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finlep [7]
3 years ago
13

What are 3 ways a car can accelerate? (CRE)

Physics
2 answers:
Nookie1986 [14]3 years ago
6 0
Answer - There are three ways an object can accelerate: a change in velocity, a change in direction, or a change in both velocity and direction.
satela [25.4K]3 years ago
5 0

Answer:

change a velocity

Explanation:

kkfjfjdkdjcnfkdjxbcbdkgkg

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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.

\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
3 years ago
What is the difference between Is it resource that is limited and one that is not limited? Give an example of each
sveticcg [70]
Limited resources: resources that take a long time to replenish
Example: coal, oil, nuclear gas

Non- limited resource: resources that are constantly being replenished
Example: soil, wind, water

5 0
3 years ago
An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9∘ above the horizontal. part a determine x-v
ankoles [38]
Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal

Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s

The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s

The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.

Answer:

  t, s    x, m
------  --------
     0   0
     1   39.98
     2   79.79
     3   112.68
     4   159.58
     5   199.47
     6   239.37

5 0
3 years ago
Read 2 more answers
Hurry need an answer
notsponge [240]

Answer:

i think b

Explanation:

6 0
3 years ago
What is the impulse of a 2 kg object that starts at 10 m/s and comes to a stop over 0.4 seconds?
gizmo_the_mogwai [7]

Answer: Impulse = 20 Ns

Explanation:

Impulse is the product of force and time

Also impulse = momentum

Where momentum is the product of mass and velocity.

Given that

M = 2kg

V = 10 m/s

Impulse = MV = 2 × 10 = 20 Ns

8 0
2 years ago
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