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OleMash [197]
3 years ago
14

What will be the time period of simple pendulum at the center of earth​

Physics
1 answer:
Furkat [3]3 years ago
7 0

Answer:

As g=0 at the centre of earth the time period becomes infinite as T=2pi/underoot g. At center of earth, r = 0 & hence g = 0. Time period of a simple pendulum is T = 2π sqrt (1/g).

Explanation:

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What is the magnitude of the gravitational force acting on the sun due to the earth?The earth does not exert any gravitational f
Sholpan [36]

I haven't worked on Part-A, and I don't happen to know the magnitude of the gravitational force that the Sun exerts on the Earth.  

But whatever it is, it's exactly, precisely, identical, the same, and equal to the magnitude of the gravitational force that the Earth exerts on the Sun.

I think that's the THIRD choice here, but I'm not sure of that either.

4 0
3 years ago
water is pumped from a stream at the rate of 90kg every 30s and sprayed into a farm at a velocity of 15m/s. Calculate the power
nikitadnepr [17]

Answer:

340 W

Explanation:

Power = change in energy / change in time

P = ΔKE / Δt

P = ½ mv² / Δt

P = ½ (90 kg) (15 m/s)² / (30 s)

P = 337.5 W

Rounded to 2 significant figures, the power is 340 W.

7 0
3 years ago
Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da
valentinak56 [21]

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

k = 3906250 N/m

Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

3 0
3 years ago
a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after
STatiana [176]

a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )

2 = ( v(2) - 10 ) ÷ ( 6 - 0 )

2 × 6 = v(2) - 10

v(2) = 12 + 10

v(2) = 22 m/s

7 0
3 years ago
A car of mass 750 kg accelerates away from traffic lights. At the end of the first 100 m it has reached a speed
malfutka [58]

the work done on the car by the force of its engine is 78,000 J.

" The work done on the car by the force of friction is 24,000 J.

Increasing the car's kinetic energy at the end of the first 100 m is 54,000J

a. Completed work = force x distance. Engine output = 780 N, that is,

780 N x 100 m = 78,000 J.

b. Completed work = force x distance. Friction force = 240 N, that is,

240 N x 100 m = 24,000 J.

c. Kinetic energy = 1 \ 2 x m x v2

= 1 \ 2 x 750 kg x 12 squared = 375 x 144 = 54,000 J.

<h3>How powerful is the engine of a car? </h3>

Mainstream car and truck engines typically produce 100-400 pounds. -Torque feet. This torque is generated by the engine piston as it moves up and down on the engine crankshaft, causing the engine to rotate (or twist) continuously.

Learn more about work done here:  brainly.com/question/25573309

#SPJ10

5 0
2 years ago
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