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Alexus [3.1K]
3 years ago
9

A joule equals the amount of work done on an object when a force _______ displaces that object ______.

Physics
1 answer:
Anastaziya [24]3 years ago
3 0

Answer:

A joule equals the amount of work done on an object when a force <u>of</u><u> </u><u>1</u><u> </u><u>Newton</u><u> </u> displaces that object <u>1</u><u> </u><u>metre</u>

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Kristen has two identical sized cubes, one is lead (Pb) and one is copper (Cu). Kristen is determining the density of each cube.
natta225 [31]

Answer:

volume

Explanation:

Identical size means volume will be the same in each calculation of

density =   mass / <u>volume</u>

3 0
2 years ago
Describe why drawing a line of best fit is useful.
Amiraneli [1.4K]

Answer:

Cause life

Explanation:

Cause life

8 0
3 years ago
A skydiver jumps out of a plane and immediately begins falling toward the Earth.
worty [1.4K]

Answer:

5.2m/s^2

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t (1)\\{Vf^{2}-Vo^2}/{2.a} =X(2)\\X= VoT+0.5at^{2} (3)\\X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 4 above equations and use algebra to solve

for this case we can use the ecuation number 3

x=100m

t=6.2s

Vo=0m/s

X= VoT+0.5at^{2}\\X= 0.5at^{2}\\a=\frac{X}{0.5t^2} \\a=\frac{100}{0.5(6.2)^2}=5.2m/s^2

7 0
3 years ago
An internal combustion engine does 356 kJ of useful work using 946 kJ of thermal energy from the gasoline consumed in the engine
OlgaM077 [116]

Answer:

The efficiency of this engine is 37.63 %.

Explanation:

Given;

useful output work done by the combustion engine, = 356 kJ

input work, = 946 kJ

The efficiency of the combustion engine is calculated as;

Efficiency = \frac{0utput \ work}{1nput \ work} \times 100\%\\\\Efficiency = \frac{356 \ \times \ 10^3}{946 \ \times \ 10^3} \times 100\%\\\\Efficiency = 37.63 \ \%

Therefore, the efficiency of this engine is 37.63 %.

3 0
3 years ago
the dog has a momentum of 60 kilogram meters per second west. the dog has a velocity of 3 meters per second west. what is the ma
lianna [129]
P=mv=>m=p/v=60/3=20kg
5 0
4 years ago
Read 2 more answers
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