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emmasim [6.3K]
3 years ago
15

A 2 kg ball if at rest. If the ball accelerates to 20 m/s, what is the change in momentum?

Physics
1 answer:
katrin [286]3 years ago
4 0

Answer:

change in momentum

20m/s x2=40kg/m

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n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How f
pashok25 [27]

Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

v = \sqrt{\frac{2K.E}{m} } \\\\v =  \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.

8 0
4 years ago
Kyle has a mass of 54 kg and is jogging at a velocity of 3 m/s. What is Kyle’s kinetic energy?
marissa [1.9K]
Kinetic energy can be solved by using the following formula: Kinetic energy = (1/2)*m*v^2

Where:

m = mass = 54 kg
v = velocity = 3 m/s

Since we are already given the needed values in the problem, direct substitution will be done:

KE =  <span>(1/2)*m*v^2
</span>KE =  <span>(1/2) * 54 * (3)^2
</span>KE = 243 J
4 0
3 years ago
Read 2 more answers
What is the frequency of a pendulum that is moving at 30<br> m/s with a wavelength of 0.32 m?
forsale [732]

Answer:

85.7Hz

Explanation:

v=f*λ

f = frequency

λ = wavelength

3 0
3 years ago
A tuning fork vibrates at 15,660 oscillations every minute. What is the period (in seconds) of one back and forth vibration of t
Ierofanga [76]

<u>We are given:</u>

The tuning fork vibrates at 15660 oscillations per minute

<u>Period of one back-and forth movement:</u>

the given data can be rewritten as:

1 minute / 15660 oscillations

60 seconds / 15660 oscillations          <em>(1 minute  = 60 seconds)</em>

<em>dividing the values</em>

0.0038 seconds / Oscillation

Therefore, one back and forth vibration takes 0.0038 seconds

4 0
3 years ago
Match each simple machine with its description.
kipiarov [429]

Answer:

inclined plane- 3

wheel and axel-  1

pulley- 5

lever- 4

wedge- 2

Explanation:

got it right on edge :)

8 0
3 years ago
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