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garri49 [273]
3 years ago
5

The net force of a heavy, stationary box is?

Physics
1 answer:
Leto [7]3 years ago
7 0

Answer:

gravity

Explanation:

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Jill can use a force of 12.0 N to lift a single box up 5.00 m. How many of these boxes must she lift it in a minute to use 60.0
Mama L [17]

Answer:

60 boxes

Explanation:

The work done by lifting a single box is equal to the force applied (the weight of the box) times the displacement of the box:

W_1 = Fd=(12.0 N)(5.00 m)=60 J

Power is related to the work done by the equation:

P=\frac{W}{t}

where W is the work done and t is the time. In this problem, we are told that the power used is P=60.0 W, while the time taken is t = 1 min = 60 s, so the total work done must be

W=Pt=(60.0 W)(60 s)=3600 J

Therefore, the number of boxes that she has to lift in order to use this power is the total work divided by the work done in lifting each box:

N=\frac{W}{W_1}=\frac{3,600 J}{60 J}=60

5 0
3 years ago
1.) A stone falls from rest from the top of a cliff.
KengaRu [80]

Answer:

Explanation:

ignore air resistance

Let t be the time of fall for the dropped stone.

½(9.8)t² = 43.12(t - 2.2) + ½(9.8)(t - 2.2)²

4.9t² = 43.12t - 94.864 + 4.9(t² - 4.4t + 4.84)

4.9t² = 43.12t - 94.864 + 4.9t² - 21.56t + 23.716

     0 = 21.56t - 71.148

t = 71.148/21.56 = 3.3 s

h = ½(9.8)3.3² = 53.361 = 53 m

or

h = 43.12(3.3 - 2.2) + ½(9.8)(3.3 - 2.2)² = 53.361 = 53 m

4 0
3 years ago
the statement that current is equal to the voltage difference divided by the resistance is known as what
dangina [55]

Answer:it is known as (Ohm's law)

Explanation:

3 0
3 years ago
After the collision, car 1 has a final velocity of 6m/s south. What is the final velocity of car2?
dedylja [7]
I’m almost 100 percent sure it’s the first one
4 0
3 years ago
Four velcro-lined air-hockey disks collide with each other in a perfectly
Reil [10]

Answer:

The magnitude of the final velocity is approximately 0.526 m/s in approximately the direction of 8.746° East of South

Explanation:

The given collision parameters are;

The kind of collision experienced by the four velcro-lined air-hockey disk = Inelastic collision

The mass of the first disk, m₁ = 50.0 g

The velocity of the first disk, v₁ = 0.80 m/s West = -0.8·i

The mass of the second disk, m₂ = 60.0 g

The velocity of the second disk, v₂ = 2.50 m/s North = 2.5·j

The mass of the third disk, m₃ = 100.0 g

The velocity of the third disk, v₃ = 0.20 m/s East = 0.20·i

The mass of the fourth disk, m₄ = 40.0 g

The velocity of the fourth disk, v₄ = 0.50 m/s South = -0.50·j

Therefore, the total initial momentum of the four velcro-lined air-hockey disk, \Sigma P_{initial} is given as follows;

\Sigma P_{initial} = m₁·v₁ + m₂·v₂ + m₃·v₃ + m₄·v₄ = 50.0×(-0.80·i) + 60.0×(2.50·j) + 100 × (0.20·i) + 40.0 × (-0.50·j)

∴ \Sigma P_{initial} = -40·i + 150·j + 20·i - 20·j = -20·i + 130·j

∴ \Sigma P_{initial} = -20·i + 130·j

By the law of conservation of linear momentum, we have;

\Sigma P_{initial} = \Sigma P _{final} = -20·i + 130·j

Therefore, given that the collision is perfectly inelastic, the disks move as one after the collision and the four masses are added to form one mass, "m", m = m₁ + m₂ + m₃ + m₄ = 50.0 + 60.0 + 100.0 + 40.0 = 250.0

∴ m = 250.0 g

Let, "v" represent the final velocity of the four disks moving as one after the collision

We have;

\Sigma P _{final} = m × v = 250.0 × v = -20·i + 130·j

∴ v = -20·i/250 + 130·j/250 = -0.08·i + 0.52·j

The final velocity of the four disks after collision, v = -0.08·i + 0.52·j

The magnitude of the final velocity, \left | v \right | = √((-0.08)² + (0.52)²) ≈ 0.526

\left | v \right | ≈ 0.526 m/s

The direction of the final velocity, θ = arctan(0.52/(-0.08)) ≈ -81.254°

The direction of the final velocity, θ ≈ -81.254° which is 8.746° East of South

4 0
3 years ago
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