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sertanlavr [38]
4 years ago
9

You toss a racquetball directly upward and then catch it at the same height you released it 1.82 s later. assume air resistance

is negligible. (a) what is the acceleration of the ball while it is moving upward?
Physics
1 answer:
Anton [14]4 years ago
5 0
Neglecting air resistance, the acceleration of the ball is
the acceleration of gravity ... 9.8 m/s² downward.

It doesn't matter what you toss, what it's mass is, what it weighs,
what color it is, how much it cost, what its shape or size is, how
fast you toss it, in what direction, or how long it's in the air.

Its horizontal acceleration is zero and its vertical acceleration
is 9.8 m/s² downward, from the moment it leaves your hand
until the moment somebody catches it or it hits the ground.
 
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Nataliya [291]
I think is between A&B
I think I would answer B
6 0
3 years ago
A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and th
gizmo_the_mogwai [7]

Answer:

λ = 5.2 x 10⁻⁷ m = 520 nm

Explanation:

From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:

Δx = λL/d

where,

Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m

L = Distance between slits and screen = 3.1 m

d = Separation between slits = 0.0005 m

λ = wavelength of light = ?

Therefore,

0.00322 m = λ(3.1 m)/(0.0005 m)

λ = (0.00322 m)(0.0005 m)/(3.1 m)

<u>λ = 5.2 x 10⁻⁷ m = 520 nm</u>

5 0
4 years ago
If you run 12 m/s for 15 minutes, how far will you go?
vfiekz [6]
10800 m = 10.8 km should be the answer if I am correct
3 0
3 years ago
Read 2 more answers
What weighs more a mole of feathers or bricks?
harkovskaia [24]
Avogadro's mole is a number, like 6.023 *10^23

your question is the equivalent of asking if 100 feathers or 100 bricks weigh more

the mole of bricks would weigh more
 
6 0
3 years ago
john is using a pulley to lift the sail on his sailboat. the sail weighs 150N and he must lift it at 4.0m. how much work must be
gogolik [260]

The work done on the sail is 600 J

Explanation:

The work done to lift the sail is equal to the gain in gravitational potential energy of the sail, therefore is:

W=mg\Delta h

where

m is the mass of the sail

g is the acceleration of gravity

(mg) is the weight of the sail

\Delta h is the change in height of the sail

In this problem we have

mg = 150 N (weight)

\Delta h = 4.0 m

Substituting, we find the work done:

W=(150)(4.0)=600 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

5 0
3 years ago
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