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Aliun [14]
3 years ago
11

Hi :) In a pendulum experience , what affects the period & explanation too? Thank you sm :))

Physics
1 answer:
stich3 [128]3 years ago
4 0

Answer:

T=2π\sqrt{l/g}  (simple pendulum)

length of the string  and gravity mainly effect the pendulum period

Explanation:

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Using the picture below, what is the displacement of the triangle?
zavuch27 [327]

Answer:

i think it is 40 kilometers in the positive direction... if not im sorry

Explanation:

7 0
3 years ago
A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu
dalvyx [7]

Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 =  0.341 \ kg

   The mass attached to the other end of the string is  m_2 =  0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 =  m_1 *  g

substituting values

      T_1 =  0.341 * 9.8

      T_1 =  3.342 \ N

At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 =  m_2 *  g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

     \tau  = 0.2505 \ N \cdot m

5 0
3 years ago
The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm
Alika [10]

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force,  F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by F=kx

So 63.5=k\times 0.0531

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J

8 0
3 years ago
Question about generators and electrical currents. Please, no absurd answers! Thank you
Kamila [148]
The first answer should be correct if not then the second one
3 0
3 years ago
Read 2 more answers
A large container contains a large amount of water. A hole is drilled on the wall of the container, at a vertical distance h = 0
barxatty [35]

Answer:

Velocity = 3.25[m/s]

Explanation:

This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.

In point 1, (surface of the water) we have the atmospheric pressure and at point 2 the water is coming out also at atmospheric pressure, therefore this members in the Bernoulli equation could be cancelled.

The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"

h2 is located at point 2 and it will be zero.

(P_{1} +\frac{v_{1}^{2} }{2g} +h_{1} )=(P_{2} +\frac{v_{2}^{2} }{2g} +h_{2} )\\P_{1} =P_{2} \\v_{1}=0\\h_{2} =0\\v_{2}=\sqrt{0.54*9.81*2}\\v_{2}=3.25[m/s]

4 0
3 years ago
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