What is the rate at which energy is transformed from one form to another called?

If the vertical axis and time on the horizontal axis if the speed is steadily increasing could the speef line ever become perfec

Phantasy [73]

A graph of real speed can have a section that's as steep as you want,
but it can never be a perfectly vertical section.

Any vertical line on a graph, even it it's only a tiny tiny section, means
that at that moment in time, the speed had many different values.

It also means that the speed took no time to change from one value to
another, and THAT would mean infinite acceleration.

8 0

3 years ago

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

3 years ago

Answer ?

morpeh [17]

Answer:

2.75 m/s^2

Explanation:

The airplane's acceleration on the runway was 2.75 m/s^2

We can find the acceleration by using the equation: a = (v-u)/t

where a is acceleration, v is final velocity, u is initial velocity, and t is time.

In this case, v is 71 m/s, u is 0 m/s, and t is 26.1 s Therefore: a = (71-0)/26.1

a = 2.75 m/s^2

5 0

2 years ago

Why is accuracy important in dodgeball?

____ [38]

Accuracy and power are the only things that matter in dodge ball. If you are not accurate, then your target does not get hit.

8 0

3 years ago

1) Write Newton's three laws of motion (20 points). 2) Define acceleration (20 points). 3) Write the Month and Day of the Assign

vaieri [72.5K]

Answer:

Hey

I have no idea when YOUR assignment is due.

Newtons 1rst law:

An object that has constant motion will remain at that speed unless acted on by an external force.

Newtons 2nd law:

F=ma (force=mass*acceleration)

Newtons 3rd law:

when a force is applied to an object, there will be an opposite but equal reaction.

Acceleration:

How much your speed increases/decreases per unit of time.

I wrote all that^

4 0

2 years ago