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BabaBlast [244]
3 years ago
9

I WILL GIVE BRAINLIEST....Determine the value of F...​

Physics
1 answer:
aleksley [76]3 years ago
6 0
F should be 10. If The Whole top is 50cm, then we should subtract 10n and 30n which gives us 10.

Or it could be 15 if both top and bottom are 25. 10+15= 25.
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_____________ circular motion occurs when an object is traveling with constant speed in a circle.
emmasim [6.3K]
Your answer is ''Uniform''.

Hope this helps :)
7 0
3 years ago
Alexandra drops an egg from 30 m above the ground to hit Vanessa on the head. If Vanessa stands 3.5 meters tall. How fast is the
I am Lyosha [343]

This is problem of free falling objects, which can be solved using the formula:

V = sqrt(2gy)

Where v is the velocity upon impact

G is the acceleration due to gravity ( 9.81 m/s2)

Y is the height

Since Venessa is 3.5 m

Y = 30 -3.5 = 26.5 m

V = sqrt(2 (9.81 m/s2) ( 26.5 m))

<span>V = 22.8 m/s</span>

7 0
3 years ago
Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p
ira [324]

Answer:

The force exerted on the q_1 is  F =  2.25*10^{3} \ N

Explanation:

From the question we are told that

     The area is  A =  2.34*10^{-3} \ m^2

     The magnitude of charge placed on them is  q =  7.07 * 10^{-7} C

     The charge placed between the plate is q_1 = 6.62 *10^{-5} C

   

The electric field generated around the plate  is mathematically represented as

           E =  \frac{q}{A \epsilon_o}

Substituting values

          E =  \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}

         E = 34*10^{6} \ V/m

The force exerted the charge q_1 is  mathematically represented as

        F =  q_1 * E

Substituting values  

        F =  6.62 *10^{-5} * 34*10^{6}

        F =  2.25*10^{3} \ N

8 0
3 years ago
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
3 years ago
A falling raisin will have<br> whale. (more/less)<br> momentum than a falling blue
Natalka [10]

Answer:less

Explanation:thats the answer

3 0
2 years ago
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