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3 years ago

17. For how long should a force of 130 N be applied to an object of mass 50 kg to change its speed from 20 m/s to 60 m/s?

Physics

1 answer:

ohaa [14]3 years ago

5 0

Answer:

c. 15.4 s

Explanation:

Given the following data;

Mass, m = 50kg

Force, F = 130N

Initial velocity, u = 20m/s

Final velocity, v = 60m/s

To find the time;

First of all, we would solve for acceleration using the formula below;

Force = mass * acceleration

130 = 50*acceleration

Acceleration = 130/50

Acceleration = 2.6m/s²

Now, we would use the first equation of motion to find the time.

V = U + at

60 = 20 + 2.6t

2.6t = 60 - 20

2.6t = 40

t = 40/2.6

Time, t = 15.39 ≈ 15.4 seconds.

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3 years ago

3. A 900N mountain climber scales a

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2 years ago

How is the surface area of an average person is 2m^2 in the chapter pressure

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Now to calculate surface area we can find it from $pressure=\frac{force}{area}$ and re-arranging it to.

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So plugging the values,

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3 years ago

Read 2 more answers

A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre

AleksandrR [38]

Answer:

A) 1568.60 Hz

B) 1437.15 Hz

Explanation:

This change is frequency happens due to doppler effect

The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

$f_(observed)=\frac{(c+-V_r)}{(C+-V_s)} *f_(emmited)\\$

where

C = the propagation speed of waves in the medium;

Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;

Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.

A) Here the Source is moving towards the receiver(C-Vs)

and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

$f_(observed)=\frac{C}{C-V_s} *f_(emmited)\\=\frac{343}{343-15}*1500\\ =1568.60 Hz$

B)Here the Source is moving away the receiver(C+Vs)

and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser

$f_(observed)=\frac{C}{C+V_s} *f_(emmited)\\=\frac{343}{343+15}*1500\\ =1437.15 Hz$

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3 years ago