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3 years ago

60 POINTS!!

1 answer:

8 0

How can one explain and predict the interactions between objects and within a system of objects? ... through electric or magnetic fields to illustrate the forces between objects and the changes in energy of the objects due to the interaction. ... Forces at a distance are explained by fields (gravitational, electric, and magnetic) ...

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A 24 kg child sits on a 2.0-m-long rope swing. You are going to give the child a small, brief push at regular intervals.

Bess [88]

time should you wait between pushes is 2.83 sec.

the question is incomplete, full statement is-

A 24 kg child sits on a 2.0-m-long rope swing. You are going to give the child a small, brief push at regular intervals. If you want to increase the amplitude of her motion as quickly as possible, how much time should you wait between pushes?

<h3>What is Amplitude?</h3>

In physics, amplitude refers to the greatest displacement or distance that a point on a vibrating body or wave may move relative to its equilibrium location. It is equivalent to the vibration path's half-length.

regular interval - at similarly spaced intervals: having the same interval of time between occurrences From 4 a.m. to midnight, the buses operate at regular intervals. The boards are positioned at regular intervals, with an equal amount of space between each.

The length of swing, l = 2.1 m

The time between the pushes is nothing but the Time period

and is given by the formula,

$T = 2\pi ( \frac{l}{g} )^{\frac{1}{2} }$

= 2 * 3.14 ( 2.0/ 9.8 ) ^ (1/2)

= 2.83 sec

to learn more about Amplitude go to - brainly.com/question/3613222

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3 0

1 year ago

A 53-N force is needed to keep a 50.0-kg box sliding across a flat surface at a constant velocity. What is the coefficient of ki

earnstyle [38]

The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.

If the box is sliding at constant speed, and not speeding up or slowing down,
that means that the horizontal forces on it add up to zero.

Since you're pushing on it with 53N in that direction, friction must be pulling
on it with 53N in the other direction.

The 53N of friction is (the weight) x (the coefficient of kinetic friction).

53N = (490N) x (coefficient).

Divide each side by 490N : Coefficient = (53N) / (490N) = 0.1082 .

Rounded to the nearest hundredth, that's 0.11 . (choice 'd')

5 0

3 years ago

Read 2 more answers

An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the

nydimaria [60]

Answer: a) the greater speed for the ball is getting with the large radius of the circle. b) 1.68* 10 ^3 m/s^2 c) 1.25*10^3 m/s^2

Explanation: In order to solve this problem firstly we have to consider that speed in a of the circular movement is directly the angular rotation multiply the radius of the circle so by this we found that the second radius get large speed.

Secondly to calculate the centripetal acceleration for the ball we have to considerer the relationship given by:

acceleration in a circular movement= ω^2*r

a1= (8.44 *2*π)^2*r1=1.68 *10^3 m/s^2

a2= (5.95*2*π)^2*r2=1.25*10^3 m/s^2

3 0

3 years ago

a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after

STatiana [176]

a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )

2 = ( v(2) - 10 ) ÷ ( 6 - 0 )

2 × 6 = v(2) - 10

v(2) = 12 + 10

v(2) = 22 m/s

7 0

3 years ago

Find the final temperature of 375 grams of tea (c = 4.184 J/g°C) if its initial temperature is 95°C just before it is placed in

Minchanka [31]

Answer:

the final temperature of the tea is 7.39⁰C.

Explanation:

Given;

mass of the tea, m = 375 g

specific heat capacity of the tea, C = 4.184 JJ/g°C

initial temperature of the tea, t₁ = 95°C

the final temperature of the tea, t₂ = ?

Energy lost by the refrigerator, Q = 137,460 J

The energy lost by the refrigerator is given by the following formula;

-Q = mc(t₂ - t₁)

-137,460 =375 x 4.184(t₂ - 95°C)

-137,460 = 1569(t₂ - 95°C)

$t_2-95 = \frac{-137,460}{1569} \\\\t_2-95 = -87.61\\\\t_2 = -87.61 + 95\\\\t_2 = 7.39 \ ^0 C$

Therefore, the final temperature of the tea is 7.39⁰C.

4 0

3 years ago