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algol [13]
3 years ago
15

Ball B is suspended from a cord of length l attached to cart A, which can roll freely on a frictionless, horizontal track. The b

all and the cart have the same mass m. If the cart is given an initial horizontal velocity v0 while the ball is at rest, describe the subsequent motion of the system, specifying the velocities of A and B for the following successive values of the angle θ (assume positive counterclockwise) that the cord will form with the vertical:
(a) θ = θmax
(b) θ = 0
(c) θ = θmin
Physics
1 answer:
Kazeer [188]3 years ago
7 0

Answer:

a.Velocity of A and Velocity of B will be opposite in direction.

b.Both velocities have the same direction.

c. The two objects will move in a direction of some angle α.  

Explanation:

When the angle gets maximum the velocities of the two objects pull each other and move in opposite direction. But when the two objects make an angle zero degree with each other their motion is in the same direction.  

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A rock is thrown downward from the top of a cliff with an initial speed of 20 m/s. If the rock hits the ground after 5.0 s, what
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Answer

I hope it's helps you

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A capacitor is connected to an AC generator. As the generator's frequency is increased, what happens to the current in the capac
Licemer1 [7]

Answer:

The current increases.

Explanation:

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8 0
3 years ago
Two identical strings, of identical lengths of 2.00 m and linear mass density of μ=0.0065kg/m, are fixed on both ends. String A
kolezko [41]

Answer:

beat frequency = 13.87 Hz

Explanation:

given data

lengths l = 2.00 m

linear mass density μ = 0.0065 kg/m

String A is under a tension T1 = 120.00 N

String B is under a tension T2 = 130.00 N

n = 10 mode

to find out

beat frequency

solution

we know here that length L is

L = n × \frac{ \lambda }{2}      ........1

so  λ = \frac{2L}{10}  

and velocity is express as

V = \sqrt{\frac{T}{\mu } }    .................2

so

frequency for string A = f1 = \frac{V1}{\lambda}

f1 = \frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}

f1 = \frac{10}{2L} \sqrt{\frac{T1}{\mu } }      

and

f2 = \frac{10}{2L} \sqrt{\frac{T2}{\mu } }

so

beat frequency is = f2 - f1

put here value

beat frequency = \frac{10}{2*2} \sqrt{\frac{130}{0.0065}}  - \frac{10}{2*2} \sqrt{\frac{120}{0.0065} }

beat frequency = 13.87 Hz

6 0
4 years ago
Material A is less optically dense than material B. How can you tell from the diagram above?
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Magnetic poles that are alike a. attract each other. b. repel each other. c. do not interact. d. have the same shape.
Vera_Pavlovna [14]
They repel each other.
4 0
3 years ago
Read 2 more answers
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