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2 years ago

When you push an object, it pushes back with an equal and opposite force

Physics

1 answer:

Ksju [112]2 years ago

5 0

Answer:

I would say false

Explanation:

Because friction slows it down which means when the object bounces back it will not have the same energy or force that was applied before.

You might be interested in

How would gravity cause planets to move if they did not have inertia?

daser333 [38]

I’d think the answer would be C. i’m just kinda guessing but my thought process is this (as simply as i can put it because physics is confusing):

so for example say you throw a ball across a flat surface. inertia is what keeps the ball rolling straight in a line, so unless you were to maybe put your hand in front of the ball or something, it would just go straight forever.

this is what happens with the planets. they go in a straight line, but since there’s gravity, the planets are also being pulled towards the sun. so gravity and inertia are why the planets orbit in the circle pattern they do. so when we remove inertia, we’re removing the state in which the planets keep going straight while being pulled towards a center point (the sun). this causes gravity to be the only factor in the planets orbiting. so that being said, the planets would just be pulled towards the sun. :)

4 0

3 years ago

Read 2 more answers

3. what could you do to increase the precision of your measurements? What factors limit the

Vika [28.1K]

Answer:

Use specific tools built specifically for that specific measurement.

Explanation:

8 0

3 years ago

The car salesman tells you that the car can go from a stop position to 60 mph in six seconds is giving you the car’s rate of

slava [35]

The salesman is telling you the average magnitude of the car's acceleration.

| Acceleration | = (change in speed) / (time for the change)

| Acceleration | = (60 mi/hr) / (6 sec)

| Acceleration | = 10 miles/hr-sec

That would be 36,000 miles per hour squared,
or 0.0028 mile per second squared.

5 0

3 years ago

A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied

Vika [28.1K]

We have that for the Question "A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied on the book, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that the minimum force that must be applied on the book is

F=44N

From the question we are told

A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied on the book, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the Force is mathematically given as

$F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\$

F=44N

Therefore

the minimum force that must be applied on the book is

F=44N

For more information on this visit

brainly.com/question/23379286

8 0

2 years ago

You and a partner sit on the floor and stretch out a coiled spring to a length of 7.2 meters. You shake the coil so you

vekshin1

Answer:

Approximately $5.9\; {\rm m\cdot s^{-1}}$ (assuming that the partner is holding the other end of the coil stationary.)

Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.

There would be a node between:

the antinode at the end of the spring that is being shaken, and
the antinode between the two ends of this spring.

Overall, the nodes and antinodes on this spring would be:

node at the end that is being held still,
antinode (as mentioned in the question),
node (inferred, not mentioned in the question), and
antinode at the end that is being shaken.

The distance between two adjacent nodes is equal to one-half (that is, $(1/2)$ ) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is, $(1/4)$ ) of the wavelength of the wave.

Thus, if the wavelength of the wave in this question is $\lambda$ , the length of this spring would be:

$\displaystyle \frac{1}{2}\, \lambda + \frac{1}{4}\, \lambda = \frac{3}{4}\, \lambda$ .

The question states that the length of this coiled spring is $7.2\; {\rm m}$ . In other words, $(3/4) \, \lambda = 7.2\; {\rm m}$ . The wavelength of this wave would be $(7.2\; {\rm m}) / (3/4) = 9.6\; {\rm m}$ .

The frequency $f$ of this wave is the number of cycles in unit time:

$\begin{aligned} f &= \frac{10}{16.3\; {\rm s}} \approx 0.613\; {\rm s^{-1}}\end{aligned}$ .

Hence, the speed $v$ of this wave would be:

$\begin{aligned} v &= \lambda\, f \\ &=9.6\; {\rm m} \times 0.613\; {\rm s^{-1}} \\ &\approx 5.9\; {\rm m \cdot s^{-1}}\end{aligned}$ .

3 0

2 years ago