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Brilliant_brown [7]
3 years ago
7

How would the period Of the pendulum different from an equivalent one on earth​

Physics
1 answer:
il63 [147K]3 years ago
5 0

Answer:

If you know that that free fall acceleration g on the Moon is about 6 times less than on the Earth, it gives you the answer: on the Moon the same pendulum will have a period about √6≈2.45 longer than on the Earth.

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I need DISPLACEMENT please.<br><br>I have Average speed, i need Displacement.
Lapatulllka [165]
Define displacement and I'll help you
4 0
3 years ago
What happens if :<br> . The test charge is not tiny.
docker41 [41]

The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.

<h3>How does test charge affect electric field?</h3>

As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.

Adjusting the amount of charge on the test charge will not change the electric field force.

<h3>What is a test charge used for?</h3>

The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.

To learn more about test charge, refer

brainly.com/question/16737526

#SPJ9

3 0
2 years ago
un tanque de gasolina de 40 litros fue llenado por la noche, cuando la temperatura era de 68 grados farenheit. Al dia siguiente
Sedaia [141]

Answer:

Volume of gasoline that expands and spills out is 1.33 ltr

Explanation:

As we know that when temperature of the liquid is increased then its volume will expand and it is given as

\Delta V = V_o\gamma \Delta T

here we know that

V_o = 40 Ltr

volume expansion coefficient of the gasoline is given as

\gamma = 950 × 10^{–6}

change in temperature is given as

\Delta T = (131 - 68) \times \frac{5}{9}

\Delta T = 35 ^oC

Now we have

\Delta V = 40(950 \times 10^{-6})(35)

\Delta V = 1.33 Ltr

3 0
3 years ago
A car travelling at 5.0 m/s starts to speed up. After 3.0 s its velocity has increased to 11 m/s. a) What is its acceleration? (
lions [1.4K]

initially, the car is traveling at 5.0 m/s.

so, we know acceleration for changing velocity is :

a = (v-v_{o})/t ..........(i)

where v is the final velocity

v_{o} is the initial velocity

t is the time taken to change velocity

Now, as per the question :

initial velocity, v_{o}=5.0 m/s

final velocity, v =11 m/s

time taken, t = 3 s

putting the values in equation (i),

a = ( 11-5 )/3

a = 2 m/s²

Therefore, a, after 3 s, is  <em>2 m/s².</em>

4 0
1 year ago
Assume that you have a rectangular tank with its top at ground level. The length and width of the top are 14 feet and 7 feet, re
ch4aika [34]

To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of 62 lb / ft ^ 3 (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as

W = \gamma A * \int_0^15 dy

Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing

W = (62)(14*7)\int^{15}_0 (15-y)dy

W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0

W = (14*7*62)[15(15)-\frac{(15)^2}{2}]

W = 683550ft-lbs

Therefore the total work in the system is 683550ft-lbs

6 0
3 years ago
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