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2 years ago

8

Helpp pleaseeeeee …..

1 answer:

Lilit [14]2 years ago

8 0

Answer:

300 cos 30 = 40 a + 40 * .2 * 10

Total force = mass * acceleration + frictional force

260 = 40 a + 80

a = 180 / 40 = 4.5 m/s^2

Check:

15 a + 15 * 10 * .2 = T acceleration of 15 kg block (assuming a = 4.5)

T = 15 (4.5) + 30 = 97.5 force required to accelerate 15 kg block

260 - 97.5 = 162.5 net force on 25 kg block

162.5 = 4.5 (25) + 25 * 10 * .2

162.5 = 112.5 + 50 = 162.5

4.5 m/s^2 checks out as correct

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A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current

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Answer: $410.90\times 10^{-6} V$

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3 years ago

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3 years ago

tatuchka [14]

1) 9.57 N

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- The force exerted by the wind, in the horizontal direction (to the right):

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The two forces are perpendicular to each other, so we can find the magnitude of the net force by using Pythagorean's theorem.

Therefore, we have:

$F=\sqrt{W^2+F_w^2}=\sqrt{(9.42)^2+(1.68)^2}=9.57 N$

2) $10^{\circ}$

The direction of the net external force, measured from the downward vertical, can be measured using the following formula:

$\theta = tan^{-1}(\frac{F_x}{F_y})$

where

$F_x$ is the force in the horizontal direction

$F_y$ is the force in the vertical direction

In this problem,

$F_x = F_w = 1.68 N$

$F_y = W = 9.42 N$

and so we find:

$\theta = tan^{-1}(\frac{1.68}{9.42})=10^{\circ}$

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2 years ago