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Charra [1.4K]
3 years ago
7

If another vehicle is trying to pass your vehicle,

Physics
1 answer:
Vladimir [108]3 years ago
8 0

Answer:

Reduce you're speed, and let the other vehicle pass you

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Electrical forces travel far through the universe
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2 questions, brainliest for both correct! please: only answer if you are sure of the answer!! :)
Pani-rosa [81]

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Explanation:

1. B

2.D

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An airplane flying at 119 m/s is accelerated uniformly at the rate of 0.5 m/s2 for 10
miss Akunina [59]

Answer:

124m/s

Explanation:

v=u+at

v=119+(0.5×10)

v=119+5

v=124m/s

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3 years ago
7. A roller coaster’s velocity at the top of a hill is 10 m/s. Two seconds later it reaches the bottom ofthe hill with a velocit
zheka24 [161]

Answer:

7) a=8m/s^2

8) a=-5m/s^2

9) a=-1.5m/s^2

10) v=29.4m/s

Explanation:

For the problems 7, 8 and 9 we just apply the definition of acceleration, since no more information is given, which is:

a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t_f-t_i}

So for each problem we will have:

7) a=\frac{26m/s-10m/s}{2s}=8m/s^2

8) a=\frac{10m/s-25m/s}{3s}=-5m/s^2

9) a=\frac{0m/s-15m/s}{10s}=-1.5m/s^2

For the problem 10, we use the equation of velocity in accelerated motion:

v=v_0+at

Since the ball starts from rest and the acceleration is that of gravity (we take the downward direction positive), we have:

v=(0m/s)+(9.8m/s)(3s)=29.4m/s

5 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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