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Ludmilka [50]
3 years ago
11

Mary has a mass of 40 kg and sprints at 1 m/s. How much kinetic energy does she have?

Physics
1 answer:
Sav [38]3 years ago
8 0

Answer:

20 J

Explanation:

Kinetic energy is given as half of the product of mass and the square of velocity of an object:

KE = \frac{1}{2}mv^2

where m = mass = 40 kg

v = velocity = 1 m/s

Hence, Mary's kinetic energy is:

KE = \frac{1}{2} * 40 * 1^2

KE = 20 * 1 = 20 J

She has a kinetic energy of 20 J.

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An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o
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Answer:

0.53m/s^2

Explanation:

We are given that

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Using the formula

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8 0
4 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
3 years ago
A weight suspended from a spring is seen to bob up and down over a total distance of 20 centimeters twice each second.
Zielflug [23.3K]

<u>Answer</u>

8. 2 Hz

9. 0.5 seconds

10. 20 cm


<u>Explanation</u>

<u>Q 8</u>

Frequency is the number of oscillation in a unit time. It is the rate at which something repeats itself in a second.

In this case, the spring bob up and down 2 times per second.

∴ Frequency = 2 Hz

<u>Q 9</u>

Period is the time taken to complete one oscillation.

2 oscillations takes 1 second

1 oscillation = 1/2 seconds.

∴ Period = 0.5 seconds


<u>Q 10</u>

Amplitude is the the maximum displacement of the spring.

In this case the spring bob up 20 cm. This is it's displacement.

∴ Amplitude = 20 cm

5 0
3 years ago
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