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hodyreva [135]
3 years ago
8

A mass m = 75 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.2 m and finally

a flat straight section at the same height as the center of the loop (19.2 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)
Physics
1 answer:
dalvyx [7]3 years ago
4 0

Answer:

The velocity is 19.39 m/s

Solution:

As per the question:

Mass, m = 75 kg

Radius, R = 19.2 m

Now,

When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.

F_{C} = F_{G}

\frac{mv^{2}}{R} = mg

where

v = velocity

g = acceleration due to gravity

v = \sqrt{2gR} = \sqrt{2\times 9.8\times 19.2} = 19.39\ m/s

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Answer:

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You push a desk with 245 N, but the desk doesn't move due to its friction with the ground. What is the magnitude of the friction
const2013 [10]

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7 0
3 years ago
A coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to i = 1680 mA2 cos [πt/(0.
zhannawk [14.2K]

Answer:

(a) 1.58 V

(b) 0.0126 Wb

(c) 0.0493 V

Solution:

As per the question:

No. of turns in the coil, N = 400 turns

Self Inductance of the coil, L = 7.50 mH = 7.50\times 10^{- 3}\ H

Current in the coil, i = 1680cos[\frac{\pi t}{0.0250}] A

where

i_{max} = 1680\ mA = 1.680\ A

Now,

(a)  To calculate the maximum emf:

We know that maximum emf induced in the coil is given by:

e = \frac{Ldi}{dt}

e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]

e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]

For maximum emf, sin\theta should be maximum, i.e., 1

Now, the magnitude of the maximum emf is given by:

|e| = 7.50\times 10^{- 3}\times 1680\times 10^{- 3}\times \frac{\pi}{0.0250} = 1.58\ V

(b) To calculate the maximum average flux,we know that:

\phi_{m, avg} = L\times i_{max} = 7.50\times 10^{- 3}\times 1.680 = 0.0126\ Wb

(c) To calculate the magnitude of the induced emf at t = 0.0180 s:

e = e_{o}sin{\pi t}{0.0250}

e = 7.50\times 10^{- 3}\times sin{\pi \times 0.0180}{0.0250} = 2.96\times 10^{- 4} =0.0493\ V

7 0
3 years ago
Estimate the smallest possible period of a satellite in a circular orbit around earth. (mass and radius of earth is 5.98 x 1024
myrzilka [38]
Given: Mass of earth Me = 5.98 x 10²⁴ Kg

           Radius of earth r = 6.37 x 10⁶ m

            G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: Smallest possible period T = ?

Formula: F = ma;  F = GMeMsat/r²     Centripetal acceleration ac = V²/r

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equate T from all equation.

F = ma

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GMe = 4π²r³/T²

T² = 4π²r³/GMe  

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T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 25,563,909.77 s²

T = 5,056.08 seconds    or around   1.4 Hour

3 0
3 years ago
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