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Ray Of Light [21]
3 years ago
10

A cord is attached to the box and run through a pulley directly above the box, so that the cord is vertical. The free end of the

cord is then connected to a hanging weight. Calculate the force exerted by the ground on the box (in lb), if the hanging weight is 10.5 lb.

Physics
1 answer:
Harman [31]3 years ago
4 0

Answer:

The answer is given here would be a simplified equation, seeing as there are some missing variables in the question.

<u>F1 = T- 46, 674.656 gm/s² </u>

Explanation:

<em>Note: Once we have the mass of the second object and/or acceleration of the cord, we can solve for the force of the ground acting on the box.</em>

To calculate the force caused by gravity on the basic pulley system we use the following equation:

F2 = M2 x g; where g= gravitational acceleration (a constant equal to 9.8 m/s²). The mass M2 = 10.5 lb = 4762.72g

∴ F2 = 4762.72g x 9.8 m/s²

= 46, 674.656 gm/s² or 46, 674.656 N

But since this F2 is acting in a downlowrd direction, it would be negative.

Tension of the cord, T = Mass, x × acceleration. ( x is in the pulley diagram)

⇒ F1 = T - F2

<u>F1 = T- 46, 674.656 gm/s² </u>

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Answer:

I think the answer is C.

Explanation:

A primary source is a first hand account of an event while a secondary source is a retelling or second hand account meaning as many details will be prevalent.

8 0
2 years ago
Suppose a ball is dropped from shoulder height, falls, makes a perfectly elastic collision with the floor, and rebounds to shoul
Bezzdna [24]

Answer:Same magnitude

Explanation:

When ball is dropped from shoulder height h then velocity at the bottom is given by

v_1=\sqrt{2gh}

if it makes elastic collision then it will acquire the same velocity and riser up to the same height

If m is the mass of ball then impulse imparted is given by

J=m(v_2-v_1)

J=2m\sqrt{2gh}

Thus impulse imparted by gravity and Floor will have same magnitude of impulse but direction will be opposite to each other.

7 0
4 years ago
oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student
puteri [66]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0
3 years ago
Which change would cause the largest increase in a mountain climber’s gravitational potential energy?
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4 years ago
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The rotating loop in an AC generator is a square 10.0 cm on each side. It is rotated at 60.0 Hz in a uniform field of 0.800 TO.
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Answer:

Explanation:

Given a square side loop of length 10cm

L=10cm=0.1m

Then, Area=L²

Area=0.1²

Area=0.01m²

Given that, frequency=60Hz

And magnetic field B=0.8T

a. Flux Φ

Flux is given as

Φ=BA Sin(wt)

w=2πf

Φ=BA Sin(2πft)

Φ=0.8×0.01 Sin(2×π×60t)

Φ=0.008Sin(120πt) Weber

b. EMF in loop

Emf is given as

EMF= -N dΦ/dt

Where N is number of turns

Φ=0.008Sin(120πt)

dΦ/dt= 0.008×120Cos(120πt)

dΦ/dt= 0.96Cos(120πt)

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Emf=-0.96NCos(120πt). Volts

c. Current induced for a resistance of 1ohms

From ohms law, V=iR

Therefore, Emf=iR

i=EMF/R

i=-0.96NCos(120πt) / 1

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d. Power delivered to the loop

Power is given as

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P=-0.96NCos(120πt)•-0.96NCos(120πt)

P=0.92N²Cos²(120πt) Watt

e. Torque

Torque is given as

τ=iL²B

τ=-0.96NCos(120πt)•0.1²×0.8

τ=-0.00768NCos(120πt) Nm

8 0
3 years ago
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