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Westkost [7]
2 years ago
7

What force is used to make the back-and-forth motion of a swing higher?

Physics
2 answers:
Eddi Din [679]2 years ago
5 0

Answer:

its a

Explanation:

scrape - push or pull a hard or sharp implement across (a surface or object) so as to remove dirt or other matter.

Inessa [10]2 years ago
5 0

<em>Answer:</em>

<em>Scrape</em>

<em>Hope you like my Answer!</em>

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3 years ago
13. An object with a mass of 2.0 kg has a force of 4.0 newtons applied to it.
julsineya [31]

Answer:2m/s²

Explanation: Well F=MA so sice F=4N and M=2kg let's plug in the values

4N=2KG*A

A=4N/2KG

A=2m/s²

7 0
3 years ago
In the simulation, there are three balls on the floor. Drag each of them up off the floor, and then let go. See what happens to
Vlad1618 [11]

Answer:

I hope this helps and I'm not to late

A way the balls behave the same way is by bouncing about 1 time after throwing the balls up. A way the balls act differently is the blue ball is bouncier than all the balls, the red ball bounces about 2 times before stopping, and the green ball doesn’t really bounce except for one time.

Explanation:

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7 0
2 years ago
A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m
Marrrta [24]

Answer:

-1.47 m/s^2

Explanation:

We can use the following SUVAT equation to solve the problem:

v^2 - u^2 = 2ad

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2

4 0
3 years ago
~~~NEED HELP ASAP~~~<br>Please solve each section and show all work for each section.
anastassius [24]

Explanation:

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass m_A as

x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)

y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)

Substituting (2) into (1), we get

\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)

where f_N= \mu_kN, the frictional force on m_A. Set this aside for now and let's look at the forces on m_B

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>

Let the x-axis be (+) up along the inclined plane. We can write the forces on m_B as

x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)

y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)

From (5), we can solve for <em>N</em> as

N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)

Substituting (7) into (4) we get

m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba

Collecting similar terms together, we get

(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag

or

a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)

Putting in the numbers, we find that a = 1.4\:\text{m/s}. To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get T = 21.3\:\text{N}. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get N = 50.9\:\text{N}

8 0
3 years ago
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