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alisha [4.7K]
2 years ago
8

M=3000km v=25m/s what’s the momentum

Physics
1 answer:
valina [46]2 years ago
8 0
<h3>Answer:</h3>

Momentum of the given body will be : 75000 Kg m/s

<h3>Explanation:</h3>

According to Newton's first law of motion, all bodies continue to be in the state of rest or motion unless an external force is applied on the body. We can use this in the case of momentum also

The formula of momentum is given by :

:\implies \sf\quad \sf \:  P = mv

Here, we are given the mass of the body ( m ) as 3000kg and the velocity of the body ( v ) as 25 m/s. On putting the values in the formula:

\begin{aligned}&:\implies \sf\quad \sf \:  P = mv \\& :\implies \sf\quad \sf \:  P = 3000 \times 25 \\ & :\implies \sf\quad \sf \:   \boxed{ \sf \: P = 75000kgm {s}^{ - 1} } \end{aligned}

Momentum is associated with the mass of the moving body and can be defined as the quantity of motion measured as a product of mass and velocity.

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A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

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We have: I=\frac{U}{R}=\frac{120}{20}=6A

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