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svet-max [94.6K]
2 years ago
14

What is the average speed of a car that travels 30 m in the 3 secs and 2 m 4 secs and 20 m 5 secs

Physics
1 answer:
Doss [256]2 years ago
6 0

Answer:

4.3333333

Explanation:

total distance / total time taken

30 + 2 + 20 / 3 + 4 + 5

52 / 12

4.3 recurring

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A device that does work with one movement and changes the size or direction of a force is a simple machine.
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Two cars are travelling in the same direction on a road. The blue car is travelling at 50 m/s in front of the red car, which is
Anestetic [448]

Explanation:

The speed of the red car, relative to the blue car, is:

v = 75 m/s − 50 m/s

v = 25 m/s

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You exert 500 N of force down on the earth as you jump into the air. How much force does the earth exert back on you?
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The same 500N, is the Newton’s Third Law.
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3 years ago
1. A sprinter races in the 100 meter dash. It takes him 10 second to reach the finish line
poizon [28]

Answer:

v = 10 m/s

Explanation:

Given that,

Distance covered by a sprinter, d = 100 m

Time taken by him to reach the finish line, t = 10 s

We need to find his average velocity. We know that velocity is equal to the distance covered divided by time taken. So,

v = d/t

v=\dfrac{100\ m}{10\ s}\\\\v=10\ m/s

Hence, his average velocity is 10 m/s.

6 0
3 years ago
A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w
cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

\displaystyle P=\frac{W}{t}

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

W=F.x

The second Newton's law gives us the net force as

F=m.a

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

v_f=v_o+at

v_f=at

Solving for a

\displaystyle a=\frac{v_f}{t}={5.4}{1}

a=5.4\ m/s^2

The distance traveled in the interval is given by

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Since vo=0

\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}

x=2.7\ m

The force is given by

F=m.a

We don't know the value of m, so the force is

F=2.7m

Computing the work done by the sprinter

W=F.x=2.7m(5.4)

W=14.58m

The power is finally computed

\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}

P=14.58m

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}

a=8.8\ m/s^2

The distance is

\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}

x=3.8\ m

The net force is

F=m(8.8)=8.8m

The work done by the sprinter is now computed as

W=8.8m(3.8)=33.44m

At last, the output power is

\displaystyle P=\frac{33.44m}{0.5}=66.88m

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0
3 years ago
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