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algol [13]
4 years ago
6

A block (mass 9.0 kg) is released from rest on a frictionless slope angled at 37° to the horizontal. Draw a diagram with labelle

d vectors for forces/acceleration. What is the block's acceleration?

Physics
1 answer:
maxonik [38]4 years ago
7 0

Answer:

Force diagram is attached here.

Acceleration of the mass down the inclined plane is independent of its mass.

ma = mg sin 37

or Acceleration = a = g sin 37 = 5.9 m/s/s.

Explanation:

The object sliding down the inclined plane has several forces acting on it.

Forces are labelled here as 1, 2, 3, 4.

Force 1 = Normal force that acts at right angles to the inclined surface.

Force 2 : Parallel component of the gravitational force

                = mg sin 37.

Force 3: Mass x Acceleration due to gravity = Weight

This is the gravity force that acts perpendicular to the  

floor or ground.  

Force 4 : Perpendicular component of the gravitational force  = mg cos 37.  

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8 0
3 years ago
The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

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.

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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:

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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
3 years ago
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Mars2501 [29]

Answer:

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given,

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magnitude of force = ?

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direction

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\theta = tan^{-1}(0.728)

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5 0
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