H2O), and silicate materials in many minerals and rocks.
Explanation:The Earth's troposphere, hydrosphere, and lithosphere all contain large amounts of the element oxygen. The troposphere, part of the atmosphere, contains oxygen in the air we breathe, while the hydrosphere, which encompasses all water on Earth, would have oxygen as a component of H2O. In the lithosphere, which covers the crust and the upper part of the mantle, oxygen is found in many minerals and rocks as part of silicate materials.
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A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules. The height (relative to the water) from which the diver dove was approximately ____ meters.
Answer:
h = 10 m
Explanation:
given,
mass of platform = 50 Kg
Kinetic energy = 5000 J
height from which the diver dove = ?
taking acceleration due to gravity = 10 m/s²
using conservation of energy
Kinetic energy is converted into mechanical energy
K.E = P.E
K.E = m g h
5000 = 50 x 10 x h
500 h = 5000
[tex]h = \dfrac{5000}{500}[/tex]
h = 10 m
The height from which the diver dove is equal to h = 10 m
The height from which the diver drove is mathematically given as
h=10m
What is the height from which the diver drove?Question Parameters:
A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules
Generally, the equation for the Kinetic energy is mathematically given as
K.E = P.E
Therefore
K.E = m g h
5000 = 50 x 10 x h
500 h = 5000
In conclusion,the height is
[tex]h = \frac{5000}{500}[/tex]
h=10m
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Through what vertical height is a 50 N object moved if 250 J of work is done lifting it against the gravitational field of Earth?
5m
Explanation:
Given parameters:
Weight of object = 50N
Work done in lifting object = 250J
Unknown:
Vertical height = ?
Solution:
The work done on an object is the force applied to lift a body in a specific direction.
Work done = force x distance
Weight is a force in the presence of gravity;
Work done = weight x height of lifting
Height of lifting = [tex]\frac{work done }{weight}[/tex]
Height of lifting = [tex]\frac{250}{50}[/tex] = 5m
The vertical height through which the object was lifted is 5m
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Final answer:
To calculate the vertical height that a 50 N object is moved with 250 J of work done against Earth's gravity, divide the work by the force. This gives a height of 5 meters.
Explanation:
To find the vertical height an object is moved, given the work done against the Earth's gravitational field, we use the formula for work: Work = Force × Distance, where the force in this context is the weight of the object due to gravity. Given the work done (250 J) and the weight of the object (50 N), we can rearrange the formula to solve for distance, which is the vertical height moved: Distance = Work / Force.
Distance = 250 J / 50 N = 5 meters.
Therefore, the object was moved through a vertical height of 5 meters. This calculation assumes that the work done is solely against the gravitational pull of the Earth, and it highlights the direct relationship between work, force, and distance in physical interactions.
Listed following are several objects in the solar system. Rank these objects from left to right based on their orbital period around the Sun from shortest to longest.
Question:
Listed following are several objects in the solar system. Rank these objects from left to right based on their orbital period around the Sun from shortest to longest.
A typical asteroid in the asteroid belt, a trojan asteroid, a typical kuiper belt object, a typical oort cloud object
Answer:
Ranking of the objects from left to right based on their orbital period around the sun from shortest to longest
a typical asteroid in the asteroid belta Trojan asteroida typical Kuiper belta typical Oort cloud objectExplanation:
Half the mass of the belt is present in the largest four asteroids, they are ceres, vesta, pallas and hygiea. The asteroid belt total mass is 4% than that of moon, 22% Pluto. The three dwarf planets are present in the Kuiper belt Pluto, Haumea and Makemake. The spherical layer of icy objects around the sun are Oort cloud. Objects in the Oort cloud are made up of water ice, ammonia, and methane. It is spherical from outer and and torous shaped from inside.
From the perspective of the cell receiving the message, the three stages of cell signaling are
A. the paracrine, local, and synaptic stages.
B. signal reception, cellular response, and cell division.
C. signal reception, nucleus disintegration, and new cell generation.
D. the alpha, beta, and gamma stages.
E. signal reception, signal transduction, and cellular response.
Answer:
E. signal reception, signal transduction, and cellular response.
Explanation:
General mechanisms of cellular communication : reception
Cellular communication can be established in different ways that include: a) paracrine communication: the signal acts on neighboring cells; b) endocrine: the signal travels through the bloodstream and reaches distant cells; c) autocrine: the signal reaches the same cell from which it came out; d) neurotransmission: the signal is released by the sending cell into the synaptic space, where it is picked up by the receiving cell; e) cell-cell contacts: the signal remains anchored to the membrane of the sending cell while interacting with the receiving cell and f) through gaps: the signal is diffused from the sending cell to the receiving cell.
Transmission of the signal inside the cell
When a ligand interacts with its membrane receptor, the signal is transmitted into the cell. A cascade of events is then triggered that includes the synthesis of second messengers and the phosphorylation of enzymes catalyzed by protein kinases.
The end of the pathway: the biological response or cellular response
Transduction of external signals produces a response from the white cell. This response is an alteration as a result of the activation or inhibition of some metabolic pathway, and of changes in the shape or movement of the cells. The role of these processes in the control of differentiation, proliferation and cell growth is essential for the normal development and functioning of the organism.
Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's modulus of the steel is Y=2.0×1011 pascals. How far ( ΔL) would such a string stretch under a tension of 1500 newtons? Use two significant figures in your answer. Express your answer in millimeters.
Answer:
[tex]\Delta l=0.015m[/tex]
Explanation:
We have given initial length of the steel guitar l = 1 m
Cross sectional area [tex]A=0.5mm^2=0.5\times 10^{-6}m^2[/tex]
Young's modulus [tex]\gamma=2\times 10^{11}Pa[/tex]
Force F = 1500 N
So stress [tex]=\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa[/tex]
We know that young's modulus [tex]=\frac{stress}{strain}[/tex]
So [tex]2\times 10^{11}=\frac{3\times 10^{9}}{strain}[/tex]
[tex]strain=1.5\times 10^{-2}=0.015m[/tex]
Now strain [tex]=\frac{\Delta l}{l}[/tex]
[tex]0.015=\frac{\Delta l}{1}[/tex]
[tex]\Delta l=0.015m[/tex]
Answer:
Explanation:
L = 1 m
A = 0.5 mm² = 0.5 x 10^-6 m²
Y = 2 x 10^11 Pa
F = 1500 N
ΔL = ?
Use the formula for the young's modulus
[tex]Y = \frac{FL}{A\Delta L}[/tex]
[tex]\Delta L = \frac{FL}{AY}[/tex]
[tex]\Delta L = \frac{1500\times 1}{0.5\times10^{-6}\times 2\times 10^{11}}[/tex]
ΔL = 0.015 m
ΔL = 0.02 m
Loss of traction between the rear wheels and road surfaces like ice, sand, or gravel results in what is called _______________.
Answer:
Fish tailing
Explanation:
Fish tailing is the vehicle handling problem which occurs due to the decrease in friction between the the rear wheels and the surface. So we mostly observe that rear wheels slide out of control when there is rain or ice covered roads, sand etc. They reduce the friction of the surface resulting in over-steering. This problem is named as fish tailing.
Fishtailing
Fishtailing is a vehicle handling issue that happens when the rear wheels lose traction and oversteer ensues. Low friction surfaces can contribute to this. Rear-drive vehicles is a term used to describe when a rear-drive vehicle with enough power loses traction on any terrain.Loss of traction between the rear wheels and road surfaces like ice, sand, or gravel results in what is called fishtailing.For more information:
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10. A chemical reaction was run in 215.5 mL of H2O. During the reaction, the temperature of the H2O changes from 34.4 oC to 57.6 oC. How much heat (in kJ) did the chemical reaction produce? For H2O, Cp= 4.184 J/(g · oC). Assume the mass of the solution is 215.5 g. Enter your answer as a positive decimal number.
Answer: 20.9 kJ
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed by water= ?
m= mass of water = 215.5 g
c = specific heat capacity = [tex]4.184J/g^0C[/tex]
Initial temperature of the water = [tex]T_i[/tex] = 34.4°C
Final temperature of the water = [tex]T_f[/tex] = 57.6°C
Change in temperature ,[tex]\Delta T=T_f-T_i=(57.6-34.4)^0C=23.2^0C[/tex]
Putting in the values, we get:
[tex]Q=215.5\times 4.184\times 23.2^0C[/tex]
[tex]Q=20918J=20.9kJ[/tex] (1kJ=1000J)
The heat absorbed by water is equal to the heat released during the reaction. Thus heat produced by chemical reaction is 20.9 kJ
The chemical reaction produced approximately 21.67 kJ of heat. This is calculated using the formula ΔQ = MCΔT, utilizing the given values for the mass of the solution, the specific heat capacity of water, and the change in temperature.
Explanation:To calculate the amount of heat produced in the chemical reaction, we use the formula ΔQ = MCΔT, where M is the mass of the solution, C is the specific heat capacity of water, and ΔT is the change in temperature. Given that M = 215.5 g, C = 4.184 J/(g·°C), and ΔT is the increase from 34.4°C to 57.6°C (which is 57.6°C - 34.4°C = 23.2°C), the formula becomes:
ΔQ = (215.5 g) x (4.184 J/g°C) x (23.2°C)
On calculating the above, the result comes out to be approximately 21669.8 Joules. Since 1 kJ = 1000 J, we need to convert 21669.8 Joules to kilojoules by dividing by 1000. Thus, the chemical reaction produces approximately 21.67 kJ of heat.
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If the mass of the earth and all objects on it were suddenly doubled, then the ________
Answer:
The force of universal gravitation between earth and all objects in it will be quadrupled
Explanation:
Newtons law of universal gravitation tell us the force of attraction between two bodies on the earth surface . This force is directly proportional to the product of the masses of the bodies and inversely proportional to their distance apart.
f is directly proportional to m1*m2
doubling these masses will be 2m1*2m2 = 4m1m2
if mas of earth is Me and mass of all objects is Mo
the f is proportional to Me*Mo
doubling the masses of the earth and all objects we have, 2Me*2Mo= 4MeMo
This means that doubling the masses of the earth and all objects on it will cause the force of gravitational attraction to be quadrupled.
A bathysphere used for deep-sea exploration has a radius of 1.50 m and a mass of 1.19 104 kg. In order to dive, this submarine takes on mass in the form of sea water. Determine the amount of mass that the submarine must take on if it is to descend at a constant speed of 1.40 m/s, when the resistive force on it is 1105 N in the upward direction. Take 1.03 103 kg/m3 as the density of seawater. answer in kg
The bathysphere cannot descend at a constant speed of [tex]\(1.40 \, \text{m/s}\)[/tex] with the given resistive force. If the resistive force were smaller, the bathysphere would need to take on approximately [tex]\(11741.33 \, \text{kg}\)[/tex] of seawater to descend at the desired speed
To find the mass of seawater that the submarine must take on in order to descend at a constant speed, we need to consider the forces acting on it.
1. The weight of the bathysphere, acting downward, is given by [tex]\( W = mg \)[/tex], where m is the mass of the bathysphere and g is the acceleration due to gravity.
2. The buoyant force acting on the bathysphere, which is equal to the weight of the water displaced by the bathysphere, is given by [tex]\( F_b = \rho V g \),[/tex] where [tex]\( \rho \)[/tex] is the density of seawater, [tex]\( V \)[/tex] is the volume of the bathysphere, and [tex]\( g \)[/tex] is the acceleration due to gravity.
3. The resistive force opposing the motion of the bathysphere, which is given as [tex]\( F_{\text{resistive}} = 1105 \, \text{N} \),[/tex] acts upward.
At constant speed, the net force on the bathysphere is zero. Therefore, the weight of the bathysphere must be balanced by the buoyant force and the resistive force. Mathematically, we can express this as:
[tex]\[ W = F_b + F_{\text{resistive}} \]\[ mg = \rho V g + F_{\text{resistive}} \][/tex]
We can rearrange this equation to solve for the mass of seawater [tex](\( m_{\text{seawater}} \))[/tex] that the bathysphere must take on:
[tex]\[ m_{\text{seawater}} = \frac{F_{\text{resistive}} - mg}{g} \][/tex]
Given:
- [tex]\( r = 1.50 \, \text{m} \)[/tex] (radius of the bathysphere),
- [tex]\( m = 1.19 \times 10^4 \, \text{kg} \)[/tex] (mass of the bathysphere),
- [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex] (acceleration due to gravity),
- [tex]\( F_{\text{resistive}} = 1105 \, \text{N} \)[/tex] (resistive force),
- [tex]\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \)[/tex] (density of seawater).
First, we need to calculate the volume of the bathysphere [tex](\( V \))[/tex]:
[tex]\[ V = \frac{4}{3} \pi r^3 \]\[ V = \frac{4}{3} \pi (1.50 \, \text{m})^3 \]\[ V = \frac{4}{3} \pi (3.375 \, \text{m}^3) \]\[ V \approx 14.137 \, \text{m}^3 \][/tex]
Now, let's calculate the mass of seawater [tex](\( m_{\text{seawater}} \))[/tex]:
[tex]\[ m_{\text{seawater}} = \frac{1105 \, \text{N} - (1.19 \times 10^4 \, \text{kg} \times 9.8 \, \text{m/s}^2)}{9.8 \, \text{m/s}^2} \]\[ m_{\text{seawater}} = \frac{1105 \, \text{N} - 116220 \, \text{N}}{9.8 \, \text{m/s}^2} \]\[ m_{\text{seawater}} = \frac{-115115 \, \text{N}}{9.8 \, \text{m/s}^2} \]\[ m_{\text{seawater}} \approx -11741.33 \, \text{kg} \][/tex]
Since mass cannot be negative, this result indicates that the resistive force is greater than the weight of the bathysphere and the buoyant force. Therefore, the bathysphere cannot descend at a constant speed of [tex]\(1.40 \, \text{m/s}\)[/tex] with the given resistive force. If the resistive force were smaller, the bathysphere would need to take on approximately[tex]\(11741.33 \, \text{kg}\)[/tex] of seawater to descend at the desired speed.
The theory of force generation proposed in the passage is best supported by which of the following observations about Amoeba locomotion?
Question:
The theory of force generation proposed is best supported by which of the following observations about Amoeba locomotion?
a. amoeboid movement stops upon expose to cytochalasins
b. amoeboid movement cannot occur if mitosis is blocked
c. moving amoeba cells produce more troponin than do stationary ones
d. the rate of movement is inversely proportional to the viscosity of the medium in the Amoeba moves
Answer:
The amoeba location can be supported by :The amoeboid movement stops upon expose to cytochalasins
Explanation:
The amoeboid movement common in eukaryotic cells. cytochalasins are the type of drug that inhibits the growth of the micro filaments, micro filaments capability to elongate are both implemented as important to make the force for movement in amoeba. further Cytochalasins are known as the fungal metabolites which can bind the actin filaments and also restrict polymerization and elongation of actin. It can also permeate cell membranes, preventing the cellular translocation and cause cells to enucleate.
Technician A says that knowing if there are any stored diagnostic trouble codes (DTCs) may be helpful when checking for related technical service bulletins (TSBs). Technician B says that only a factory scan tool should be used to retrieve DTCs. Which technician is correct?
Answer:
Technician A only
Explanation:
Technician A says that knowing if there are any stored diagnostic trouble codes (DTCs) may be helpful when checking for related technical service bulletins (TSBs). Technician B says that only a factory scan tool should be useWhich technician is correct?
DTC's, or Diagnostic Trouble Codes, are used by automobile manufacturers to detect problems related to the vehicle. ... The first number in the DTC indicates whether the code is an SAE generic code
tsb are recommended techniques for repairing a vehicle. the manufactures of vehicles make this available so the technicians can work and make reference to it
Technician A says that the connecting rods should be marked before disassembly. Technician B says that pistons are cam ground so that when operating temperature is reached, the piston will have expanded enough across the piston pin area to become oval shaped. Who is right?
A) Technician A only B) Technician B only
C) Both technicians A and B D) Neither technician A nor B
Answer:
C.
Explanation:
Both technicians A & B are right since the connecting rods must be marked to indicate which surface faces front. And a cam-ground piston machine must also be configured oval shape so that it will be able to fit the cylinder better throughout its operating temperature range.
Technician A is correct that the connecting rods should be marked before disassembly. Technician B is incorrect about the cam ground pistons. The correct answer is A) Technician A only.
Explanation:Technician A is correct that the connecting rods should be marked before disassembly. This is important in order to ensure that the rods are reinstalled in the same location and orientation as before, which helps maintain balance and proper engine function.
Technician B is incorrect. Pistons are not cam ground to become oval shaped when operating temperature is reached. Instead, pistons are typically designed with expansion clearances to account for thermal expansion, ensuring that they maintain proper shape and function at operating temperatures.
Therefore, the correct answer is A) Technician A only.
Label 1a lies on the rim of a large crater, and 1b lies on the rim of a smaller one. Which crater must have formed first?
Answer:
Label 1a
Explanation:
The craters are usually referred to as those large depressions that are created due to the impact by meteorites, asteroids or other astronomical bodies. These are of variable size and shape, which are controlled by factors such as the mass of the impact body, velocity at which the impact body strikes.
According to the rule of thumb, the geologically older surfaces are exposed to various types of impacting bodies such the meteoroids, asteroids, and comets, for a longer duration of time in comparison to the surfaces that are younger in age.
The bigger craters are formed first, and the smaller craters are formed later. Thus, the age of the craters is directly proportional to the size of the craters.
Hence, the larger crater (label 1a) was formed first.
The law of superposition implies that the larger crater, labeled as 1a, must've formed first, as new craters form on top of existing surfaces. Therefore, the rim of the larger crater existed prior to the formation of the smaller crater, 1b.
Explanation:In the study of craters, the principle that is usually applied is the Law of Superposition. This law states that in an undisturbed sequence, the oldest layers lie at the bottom and the youngest layers at the top. Applying this to your question: if a crater (labelled 1a) locates on the rim of a large crater, and another crater (labelled 1b) lies on the rim of a smaller one, the larger crater would have formed first. This is because any new crater that forms would be on top of, or superimposed on, the existing surface. Therefore, the rim of the larger crater must have been in existence for the smaller crater (1b) to have formed on it.
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Free monitoring tools are useful for:
a. calculating customer satisfaction KPI formulas.
b. qualitative measurements.
c. observing what people are saying about a brand.
d. evaluating the effectiveness of social media tactics.
e. measuring branding efforts.
Answer:
c.observing what people are saying about a brand
Explanation:
Monitoring tool systems are used to keep track of the system's condition constantly, with the goal of early warning against and strengthening errors, defects or issues. The computer, network, storage, health, results, website and Internet use and applications are controlled.
therefore, Free monitoring tools are useful for observing what people are saying about a brand.