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1 year ago

How much work is done on a pumpkin with a force of 16 newtons when you lift it 15 meters?

Physics

1 answer:

blsea [12.9K]1 year ago

3 0

Answer:

240 Joules

Explanation:

work = f x d

16 newtons is the force

15 meters is the distance

15*16= 240

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A car traveling east at 46.8 m/s passes a trooper hiding at the roadside. the driver uniformly reduces his speed to 25.0 m/s in

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(a). a=5, 89 m/s west.
(b) x=132,82 m.

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2 years ago

What is the wavelength of a wave traveling with a wave speed of 1348 m/s

saw5 [17]

Answer: f = 4 Hz i think

Explanation:

sorry i dont rember (cant type lol)

5 0

2 years ago

If an ocean wave has a frequency of 2 waves/s and a speed of

LuckyWell [14K]

Answer:2m

Explanation:

Wavelength=velocity/frequency

Wavelength=4/2=2m

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2 years ago

A 4.0 µC point charge and a 3.0 µC point charge are a distance L apart. Where should a third point charge be placed so that the

Rudik [331]

Answer:

Q must be placed at 0.53 L

Explanation:

Given data:

q_1 = 4.0 μC , q_2 = 3.0μC

Distance between charge is L

third charge q be placed at distance x cm from q1

The force by charge q_1 due to q is

$F1 = \frac{k q q_1}{x^2}$

$F1 = \frac{k q ( 4.0 μC )}{ x^2}$ ----1

The force by charge q_2 due to q is

$F2 = \frac{k q q_2}{(L-x)^2}$

$F2 = \frac{kq (3.0 μC)}{(L-x)^2}$ --2

we know that net electric force is equal to zero

F_1 = F_2

$\frac{k q ( 4.0 μC )}{x^2} =\frac{k q ( 3.0 μC )}{(l-x)^2}$

$\frac{4}{3}*(L-x)^2 = x^2$

$x = \sqrt{\frac{4}{3}*(L - x)$

$L-x = \frac{x}{1.15}$

$L = x + \frac{x}{1.15} = 1.86 x$

x = 0.53 L

Q must be placed at 0.53 L

3 0

2 years ago

A rope is wrapped three and a half times around a cylinder. Determine the range of force T exerted on the free end of the rope f

Taya2010 [7]

The range of force exerted at the end of the rope is 285.7 N to 1,000 N.

<h3>Net horizontal force of the cylinder</h3>

The net horizontal force of the cylinder when it is at equilibrium position is determined by applying Newton's second law of motion.

∑F = 0

F - μFn = 0

F - 0.2(5,000) = 0

F - 1,000 = 0

F = 1,000 N

The strength of the applied force increases as the number of turns of the rope increases.

minimum force = total force/number of turns of rope

minimum force = 1,000/3.5

minimum force = 285.7 N

Thus, the range of force exerted at the end of the rope is 285.7 N to 1,000 N.

Learn more about Newton's second law of motion here: brainly.com/question/3999427

8 0

1 year ago