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Lera25 [3.4K]
3 years ago
15

Two identical stones, A and B, are thrown from a cliff from the same height and with the same initial speed. Stone A is thrown v

ertically upward, and stone B is thrown vertically downward. Which of the following statements best explains which stone has a larger speed just before it hits the ground, assuming no effects of air friction?
a. Both stones have the same speed; they have the same change in Ugand the same Ki
b. A, because it travels a longer path.
c. A, because it takes a longer time interval.
d. A, because it travels a longer path and takes a longer time interval.
e. B, because no work is done against gravity.
Physics
2 answers:
charle [14.2K]3 years ago
6 0

Answer:

b. A, because it travels a longer path.

Explanation:

  • If the stone A is thrown is thrown vertically upwards and another stone is dropped down directly from the same height above the ground then the stone A will hit the ground with a higher speed because it falls down from a greater height above the earth surface.

<u>This can be justified by the equation of motion given below:</u>

v^2=u^2+2\times a\times s

where:

v= final velocity

u= initial velocity

a= acceleration = g (here)

s= displacement of the body

  • Now we know that at the maximum height the speed of the object will be zero for a moment. So for both the stones A and B the initial  velocity is zero, stone B is also dropped from a height with initial velocity zero.
  • Acceleration due to gravity is same for the stones so the only deciding factor that remains is s, displacement of the stones. Since stone A is thrown upwards it will attain a greater height before falling down.
Taya2010 [7]3 years ago
5 0

Answer:

Option A

Explanation:

This can be explained based on the conservation of energy.

The total mechanical energy of the system remain constant in the absence of any external force. Also, the total mechanical energy of the system is the sum of the potential energy and the kinetic energy associated with the system.

In case of two stones thrown from a cliff one vertically downwards the other vertically upwards, the overall gravitational potential energy remain same for the two stones as the displacement of the stones is same.

Therefore the kinetic energy and hence the speed of the two stones should also be same in order for the mechanical energy to remain conserved.

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Explanation:

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It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used
stepladder [879]

Answer:

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

a_{c} = rω²

ω² = a_{c} / r

ω = √( a_{c} / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a a_{c}  = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²  

ω = 0.0500647 ≈ 0.05 rad/s  

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM  

1 rpm = 60 rph  

so  

ω = 0.478082 × 60

ω = 28.6849  revolutions per hour  

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7 0
3 years ago
The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
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Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
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