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2 years ago

the force that is exerted when a shopping cart is pushed. the forces that causes a metal ball to move toward a magnet

Physics

2 answers:

alexandr402 [8]2 years ago

8 0

Answer:

The force that is exerted when a shopping cart is pushed is a type of push force, supplied by the muscles of the cart pusher's body.

The forces that causes a metal ball to move toward a magnet is a type of pull force that is as a result of the magnetic field forces.

Explanation:

Forces are divided into push forces that tends to accelerate a body away from the source of the force, and pull forces that accelerates the body towards the force source.

Examples of push forces includes pushing a cart, pushing a table, repulsion of two similar poles of a magnet etc. Examples of pull forces includes a attractive force between two dissimilar poles of a magnet, pulling a load by a rope, a dog pulling on a leash etc.

MariettaO [177]2 years ago

4 0

Answer:

contact

noncontact

Explanation:

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3 years ago

A sample of gas in a syringe releases heat to its surroundings while the

blsea [12.9K]

Answer: b

Explanation:

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2 years ago

Read 2 more answers

How much work is done? A Net Force of 9.0 N acts through a distance of 3.0 m in a time of 3.0 s. The answers are 3.0 J, 9.0 J, 2

Vadim26 [7]

If the force and the motion are along the same direction (like it is here) then work is force*distance. The time doesn't come into play until you want the power used. So here
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3 years ago

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

Known Data

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

Time of the dropped Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

Time of the Thrown Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$

Finally, we can find how much earlier does the thrown rock strike the ground, so $t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s$

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3 years ago

If a sound wave is produced with a wavelength of 1.04m what is the waves frequency

dmitriy555 [2]

You should just ask the wave

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2 years ago