A load of bricks is being lifted by a crane at a steady velocity of 5.0m/s when one bricks falls off 6.0 m above the ground

When a wheel is rotated through an angle of 21◦ , a point on the circumference travels through an arc length of 2.1 m. When the

LiRa [457]

Answer:

R = 5.73 m

Explanation:

For an angle of rotation through 21 degree we know that

arc length is given as

$angle \times Radius = Arc$

now we know that

Arc = 2.1 m

Angle = 21 degree

$Angle = 21\times \frac{\pi}{180}$

so now we have

$21\times \frac{\pi}{180} = \frac{2.1}{R}$

$R = \frac{2.1 \times 180}{21 \times \pi}$

$R = 5.73 m$

6 0

3 years ago

ASE (National Institute for Automotive Service Excellence) certified professionals must be retested every _______ years to keep

Bess [88]

Hi,

The correct answer is C. five

I hope I helped :)

4 0

3 years ago

A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What

Crank

Answer:

$\mu = \frac{r (2\pi f)^{2}}{g}$

Explanation:

$N$ = normal force acting on the coin

Normal force in the upward direction balances the weight of the coin, hence

$N = mg$

$f$ = frequency of rotation

Angular velocity of turntable is hence given as

$w = 2\pi f$

$r$ = distance from the axis of rotation

$\mu$ = minimum coefficient of static friction

static frictional force is given as

$f = \mu N\\f = \mu mg$

The static frictional force provides the necessary centripetal force , hence

Centripetal force = Static frictional force

$m r w^{2} = \mu mg\\r w^{2} = \mu g\\\\\mu = \frac{r w^{2}}{g} \\\mu = \frac{r (2\pi f)^{2}}{g}$

3 0

3 years ago

How far from the base of the platform does she land?

seropon [69]

When Janet leaves the platform, she's moving horizontally at 1.92 m/s. We assume that there's no air resistance, and gravity has no effect on horizontal motion. There's no horizontal force acting on Janet to make her move horizontally any faster or slower than 1.92 m/s.

She's in the air for 1.1 second before she hits the water.

Moving horizontally at 1.92 m/s for 1.1 second, she sails out away from the platform

(1.92 m/s) x (1.1 sec) = <em>2.112 meters</em>

3 0

3 years ago

When a cannon fires a cannonball, the cannon will recoil backward because the:

gulaghasi [49]

C) total linear momentum of the ball and cannon is conserved.

Basically it happens that in the beginning before there is a momentum acting on the two bodies, these are a unique system. Here the total momentum of the System is 0. However, when the positive momentum of the cannonball is added, the system will be immediately affected by a negative momentum which will pull back the cannon. Could this be extrapolated as a condition of Newton's third law.

4 0

3 years ago