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lisabon 2012 [21]

4 years ago

14

How much work does the electric field do in moving a proton from a point with a potential of +130 v to a point where it is -70 v

? express your answer in joules?

1 answer:

NikAS [45]4 years ago

6 0

Hi Gonnah,
W=qΔV
W= q(-70-130)
q of proton look it up and plug in and find out
it should be negative since you are doing work form greater potential to lower

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A wheel has a radius of 9 centimeters and an axle radius of 3 centimeters. The IMA of the system is ___?

otez555 [7]

Answer:

Answer is C.

Explanation:

To calculate the IMA for a wheel and axle, find the ratio of the larger radius to the smaller radius:

IMA = R/r

IMA = 9/3

IMA = 3

4 0

4 years ago

A constant amount of charge passes through a conductor. How is current affected if the same amount of charge passes in less time

baherus [9]

Answer:

B. The current increases.

Explanation:

As we know that rate of flow of charge through the conductor is known as electric current

So we have

$i = \frac{q}{t}$

here we know that charge Q flowing through the conductor is constant while the time in which it passes through it is decreased

so we can say that the ratio of charge and time will increase

so here we have

$i = increased$

So correct answer will be

B. The current increases.

4 0

4 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03 s la

Nookie1986 [14]

Answer:

$h=53.09m$ (2)

$v_{min}>5.05m/s$

$v_{max}$

Explanation:

<u>a)Kinematics equation for the first ball:</u>

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=8.9m/s$

The ball reaches the ground, y=0, at t=t1:

$0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}$

$h=1/2*g*t_{1}^{2}-v_{o}t_{1}$ (1)

Kinematics equation for the second ball:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=0$ the ball is dropped

The ball reaches the ground, y=0, at t=t2:

$0=h-1/2*g*t_{2}^{2}$

$h=1/2*g*t_{2}^{2}$ (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03 (3)

We solve the equations (1) (2) (3):

$1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}$

$g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)$

$g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)$

$-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)$

$2.06*gt_{1}-2v_{o}t_{1}=g*1.06$

$t_{1}=g*1.06/(2.06*g-2v_{o})$

vo=8.9m/s

$t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s$

t2=t1-1.03 (3)

t2=3.29sg

$h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m$ (2)

b) $t_{1}=g*1.06/(2.06*g-2v_{o})$

t1 must : t1>1.03 and t1>0

limit case: t1>1.03:

$1.03>9.81*1.06/(2.06*g-2v_{o})$

$1.03*(2.06*9.81-2v_{o})$

$20.8-2.06v_{o}$

$(20.8-10.4)/2.06$

$v_{min}>5.05m/s$

limit case: t1>0:

$g*1.06/(2.06*g-2v_{o})>0$

$2.06*g-2v_{o}>0$

$v_{o}$

$v_{max}$

8 0

4 years ago

HELP ME PLEASE AND HURRY !!!!!!

hodyreva [135]

Answer:

mass number = protons + neutrons

8 0

2 years ago

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Why is a steam burn more damaging than a burn from boiling water at the same temperature?

maksim [4K]

The phase change (from gas to liquid) that the steam undergoes right when it touches your skin releases a lot of energy right onto your skin so it burns a ton. The boiling water would just burn you until it reaches the same temperature as your skin.

4 0

3 years ago