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Free_Kalibri [48]

4 years ago

6

You have built a device that measures the temperature outside and displays it on a dial as a measure of how far away from room t

emperature outside is. The way the dial works is that a needle with a charged ball on the end is placed between two charged parallel plates. The strength of the uniform electric field between the plates is proportional to the outside temperature. Given that the charged ball on the needle has a charge of?

1 answer:

Bogdan [553]4 years ago

8 0

Answer:

m=33.734 grams

E=41435.95 N/C

Explanation:

The detailed explanation of Answer is given in the attached file.

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An 800 N man climbs 5 m up a ladder. How much gravitational potential energy does he gain?

Artemon [7]

Answer:

4000J

Explanation:

Given parameters:

Weight of the man = 800N

Height of ladder = 5m

Unknown:

Gravitational potential energy gained = ?

Solution:

The gravitational potential energy is due to the position of a body.

Gravitational potential energy = weight x height

Now insert the parameters;

Gravitational potential energy = 800 x 5 = 4000J

5 0

3 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

5 0

3 years ago

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the 14) car a deceleration of 3.50

sergij07 [2.7K]

To be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving.

First, we manipulate the one of the kinematic equations

v^2 = v0^2 + 2 (a) (x) where v = 0 since the car stopped

Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,

<span>we get v0 = sqrt (2(a)(x))

Substituting the known values,

v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s
</span>
Therefore, before stopping the car the original speed of the car would be 14.49 m/s

7 0

4 years ago

Please help help me please

vodomira [7]

The answer is either c or d but c is the best answer

4 0

3 years ago

A force vector F1 points due east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The resultant of the two

Andrew [12]

A force vector F1 points due east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The resultant of the two vectors has a magnitude of 400 newtons and points along the due east/west line. Find the magnitude and direction of F2. Note that there are two answers. <span>The given values are
F1 = 200 N</span> F2 =? Total = 400 N

Solution: F1 + F2 = T 200 N + F2 = 400N
F2 = 400 - 200
F2 = 200 N

4 0

3 years ago