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Anastasy [175]
2 years ago
7

A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end

of the wire, a pulse travels down the wire and returns 7.84 ms later, having reflected off the lower end. The stone is heavy enough to prevent the lower end of the wire from moving. What is the mass m of the stone
Physics
1 answer:
Triss [41]2 years ago
3 0

Answer: m= 35.6 kg

Explanation:

For finding the mass of the stone we have the formula

v= \sqrt{\frac{Tension}{Linear. Mass. density} }

Here, Tension= m*g = m*9.81

and linear mass density= \frac{8.25 g}{65 cm}

Linear mass density= \frac{8.25*10^-3}{65*10^-2}

Linear mass density= 0.0127 kg/m

Velocity= 2*\frac{l}{t}

Velocity= 2 * \frac{65*10^-2}{7.84}

Velocity= 165.8 m/s

So putting all these values in equation we get

v= \sqrt{\frac{Tension}{Linear. Mass. density} }

165.8= \sqrt{\frac{m*9.81}{0.0127} }

Solving we get

m= 35.58 kg

or m= 35.6 kg

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Answer:

Mass of the pull is 77 kg

Explanation:

Here we have for

Since  the rope moves along with pulley, we have

For the first block we have

T₁ - m₁g = -m₁a = -m₁g/4

T₁ =  3/4(m₁g) = 323.4 N

Similarly, as the acceleration of the second block is the same as the first block but in opposite direction, we have

T₂ - m₂g = m₂a = m₂g/4

T₂ =  5/4(m₂g) = 134.75 N

T₂r - T₁r = I·∝ = 0.5·M·r²(-α/r)

∴ M = -\frac{2}{a} (T_2-T_1)

M = -\frac{2}{2.45} (134.75-323.4) = 77 \, kg

Mass of the pull = 77 kg.

5 0
2 years ago
A balloon is at a height of 81m and is ascending upwards with a velocity of 12m/s. A body of 2kg weight is dropped from it. If g
kap26 [50]
I know you are Indian by your question, HC Verma class 9 or 11 !!

if you got any problem, comment !!

7 0
2 years ago
In an experiment, students release a block from rest at the top of an inclined plane. The block slides down the plane through a
mote1985 [20]

Answer:

B) Friction

Explanation:

The main source of error is the omission of the effect from friction between block and incline, which is directly proportional to the mass of the block. The force of gravity is constant. The friction force dissipates part of the gravitational potential energy, generating a final speed less than calculated under the consideration of a conservative system. Air resistance is neglected at low speeds like this case.

8 0
2 years ago
Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of
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Answer:

3311N

Explanation:

r = radius = 600m

V = speed = 150m/s

Mass = weight = 70kg

The weight of pilot when calculated due to circular motion

W = tv

Fv = mv²/r

Fv = 70x150²/600

Fv = 79x22500/600

= 15750000/600

= 2625N

Real Weight of the pilot = m x g

= 70 x 9.8

= 686N

The apparent Weight is calculated by

Mv²/r + mg

= 2625N + 686N

= 3311 N

Therefore the apparent Weight is 3311N

6 0
1 year ago
Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat
Alja [10]

Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

Note: the mass doesn't change, it is only the velocity that change.

Also,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Let assume that if is constant acceleration

Then, m and a is constant,

Then,

K.E is directly proportional to d

So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

K.E_1 = E_k

d_1 = d

d_2 = d/3

K.E_2 = K.E_1 / d_1 × d_2

K.E_2 = E_k × ⅓d / d

Then,

K.E_2 = ⅓E_k

So, the new kinetic energy is one third of the E_k

7 0
2 years ago
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