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3 years ago

6

The electrons that produce the picture in a

1 answer:

pantera1 [17]3 years ago

7 0

We use the formula:

a = (v^2 - u^2)/2s <span>

a = acceleration = ?
v = final velocity = 2.5x10^6
u = initial Velocity = 1x10^5
s = distance traveled. = 0.012m

To find the acceleration, we need to fill in the variables:
a = ((2.5x10^6)^2 - (1x10^5)^2)/ 2* 0.012 = 260,000,000,000,000 m/s^2

<span>Our acceleration is 2.6 × 10^14 m/s^2 </span></span>

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A truck is parked at the bottom of a hill. It moves from the bottom of a hill

Marat540 [252]

Answer:

C

Explanation:

potential energy is stored energy that an object has due to its position or chemical composition. Since the truck has the possibility to move down the hill when it is at the top of the hill, it has a great amount of potential energy, howver when its at the bottom of the hill, it doesn’t have the possibility to move or can move v little due to its positions, therefore it has little to no potienatal energy.

3 0

3 years ago

Pls help i have test

ludmilkaskok [199]

Answer:

I think the answer is a no. I guess

Explanation:

don't mind if it is write or wrong

5 0

3 years ago

Read 2 more answers

When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6

Bezzdna [24]

Answer:

Distance between the charges, r = 0.8 meters

Explanation:

Given that,

Charge 1, $q_1=+8.4\ \mu C=+8.4\times 10^{-6}\ C$

Charge 2, $q_2=+5.6\ \mu C=+5.6\times 10^{-6}\ C$

Repulsive force between charges, F = 0.66 N

Let r is the distance between charges. The formula for the electrostatic force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$r=\sqrt{\dfrac{kq_1q_2}{F}}$

$r=\sqrt{\dfrac{9\times 10^9\times 8.4\times 10^{-6}\times 5.6\times 10^{-6}}{0.66}}$

r = 0.8009 meters

or

r = 0.8 meters

So, the distance between the charges i 0.8 meters. Hence, this is the required solution.

4 0

3 years ago

An astronaut of mass m in a spacecraft experiences a gravitational force F=mg when stationary on the launchpad.

zaharov [31]

gravitational force is the attraction force of earth on an object which is near the surface of earth

It will not depend on the velocity or acceleration of earth

So it will not change while an object is stationary or it is moving with some acceleration

So here we will say that force will remain the same

$F = mg$

7 0

3 years ago

Three identical train cars, coupled together, are rolling east at speed v0. A fourth car traveling east at 2v0 catches up with t

aliina [53]

Answer: $v_o$

Explanation:

It is given that three cars has same mass m with speed $v_o$

suppose rest two cars also has same mass m

As there is no external force therefore momentum is conserved

Initial Momentum $P_i$

$P_i=3mv_0+m(2v_0)+m\times 0$

Final momentum $P_f$

$P_f=5m\times v$

where v=final velocity

$P_i=P_f$

$5mv_o=5mv$

$v=v_o$

thus final velocity is $v_o$

8 0

4 years ago