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ozzi
2 years ago
14

What force is acting on a 2 kg apple falling on the Earth (g = 10)?

Physics
1 answer:
Korvikt [17]2 years ago
5 0

Answer:

<em>D. 20 N</em>

Explanation:

<em>Force</em><em> </em><em>=</em><em> </em><em>mass(</em><em>m)</em><em> </em><em>×</em><em> </em><em>acceleration</em><em> due</em><em> to</em><em> </em><em>gravity (</em><em>g)</em>

<em>Force </em><em>=</em><em> </em><em>2</em><em>k</em><em>g</em><em> </em><em>×</em><em> </em><em>1</em><em>0</em><em>m</em><em>/</em><em>s²</em>

<em>Force</em><em> </em><em>=</em><em> </em><em>2</em><em>0</em><em>N</em>

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Answer:

the answer is equal to 2.00v

4 0
2 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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Answer:

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The theory of continental drift is most associated with the scientist Alfred Wegener. In the early 20th century, Wegener published a paper explaining his theory that the continental landmasses were “drifting” across the Earth, sometimes plowing through oceans and into each other. He called this movement continental drift. 

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2 years ago
(i) 10 m (ii) 20 m (iii) 40 m (iv) 80 m
IRINA_888 [86]

Answer:

20m

420=80m

100

increases

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6 0
2 years ago
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