________________________ is the use of circuitry to manipulate the transfer of electrical energy by controlling the closed path.

How does friction affect forward motion give an example

LenaWriter [7]

Friction is an opposing force to motion. A car driving fast, slams on the brakes, locking up and the wheels. The car skids to a stop as a result of the friction between the tires and the pavement.

8 0

3 years ago

A uniform disk a uniform hoop and a uniform sphere are released at the same time at the top of an inclined ramp. They all roll w

lara [203]

Answer:

D. The sphere the disk and the hoop

Explanation:

This is because the sphere has inertial of

2/5mR²

Disk 1/2mR²

Hope mR²

So these are moment of inertial which is resistance or opposition to rotation so since the sphere has a smaller moment to inertial it will move faster and reach the ground first then the disk then the hoop in that order

6 0

3 years ago

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr

fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

$\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}$

The values are

$\displaystyle y_o=15\ m$

$\displaystyle v_o=14\ m/s$

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

$\displaystyle y=0$

$\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0$

Multiplying by 2

$\displaystyle 2y_o+2v_ot-gt^2=0$

Clearing the coefficient of $t^2$

$\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0$

Plugging in the given values, we reach to a second-degree equation

$\displaystyle t^2-2.857t-3.061=0$

The equation has two roots, but we only keep the positive root

$\displaystyle \boxed {t=3.69\ s}$

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

$\displaystyle Y_r=15+V_r.t$