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3 years ago

You are sliding across a hockey rink and catch a hockey puck as you continue to slide. What type of collision is this?

Physics

2 answers:

nexus9112 [7]3 years ago

8 0

A contact,.... I Think!

Zepler [3.9K]3 years ago

3 0

It is a (Perfectly Inelastic Collision)

Though art welcome

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When a single source of light shines through an extremely thin rectangular slit and projects on a far away viewing screen, a sin

saveliy_v [14]

True
If it helps you plz brainlest me

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3 years ago

7. A 50 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is over the bar is 1.0 m/s. Neglect air resist

luda_lava [24]

Answer: 5.05 m

Explanation:

4 0

3 years ago

Read 2 more answers

A student (m = 68 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a

Gnesinka [82]

Answer:

5.7141 m

Explanation:

Here the potential and kinetic energy will balance each other

$mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}$

This is the initial velocity of the system and the final velocity is 0

t = Time taken = 0.04 seconds

F = Force = 18000 N

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}$

From Newton's second law

$F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}$

Squarring both sides

$112.11061=2\times 9.81\times h\\\Rightarrow h=\frac{112.11061}{2\times 9.81}\\\Rightarrow h=5.7141\ m$

The height from which the student fell is 5.7141 m

5 0

3 years ago

What is the least force you could exert on the refrigerator to move it?

Ulleksa [173]

<span>If the refrigerator weights 1365 and you are not exerting any vertical force on it, then the normal force is also 1365N. so Fn=1365

Fsf = Static frictional force = (coefficient of static friction) * (Normal force)

So the least for you could exert to move it is equal to the Fsf.
Fsf = (0.49)(1365N)</span><span>
</span>

8 0

3 years ago

One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t < 5.0 s P(t)=60 W, 0≤

OLga [1]

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$