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Leona [35]
3 years ago
11

A 65.0-kg runner has a speed of 5.20 m/s at one instant dur- ing a long-distance event. (a) What is the runner’s kinetic energy

at this instant? (b) How much net work is required to double his speed?
Physics
1 answer:
vladimir2022 [97]3 years ago
8 0

Answer:

a)KE=878.8 J

b)W=2636.4 J      

Explanation:

Given that

mass ,m = 65 kg

Initial speed ,u = 5.2 m/s

a)

We know that kinetic energy KE is given as follows

KE=\dfrac{1}{2}mu^2

m=mass

u=velocity

Now by putting the values in the above equation we get

KE=\dfrac{1}{2}\times 65\times 5.2^2\ J

KE=878.8 J

b)

We know that

Work done by all forces = Change in the kinetic energy

The final velocity , v= 2 u = 2 x 5.2 m/s

v= 10.4 m/s

W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

Now by putting the values in the above equation we get

W=\dfrac{1}{2}\times 65\times 10.4^2-\dfrac{1}{2}\times 65\times 5.2^2\ J

W=2636.4 J

a)KE=878.8 J

b)W=2636.4 J

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Daffy Duck is standing 6.8 m away from Minnie Duck. The attractive gravitational force between them is 5.4x10-8 N. If Daffy Duck
artcher [175]

Answer:

432.78 Kg

Explanation:

From the question given above, the following data were obtained:

Distance apart (r) = 6.8 m

Force of attraction (F) = 5.4×10¯⁸ N

Mass of Daffy Duck (M₁) = 86.5 kg

Mass of Minnie Duck (M₂) =?

NOTE: Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

The mass of Minnie Duck can be obtained as follow:

F = GM₁M₂ / r²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 6.8²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 46.24

Cross multiply

6.67×10¯¹¹ × 86.5 × M₂ =5.4×10¯⁸ × 46.24

Divide both side by 6.67×10¯¹¹ × 86.5

M₂ = 5.4×10¯⁸ × 46.24 / 6.67×10¯¹¹ × 86.5

M₂ = 432.78 Kg

Therefore, the mass of Minnie Duck is 432.78 Kg

8 0
3 years ago
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