We dont see objects. We see the light ___

How does the vertical acceleration at point a compare to the vertical acceleration at point c?

Anika [276]

The vertical accelerations of the point a and point c are equal. because the vertical acceleration is due to the acceleration due to gravity. Acceleration due to gravity<span> may refer to: Gravitational acceleration, the acceleration caused by the gravitational attraction of massive bodies in general, which is equal to 9.8 m/s^2</span>

4 0

3 years ago

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In the following reaction, which substance is the precipitate? (NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq) (NH4)2SO4 Ba(

Mariana [72]

I think it's Barium sulfate, the soild and percipitate

4 0

3 years ago

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A girl moves quickly to the center of a spinning merry-go-round, traveling along the radius of the merry-go-round. Which of the

DochEvi [55]

Answer:

The angular speed of the system increases.

The moment of inertia of the system decreases.

Explanation:

As we know that the girl is going towards the center of the circle so here the moment of inertia of the girl is given as

$I = mr^2$

here we know that

r = position of the girl from the center of the disc

now we know that the girl is moving towards the center so its distance will continuously decreasing

So the moment of inertia of the girl will decrease

Now we know that that with respect to the center of the disc there is no torque on the disc + girl system

So here we can use angular momentum conservation

So we have

$I\omega = constant$

since moment of inertia is decreasing for the system

so angular speed will increase

3 0

4 years ago

A 1000 kg car makes a turn on a banked curve. The radius is 300 m.

Alenkinab [10]

b) The maximum speed of the car is 47.0 m/s

a) The centripetal force on the car is 7363 N

Explanation:

b)

We have to proceed by solving part b) first. The frictional force between the tires of the car and the road provides the centripetal force that keeps the car in circular motion, therefore we can write:

$\mu mg = m\frac{v^2}{r}$