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4 years ago

If 3,600 J of work is done in 3.0 s, what is the power?

Physics

2 answers:

CaHeK987 [17]4 years ago

6 0

The power when 3600 J is done in 3 secs is 1200W

zlopas [31]4 years ago

6 0

What's the answer choices?

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Someone is trying to balance a (110cm) plank with certain forces.

Mashutka [201]

Answer:

a. i. 30 Nm ii. This moment is a clockwise positive moment.

b. i. 15 Nm ii, This moment is a counter-clockwise negative moment.

c. i. The plank will not balance. ii. The plank would tip up.

d. 150 N

Explanation:

a) Calculate the moment of the 60N force (about O), then name its type.

i. Calculate the moment of the 60N force (about O)

Since moment = Force × perpendicular distance from point of moment ,

M = Fd

Since F = 60 N and d = 50 cm = 0.5 m

M = 60 N × 0.5 m = 30 Nm

ii. Then name its type.

This moment is a clockwise positive moment.

b) Calculate the moment of the 30N force (about O), then name its type.

i. Calculate the moment of the 30N force (about O),

Since moment = Force × perpendicular distance from point of moment ,

M' = F'd'

Since F' = 30 N and d' = 50 cm = 0.5 m

M' = 30 N × 0.5 m = 15 Nm

ii. Then name its type.

This moment is a counter-clockwise negative moment.

c) Will the plank balance? If not, which way will it tip?

i. Will the plank balance?

The plank will balance if the net moment on it is zero

So net moment, M' = positive moment - negative moment = M - M' = 30 Nm - 15 Nm = 15 Nm

Since the net moment on the plank is M" = 15 Nm ≠ 0,<u>the plank will not balance.</u>

ii Which way will it tip?

The plank would tip in the direction of the greater moment since the net moment is positive. <u>This moment tilts the plank in a clockwise direction, so the plank would tip up.</u>

d) What extra force would be needed at (B) to balance the plank?

The extra force must balance the net moment,

So M" = F"d" where F" = force and d" = distance of force from O = 10 cm = 0.10 m

F" = M"/d"

= 15 Nm/0.10 m

= 150 N

4 0

3 years ago

A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio

Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to $F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N$

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

$F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3$

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be $8.21\times 10^3~N$

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0

4 years ago

Which of the following advancements could science and technology reasonably contribute to society?

Alexxx [7]

Answer:

idk search it on google chrome

Explanation:

7 0

3 years ago

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Which organelle acts like the transportation or circulatory system of the cell?

REY [17]