a box is pushed 40 m by a mover. the amount of work done was 2,240 J. How much force was exerted on the box

What is the changing of the position of an object relative to a point of reference

Bond [772]

The displacement ........................

5 0

3 years ago

How know that rocks formed from the seafloor sediments deposited in the Amadeus basin were softer than the arkose sandstone

svp [43]

The rocks formed from the seafloor sediments deposited in the Amadeus basin were softer than the Arkose sandstone because the Amadeus basin were made up of marine and non-marine sedimentary rocks which are softer compared to quarts which make up mostly the Arkose sandstone.

7 0

3 years ago

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A ball is rolling along at speed v without slipping on a horizontal surface when it comes to a hill that rises at a constant ang

MissTica

Answer:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

Explanation:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

This is because:

If we consider the ball initially at rest on a frictionless surface and a force is exerted through the centre of mass of the ball, it will slide across the surface with no rotation, and thus, there will only be translational motion.

Now, if there is friction and force is again applied to the stationary ball, the frictional force will act in the opposite direction to the force but at the edge of the ball that rests on the ground. This friction generates a torque on the ball which starts the rotation.

Therefore, static friction is infact necessary for a ball to begin rolling.

Now, from the top of the ball, it will move at a speed 2v, while the centre of mass of the ball will move at a speed v and lastly, the bottom edge of the ball will instantaneously be at rest. So as the edge touching the ground is stationary, it experiences no friction.

So friction is necessary for a ball to start rolling but once the rolling condition has been met the ball experiences no friction.

6 0

3 years ago

Assume (unrealistically) that a TV station acts as a point source broadcasting isotropically at 3.2 MW. What is the intensity of

Dahasolnce [82]

Answer:

$I=1.5\times10^{-28}W/m^2$

Explanation:

The intensity is related to the power and surface area by $I=\frac{P}{A}=\frac{P}{4\pi r^2}$ . We need to calculate the surface area of a sphere of radius r=4.3ly.

Since 4.3ly is the distance light travels in 4.3 years at 299792458m/s, we can obtain it in meters by doing:

$r=vt=(299792458m/s)(4.3\times365\times24\times60\times60s)=4.1\times10^{16}m$

So we have:

$I=\frac{P}{4\pi r^2}=\frac{3.2\times10^6W}{4\pi (4.1\times10^{16}m)^2}=1.5\times10^{-28}W/m^2$

4 0

3 years ago

You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a consta

Arte-miy333 [17]

Answer:

1) 3.66 s

2) 124.44 m

3) 3.12 s

Explanation:

Let's start by first listing down the information in the question.

Red Car : 34 m/s

Blue Car: 28 m/s

Distance between them : 22 m

The difference in speed between the cars is: 34 - 28 = 6 m/s

This means that the red car is catching up to the blue car at a speed of 6 m/s.

1) We can solve this by just dividing the distance by the difference in speed. This becomes: $\frac{Distance}{Speed}= \frac{22}{6} = 3.66$

Thus it takes 3.66 seconds for the red car to catch up to the blue car.

2) We know from (1) that it took 3.66 seconds for the red car to catch up. Since the speed it was travelling at is constant, we only need to multiply it by the time from (1) to get the distance.

This becomes: $Speed * Time = 34 * 3.66 = 124.44$

Thus the red car travels 124.44 m before catching up to the blue car.

3) If the red car starts to accelerate the moment we see it, the time taken to get to the blue car will be less than before. We can find this in a simple way.

We can use the motion equation : $s = u*t + \frac{1}{2}(a * t^2)$

Here s = 22 m

We can take u as the difference in speed. u = 6 m/s

Acceleration a = (2/3) m/s^2

Substituting the these into the equation we get:

$22 = 6t + \frac{1}{2}(\frac{2}{3}t^2)$

Solving this for the variable 't' using the quadratic formula we get the following two answers:

t1 = 3.12 s

t2 = - 21.12 s

Since t2 is not possible, the answer is t1. This means it takes 3.12 seconds for the red car to catch up to the blue.

4 0

3 years ago