A _______ is a tool often used to measure the amount of force exerted by an object

A force of 350N is applied to a body. If the work done is 40kJ, what is the distance through which the body moved?

Studentka2010 [4]

The distance covered by the body is 114.3 m

Explanation:

The work done by a force exerted on an object is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and of the displacement

For the object in this problem, we have

F = 350 N is the force applied

$W=40 kJ = 40,000 J$ is the work done

$\theta=0^{\circ}$ if we assume that the force is applied parallel to the motion of the object

Solving for d, we find the distance covered by the object:

$d=\frac{W}{F cos \theta}=\frac{40,000}{(350)(cos 0)}=114.3 m$

Learn more about work:

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#LearnwithBrainly

7 0

3 years ago

Kawas 200 osebanywhes he remembered he had to return some books they

tia_tia [17]

I think you need to add more.. but I may know where you are leading

Was he 200 m away and made the trip in 200 seconds?

If yes...
2 m/s was his speed and 0 velocity

6 0

3 years ago

The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st

Aleonysh [2.5K]

Answer:

$a = 5.83 \times 10^{-4} m/s^2$

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = $1.50 m/s^2$

now we know that

$F = ma$

$F = 70(1.50) = 105 N$

now for the space station will be same as above force

$F = ma$

$105 = 1.8 \times 10^5 (a)$

$a = \frac{105}{1.8 \times 10^5}$

$a = 5.83 \times 10^{-4} m/s^2$

3 0

3 years ago

Imagine that you are working as a roller coaster designer. You want to build a record breaking coaster that goes 70.0 m/s at the

Rzqust [24]

Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.

At the top of the first hill, the car's potential energy is

PE = (mass) x (gravity) x (height) .

At the bottom, the car's kinetic energy is

   KE = (1/2) (mass) (speed²) .

You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:

   KE = 0.9 PE

(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)

Divide each side by (mass):

   (0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)

(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)

Divide each side by (0.9):

   (0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)

Divide each side by (9.8 m/s²):

   Height = (5/9)(4900 m²/s²) / (9.8 m/s²)

      = (5 x 4900 m²/s²) / (9 x 9.8 m/s²)

      = (24,500 / 88.2) (m²/s²) / (m/s²)

      = 277-7/9 meters
      (about 911 feet)

3 0

3 years ago

I feel really isolated these days and wish I had more friends. I've chosen to put myself in new social situations to try to make

leva [86]

Behaviorist approach to psychology

3 0

4 years ago