Emigration refers to the number of

Muscles use force to move objects. When using a ramp to do work, would your muscles use more force or less force?

White raven [17]

Less because the ramp is letting off force but i does depend on the way you are going on the ramp

7 0

3 years ago

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A fold is a _ in a rock , and a fault is a _in a rock?

777dan777 [17]

A geological fold occurs when one or a stack of originally flat and planar surfaces, such as sedimentary strata, are bent or curved as a result of permanent deformation.

So A fold is a Bend? in a rock. Maybe.

A fault is a planar fracture or discontinuity in a volume of rock, across which there has been significant displacement as a result of rock-mass movement.

3 0

3 years ago

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If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w

marshall27 [118]

Answer:

$P(3600)=593.247W$

Explanation:

First, let's find the voltage through the resistor using ohm's law:

$V=IR=20*8=160V$

AC power as function of time can be calculated as:

$P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)$ (1)

Where:

$\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency$

Because of the problem doesn't give us additional information, let's assume:

$\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi$

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

$P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W$

5 0

4 years ago

What is the net force for this? (please don't forget it's direction and magnitude)

avanturin [10]

Answer:

Explanation:

As the sum of the two right directed forces match exactly the left directed force, the only unbalanced force, and thus the net force, is the upward 25 N force.

3 0

3 years ago

Very large forces are produced in joints when a person jumps from some height to the ground. Calculate the force produced if an

krok68 [10]

Answer:

A) 31 kJ

B) 1.92 KJ

C) 40 , 2.48

Explanation:

weight of person ( m ) = 79 kg

height of jump ( h ) = 0.510 m

Compression of joint material ( d ) = 1.30 cm ≈ 0.013 m

A) calculate the force

Fd = mgh

F = mgh / d

W = mg

F(net) = W + F = mg ( 1 + $\frac{h}{d} )$

= 79 * 9.81 ( 1 + (0.51 / 0.013) )

= 774.99 ( 40.231 ) ≈ 31 KJ

B) calculate the force when the stopping distance = 0.345 m

d = 0.345 m

Fd = mgh hence F = mgh / d

F(net) = W + F = mg ( 1 + $\frac{h}{d} )$

= 79 * 9.81 ( 1 + (0.51 / 0.345) )

= 774.99 ( 2.478 ) = 1.92 KJ

C) Ratio of force in part a with weight of person

= 31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40

Ratio of force in part b with weight of person

= 1920 / 774.99 = 2.48

4 0

3 years ago