Convert the following as instructed:

Two wheels with fixed hubs and radii 0.51 m and 1.9 m, each having a mass of 3 kg, start from rest. Forces 5 N and F2 are applie

Katarina [22]

Answer:

18.63 N

Explanation:

Assuming that the sum of torques are equal

Στ = Iα

First wheel

Στ = 5 * 0.51 = 3 * (0.51)² * α

On making α subject of formula, we have

α = 2.55 / 0.7803

α = 3.27

If we make the α of each one equal to each other so that

5 / (3 * 0.51) = F2 / (3 * 1.9)

solve for F2 by making F2 the subject of the formula, we have

F2 = (3 * 1.9 * 5) / (3 * 0.51)

F2 = 28.5 / 1.53

F2 = 18.63 N

Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.

3 0

3 years ago

JOSEPH JOGS FROM END A TO OTHER END B OF A straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back

zzz [600]

(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second

<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>

3 0

3 years ago

What metric unit would you use to measure the weight of an elephant

Tems11 [23]

Answer:

kilograms

Explanation:

5 0

3 years ago

The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from

grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

$\epsilon=-N\dfrac{d\phi}{dt}$

Where

$\dfrac{d\phi}{dt}$ is the rate of change of magnetic flux,

And $\phi=BA$

$\epsilon=-NA\dfrac{dB}{dt}$

$\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}$

$\epsilon=28.32\ V$

Using Ohm's law, $\epsilon=I\times R$

Induced current, $I=\dfrac{\epsilon}{R}$

$I=\dfrac{28.32}{1.5}$

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0

2 years ago

Heavy crate sits at rest on the floor of a warehouse. you push on the crate with a force of 400 n, and it doesn't budge. what is

maks197457 [2]

Heavy crate sits at rest on the floor of a warehouse. you push on the crate with a force of 400 N, and it doesn't budge. The magnitude of the friction force on the crate in Newton is 400N

This is due to Friction force, which is defined as the resisting force that acts on a body when it is at rest (Static friction) or when it is in motion (Kinetic friction).

When a force is applied on a stationary body, the force of static friction starts to act on the body which prevents any relative motion between the object and surface. The magnitude of friction increases up to μsN, where μs is the coefficient of static friction. As the crate didn't budge, it means the amount of force applied was less than μsN. Hence the force applied was canceled by an equal and opposite amount of frictional force which was equal to 400N.

Learn more about frictional force here

brainly.com/question/1714663

#SPJ4

8 0

1 year ago