All categories

2 years ago

Find the acceleration if a 32.5 N force is used on an object that has a mass of 128.6 kg.

Physics

1 answer:

Sholpan [36]2 years ago

3 0

Answer:

Acceleration=3.95

Explanation:Use the formula a=m/f

a=128.6/32.5

a=3.95

You might be interested in

Define First Condition of Equilibrium ??

spayn [35]

Answer:

First Condition of Equilibrium is that it must be experiencing no acceleration or the external forces acting on the body should be zero.

7 0

2 years ago

Read 2 more answers

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

A car traveling with constant speed travels 150 km in 7200s. What is the speed of the car?

LiRa [457]

Speed= distance/time
Speed= 150000m/7200s=20.83m/s(cor.to.2d.p.)

7 0

3 years ago

In order to start an object moving, you must first overcome the force of _________ friction.

navik [9.2K]

It would be static friction which is what you have to overcome when an object is not in motion. When you move an object friction works against it like gravity and air resistance. I hope this helps!

7 0

3 years ago

Read 2 more answers

What are the diferent units that we typically use to measure average speed

Juli2301 [7.4K]

Mph= miles per hour
kph= kilometers per hour

6 0

2 years ago