Changes in forms of energy are called energy?

In the figure, if Q = 52 µC q =10 µC and d = 55 cm, what is the magnitude of the electrostatic force on q?

GenaCL600 [577]

Answer:

F = 15.47 N

Explanation:

Given that,

Q = 52 µC

q = 10 µC

d = 55 cm = 0.55 m

We need to find the magnitude of the electrostatic force on q. The formula for the electrostatic force is given by :

$F=k\dfrac{q_1q_2}{d^2}\\\\F=9\times 10^9\times \dfrac{52\times 10^{-6}\times 10\times 10^{-6}}{(0.55)^2}\\\\F=15.47\ N$

So, the magnitude of the electrostatic force is 15.47 N.

4 0

3 years ago

Which is an example of transforming potential energy to kinetic energy? Select two options

blsea [12.9K]

Answer : 

.changing chemical energy to thermal energy

.changing nuclear energy to radiant energy

Explanation : 

<h3>Hope it helps you mate ))</h3>

3 0

3 years ago

A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

(c) 2.86s

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

$x(t)=x_{0}+v_{0}t$

Where $x(t)$ is the horizontal distance at a time $t$ , $x_{0}$ is the initial horizontal position which in this case will be zero: $x_{0}=0$ . And $v_{0}$ is the initial speed: $v_{0}=2m/s$

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: $t=0.8s$ so the horizontal distance is:

$x(0.8)=(2m/s)(0.8s)1.6m$

The answer for (a) is 1.6m

Now, to solve (b) we need the equation for vertical distance:

$y(t)=y_{0}-\frac{1}{2}gt^2$

Where $y(t)$ is the vertical distance at a time $t$ , $y_{0}$ is the initial vertical distance: $y_{0}=10m$ And $g$ is the gravitational acceleration: $g=9.81m/s^2$ }

(b) At what vertical distance above the surface of the water is the diver just then?

The time is the same as in the last question: $t=0.8s$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)$

$y(0.8)=10m-\frac{1}{2}(6.2784m)$

$y(0.8)=10m-(3.1392)$

$y(0.8)=6.86m$

the answer to (b) is 6.86m

(c) At what horizontal distance from the edge does the diver strike the water?

To solve (c) we first need to know how long it took to reach the height of the water, that is, for what time y = 0

Using $y(t)=y_{0}-\frac{1}{2}gt^2$

if $y(t)=0$

$0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t$

and since $g=9.81m/s^2$

$t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s$

At $t=1.43s$ the car hits the water, so the horizontal distance at that time is:

$x(t)=x_{0}+v_{0}t$

$x(1.43)=(2m/s)(1.43s)=2.86m$

the answer to (c) is 2.86s

6 0

3 years ago

An object is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/s2.

Alex

Answer:

12s

Explanation:

the formula is

s = u×t + 0.5×a×t²

where:

s is the displacement of the object at this time (so it's equal to 0)

u is the initial speed

t is the time you have to find

a is the acceleration due to gravity

$0 = 15 \frac{m}{s} \times t + 0.5 \times 2.5 \frac{m}{{s}^{2} } \times {t}^{2}$

$15 \frac{m}{s} \times t = 0.5 \times 2.5 \frac{m}{ {s}^{2} } \times {t}^{2}$

$15 \frac{m}{s} = 1.25 \frac{m}{{s}^{2} } \times {t}$

$t = \frac{15 \frac{m}{ {s}^{} } }{1.25 \frac{m}{ {s}^{2} } }$

$t = 12s$

7 0

2 years ago

Please help 1-9 15 points

Serjik [45]

Answer:

<h2>3. The Sputnik orbitted at approximately 577.1 km above the surface of the Earth. </h2><h2>4.First Successful U.S. Rocket Launch into Space was The Jupiter C</h2>

Explanation:

The first two are already answered 5-9 I cannot answer because I have not watched "October sky video" and I don't know who homer and the boys are.

I Hope This helps

5 0

3 years ago