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Yuri [45]
2 years ago
15

Calculate the acceleration for the following data: Initial Velocity=20 m/s, Final

Physics
1 answer:
Korolek [52]2 years ago
7 0

Answer:

a = 3 [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 = 60 [m/s]

Vo = initial velocity = 20 [m/s]

a = acceleration [m/s²]

t = time = 10 [s]

Now replacing these values we have:

60 = 20 + (a*10)

60 - 20 = 10*a

30 = 10*a

a = 3 [m/s²]

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The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 651 nm in vacuum. This light propag
sukhopar [10]

Answer:

v=2.58\times10^8m/s

Explanation:

The index of refraction is equal to the speed of light c in vacuum divided by its speed v in a substance, or n=\frac{c}{v}. For our case we want to use v=\frac{c}{n}, which for our values is equal to:

v=\frac{c}{n}=\frac{299792458m/s}{1.16}=258441774.138m/s

Which we will express with 3 significant figures (since a product or quotient must contain the same number of significant figures as the measurement with the  <em>least</em> number of significant figures):

v=2.58\times10^8m/s

4 0
3 years ago
Guys please help me out I’ll give extra points
zvonat [6]

Answer: h = 3.34 m

Explanation:

If the hat is thrown straight up, then at its highest point it has no motion and no kinetic energy. All energy is potential energy

PE = mgh

h = PE/mg = 4.92 / (0.150(9.81)) = 3.34352... ≈3.34 m

8 0
3 years ago
The movement of an object at a constant speed around a circular radius is known as
forsale [732]

Answer:

Uniform Circular Motion

Explanation:

Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. At all instances, the object is moving tangent to the circle.

8 0
3 years ago
A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0
3 years ago
Can anyone help me with this??? Anyone familiar with this sorta project??
SIZIF [17.4K]

yah set up an experiment do u have the rocks with u?


7 0
2 years ago
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