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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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. What is the energy of an object if the height is 20 m and the mass is 2kg?
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A car traveling in a straight line has a velocity of 5.0 m/s. After an acceleration of 0.75 m/s/s, the cars velocity is 8.0. In
Bogdan [553]
Vs - velocity on beginning
ve - velocity on ending. You've got:
v_s = 5 \frac{m}{s} \\ v_e=8 \frac{m}{s} \\ \hbox{Then:} \\ \Delta v=v_e - v_s = 8 \frac{m}{s} - 5\frac{m}{s}=3 \frac{m}{s} \\ a=0,75 \frac{m}{s^2} \\ \hbox{And from formula:} \\ a=\frac{\Delta v}{\Delta t} \qquad \Rightarrow \qquad  \Delta t= \frac{\Delta v}{a} \\ \hbox{Substitute:} \\ \Delta t=\frac{3\frac{m}{s}}{0,75 \frac{m}{s^2}}= \frac{3}{\frac{3}{4}} s= 3 \cdot \frac{4}{3} s= 4 s
So he needed  4 second. 

3 0
3 years ago
Two children (each having a mass of 60 kg) are standing on the edge a merry-go-round (mass of 140 kg) as it spins with an angula
poizon [28]

Answer:

The angular velocity after the children jump off is approximately 1.4 rad/s

Explanation:

The given parameters are;

The masses of each child, m₁, and m₂ = 60 kg

The mass of the merry-go-round, m₃ = 140 kg

The initial angular velocity, \omega_i = 0.75 rad/s

The angular velocity after the children jump off = \omega_f  

According to the principle of conservation of angular momentum

The angular momentum = I × ω

The moment of inertia, I = m × R²

The total initial angular momentum = I_i \times \omega_i = m_i \times R^2 \times \omega_i

The total angular momentum after the children jump off = I_f \times \omega_f = m_f \times R^2 \times \omega_f

The initial mass, m_i = m₁ + m₂ + m₃ = 60 kg + 60 kg + 140 kg = 260 kg

The final mass, m_f = m₃ = 140 kg

According to the principle of conservation of linear momentum, we have;

I_i \times \omega_i = I_f \times \omega_f

Therefore;

260 kg × R² × 0.75 rad/s = 140 kg × R² × \omega_f

∴ \omega _f = (260 kg × R² × 0.75 rad/s)/(140 kg × R²) = 1.39285714 rad/s. ≈ 1.4 rad/s

The angular velocity after the children jump off, \omega _f ≈ 1.4 rad/s.

7 0
3 years ago
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