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Gnom [1K]
3 years ago
6

The center of a long frictionless rod is pivoted at the origin and the rod is forced to rotate at a constant angular velocity Q

in a horizontal plane. Write down the equation of motion for a bead that is threaded on the rod, using the coordinates x and y of a frame that rotates with the rod (with x along the rod and y perpendicular to it). Solve for x (t). What is the role of the centrifugal force

Physics
1 answer:
Doss [256]3 years ago
7 0

Answer:

The answers can be found attached.

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A 6 kg block is attached to a massless rope that passes over a frictionless pulley. The pulley can be modeled as a uniformly den
scoundrel [369]

The effective mass of the system is 12 kg.

<u>Explanation:</u>

  • The effective mass of the system is given by the expression,

                    = mₓ + 1/2 m₁ + 2/5 m₂

  • In the above equations the subscript x denotes block B and the subscript 2 denotes the mass of the solid sphere and the subscript 1 denotes the pulley.
  • Expressing the above functions with the data given below. we have,

           = 6 kg + 1/2 (4 kg) + 2/5 (10 kg)

           =  12 kg.

with the findings above we can conclude that the effective mass of the system is 12 kg

6 0
3 years ago
Is there any change in mass of substance after it changes its state? Explain with an example​
Flura [38]

Answer:

No

Explanation:

When a substance changes its state , there will not be any change in the mass of substance . For example , if we change 100 grams of water (liquid) to ice(solid) , the mass of ice will be same i.e., 100 grams . This shows there is no change in the mass even if a substance changes its state.

MARK ME AS BRAINLIST

3 0
2 years ago
What is the magnetic field strength, in units of microTesla (T), at a distance of 6.3 cm from a long straight wire carrying a cu
bezimeni [28]

Answer:

2.1 \mu T

Explanation:

The magnetic field strength is given by:

B=\frac{\mu_0 I}{2 \pi r}

where

\mu_0 is the vacuum permeability

I is the current

r is the distance from the wire

In this problem,

I = 6.5 A

r = 6.3 cm = 0.063 m

So, the magnetic field strength is

B=\frac{(1.256\cdot 10^{-7} H/m)(6.5 A)}{2 \pi (0.063 m)}=2.1\cdot 10^{-6} T=2.1 \mu T

7 0
3 years ago
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