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inysia [295]
3 years ago
7

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

e goalie and the puck have an elastic collision, and the puck is reflected back in the direction from which it came. What would the final velocities of the goalie and the puck be in this case? Assume that the collision is completely elastic.v goalie = _____ m / sv puck = ______ m / s
Physics
1 answer:
NISA [10]3 years ago
4 0

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

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A 800 kg safe is 2.1 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses
stellarik [79]

Answer:

k = 17043.5 N/m = 17.04 KN/m

Explanation:

First we need to find the force applied by safe pn the spring:

F = Weight of Safe

F = mg

where,

F = Force Applied by the safe on the spring = ?

m = mass of safe = 800 kg

g = 9.8 m/s²

Therefore,

F = (800 kg)(9.8 m/s²)

F = 7840 N

Now, using Hooke's Law:

F = kΔx

where,

K = Spring Constant = ?

Δx = compression = 46 cm = 0.46 m

Therefore,

7840 N = k (0.46 m)

k = 7840 N/0.46 m

<u>k = 17043.5 N/m = 17.04 KN/m</u>

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4 years ago
The value of a scientific variable ____. A) can change
gregori [183]
The answer is A: can change
5 0
3 years ago
Read 2 more answers
Assertion – Reason
Norma-Jean [14]

u have not displayed the question fella but i can guess the answer

Explanation:

the answer is (i)

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3 0
3 years ago
Water drips from a shower head (the sprayer at the top of the shower) and falls onto the floor 2.4 m below. The droplets are fal
expeople1 [14]

Answer:

The third drop is 0.26m

Explanation:

The drop 1 impacts at time T is given by:

T=sqrt(2h/g)

T= sqrt[(2×2.4)/9.8]

T= sqrt(4.8/9.8)

T= sqrt(0.4898)

T= 0.70seconds

4th drops starts at dT=0.70/3= 0.23seconds

The interval between the drops is 0.23seconds

Third drop will fall at t= 0.23

h=1/2gt^2

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4 0
3 years ago
Torque can cause the angular momentum vector to rotate in UCM. This motion is called ___________.
emmainna [20.7K]

Torque can cause the angular momentum vector to rotate in UCM. This motion is called _Conservation of Angular momentum__________.

Answer:

Conservation of Angular momentum

Explanation:

The motion of an object in a circular path at constant speed is known as uniform circular motion (UCM). An object in UCM is constantly changing direction, and since velocity is a vector and has direction, you could say that an object undergoing UCM has a constantly changing velocity, even if its speed remains constant.

The law of conservation of angular momentum states that when no external torque acts on an object, no change of angular momentum will occur.

Key Points

When an object is spinning in a closed system and no external torques are applied to it, it will have no change in angular momentum.

The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation.

If the net torque is zero, then angular momentum is constant or conserved.

Angular Momentum

The conserved quantity we are investigating is called angular momentum. The symbol for angular momentum is the letter L. Just as linear momentum is conserved when there is no net external forces, angular momentum is constant or conserved when the net torque is zero. We can see this by considering Newton’s 2nd law for rotational motion:

τ→=dL→dt, where  

τ is the torque. For the situation in which the net torque is zero,  

dL→dt=0.

If the change in angular momentum ΔL is zero, then the angular momentum is constant; therefore,

⇒

L  =constant

L=constant (when net τ=0).

This is an expression for the law of conservation of angular momentum.

Example and Implications

An example of conservation of angular momentum is seen in an ice skater executing a spin,  The net torque on her is very close to zero,

because (1) there is relatively little friction between her skates and the ice, and (2) the friction is exerted very close to the pivot point.

Conservation of angular momentum is one of the key conservation laws in physics, along with the conservation laws for energy and (linear) momentum. These laws are applicable even in microscopic domains where quantum mechanics governs; they exist due to inherent symmetries present in nature.

7 0
3 years ago
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