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Keith_Richards [23]
3 years ago
14

A pendulum can be formed by tying a small object, like a tennis ball, to a string, and then connecting the other end of the stri

ng to the ceiling. Suppose the pendulum is pulled to one side and released at t1. At t^2, the pendulum has swung halfway back to a vertical position. At t^3, the pendulum has swung all the way back to a vertical position. Rank the three instants in time by the magnitude of the centripetal acceleration, from greatest to least. Most of the homework activities will be Context-rich Problems.
Physics
1 answer:
Usimov [2.4K]3 years ago
5 0

Answer:

1- t^3

2- t^2

3- t1

Explanation:

The acceleration produced in a body, while travelling in a circular motion, due to change in direction of motion is called centripetal acceleration. The formula of the centripetal acceleration is as follows:

ac = v²/r

where,

ac = centripetal acceleration

v = speed

r = radius

for a constant radius the centripetal acceleration will be directly proportional to the speed of object. The speed of pendulum will be lowest at t1 due to zero speed initially. Then the speed will increase gradually having greater speed at t^2 and the highest speed and centripetal acceleration at t^3. Therefore, the three instants in tie can be written in following order from greatest centripetal acceleration to lowest:

<u>1- t^3</u>

<u>2- t^2</u>

<u>3- t1</u>

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Answer:

Pascal is a derived unit because <u>it</u><u> </u><u>cannot</u><u> </u><u>be</u><u> </u><u>expressed</u><u> </u><u>in</u><u> </u><u>any</u><u> </u><u>physics</u><u> </u><u>terms</u><u>,</u><u> </u><u>but</u><u> </u><u>it</u><u> </u><u>is</u><u> </u><u>an</u><u> </u><u>expression</u><u> </u><u>of</u><u> </u><u>fundamental</u><u> </u><u>quantities</u><u>.</u>

Explanation:

{ \sf{Pasacal \: ( Pa) =  \frac{newtons}{metres {}^{2} } }} \\  \\ { \sf{Pasacal  \: (Pa) =  \frac{kg \times  {ms}^{ - 2} }{ {m}^{2} } }}

4 0
3 years ago
ehicle collisions involve huge amounts of energy. How does the crashworthiness of a car affect the transfer and transformations
Angelina_Jolie [31]

When the two-vehicle collides transformation of the energy is done in terms of kinetic energy.

<h3>What is the law of conservation of energy?</h3>

According to the Law of Conservation of Energy, energy can neither be created nor destroyed, but it can be transferred from one form to another.

The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

In the above problem, the Vehicle get collides so that the kinetic energy of the vehicle is converted into the kinetic energy of another vehicle the speed of the vehicle will reduce when they collide.

Momentum also gets conserved when the two vehicles collide.

Hence, the transformation of the energy is done in terms of kinetic energy.

To learn more about the law of conservation of energy, refer to brainly.com/question/2137260.

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8 0
2 years ago
Which of the following equations can be used to directly calculate an object's impulse?
Iteru [2.4K]
I believe the answer is B
7 0
4 years ago
A particle with a charge of 5 × 10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a sta
creativ13 [48]

Answer:

r = 0.22m

Explanation:

To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.

Then, you have:

F_c=F_e=ma_c      (1)

m: mass of the particle = 20g = 20*10-3 kg

ac: centripetal acceleration = ?

q: charge of the particle = 5*10^-6C

Fe: electric force between the charges

The electric force is given by:

F_e=k\frac{qq'}{r^2}             (2)

r: radius of the orbit

q': charge of the particle at the center of the orbit = -5*10^-6C

Furthermore, the centripetal acceleration is:

a_c=\frac{v^2}{r}                 (3)

v: speed of the particle = 7m/s

You replace the expressions (2) and (3) in the equation (1) and solve for r:

k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}

Finally, you replace the values of all parameters in the previous expression:

r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m

The radius of the circular trajectory is 0.22m

5 0
3 years ago
A small, rigid object carries positive and negative 3.00 nC charges. It is oriented so that the positive charge has coordinates
enot [183]

Answer:

Umax = 105.8nJ

Umin =-105.8nJ

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Explanation:

8 0
3 years ago
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