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Evgen [1.6K]
3 years ago
7

HELP PLEASE BRAINLIEST!!!

Physics
2 answers:
Dmitrij [34]3 years ago
6 0
<span>The potential energy decreases liquid particles that are attracted to one another move closer together to form a solid. Temperature remains constant because kinetic energy remains constant</span>
irina1246 [14]3 years ago
5 0
Was haing trouble with a similar wuestion
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An object is at rest in front of a compressed spring. It travels over a surface that exerts a kinetic frictional force on it and
PolarNik [594]

Answer:

the object will travel 0.66 meters before to stop.

Explanation:

Using the energy conservation theorem:

E_i+K_i+W_f=K_f+U_f

The work done by the friction force is given by:

W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]

so:

\frac{1}{2}1800*(10*10^{-2})+0-13.7d=0+0\\d=0.66m

3 0
3 years ago
A sand mover at a quarry lifts 2,000 kg of sand per minute a vertical distance of 12 m. The sand is initially at rest and is dis
Marianna [84]

Answer:

<h2>E. 3.95kW</h2>

Explanation:

Power is defined as the rate of workdone.

Power = Workdone/time taken

Given Workdone = Force * distance

Power = Force * distance/time taken

Power = mgd/t (F = mg)

m = mass of the sand in kg

g = acceleration due to gravity in m/s²

d = vertical distance covered in metres

t = time taken in seconds

Given m = 2000kg, d = 12m, t = 1min = 60secs, g = 9.8m/s²

Power = 2000*9.8*12/60

Power = 3920Watts

Minimum rate of power that must be supplied to this machine is 3920Watts or 3.92kW

5 0
3 years ago
Why are chloroplasts essentially to the process of photosynthesis to occurs
lesya [120]
 they absorb sunlight and turn it into glucose I think.
8 0
3 years ago
Read 2 more answers
How much work would it take to push two protons very slowly from a separation of 2.00×10−10m (a typical atomic distance) to 3.00
laiz [17]

Answer:

Work= -7.68×10⁻¹⁴J

Explanation:

Given data

q₁=q₂=1.6×10⁻¹⁹C

r₁=2.00×10⁻¹⁰m

r₂=3.00×10⁻¹⁵m

To find

Work

Solution

The work done on the charge is equal to difference in potential energy

W=ΔU

Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\  Work=-7.68*10^{-14}J

4 0
3 years ago
Two astronauts in space with a baseball decide to play catch to pass the time. In the language of conservation of momentum, desc
Anna35 [415]
As the first astronaut throws the ball, lets assume it goes with v velocity and the mass of the ball be m
the momentum comes out be mv, thus to conserve that momentum the astronaut will move opposite to the direction of the ball's motion with the velocity mv/M (where M is the mass of the astronaut).
4 0
3 years ago
Read 2 more answers
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