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dmitriy555 [2]
3 years ago
15

How much work must be done to raise a 1100kg car 2m above the ground?​

Physics
2 answers:
N76 [4]3 years ago
6 0
21560J? i dont know
Xelga [282]3 years ago
4 0

Answer:

21560 J

Explanation:

Work = mg*h = 1100*9.8*2 = 21560 J

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There is a filing cabinent that is 52 in. tall and 15 in. wide. The center of gravity of the cabinet is right at the center. To
erastova [34]

Answer:

16 degrees

Explanation:

The tipping point of the cabinet is sketched below.

5 0
1 year ago
The amount of space between two points is measure in unit?
Andreas93 [3]

Answer:

meters

Explanation:

I'm not positive if this is correct, your teacher may be looking for a broader answer so possibly just 'distance'. Hope this helps! <3

3 0
3 years ago
Read 2 more answers
A swan on a lake gets airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.40
Veseljchak [2.6K]

Answer:

53.895 m.

Explanation:

Using the equation of motion,

v² = u² + 2as .............. Equation 1

Where v = final velocity of the swan, u = initial velocity of the swan, a = acceleration of the swan, s = distance covered by the swan.

make s the subject of the equation,

s = (v² - u²)/2a----------- Equation 2

Given: v = 6.4 m/s, u = 0 m/s ( from rest)  a = 0.380 m/s².

Substitute into equation 2

s = (6.4²-0²)/(2×0.380)

s = 40.96/0.76

s = 53.895 m.

Hence the swan will travel 53.895 m before becoming airborne.

6 0
4 years ago
Even if there were some friction on the ice, it is still possible to use conservation of momentum to solve this problem, but you
hjlf

The problem referred to in this question is missing and it is;

Two hockey pucks of identical mass are on a flat, horizontal ice hockey rink. The red puck is motionless; the blue puck is moving at 2.5 m/s to the left. It collides with the motionless red puck. The pucks have a mass of 15 g. After the collision, the red puck is moving at 2.5 m/s, to the left. What is the final velocity of the blue puck?

Answer:

The condition is that p_f - p_i which is the change in momentum will not be equal to zero but equal to the impulse (Ft).

Explanation:

In the problem described, by inspection, we can say that since there is no friction, we have a closed system and thus momentum is conserved.

Since momentum is conserved, we can say that;

Initial momentum(p_i) = final momentum(p_f)

Now, in this question we are told that some friction wants to be introduced on the ice and it's possible to still use conservation of momentum.

From impulse - momentum theory, we know that;

Impulse = change in momentum

Impulse is zero when no force is acting on the ice and we have; 0 = p_f - p_i

This will yield initial momentum = final momentum.

Now, since a force is applied, we know that impulse is; J = F × t

Thus;

Ft = p_f - p_i

Where F is the force due to friction.

Thus, the condition is that p_f - p_i will not be equal to zero

6 0
3 years ago
How do you calculate energy lost due to friction in an experiment?
Snowcat [4.5K]

Answer:

A treadmill get it? but its   Ff * d cos theta

Explanation:

6 0
4 years ago
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