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bonufazy [111]
2 years ago
6

A golf ball starts with an initial velocity of 100m/s and slows at a constant rate of -8m/s2, what is its final velocity after 2

s?
Vf = Vi+at

A. 20 m/s

B. 84 m/s

C. 16 m/s
Physics
1 answer:
Illusion [34]2 years ago
8 0

Answer:

<em>Answer: B. 84 m/s</em>

Explanation:

<u>Uniform Speed Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time thus, the acceleration is constant.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the final speed is calculated as follows:

v_f=v_o+at

The golf ball starts with an initial speed of 100 m/s and slows down at a=-8\ m/s^2. We are required to find the final speed at t=2 s:

v_f=100-8*2=100-16

v_f=84\ m/s

Answer: B. 84 m/s

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Two laws are described below:
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A block of mass 0.510 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T
maria [59]

Answer:

x=0.46m, speed=7.9m/s

Explanation:

Using the concept of conservation of energy:

1. kinetic energy of mass m and velocity v: E_k=\frac{1}{2}mv^2

2. gravitational potential energy of mass m, grav. acc. g and height h: E_g=mgh

3. potential energy in a spring with spring constant k and displacement from equilibrium x: E_s=\frac{1}{2}kx^2

Calculating x:

\frac{1}{2}mv_a^2=\frac{1}{2}kx^2

x=\sqrt{\frac{m}{k}}v_a

Calculating the speed:

\frac{1}{2}mv_a^2 +mgh_a=\frac{1}{2}mv_b^2+mgh_b + W_{friction}

h_a=0, h_b=2R,W_{friction}=F_{friction}\times distance=7\pi R

\frac{1}{2}mv_a^2=\frac{1}{2}mv_b^2+2mgR+7\pi R

Solving for v_b:

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7 0
3 years ago
a body thrown vertically upwards from grounf with inital vel 40m/s then time taken by it to reach max hieght is?
777dan777 [17]

Answer:

t = 4.08 s

Explanation:

if the body is thrown upward, it has negative gravity. Knowing through the International System that the earth's gravity is 9.8 m/s²

Data:

  • Vo = 40 m/s
  • g = -9.8 m/s²
  • t = ?

Use formula:

  • \boxed{\bold{t=\frac{-(V_{0})}{g}}}

Replace and solve:

  • \boxed{\bold{t=\frac{-(40\frac{m}{s})}{-9.8\frac{m}{s^{2}}}}}
  • \boxed{\boxed{\bold{t=4.08\ s}}}

Time taken by it to reach max height is <u>4.08 seconds.</u>

Greetings.

4 0
2 years ago
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