Answer:
E = 307667 N/C
Explanation:
Since the object's mass is 1 g, then its weight in newtons is 0.001 * 9.8 = 0.0098 N.
This weight should have the same magnitude of the vertical component of the tension T of the string (T * cos(37)) so we can find the magnitude of the tension T via:
0.0098 N = T * cos(37)
then T = 0.0098/cos(37) N = 0.01227 N
Knowing the tension's magnitude, we can find its horizontal component:
T * sin(37) = 0.007384 N
and now we can obtain the value of the electric field since we know the charge of the ball to be: -2.4 * 10^(-8) C:
0.007384 N = E * 2.4 * 10^(-8) C
Then E = 0.007384/2.4 * 10^(-8) N/C
E = 307667 N/C
Answer:
A) d_o = 20.7 cm
B) h_i = 1.014 m
Explanation:
A) To solve this, we will use the lens equation formula;
1/f = 1/d_o + 1/d_i
Where;
f is focal Length = 20 cm = 0.2
d_o is object distance
d_i is image distance = 6m
1/0.2 = 1/d_o + 1/6
1/d_o = 1/0.2 - 1/6
1/d_o = 4.8333
d_o = 1/4.8333
d_o = 0.207 m
d_o = 20.7 cm
B) to solve this, we will use the magnification equation;
M = h_i/h_o = d_i/d_o
Where;
h_o = 3.5 cm = 0.035 m
d_i = 6 m
d_o = 20.7 cm = 0.207 m
Thus;
h_i = (6/0.207) × 0.035
h_i = 1.014 m
this process is called parellelogram method of resolving vectors.
Answer:

Explanation:
As we know that resistance of one copper wire is given as

here we know that

now we have


now we know that such 17 resistors are connected in parallel so we have


Now if a single copper wire has same resistance then its diameter is D and it is given as

now from above two equations we have


now we have

The acceleration of the car would be 0.33 first and then it would be 0.17.
<u>Explanation:</u>
An applied force is a force that is applied to an object by an individual or another item. On the off chance that an individual is pushing a work area over the room, at that point there is an applied power following up on the article. The applied power is the power applied on the work area by the individual.
The net force applied to the object rises to the mass of the article increased by the measure of its acceleration. The net power following up on the soccer ball is equivalent to the mass of the soccer ball duplicated by its adjustment in speed each second (its acceleration).