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mixas84 [53]
3 years ago
5

A block oscillating up and down on a spring comes to the top of its path, where it is momentarily at rest. At the instant it is

at rest, what is the direction of the rate of change of its momentum?
Physics
1 answer:
True [87]3 years ago
7 0

Opposite to the direction of the velocity which led it to its current position.

Explanation:

The direction of momentum when a vertically oscillating block comes to the rest momentarily will be opposite to the direction of the velocity that it has just followed to reach reach its current position.

The direction of change in momentum at the bottom will be upwards and at the top will be downwards.

The change in momentum is mathematically defined as:

\Delta P=m.v_f-m.v_i

where:

m= mass of the block

v_f= final velocity of the block

v_i= initial velocity of the block

When the block comes to rest it is due to the result of continuously decreasing velocity.

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A bullet with a mass 2.25g is fired up into the air with a velocity of 187.5 m/s. What is the maximum height of the bullet
Ksivusya [100]

Answer:

1793.7m

Explanation:

From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

Now the kinetic energy; is

K.E = 1/2 × m × v2

Where m is mass

v is velocity

Hence.

K.E = 1/2 × 2.25 × (187.5)^2

Now this should be same with the potential energy which is given as;

P.E = m× g× h

Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P.E = 2.25 × 9.8 × h

From the foregoing analysis of energy conversation it implies;

1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h

=> 1/2 × (187.5)^2 = 9.8 × h

=>1/2 × (187.5)^2 / 9.8 = h

=> 1793.69m = h

h= 1793.69m

h =1793.7m to 1 decimal place

3 0
3 years ago
In one of the original Doppler experiments, a tuba was played at a frequency of 64.0 Hz on a moving flat train car, and a second
wolverine [178]

Answer:

 f_{beat} = 1.64\ Hz

Explanation:

given,

frequency of tuba.f = 64 Hz

Speed of train approaching, v = 8.50 m/s

beat frequency = ?

using Doppler's effect formula

 f' = f(\dfrac{v}{v-v_s})

v_s is the velocity of the source

v is the speed of sound, v = 340 m/s

now,

 f' = 64\times (\dfrac{340}{340 - 8.50})

       f' = 65.64 Hz

now, beat frequency is equal to

 f_{beat} = f' - f

 f_{beat} = 65.64 - 64

 f_{beat} = 1.64\ Hz

hence, beat frequency is equal to 1.64 Hz

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