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Schach [20]
3 years ago
7

A constant net force of 20 N is applied to a 32kg car, causing it to speed up from 4.0 to 9.0 m/s. How long is the force applied

Physics
1 answer:
pishuonlain [190]3 years ago
8 0

Answer:

8 seconds

Explanation:

From Newton's second law;

Ft = m(v-u)

F = Force applied

t = time taken

v = final velocity

u  = initial velocity

20 * t = 32 (9 - 4)

20t = 32 * 5

t = 32 * 5/ 20

t = 8 seconds

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dybincka [34]

Given,

A player kicks a soccer hits at an angle of 30° at a speed of 26 m/s

We can resolute the trajectory of soccer into horizontal and vertical components.(Please see the attached file)

We can have,

Horizontal velocity component of ball= 26cos(30°)  = 26×(√3÷2) = 22.51 m/s

And vertical velocity component of ball = 26sin(26°) = 26×(1÷2) = 13 m/s


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3 years ago
a pelican flying along a horizontal path drops a fish from a height of 5.4m. the fish travels 8.0m horizontally before it hits t
oksian1 [2.3K]

Answer:

7.0 m/s

Explanation:

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w = mg

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3 0
2 years ago
Read 2 more answers
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 depends on public opinion and support?
Lelechka [254]
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