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never [62]
3 years ago
15

When you push a child on a swing, you are doing work on the child because _____.

Physics
2 answers:
max2010maxim [7]3 years ago
8 0

ur answer is A or also known as

When you push a child on a swing, you are doing work on the child because  you are pushing against the force of gravity

hope this helps :)

julsineya [31]3 years ago
3 0

Thinking it’s A! Sorry if I’m wrong! :)

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Identify five common business uses for electronic spreadsheets
Sloan [31]


For the answer to the question above
Forecasting how a business might do in the future. 
Calculating tax. 
Doing basic payrolls. 
Calculating Revenues. 
Producing charts. 

--Going past 5-- 

Inventory tracking 
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I hope my answer helped you. 
4 0
3 years ago
Read 2 more answers
A ball is thrown vertically upward with an initial velocity
ivolga24 [154]

Answer:

D

Explanation:

First we define our variables

V0=29.4

a=-9.8

V=0

We have to find the maximum displacement , which I will define as X

We use formula v^2=v0^2+2aX

All we do is substitute our values

0=29.4^2-19.6X

29.4^2=19.6X

X=29.4^2/19.6=44.1

5 0
2 years ago
Compare and contrast the particle motion in each state of matter
Vladimir79 [104]

Answer:

Gases, liquids and solids are all made up of microscopic particles, but the behaviors of these particles differ in the three phases. ... gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

<h3>Hope this is fine for you</h3>
6 0
3 years ago
A 5.00 kg object oscillates back and forth at the end ofa spring whose spring constant is 49.3 N/m. An obersever istraveling at
defon

Answer:

5.571 sec

Explanation:

angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s

Period To = 2π / angular frequency

Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got

T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec

t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been  ( To / (√ (1 - (v²/c²))).  where To = 2.00 sec

8 0
3 years ago
What is the speed of an object after falling (from rest) a distance of 9.80 meters near the surface of the Earth? Assume air
madam [21]

Hi there!

We can use the kinematic equation:

v_f^2 = v_i^2 + 2ad

vf = Final velocity (? m/s)

vi = initial velocity (0 m/s, dropped from rest)

a = acceleration (due to gravity, 9.8 m/s²)

d = distance (9.8 m)

Simplify the equation to solve for vf:

v_f^2 = 0 + 2ad\\\\v_f = \sqrt{2ad}

Substitute in the given values:

v_f = \sqrt{2(9.8)(9.8)} = \boxed{13.86 m/s}

8 0
2 years ago
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