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meriva
3 years ago
6

What happens when velocity and acceleration are at right angles to each other

Physics
1 answer:
photoshop1234 [79]3 years ago
7 0
You get circular motion, where the acceleration is pointing towards the center of the circle, as long as they are constant, and not fluctuating.
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The superheroine Xanaxa, who has a mass of 68.1 kg , is pursuing the 75.3 kg archvillain Lexlax. She leaps from the ground to th
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The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as

U = mg\Delta h

Here,

\Delta h = Change in height

m = mass of super heroine

g = Acceleration due to gravity

The change in height will be,

\Delta h = h_f - h_i

The final position of the heroin is below the ground level,

h_f = -16.1m

The initial height will be the zero point of our system of reference,

\Delta h = -16.1m-0m

\Delta h = -16.1m

Replacing all this values we have,

U = mg\Delta h

U = (68.1kg)(9.8m/s^2)(-16.1m)

U = -10744.81J

Since the final position of the heroine is located below the ground, there will net loss of gravitational potential energy of 10744.81J

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mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan
Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, v_e, for the planet is v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }

Where;

v_e = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, M_E

r = The radius of the planet = 3 × The radius of Earth, R_E

The escape velocity for Earth, v_e_E, is therefore given as follows;

v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }

\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } =  \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E

v_e = 16 × v_e_E

Given that the escape velocity for Earth, v_e_E ≈ 11,186 m/s, we have;

The escape velocity on the planet = v_e ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

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3 years ago
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