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Klio2033 [76]
4 years ago
15

How is a net electric charge produced​

Physics
1 answer:
Vlad [161]4 years ago
4 0

If an object has more protons than electrons, then the net charge on the object is positive. If there are more electrons than protons, then the net charge on the object is negative. If there are equal numbers of protons and electrons, then the object is electrically neutral.

Source: https://www.khanacademy.org/science/ap-physics-1/ap-electric-charge-electric-force-and-voltage/electric-charge-ap/a/electric-charge-ap1

Hope this helps :)

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Question 1 (1 point)
deff fn [24]

Answer:

Travelled 18 km, they are 6 km from home.

Explanation:

12/2 (halfway) is 6km. So, 6 + 12 would be 18 km, total amount travelled. The total distance of the trip would be 24 km (12 km out, 12km back) if they travelled 12+6 (18km) then they only have 6 km more to go.

5 0
3 years ago
How do you know or find out if an element is a solid, liquid or gas?
lubasha [3.4K]
Liquid is not a solid object so make sure you define them
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3 years ago
A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr
fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}

The values are

\displaystyle y_o=15\ m

\displaystyle v_o=14\ m/s

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

\displaystyle y=0

\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0

Multiplying by 2

\displaystyle 2y_o+2v_ot-gt^2=0

Clearing the coefficient of t^2

\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0

Plugging in the given values, we reach to a second-degree equation

\displaystyle t^2-2.857t-3.061=0

The equation has two roots, but we only keep the positive root

\displaystyle \boxed {t=3.69\ s}

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

\displaystyle Y_r=15+V_r.t

\displaystyle Y_r=15+14\times3.69

\displaystyle \boxed{Y_r=66.62\ m}

3 0
3 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
kirill115 [55]

Answer:

Part A

it would take 6 sec

it would take 3 sec

Explanation:

We are told that the power supplied to the wheel is constant which means that the sport car is gaining energy i.e

                           power \ \alpha  \ energy

Hence if power is constantly supplied energy constantly increase

From the formula of the Kinetic energy

                       KE  = \frac{1}{2} mv^2

we can see that as the speed doubles from 29 mph  to 58 mph  the energy needed is 2^2 = 4 times of the energy from the formula

   Also the time needed would also be 4 times because energy i directly proportional to time

       Hence to reach 58mph the time that it would take is

                          = 4* 1.5sec = 6sec

     

We are told that the ground pushes the car  with a constant force and

                 F = ma

this means that the acceleration is also constant

             now from newtons law

     v = u +at  

 Looking at it we see that final velocity is directly proportional with time

hence it would take twice the time to reach twice the final velocity

        Time to reach 58mph = 3 s

        since time to reach 29 mph(\frac{1}{2} \ of \ [58mph]) =( \frac{1}{2} \ of \ 3sec )1.5 s

6 0
3 years ago
What happens to the Total Energy as the spring bounce?
fiasKO [112]
Total energy will remain the same
4 0
3 years ago
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