Ask question

Ask question

All categories

3 years ago

14

Bernie, whose mass is 70.0 kg, leaves a ski jump with a velocity of 21.0 m/s.What is Bernie’s momentum (kg*m/s) as he leaves the

ski jump?

1 answer:

zheka24 [161]3 years ago

6 0

Momentum = mass * velocity
Momentum = 70 * 21
Momentum = 1,470

Therefore, Bernie’s momentum was 1,470kg*m/s as he leaves the ski jump. :)

You might be interested in

Why collaboration in science is critical to success in the scientific community

Marianna [84]

Collaboration in science is important because if only one scientist does an experiment, and gets a result, he/she could have messed. So this is where collaboration comes in. A few other scientists could try the experiment, and if they get the same answers, the result may be proven correct.

Hope this helped!

5 0

3 years ago

The fluid inside the hydraulic jack has a pressure of 30,000 Pa. If the surface of

aleksley [76]

Explanation:

p = F /A

F = P×A

F = 30,000 Pa / 0.1 m²

F = 300,000 N

5 0

2 years ago

You have learned in biology class that a substance called chlorophyll makes leaves green. You guess that perhaps leaves change c

Shkiper50 [21]

You have formed a hypothesis.

7 0

3 years ago

Read 2 more answers

A coin dropped in the lift it takes time 0.5 s to reach the floor when lift is staionary it takes time t when lift is moving up

BARSIC [14]

Answer:

t₁ > t₂

Explanation:

A coin is dropped in a lift. It takes time t₁ to reach the floor when lift is stationary. It takes time t₂ when lift is moving up with constant acceleration. Then t₁ > t₂, t₁ = t₂, t₁ >> t₂ , t₂ > t₁

Solution:

Newton's law of motion is given by:

s = ut + (1/2)gt²;

where s is the the distance covered, u is initial velocity, g is the acceleration due to gravity and t is the time taken.

u = 0 m/s, t₁ is the time to reach ground when the light is stationary and t₂ is the time to reach ground when the lift is moving with a constant acceleration a.

hence:

When stationary:

$s=\frac{1}{2}gt_1^2\\\\t_1^2=\frac{2s}{g} \\\\When\ moving\ with\ acceleration(a):\\\\s=\frac{1}{2}(g+a)t_2^2\\\\t_2^2=\frac{2s}{g+a}$

Hence t₂ < t₁, this means that t₁ > t₂.

4 0

3 years ago

A 6.7 kg block is released from rest on a frictionless inclined plane making an angle of 25.7° with the horizontal. Calculate th

yulyashka [42]

Answer:

So time taken by block to travel 0.5 m from rest will be 0.4853 sec

Explanation:

We have given mass of the block m = 6.7 kg

angle of inclination $\Theta =25.7^{\circ}$

Distance traveled s = 0.5 m

Initial velocity u = 0 m/sec

Acceleration of the block $a=gsin\Theta =9.8\times sin25.7^{\circ}=4.243m/sec^2$

From second equation of motion we know that $S=ut+\frac{1}{2}at^2$

So $0.5=0\times t+\frac{1}{2}\times 4.243\times t^2$

$t^2=0.2356$

t = 0.4853 sec

So time taken by block to travel 0.5 m from rest will be 0.4853 sec

5 0

3 years ago