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Crank
4 years ago
6

How much tension must a rope withstand if it is used to accelerate a 960kg cat horizontally along a frictionless surface at 1.20

m/s^2 ?
Physics
1 answer:
qaws [65]4 years ago
5 0
Tension is a type of force and
F = m*a (force is equal to mass times acceleration)
in this case 
T = 960 * 1.20 m/s^2 = 1152 N 
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A single loop of area 0.193 m2 is placed perpendicular to a magnetic field of 0.374 T. The loop is rotated about an axis lying i
DENIUS [597]

Answer: The angular speed w of the loop = 181 rad/s

Explanation:

Given that;

Area A = 0.193 m2 

Magnetic field B = 0.374 T

E.m.f = 9.24v

Ø = 45 degree

According to Faraday's law when the magnetic flux linking a circuit changes, an electromotive force is induced in the circuit proportional to the rate of change of the flux linkage.

Using the formula

E.m.f = NABwcosØ

Where w = angular velocity.

Let assume that N = 1 then,

9.24 = 0.193 × 0.374 × w × cos45

9.24 = 0.051w

w = 9.24/0.051

w = 181 m/s

Therefore, the angular speed w of the loop = 181 rad/s

4 0
3 years ago
Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuvers. A trained, fit person we
jeka57 [31]

(a) 4.14 rad/s^2

The relationship beween centripetal acceleration and angular speed is

a=\omega^2 r

where

\omega is the angular speed

r is the radius of the circular path

Here we gave

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r = 5.15 m is the radius

Solving for \omega, we find:

\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{88.2 m/s^2}{5.15 m}}=4.14 rad/s^2

(b) 21.3 m/s

The relationship between the linear speed and the angular speed is

v=\omega r

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v is the linear speed

\omega is the angular speed

r is the radius of the circular path

In this problem we have

\omega=4.14 rad/s

r = 5.15 m

Solving the equation for v, we find

v=(4.14 rad/s)(5.15 m)=21.3 m/s

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3 years ago
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D=v×t
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Who was Christa McAuliffe and what happened this year to make her dream come true?
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