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frez [133]
3 years ago
6

A 0.111 kg hockey puck moving at 55 m/s is caught by a 80 kg goalie at rest. With what speed does the goalie slide on the (frict

ionless) ice? a 3.78 m/s b 2 m/s C 0.076 m/s d 1.056 m/s​
Physics
1 answer:
Andrews [41]3 years ago
6 0

Answer: 0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

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A proton is moved so that its electric potential energy increases from 4.0 × 10-14 J to 9.0 × 10-14 J. The magnitude of the char
Kamila [148]

Answer:

B. 3.1 × 10^5 V

Explanation:

8 0
2 years ago
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Place the following structures in the appropriate category based on their location in a cell
Studentka2010 [4]

Answer:

Cell surface.

  • Cilium
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Nucleus

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3 0
3 years ago
The acceleration of gravity is -9.8 m/s2. A sling shot fires a rock straight up into the air with a speed of +39.2 m/s. 1. what
Vanyuwa [196]

Given that,

The acceleration of gravity is -9.8 m/s²

Initial velocity, u = 39.2 m/s

Time, t = 2 s

To find,

The final velocity of the shot.

Solution,

Let v is the final velocity of sling shot. Using first equation of motion to find it.

v = u +at

Here, a = -g

v = u-gt

v = (39.2)-(9.8)(2)

v = 19.6 m/s

So, its velocity after 2 seconds is 19.6 m/s.

4 0
2 years ago
A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
Irina-Kira [14]

Answer:

part (a) \alpha\ =\ -2.5\ rad/s^2

part (b) N = 79.61 rev

part (c) \tau\ =\ 23.54\ Nm

Explanation:

Given,

  • Initial speed of the wheel = w_o\ =\ 50.0\ rad/s
  • total time taken = t = 20.0 sec

part (a)

Let \alpha be the angular acceleration of the wheel.

Wheel is finally at the rest. Hence the final angular speed of the wheel is 0.

\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2

part (b)

Let \theta be the total angular displacement of the wheel from initial position till the rest.

\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad

We know,  1 revolution = 2\pi rad

Let N be the number of revolution covered by the wheel.

\therefore N\ =\ \dfrac{\theta}{2\pi}\\\Rightarrow N\ =\ \dfrac{500}{2\times 3.14}\\\Rightarrow N\ =\ 79.61\ rev

Hence the 79.61 revolution is covered by the wheel in the 20 sec.

part (c)

Given,

  • Mass of the pole = m = 4 kg
  • Length of the pole = L = 2.5 m
  • Angle of the pole with the horizontal axis = \theta\ =\ 60^o

Now the center of mass of the pole = d\ =\ \dfra{L}{2}\ =\ \dfrac{2.5}{2}\ =\ 1.25\ m

Weight component of the pole perpendicular to the center of mass = F\ =\ mgcos\theta

\therefore \tau\ =\ F\times d\\\Rightarrow \tau\ =\ 4\times 9.81\times cos60^o\times 1.25\\\Rightarrow \tau\ =\ 23.54\ Nm

3 0
3 years ago
The force of gravity acting on an object is 300 N. What is the mass of the object?btw u’ll get 25 points if u answer
Nikolay [14]

Answer:

m=30.581kg

<em>If you use 10m/s/s for your value of g then the answer is 30kg)</em>

Explanation:

Assuming this is happening on earth, acceleration due to gravity (g) is 9.81m/s/s.

Force of gravity is the acceleration due to gravity multiplied by its mass:

F=mg

300=(9.81)m

m=30.581kg

<em>If you use 10m/s/s for your value of g then the answer is 30kg)</em>

8 0
3 years ago
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