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Ann [662]
3 years ago
14

A hockey player uses a 400- gram stick to apply 50 newtons of force to a 170-gram puck, causing a momentum of change 5kg x m/s.

What is the impulse experienced by the hockey puck?
Physics
1 answer:
gavmur [86]3 years ago
4 0

Answer: 5 N•s

Explanation:

An impulse will result in a change of momentum.

If the momentum changed 5 kg•m/s, you can be certain that an impulse of 5 N•s has been applied.

The units for Impulse and momentum are identical

kg•m/s = N•s

if we divide each side by s

kg•m/s/s = N•s/s

kg•m/s² = N

which is probably familiar already

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A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou
koban [17]

Answer:

The value is  f  =  0.05 \ Hz

Explanation:

From the question we are told that

     The mass of the car is  m  =  615 \  kg

      The period of the circular motion is  T  =  20 \  s

      The radius  is r =  80 \  m/s

Generally the frequency of the circular motion is  

       f  =   \frac{1}{T }

=>    f  =   \frac{1}{ 20  }

=>    f  =  0.05 \ Hz

3 0
3 years ago
A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc
Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0
3 years ago
I forgot this someone help
dexar [7]

Answer:

A. water + carbon dioxide + sunlight = oxygen + <u>glucose </u><u>!</u>

6 0
2 years ago
What is the final velocity of a car that starts at 22 m/s and accelerates at 3.78 m/s for distance of 45 m
Pepsi [2]

v^2 = v0^2 +2ad v^2 = 22^2 + 2*3.78*45 = 824.2 v= √824.2 = 28.7 m/s

5 0
4 years ago
Another athlete using a different spring exerts an average force of 400 N to enable her
babymother [125]

Answer:84Nm

Explanation:

force=400N

Distance=0.210m

Workdone=force x distance

Workdone=400 x 0.210

Workdone=84Nm

6 0
3 years ago
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