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nikitadnepr [17]
3 years ago
12

You're in the gym lifting weights up and down above your head. When are you doing (positive) work on the weights?(A) When you ho

ld them still but not when you raise them up or lower them down. (B) When you raise them up but not when you hold them still or lower them down. (C) When you raise them up or lower them down but not when you hold them still. (D) When you lower them down but not when you hold them still or raise them up.
Physics
1 answer:
a_sh-v [17]3 years ago
8 0

Answer:

(B) When you raise them up but not when you hold them still or lower them down.

Explanation:

Work is defined as:

W=Fd cos \theta

where

F is the force applied

d is the displacement of the weight

\theta is the angle between the direction of the force and the displacement

Let's analyze the three situations:

- When you are raising the weight up: in this case, the work done is positive, because the force (upward) and the displacement of the weight (up) are in the same direction, so \theta=0^{\circ}, cos \theta=1 and the work is positive.

- When you are lowering the weight down: in this case, the work done is negative, because the force (upward) and the displacement of the weight (down) are in the opposite direction, so \theta=180^{\circ}, cos \theta=-1 and the work is negative.

- When you are holding the weight still: in this case, the work done is zero, because the displacement of the weight is zero: d=0, so the work done is also zero.

Therefore, the only correct choice is

(B) When you raise them up but not when you hold them still or lower them down.

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Answer:

a).β=0.53x10^{-3} T

a).β=0.40 x10^{-4} T

Explanation:

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a).

N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A}  \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T

b).

The distance is the radius add the cross section so:

r_{1}=15cm+5cm\\r_{1}=20cm

r_{1} =20cm*\frac{1m}{100cm}=0.20m

\beta =\frac{u_{o}*I*N}{2\pi*r1}

\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T

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The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo
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Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

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the weight of the hyppo is

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m is the mass of planet
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So, if they have the same a:

m₁/r₁² = m₂/r₂²
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m/r₁² = m₂/(2r₁)²
m₂ = 4m

<em>Thus, the answer is D.</em>
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<h3><u>Answer and explanation;</u></h3>
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