Ask question

Ask question

All categories

jarptica [38.1K]

3 years ago

5

If an object is rolling without slipping, how does its linear speed compare to its rotational speed? If an object is rolling wit

hout slipping, how does its linear speed compare to its rotational speed? v = Rω v = ω/R They are unrelated. v = ω

1 answer:

salantis [7]3 years ago

6 0

Answer:

$v = r\omega$

Explanation:

If the object is rolling without slipping, every unit of rotated angle equals to a distance perimeter rotated.

Suppose the object complete 1 revolution within time t. The angular distance is 2π rad. Its angular velocity is 2π/t

The distance it covered is its circumference, which is 2πr, and so the speed is 2πr/t

So the linear speed compared to angular speed is

$\frac{v}{\omega} = \frac{2\pi r/t}{2\pi /t} = r$

$v = r\omega$

You might be interested in

What happens to ar object in free fall?

Inessa05 [86]

Energy from the gravitational potential store in converted to kinetic energy. Air friction acts against the object, dissipating some energy as heat or sound. The object will continuously accelerate until the acceleration is equal to the air friction acting against it. This is when it reaches terminal velocity

5 0

2 years ago

What do all wetlands have in common?

Ksivusya [100]

They are all coverd in water 24/7 they never clear up

4 0

3 years ago

Read 2 more answers

Classify the following situations into contact and non-contact forces.

NARA [144]

Answer: Contact force

a. Applying break in a vehicle.

d. The speed of ball rolling on ground is reduced

Non contact force

b. A coconut falling from a coconut tree.

c. The planets revolving around the sun.

Explanation:

The contact force is the force which exerts when one object or entity comes in contact with other object or entity. For example, on application of break the vehicle stops, the force is applied on the breaks to stop the vehicle. The ball rolling on the ground the speed reduces so the application of force on the ground also reduces.

The non contact force is the force one object exerts on the other without coming in direct contact with the other object. The force exerted by one object on other due to gravity is a non contact force. The coconut falling on the ground and planets revolving around the sun are examples of non contact force due to gravity.

8 0

2 years ago

An object weighs 79.1 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 21.8 N.

Free_Kalibri [48]

Answer:

(a). The density of the object is 1382 kg/m³.

(b). The density of the oil is 536.4 kg/m³.

Explanation:

Given that,

Weight in air = 79.1 N

Weight in water = 21.8 N

Weight in oil = 48.4 N

We need to calculate the volume of object

Using formula of buoyant force

$F_{b}=W_{air}=W_{water}$

$F_{b}=79.1-21.8$

$F_{b}=57.3\ N$

$F_{b}=\rho g h$

Put the value into the formula

$57.3=1000\times V\times 9.8$

$V=\dfrac{57.3}{1000\times9.8}$

$V=5.84\times10^{-3}\ m^3$

We need to calculate the density

Using formula of buoyant force

$F_{b}=\rho Vg$

$79.1=\rho\times5.84\times10^{-3}\times9.8$

$\rho=\dfrac{79.1}{5.84\times10^{-3}\times9.8}$

$\rho=1382\ kg/m^3$

The density of the object is 1382 kg/m³.

(b). We need to calculate the volume of object

Using formula of buoyant force

$F_{b}=W_{air}=W_{oil}$

$F_{b}=79.1-48.4$

$F_{b}=30.7\ N$

We need to calculate the density

Using formula of buoyant force

$F_{b}=\rho_{oil} Vg$

$30.7=\rho_{oil}\times5.84\times10^{-3}\times9.8$

$\rho_{oil}=\dfrac{30.7}{5.84\times10^{-3}\times9.8}$

$\rho=536.4\ kg/m^3$

The density of the oil is 536.4 kg/m³.

Hence, This is the required solution.

4 0

3 years ago

Identify one force acting on the cart.

Artist 52 [7]

Answer:

pulling force

Explanation:

because the person is pulling that cart.

5 0

3 years ago