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2 years ago

What is the law of variation of the period T of a simple pendulum

1 answer:

3 0

The period of a simple pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the length of the pendulum and $g=9.81 m/s^2$ is the gravitational acceleration. As we can see, the period of a simple pendulum depends only on its length.

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Michael has a substance that he puts in Container 1. The substance has a volume of 5 cubic meters. He then puts the substance in

m_a_m_a [10]

Gas because liquids and solids volumes don't change from switching containers.

3 0

2 years ago

Suppose you have thrown a rock nearly straight up at a coconut in a palm tree,and the rock misses on the way up but hits the coc

Margarita [4]

Most likely it would dislodge the coconut on the way down due to gravity because on the way up the gravity would slow down the rock but on they down the gravity pulls the rock

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2 years ago

A car is moving at 10 m/s to the right. It accelerates for 10 s after which it is moving at 5 m/s to the left. What was the car'

NNADVOKAT [17]

Answer:

Acceleration, $a=-1.5\ m/s^2$

Explanation:

It is given that,

Initial velocity of the car, u = 10 m/s (in right)

Final velocity of the car, v = -5 m/s (in left)

Time taken, t = 10 s

Let a is the acceleration of the car. It can be calculated using the equation of kinematics. The equation is as :

$v=u+at$

$a=\dfrac{v-u}{t}$

$a=\dfrac{-5-10}{10}$

$a=-1.5\ m/s^2$

So, the acceleration of the car is $-1.5\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

A proton moving at 8.9 × 106 m/s through a magnetic field of 0.96 T experiences a magnetic force of magnitude 3.8 × 10−13 N. Wha

goblinko [34]

Answer: 15.66 °

Explanation: In order to solve this proble we have to consirer the Loretz force for charge partcles moving inside a magnetic field. Thsi force is given by:

F=q v×B = qvB sin α where α is teh angle between the velocity and magnetic field vectors.

From this expression and using the given values we obtain the following:

F/(q*v*B) = sin α

3.8 * 10^-13/(1.6*10^-19*8.9*10^6* 0.96)= 0.27

then α =15.66°

8 0

3 years ago

A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?

garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

$U_i +K_i = U_f + K_f$