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kozerog [31]
3 years ago
15

Identify the action and reaction forces in the following situations: a) Earth attracts the Moon, b) a boy kicks a football, c) a

rocket accelerates upward, d) a car accelerates forward, e) a high jumper leaps, and f) a bullet is shot from a gun.
Physics
1 answer:
jeka943 years ago
7 0

Answer:

#See solution for details

Explanation:

a. Action: Earth pulls on the Moon, reaction: Moon pulls on Earth;

b. Action: foot applies force to ball, reaction: ball applies force to foot;

c. Action: rocket pushes on gas, reaction: gas pushes back on rocket;

d. Action: car tires push backward on road, reaction: road pushes forward on tires;

e. Action: jumper pushes down on ground, reaction: ground pushes up on jumper;

f. Action: gun pushes forward on bullet, reaction: bullet pushes backward on gun

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A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.
Katen [24]

Answer:

9.4 m/s

Explanation:

The work-energy theorem states that the work done on an object is equal to the change in kinetic energy of the object.

So we can write:

W=K_f - K_i

where in this problem:

W = -36.733 J is the work performed on the car (negative because its direction is opposite to the motion of the car)

K_i = 66,120 J is the initial kinetic energy of the car

K_f is the final kinetic energy

Solving for Kf,

K_f = W+K_i = -36,733+66,120=29,387 J

The kinetic energy of the car can be also written as

K_f = \frac{1}{2}mv^2

where:

m = 661 kg is the mass of the car

v is its final speed

Solving, we find

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s

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3 years ago
What is the moment of inertia of a disc of mass 5kg and radius 10cm?
vodka [1.7K]

Answer:

500

Explanation:

6 0
3 years ago
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As rotational speed increases, thrust____?
never [62]
Increases exponentially is your correct answer
6 0
3 years ago
How can accuracy be limited?
lozanna [386]

Answer:

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A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra
blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

y=\sqrt{1+x^3}

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

Now at (2,3)

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s

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2 years ago
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