All categories

2 years ago

A roller of radius 12.5 cm turns at 14 revolutions per second. What is the linear velocity of the roller in meters per second?

Physics

2 answers:

Burka [1]2 years ago

8 0

Answer:

Linear velocity of the roller, v = 11 m/s

Explanation:

It is given that,

Radius of roller, r = 12.5 cm = 0.125 m

Angular velocity of the roller, $\omega=14\ rev/s$

Firstly, we will convert revolution per second to radian per second i.e.

Angular velocity, $\omega=87.96\ rad/s$

We need to find the linear velocity of the roller. It can be calculated by taking the product of angular velocity and the radius of roller.

$v=r\times \omega$

$v=0.125\ m\times 87.96\ rad/s$

v = 10.995 m/s

v = 11 m/s

So, the linear velocity of the roller is 11 m/s. Hence, this is the required solution.

Firdavs [7]2 years ago

7 0

12.5 times 14 and convert to meters its 1.75 meters per second

You might be interested in

A dolphin is able to tell in the dark that the ultrasound echoes received from two sharks come from two different objects only i

Vlad1618 [11]

Answer:

a) Wavelength of the ultrasound wave = 0.0143 m <<< 3.5m, hence its ability is not limited by the ultrasound's wavelength.

b) Minimum time difference between the oscillations = Period of oscillation = 0.00952 ms

Explanation:

The frequency of the ultrasound wave = 105 KHz = 105000 Hz. The speed of ultrasound waves in water ≈ 1500 m/s. Wavelength = ?

v = fλ

λ = v/f = 1500/105000 = 0.0143 m <<< 3.5m

This value, 0.0143m is way less than the 3.5m presented in the question, hence, this ability is not limited by the ultrasound's wavelength.

b) Minimum time difference between the oscillations = The period of oscillation = 1/f = 1/105000 = 0.00000952s = 0.00952 ms

Hope this helps!

6 0

2 years ago

A 1 mg ball carrying a charge of 2 x 10-8 C hangs from a

Fed [463]

Answer:

σ = 0.255*10^-3 C/m²

Explanation:

The Electric field Intensity act due to plate = σ/ε₀, where σ is surface charge density of plate.

At equilibrium ,

Upward force = downward force

Tcosθ = mg ----(1)

Assuming that the Forward force = backward force, then

Tsinθ = σq/ε₀

[ ∵ F = qE , ∴ F = qσ/ε₀ ] -----(2)

Dividing equation (2) by (1)

Tsinθ/Tcosθ = qσ/ε₀mg

⇒Tanθ = qσ/ε₀mg

σ = ε₀mg tanθ/q

Now substituting the values of

σ = (8.85*10^-12 * 1 * tan 30) / 2*10^-8

σ = (8.85*10^-12 * 0.5774) / 2*10^-8

σ = 5.11*10^-12 / 2*10^-8

σ = 0.255*10^-3 C/m²

7 0

2 years ago

Two roads intersect at right angles, one going north-south, the other east-west. an observer stands on the road 60 meters south

Sloan [31]

observer is standing at distance d = 60 m south from the intersection

cyclist is travelling at speed v = 10 m/s

now after t = 8 s its displacement from intersection is given by

$x = 10*8 = 80 m$

so the position of cyclist makes an angle with the observer

$\theta = tan^{-1}\frac{80}{60} = 53 degree$

now the component of velocity of cyclist along the line joining its position with the observer is given as

$v = v_o cos\phi$

here

$\phi = 90 -\theta$

$\phi = 90 - 53 = 37 degree$

$v = 10 cos37 = 8 m/s$

so at this instant cyclist is moving away with speed 8 m/s

7 0

2 years ago

Read 2 more answers

A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d

marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 35.4}$

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0

2 years ago

PLEASE HELP!!!!!! WILL GIVE BRAINLIEST TO WHOEVER ANSWERS WITH THE RIGHT ANSWER !!!!!!!!

Solnce55 [7]

It would be B and D your welcome

7 0

2 years ago

Read 2 more answers