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Degger [83]
2 years ago
8

Two hockey pucks are sliding along the ice, towards the same goal, one directly in front of the other. The hockey puck in front

has a mass of 0.170 kg, and is moving towards the goal at a speed of 6.0 m/s. The hockey puck in the rear has a mass of 0.155 kg, and is moving towards the goal at a speed of 11 m/s. After the rear puck collides with the front puck, the rear puck has a speed of 7.5 m/s. What is the speed of the front puck after the collision?
A. 9.2 m/s
B. 2.7 m/s
C. 9.5 m/s
D. 1.6 m/s
Physics
1 answer:
grandymaker [24]2 years ago
6 0

Answer: A.9.2m/s

Explanation:

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0.5 , 54.5

Explanation:

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a(0)=0.5

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8 0
3 years ago
Which characeristic makes omasis different from diffusion.
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3 years ago
Explain how columns can be used and set up to increase the effectiveness of business documents.
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3 years ago
Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This
vova2212 [387]

Answer:

1.554\times 10^{32}\ \text{kg}

Explanation:

M = Mass of each star

T = Time period = 15.5 days

v = Orbital velocity = 230 km/s

G = Gravitational constant = 6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2

Radius of orbit is given by

R=\dfrac{vT}{2\pi}

We have the relation

\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}

The mass of each star is 1.554\times 10^{32}\ \text{kg}

6 0
2 years ago
Pa help po science po yan pang grade 7
AveGali [126]

Answer:

table 1:

1. 100/15 = 6.7 m/s

2. 100/12 = 8.3 m/s

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3. 100/4 = 25 m/s

Explanation:

8 0
2 years ago
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