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CaHeK987 [17]
2 years ago
5

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?​
Physics
1 answer:
madreJ [45]2 years ago
8 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 glider moving with a speed of 200 kilometers/hour experiences a cross wind of 30 kilometers/hour. What is the resultant speed
liraira [26]
You didn't say so, but we must assume that the "200 km/hr" is
the glider's air-speed, that is, speed relative to the air. 

If the air itself is moving at 30 km/hr relative to the ground and
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                 √(200² + 30²)

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2 years ago
A pitcher exerts a force on a baseball that is 30 times the balls weight. How fast is the pitcher accelerating the ball?
iVinArrow [24]

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 Pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

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  Force applied on baseball = 30 times weight of the ball.

   Weight of ball = mg, where m is the mass of ball and g is acceleration due to gravity value.

  We have force applied is also equal to product of mass and acceleration.

                            F = ma = 30 x mg

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Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe
AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

Mm=44.0 g/mol is the molar mass of the carbon dioxide

So, the number of moles of the gas is

n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol

Now we can find the pressure of the gas by using the ideal gas equation:

pV=nRT

where

p is the pressure

V=1.00 L = 0.001 m^3 is the volume

n = 0.023 mol is the number of moles

R=8.314 J/mol K is the gas constant

T=25.0^{\circ}+273=298 K is the temperature of the gas

Solving the equation for p, we find

p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa

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