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FinnZ [79.3K]
4 years ago
7

Which property of potential energy distinguishes it from kinetic energy?

Physics
2 answers:
dimaraw [331]4 years ago
6 0
There are several differences, however a big property that distinguishes the 2 is that potential energy cannot be transferred, whereas kinetic energy can.
We can see this for example in a bow, when you pull it back it has elastic potential energy because of the strain in the bow, however you can't transfer this potential energy to another object, whereas the flying arrow can transfer it's KE to the target it hits as several things including thermal and sound energy.
kobusy [5.1K]4 years ago
5 0

Answer:

mass

Explanation:

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A particle with a mass of 0.01 kg and a net charge of -0.05 C accelerates from rest through a uniform electric field. If the str
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Answer:

The speed of the particle after travelling for 0.5 m is 100 m/s.                            

Explanation:

It is given that,

Mass of the particle, m = 0.01 kg

Net charge on the particle, q = -0.05 C  

Electric field strength, E = 2000 V/C

Distance travelled by the particle, d = 0.5 m

The work done due to motion of the particle is balanced by the change in kinetic energy as :

Fd=\dfrac{1}{2}mv^2

v is the speed of the particle

F is the electric force

qEd=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{2qEd}{m}}

v=\sqrt{\dfrac{2\times 0.05\times 2000\times 0.5}{0.01}}

v = 100 m/s

So, the speed of the particle after travelling for 0.5 m is 100 m/s. Hence, this is the required solution.

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In what setting do all frequencies of electromagnetic waves travel at the same speed?
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3 years ago
Read 2 more answers
A projectile is fired with an initial speed of 100 m/s and angle of elevation 30 degrees. The projectile eventually hits the gro
zaharov [31]

Answer:

Explanation:

Given

launch velocity u=100\ m/s

Launch angle \theta =30^{\circ}

Range of Projectile R=\frac{u^2\sin 2\theta }{g}

R=\frac{100^2\times \sin (2\times30)}{9.8}

R=883.699\approx 883.7

Horizontal velocity remain same and only vertical velocity changes.

Initially  vertical  velocity is in upward direction but as soon as it reaches the ground its direction change but magnitude remain same.

u_y=100\sin (30)=50\ m/s

u_x=100\cos (30)=86.60\ m/s

u_{net}=\sqrt{(50)^2+(86.60)^2}

u_{net}=99.99\approx 100\ m/s      

7 0
3 years ago
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