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Mandarinka [93]
3 years ago
5

A swimmer speeds up from 1.1 m/s to 3.2 m/s during the last 13.0 seconds of the race. What is the acceleration of the swimmer?​

Physics
1 answer:
rodikova [14]3 years ago
7 0

Answer:

a = 0.16 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f}=v_{o}+a*t\\

where:

Vf = final velocity = 3.2 [m/s]

Vo = initial velocity = 1.1 [m/s]

t = time = 13 [s]

a = acceleration [m/s²]

Now replacing:

3.2=1.1+a*13\\2.1=13*a\\a=0.16[m/s^{2} ]

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The energy transfer diagram represents the energy of a person diving into a pool.
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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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