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jolli1 [7]
2 years ago
13

A net force of 15 N is applied to a cart with a mass of 2.1 kg. What is the acceleration of the cart?

Physics
1 answer:
Shkiper50 [21]2 years ago
5 0
Force = mass * acceleration
15 = 2.1 * a
a = 15/2.1 = 7.143 m/s^2
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Assume that a 15-kg ball moving at 8 m/s strikes a wall perpendicularly and rebounds elastically at the same speed. What is the
Aleks [24]
I think u should follow the formulae F=MA. So I think the answer is 120N
5 0
3 years ago
What is the frequency of a clock waveform whose period is 750 microseconds?
Allushta [10]
Use this formula to find your answer...

Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms

frequency (f)=1/( Time period).

Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.

Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.

6 0
3 years ago
Describe the evolution of a pulsar over time, in particular how the rotation and pulse signal changes over time.
madam [21]

Answer:

As beams of particles and their associated energy are given off, the pulsar will lose energy slowly, which will decrease the rate of its rotation. The frequency of pulses would therefore decrease, so that fewer pulses are observed in a given time span. The strength of the pulse signal will also decrease so the pulses will become fainter. Eventually, the pulsar should rotate so slowly and have such a low emission of radiation that it would no longer be observable.

3 0
3 years ago
A small cart travels 20 meters to the right in 10 seconds calculate the velocity
stira [4]
The velocity would be 10
3 0
3 years ago
A meteoroid is moving towards a planet. It has mass m = 0.78×109 kg and speed v1 = 4.1×107 m/s at distance R1 = 2.8×107 m from t
wariber [46]

Answer:

PE=81.755\, J

Explanation:

Given that:

  • mass of meteoroid, m=7.8\times 10^8 \,kg
  • radial distance from the center of the planet, R= 2.8\times 10^7 m
  • mass of the planet, M=4.4\times 10^{25}\, kg

<u>For gravitational potential energy we have:</u>

PE=G\frac{M.m}{R}

substituting the respective values:

PE=6.67\times 10^{-11}\times \frac{4.4\times 10^{25}\times 7.8\times 10^8}{2.8\times 10^7}

PE=81.755\, J

5 0
3 years ago
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