All categories

4 years ago

Explain why a Merry-Go-Round and a Ferris Wheel have a constant acceleration when they are moving.

Physics

1 answer:

luda_lava [24]4 years ago

5 0

Explanation:What is centripetal acceleration?

Can an object accelerate if it's moving with constant speed? Yup! Many people find this counter-intuitive at first because they forget that changes in the direction of motion of an object—even if the object is maintaining a constant speed—still count as acceleration.

Acceleration is a change in velocity, either in its magnitude—i.e., speed—or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the speed might be constant. You experience this acceleration yourself when you turn a corner in your car—if you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion. What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we'll examine the direction and magnitude of that acceleration.

The figure below shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation—the center of the circular path. This direction is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration

a, start subscript, c, end subscript; centripetal means “toward the center” or “center seeking”.

You might be interested in

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit

Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

$\Delta KE = \Delta PE$

$\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1$

Where,

m = mass

$v_{f,i}$ = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

$\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2$

$\frac{1}{2} (v_f^2-v_0^2) = gh_2$

$(v_f^2-v_0^2) = 2gh_2$

$v_f = \sqrt{2gh_2+v_0^2}$

$v_f = \sqrt{2(9.8)(23.5)+13.6^2}$

$v_f = 25.4m/s$

7 0

3 years ago

The temperature of a gas is _____.

kykrilka [37]

Directly proportional to pressure

8 0

4 years ago

Read 2 more answers

A closed curve encircles several conductors. The line integral around this curve is (image attached below)

Masteriza [31]

The net current in the conductors and the value of the line integral

$I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A$
The resultant remains same 3.2 *10^4 Tm

This is further explained below.

<h3>What is the net current in the conductors?</h3>

Generally,

To put it another way, the total current In flowing across a surface S (contained by C) is proportional to the line integral of the magnetic B-field (in tesla, T).

$\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \iint_S \mathbf{J} \cdot \mathrm{d}\mathbf{S} = \mu_0I_\mathrm{enc}$

$I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A$

In conclusion, It is possible for the line integral to go around the loop in either direction (clockwise or counterclockwise), the vector area dS to point in either of the two normal directions and Ienc, which is the net current passing through the surface S, to be positive in either direction—but both directions can be chosen as positive in this example. The right-hand rule solves these ambiguities.

The resultant remains the same at 3.2 *10^4 Tm