PLEASE HELP <br> How much work is required to raise a 1200 kg rock 15 m?

Two identical wind-up cars A and B are released. Car B has a 2 kilogram weight strapped to the back of the car. Which will have

koban [17]

Answer:

Car A would have a better average speed

Explanation:

added weight to a object that is self propelled will be slower than a identical object with no added weight

7 0

3 years ago

The angular momentum of a flywheel having a rotational inertia of 0.140 kg ·m2 about its central axis decreases from 3.00 to 0.8

Rasek [7]

Answer

given,

I = 0.140 kg ·m²

decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.

a) $\tau = \dfrac{\Delta L}{\Delta t}$

$\tau = \dfrac{0.8-3}{1.5}$

τ = -1.467 N m

b) angle at which fly wheel will turn

$\theta= \omega t +\dfrac{1}{2}\alpha t^2$

$\theta= \dfrac{L}{I} t +\dfrac{1}{2}\dfrac{\tau}{I}t^2$

$\theta= \dfrac{3}{0.14}\times 1.5+\dfrac{1}{2}\dfrac{-1.467}{0.14}\times 1.5^2$

θ = 20.35 rad

c) work done on the wheel

W = τ x θ

W = -1.467 x 20.35 rad

W = -29.86 J

d) average power of wheel

$P_{av} =-\dfrac{W}{t}$

$P_{av} =-\dfrac{(-29.86)}{1.5}$

$P_{av} =19.91\ W$

7 0

3 years ago

Read 2 more answers

Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale?

lesantik [10]

The equilibrium conditions allow to find the results for the balance forces are:

F₁ = 225.4 N

F₂ = 558.6 N

When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.

∑ F = 0

∑ τ = 0

Where F are the forces and τ the torques.

The torque is the product of the force and the perpendicular distance to the point of support,

The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.

We write the translational equilibrium condition.

F₁ - W₁ - W₂ + F₂ = 0

We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.

F₂ 2 - W₁ 1 - W₂ 1.5 = 0 $\frac{W_1 \ 1 + W_2 \ 1.5}{2}$

Let's calculate F₂

F₂ = $\frac{W_1 \ 1 + W_2 \ 1.5 }{2}$

F₂ = (m g + M g 1.5)/ 2

F₂ = $\frac{(12 + 68 \ 1.5 ) \ 9.8}{2}$

F₂ = 558.6 N

We substitute in the translational equilibrium equation.

F₁ = W₁ + W₂ - F₂

F₁ = (m + M) g - F₂

F₁ = (12 +68) 9.8 - 558.6

F₁ = 225.4 N

In conclusion using the equilibrium conditions we can find the forces of the balance are:

F₁ = 225.4 N

F2 = 558.6 N

Learn more here: brainly.com/question/12830892

5 0

2 years ago

How can you tell one element from another?

3241004551 [841]

Their atomic number?? (easy guess)

6 0

3 years ago

How is heat transfer involved in the movement of tectonic plates?

Simora [160]

Mantle convection

Explanation:

The driver of moving plates is the mantle convection.

Convection is a heat transfer process that involves the movement of fluids from one place to another.

How does convection occur in the mantle?

Part of the mantle known as the asthenosphere is a hot liquid. It moves slowly and does not readily flow.
Over riding the asthenosphere is the brittle lithosphere made up of the upper mantle and the entirety of the crust.
The asthenosphere is at a higher temperature compared to the lithosphere on top.
In order to achieve stability and thermal equilibrium, the asthenosphere rises to the surface whereas the dense and cold lithosphere sinks down.
This way a convective cell is set up.
Hot materials rises to the surface because they are light whereas cold materials on the surface sinks into the asthenosphere.
Through this the plates moves.

Learn more:

Mantle brainly.com/question/9582362

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8 0

3 years ago

PLEASE HELP How much work is required to raise a 1200 kg rock 15 m?