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Dmitry [639]
3 years ago
10

Water can be used to create electricity. this type of energy is called _____. nuclear power hydroelectric power electrolysis cur

rents
Physics
1 answer:
ioda3 years ago
8 0
Hydroelectric power. That's just the term for electricity gained by using water, think of a water mill.
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Even if there were some friction on the ice, it is still possible to use conservation of momentum to solve this problem, but you
hjlf

The problem referred to in this question is missing and it is;

Two hockey pucks of identical mass are on a flat, horizontal ice hockey rink. The red puck is motionless; the blue puck is moving at 2.5 m/s to the left. It collides with the motionless red puck. The pucks have a mass of 15 g. After the collision, the red puck is moving at 2.5 m/s, to the left. What is the final velocity of the blue puck?

Answer:

The condition is that p_f - p_i which is the change in momentum will not be equal to zero but equal to the impulse (Ft).

Explanation:

In the problem described, by inspection, we can say that since there is no friction, we have a closed system and thus momentum is conserved.

Since momentum is conserved, we can say that;

Initial momentum(p_i) = final momentum(p_f)

Now, in this question we are told that some friction wants to be introduced on the ice and it's possible to still use conservation of momentum.

From impulse - momentum theory, we know that;

Impulse = change in momentum

Impulse is zero when no force is acting on the ice and we have; 0 = p_f - p_i

This will yield initial momentum = final momentum.

Now, since a force is applied, we know that impulse is; J = F × t

Thus;

Ft = p_f - p_i

Where F is the force due to friction.

Thus, the condition is that p_f - p_i will not be equal to zero

6 0
2 years ago
A rocket moves upward from rest with an acceleration of 40 m/s2 for 5 seconds. It then runs out of fuel and continues to move up
Snezhnost [94]

Answer:

Maximum height of rocket  = 2538.74 m

Explanation:

We have equation of motion s = ut + 0.5 at²

For first 5 seconds

          s = 0 x 5 + 0.5 x 40 x 5² = 500 m

Now let us find out time after 5 seconds rocket move upward.

We have the equation of motion v = u + at

After 5 seconds velocity of rocket

         v = 0 + 40 x 5 = 200 m/s

After 5 seconds the velocity reduces 9.8m/s per second due to gravity.

Time of flying after 5 seconds

          t=\frac{200}{9.81}=20.38s

Distance traveled in this 20.38 s

          s = 200 x 20.38 - 0.5 x 9.81 x 20.38² = 2038.74 m

Maximum height of rocket = 500 +2038.74 = 2538.74 m

6 0
3 years ago
Can you guys please help me with this science quiz
Xelga [282]

Answer:

1 is 90, 2 is 200 and 3 is 5

Explanation:

im big brain so i know lol

8 0
3 years ago
Review. From a large distance away, a particle of mass 2.00 g and charge 15.0σC is fired at 21.0 i^ m/s straight toward a second
MissTica

(a)

Determine the system's initial configuration at ri = infinite particle separation and the system's final configuration at the point of closest approach.

Since the two-particle system is not being affected by any outside forces, we may treat it as an isolated system for momentum and use the momentum conservation law.

m1v1 + m1v2 = (m1+m2)v

The second particle's starting velocity is zero, so:

m1v1  = (m1+m2)v

After substituting the values we get,

v = 6i m/s

(b)

Since the two particle system is also energy-isolated, we may use the energy-conservation principle.

dK + dU = 0

Ki +Ui = Kf + Uf

Substituting the values,

1/2m1v1^2i + 1/2 m2v2^2i + 0 = 1/2m1v1^2f + 1/2m2v2^2f +ke q1q2/rf

The second particle's initial speed is 0 (v2 = 0). Additionally, both the first and second particle's final velocity have the same value, v. Put these values in place of the preceding expression:

1/2m1v1^2i  = 1/2m1v1^2 + 1/2m2v2^2 +ke q1q2/rf

After solving we get,

rf = 2ke q1q2 / m1v1^2 - (m1+m2)v^2

Substituting the values we get,

rf = 3.64m

(c)

v1f = (m1-m2 / m1 + m2) v1i

v1f  = -9i m/s

(d)

v2f =  (2m1/ m1 +m2) v1i

After substituting the values,

v2f = 12i m/ s

Question :

Review. From a large distance away, a particle of mass 2.00 g and charge 15.0 \muμC is fired at 21.0 m/s straight toward a second particle, originally stationary but free to move, with mass 5.00 g and charge 8.50 \muμC. Both particles are constrained to move only along the x axis. (a) At the instant of closest approach, both particles will be moving at the same velocity. Find this velocity. (b) Find the distance of closest approach. After the interaction, the particles will move far apart again. At this time, find the velocity of (c) the 2.00-g particle and (d) the 5.00-g particle. \hat{i}

To learn more about  momentum conservation law click on the link below:

brainly.com/question/7538238

#SPJ4

5 0
2 years ago
a 74.9 kg person sits at rest on an icy pond holding a physics book. he throws the physics book west at 8.25 m/s and he recoils
kifflom [539]

Answer:

1.95 kg

Explanation:

Momentum is conserved.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

0 = (74.9) (-0.215) + m (8.25)

m = 1.95

3 0
3 years ago
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