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Fofino [41]
3 years ago
9

A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of

friction of 2.7 N opposing the motion. what is the acceleration of the hockey puck?
Choices:
0.0054 m/s^2
0.005 m/s^2
200 m/s^2
190 m/s^2
Physics
1 answer:
GREYUIT [131]3 years ago
5 0

Answer:

a=190\ m/s^2

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2

So, the acceleration of the hockey puck is 190\ m/s^2.

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The terminal velocity is not dependent on which one of the following properties? the drag coefficient 1 the force of gravity 2 c
ahrayia [7]
<h2>Answer: the falling time</h2>

Explanation:

When a body or object falls, basically two forces act on it:  

1. The force of air friction, also called<em> </em><u><em>"drag force"</em></u> D:  

D={C}_{d}\frac{\rho V^{2} }{2}A  (1)

Where:  

C_ {d} is the drag coefficient  

\rho is the density  of the fluid (air for example)

V is the velocity  

A is the transversal area of the object

So, this force is proportional to the transversal area of ​​the falling element and to the square of the velocity.  

2. Its <u>weight </u>due to the gravity force W:  

W=m.g

(2)

Where:  

m is the mass of the object

g is the acceleration due gravity  

So, at the moment <u>when the drag force equals the gravity force, the object will have its terminal velocity:</u>

D=W (3)

{C}_{d}\frac{\rho V^{2} }{2}A=m.g  (4)

V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}}  (5) This is the terminal velocity

As we can see, there is no "falling time" in this equation.

Therefore, the terminal velocity is not dependent on the falling time.

6 0
2 years ago
During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 33.5 rad/s. Find the angular displ
zavuch27 [327]

Answer: angular displacement in rad = 3038.45 rad

angular displacement in rev = 483.589 rev

Explanation: mathematically

Angular velocity = angular displacement / time taken.

Angular velocity = 33.5 rad/s, time taken = 90.7s

33.5 = angular displacement /90.7

Angular displacement = 33.5 * 90.7 = 3038.45 rad

But 1 rev =2π

Hence 3038.45 rad to rev is

3038.45/2π = 483.599 rev

7 0
3 years ago
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What is the mass of a dog that weighs 382 N?(unit=kg)
Luba_88 [7]

Answer:

The answer to your question is: mass = 38.93 kg

Explanation:

Data

mass = ?

Weight = 382 N

gravity = 9.81 m/s2

Formula

Weight = mass x gravity

mass = weight / gravity

mass = 382 / 9.81         substitution

mass = 38.93 kg           result

6 0
3 years ago
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____ [38]

What is your theory on frame of reference?

Answer:

It is the idea of seeing something from a different perspective

Explanation:

Hope this helps!

8 0
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Radda [10]
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5 0
3 years ago
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