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Murljashka [212]
3 years ago
15

A hockey puck with a mass of 0.175 kg slides over the ice. The puck initially slides with a speed of 5.25 m/s, but it comes to a

rough patch in the ice which slows it down to a speed of 2.85 m/s. How much energy is dissipated as the puck slides over the rough patch?
Physics
1 answer:
Neko [114]3 years ago
7 0

Answer:

1.70 J

Explanation:

The heat dissipated is the difference in the kinetic energies.

This is given by

E = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

v_i and v_f are the initial and final velocities.

With <em>m</em> = 0.175 kg,

E = \frac{1}{2}\times0.175(2.85^2 - 5.25^2) = -1.701\text{ J}

The negative sign appears because energy is lost.

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A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (
SIZIF [17.4K]

The solution to the questions are given as

  • t=40.39 \mathrm{sec}
  • \varepsilon &=(0.12v)e^{0.057t}
  • the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}

\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$

\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}

\varepsilon &=(0.12v)e^{0.057t}

(b) Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$

\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}

c)

In conclusion, the direction of the induced current will be Counterclockwise.

Read more about current

brainly.com/question/13076734

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4 0
2 years ago
Calculate the change in length of a Pyrex glass dish (Coefficient of linear expansion for Pyrex is 3 * 10-6 / oC) that is 0.25 m
DIA [1.3K]

9*10^{-5} m

Explanation:

Step 1:

We are given the initial length of the Pyrex glass dish at a particular temperature and need to calculate the change in the length when the temperature changes by 120° C. The coefficient of linear expansion of Pyrex is provided.

Step 2:

Change in length = Coefficient of linear expansion * Change in temperature * Initial length

Step 3:

Coefficient of linear expansion = 3*10^{-6} /°C

Change in temperature = 120°C = 120 K

Initial length = 0.25 m

Step 4:

Change in length = 3*10^{-6} * 120 * 0.25 = 9*10^{-5} m

8 0
3 years ago
A 3.0 kg rifle is held firmly by a 50.0 kg woman, initially standing still. A 0.06 kg bullet leaves the rifle muzzle with a velo
ohaa [14]

Answer:

48kg

Explanation:

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7 0
3 years ago
Why do scientists believe that light is made of streams of particles?
irina [24]
I just need points to ask my own question.
5 0
3 years ago
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During what time interval was the object traveling most quickly?
anygoal [31]

ANSWER

0s to 3s

EXPLANATION

To find the time interval where the object was traveling more quickly, we have to find the velocity of the object during each interval.

From a position-time graph, we can obtain the velocity of the object from the slope of the graph in each interval. To find those slopes, we just have to divide the vertical difference by the time interval.

The interval from 0s to 3s:

v=\frac{7-3}{0-3}=-\frac{4}{3}m/s

The interval from 3s to 5s and the interval from 7s to 8s have a horizontal line, so the slope is zero and therefore the velocity is zero - meaning that the object was not moving during these periods.

The interval from 5s to 7s,

v=\frac{5-3}{7-5}=\frac{2}{2}=1m/s

And the interval from 8s to 12s,

v=\frac{0-5}{12-8}=-\frac{5}{4}m/s

Two of these three velocities are negative. Negative velocity indicates that the object is moving backward.

From these velocities, the greatest one, in absolute value, is the one between 0s to 3s. During this interval, the object is moving backward but at the greatest velocity.

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