Ask question

Ask question

All categories

3 years ago

5

On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a rough patch that reduces her speed

by 45% due to a friction force that is 25% of her weight. Use the work‑energy theorem to find the length of this rough patch.

1 answer:

JulsSmile [24]3 years ago

7 0

Answer: 1.25 m

Explanation:

Given

initial velocity $(v_i )=3 m/s$

Final velocity $(v_f)=0.55\times 3=1.65 m/s$

Change in kinetic Energy =work done by Friction

change in Kinetic Energy $=\frac{m}{2}\left ( v_i^2-v_f^2\right )$

work done by friction $=\mu mgL$

$\frac{m}{2}\left ( 3^2-1.65^2\right )=0.25\cdot mg\times L$

$3.135=0.25\times 9.8\times L$

L=1.25

You might be interested in

Am 1 years old UwU (JK 15)

igomit [66]

Answer:

so am i ._. 15 not 1 ofc lol

Explanation:

7 0

3 years ago

Which type of light ray can produce identical but reversed

barxatty [35]

Answer: Identical rays are refracted upon entering

Explanation:

4 0

3 years ago

Read 2 more answers

The cube with 2.00m wide and 2.00m long and 2.00m high has a weight of 960.00 N what pressure does it exert

andre [41]

2m X 2m = 4m ^2
960n / 4 = 240pa

3 0

3 years ago

A computer hard disk starts from rest, then speeds up with an angular acceleration of 190 rad/s2rad/s2 until it reaches its fina

Otrada [13]

Answer:

962 rpm.

Explanation:

given,

angular acceleration = 190 rad/s²

initial angular speed = 0 rad/s

final angular speed = 7200 rpm

= $7200\times\dfrac{2\pi}{60}$

= $754\ rad/s$

we need to calculate the revolution of disk after 10 s.

time taken to reach the final angular velocity

using equation of angular motion

$\omega_f - \omega_i = \alpha t$

$754 - 0 =190\times t$

t = 4 s

rotation of wheel in 4 s

$\theta =\omega_i t+ \dfrac{1}{2}\alpha t$

$\theta = \dfrac{1}{2}\alpha t^2$

$\theta = \dfrac{1}{2}\times 190 \times 4^2$

θ = 1520 rad

$\theta = \dfrac{1520}{2\pi}$

$\theta =242\ rev$

now, revolution of the disk in next 6 s

angular velocity is constant

$\omega_f = \dfrac{\theta_f-\theta_i}{t_f-t_i}$

$754 = \dfrac{\theta_f-1520}{10-4}$

θ_f = 6044 rad

θ_f = $\dfrac{6044}{2\pi}$

revolution of the computer hard disk

θ_f = 962 rpm.

total revolution of the computer disk after 10 s is equal to 962 rpm.

3 0

3 years ago

An engine is designed to obtain energy from the temperature gradient of the ocean. What is the thermodynamic efficiency of such

rewona [7]

Answer:

0.035 (3.5 %)

Explanation:

The thermodynamic efficiency is given by:

$\eta = 1 - \frac{T_C}{T_H}$

where

$T_C$ is the cold temperature

$T_H$ is the hot temperature

In this problem we have

$T_C = 5 ^{\circ}C+ 273 = 278 KT_H = 15^{\circ}C+273 = 288 K$

So the efficiency is

$\eta = 1 - \frac{278 K}{288 K}=0.035$

6 0

3 years ago