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3 years ago

Which requires more work, lifting a 10.0kg load a vertical distance of 2m or lifting a 5.0kg load a distance of 4m?

1 answer:

6 0

For each load, Work = (mass) x (gravity) x (distance .

Bigger load: Work = (10 kg) x (9.8 m/s²) x (2 m) = 196 joules .

Smaller load: Work = (5 kg) x (9.8 m/s²) x (4 m) = 196 joules.

The work required is equal in both cases.

The mass ratio of 2:1 is exactly balanced by
the height ratio of 1:2 .

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A 7.00-kg object accelerates from rest to a final velocity in 55 seconds. If the magnitude of the

Len [333]

Answer:

The final velocity of the object is 330 m/s.

Explanation:

To solve this problem, we first must find the acceleration of the object. We can do this using Newton's Second Law, given by the following equation:

F = ma

If we plug in the values that we are given in the problem, we get:

42 = 7 (a)

To solve for a, we simply divide both sides of the equation by 7.

42/7 = 7a/7

a = 6 m/s^2

Next, we should write out all of the information we have and what we are looking for.

a = 6 m/s^2

v1 = 0 m/s

t = 55 s

v2 = ?

We can use a kinematic equation to solve this problem. We should use:

v2 = v1 + at

If we plug in the values listed above, we should get:

v2 = 0 + (6)(55)

Next, we should solve the problem by performing the multiplication on the right side of the equation.

v2 = 330 m/s

Therefore, the final velocity reached by the object is 330 m/s.

Hope this helps!

7 0

3 years ago

When you push on an object such as a wrench, a steel pry bar, or even the outer edge of a door, you produce a torque equal to th

Korvikt [17]

Since youc oncetrate all your force directly towards the moment arm it means that you push it at an angle of your force is directed to the left or the right and I bet that it must be 90 degrees to the bar. Obviuosly, if you are about to push it you will do it straight up but not in a zig zag way. In other words, it should be perpendicular to the arm because the torque can be produced only if force is applied at a constant index (90).
Hope that helps! Regards.

5 0

3 years ago

A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the

WITCHER [35]

Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\$

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

$(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\$

v_f = 10.85 m/s

8 0

3 years ago

Need help with this question. Thirty points.

Anon25 [30]

$1)\\\\\text{Kinetic energy,}\\\\E_k = \dfrac 12 mv^2 \\\\\implies m = \dfrac{2E_k}{v^2} = \dfrac{2\times 4500}{30^2}=10~ kg\\ \\2)\\\\\text{Kinetic energy,}\\\\E_k = \dfrac 12mv^2 \\\\\implies v^2 = \dfrac{2E_k}m\\\\\implies v = \sqrt{\dfrac{2E_k}{m}}=\sqrt{\dfrac{2 \times 320}{\dfrac{20}{9.8}}}= 17.709 ~ ms^{-1}\\\\3)\\\\\text{Kinetic energy,}\\\\E_k=\dfrac 12 mv^2=\dfrac 12 \times 50 \times 10^2=2500 ~J\\\\\\$

$4)\\\\\text{Potential energy,}\\\\E_p =mgh = 5\times 9.8\times 1.5 = 73.5 ~J$

5 0

2 years ago

There are 15 water molecules in the air, and the relative

AlexFokin [52]

30 30 30 30 30 30 30 30 30 30

6 0

2 years ago