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Vedmedyk [2.9K]
3 years ago
5

A kid on a playground swing makes a complete to-and-fro swing each 2 seconds.

Physics
1 answer:
sashaice [31]3 years ago
4 0

Answer:

The frequency of the swing: 1/2 Hertz

The period is: 2 Seconds

Explanation:

The time the kid takes to make a complete to-and-fro swing = 2 seconds

The period, T, is the time it takes to make one complete cycle of an oscillatory motion, therefore, we have;

The frequency, f, is the number of cycles completed each second, therefore, we have;

The time for 1 cycle = 2 seconds

2 seconds = 1 cycle

Dividing both sides by 2 gives;

2/2 seconds = 1/2 cycles

2/2 = 1

In 2/2 = 1 seconds The number of cycles completed = 1/2 cycles

Therefore, the number of cycles completed per (one) second = 1/2 cycles

Therefore the frequency of the swing, f = 1/2 cycle/seconds = 1/2 Hertz

The period, T, is the time it takes to complete one to-and-fro swing which is one cycle which is 2 seconds

Therefore, the period is 2 Seconds.

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A 1000 kg car accelerates from rest at a rate of 10 m/s² for 3 seconds. A) what is the final velocity of the car?
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Answer:

Refer to the attachment!~

5 0
3 years ago
How much elastic potential energy is stored in a bungee cord with a spring constant of 10.0 N/m when the cord is stretched 2.00
Aleonysh [2.5K]

Answer:

<em> The elastic potential energy stored in the bungee cord = 20 J</em>

Explanation:

potential energy: This is the energy possessed by a body due to its position. The S.I unit of energy is Joules. The mathematical expression for elastic potential energy is given below

E = 1/2ke²................ Equation 1

Where E = elastic potential energy of the spring, k = force constant of the spring, e = extension

<em>Given: K = 10 N/m, e = 2.00 m</em>

<em>Substituting these values into Equation 1</em>

<em>E = 1/2(10)(2)²</em>

<em>E = 5×4</em>

<em>E = 20 Joules.</em>

<em>Therefore the elastic potential energy stored in the bungee cord = 20 J</em>

<em></em>

6 0
3 years ago
If we want to know the velocity that an object is traveling, we must know the
Serga [27]
If my memory serves me well, if we want to know the velocity that an object is traveling, we must know the <span>direction and speed. Velocity includes two these points listed in the previous sentence which means the answer is D.</span>
4 0
3 years ago
Read 2 more answers
The flow control circuit design that can best control an overrunning load is the ____ circuit.
lara31 [8.8K]

The meter out circuit is the flow control circuit design that can most effectively control an overrunning load.

The meter-out circuit can be very accurate, but are not efficient. The meter-out circuit can control overrunning as well as opposing loads while the other one method must be used with opposing loads only. The choice of flown control valve method and the location of the flow control in the circuit are dependent on the type of application being controlled.

<h3>What is a Circuit ?</h3>

In electronics, a circuit is a complete circular conduit through which electricity flows. A simple circuit consists of conductors, a load, and a current source. The term "circuit" broadly refers to any continuous path via which electricity, data, or a signal might flow.

  • The directional valve shifts, causing the actuator to move faster than pump flow can fill it due to an overrunning load. Oil is leaking from one side, whereas there is none on the other.

Hence, flow control circuit design that can best control an overrunning load is the opposing circuit

Learn more about Circuit here:

brainly.com/question/26064065

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3 0
2 years ago
A 50.0-turn circular coil of radius 5.00 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0
Blababa [14]

Answer:

The maximum torque in the coil is 4.9\times 10^{-3}\ N-m.

Explanation:

Given that,

Number of turns in the circular coil, N = 50

Radius of coil, r = 5 cm

Magnetic field, B = 0.5 T

Current in coil, I = 25 mA

We need to find the magnitude of the maximum possible torque exerted on the coil. The magnetic torque is given by :

\tau=NIAB\ \sin\theta

For maximum torque, \theta=90^{\circ}

\tau=NIAB\\\\\tau=50\times 25\times 10^{-3}\times \pi (0.05)^2\times 0.5\\\\\tau=4.9\times 10^{-3}\ N-m

So, the maximum torque in the coil is 4.9\times 10^{-3}\ N-m.

4 0
3 years ago
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