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Lubov Fominskaja [6]
3 years ago
10

If a 2 kg ball has 120 J of PE, then how high off the ground is it?

Physics
1 answer:
Yakvenalex [24]3 years ago
7 0
I think 6.1m i’m not sure sorry if i’m wrong
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A very small sphere with positive charge 5.00uC is released from rest at a point 1.20cm from a very long line of uniform linear
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Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
5 0
3 years ago
A car battery can deliver up to 75 amps of current at a voltage of 12 volts. How much power is this?
Alex_Xolod [135]
A.900 watts That would be your correct answer 
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Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
IRINA_888 [86]

Answer:

Time : <u>7.96 s</u>

Distance Traveled : <u>357.8 m</u>  

Explanation:

In order to solve this problem, we first consider the accelerated motion of rocket. We will be using the subscript 1 for accelerated motion.

So, for accelerated motion, we have:

Acceleration = a₁ = 14.5 m/s²

Time Period = t₁ = 3.1 s

Initial Velocity = Vi₁ = 0 m/s    (Since, it starts from rest)

Final Velocity = Vf₁

Distance covered by sled during acceleration motion = s₁

Now, using 1st equation of motion:

Vf₁ = Vi₁ + (a₁)(t₁)

Vf₁ = 0 m/s + (14.5 m/s²)(3.1 s)

Vf₁ = 44.95 m/s

Now, using 2nd equation of motion:

s₁ = (Vi₁)(t) + (0.5)(a₁)(t₁)

s₁ = (0 m/s)(3.1 s) + (0.5)(14.5 m/s²)(3.1 s)

s₁ = 22.5 m

Now, we first consider the decelerated motion of rocket. We will be using the subscript 2 for decelerated motion.

So, for accelerated motion, we have:

Deceleration = a₂ = - 5.65 m/s²

Time Period = t₂ = ?

Initial Velocity = Vi₂ = Vf₁ = 44.95 m/s    (Since, decelerate motion starts, where accelerated motion ends)

Final Velocity = Vf₂ = 0 m/s    (Since, rocket will eventually stop)

Distance covered by sled during deceleration motion = s₂

Now, using 1st equation of motion:

Vf₂ = Vi₂ + (a₂)(t₂)

0 m/s = 44.95 m/s + (- 5.65 m/s²)(t₂)

t₂ = (44.95 m/s)/(5.65 m/s²)

<u>t₂ = 7.96 s</u>

Now, using 2nd equation of motion:

s₂ = (Vi₂)(t₂) + (0.5)(a₂)(t₂)

s₂ = (44.95 m/s)(7.96 s) + (0.5)(- 5.65 m/s²)(7.96 s)

s₂ = 357.8 m - 22.5 m

s₂ = 335.3 m

Thus, the total distance covered by sled will be:

Total Dustance = S = s₁ + s₂

S = 22.5 m + 335.3 m

<u>S = 357.8 m</u>

7 0
3 years ago
How do the units of work and power compare?
ki77a [65]

Answer:

Option (d) is correct.

Explanation:

Work done is given by :

W = Fd, where F is force and d is displacement

Unit of work done :

The SI unit of force is Newton (N) and that of displacement is meter (m). So, the unit of work done is N-m. It is call Joule. It means that the unit of work done is Joule.

Power is given by rate at which the work is done. It is given by :

P = W/t, W is work done and t is time

Unit of power:

Unit of work is Joule (J) and that of time is second (s). It means that the unit of power is Watt and it is equal to Joule/second

Hence, the correct option is (d) "The unit for work is a joule. The unit for power is a watt, which is a joule per second".

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3 years ago
Which statement is TRUE concerning the particle arrangement of a solid? A) the particles have no definite shape B) the particles
Anarel [89]

C is the correct answer.

If you find my answer helpful, please brainliest me!

8 0
3 years ago
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