The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.
<h3>How does test charge affect electric field?</h3>
As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.
Adjusting the amount of charge on the test charge will not change the electric field force.
<h3>What is a test charge used for?</h3>
The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.
To learn more about test charge, refer
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Answer: 0.5 m/s
Explanation:
Given
Speed of the sled, v = 0.55 m/s
Total mass, m = 96.5 kg
Mass of the rock, m1 = 0.3 kg
Speed of the rock, v1 = 17.5 m/s
To solve this, we would use the law of conservation of momentum
Momentum before throwing the rock: m*V = 96.5 kg * 0.550 m/s = 53.08 Ns
When the man throws the rock forward
rock:
m1 = 0.300 kg
V1 = 17.5 m/s, in the same direction of the sled with the man
m2 = 96.5 kg - 0.300 kg = 96.2 kg
v2 = ?
Law of conservation of momentum states that the momentum is equal before and after the throw.
momentum before throw = momentum after throw
53.08 = 0.300 * 17.5 + 96.2 * v2
53.08 = 5.25 + 96.2 * v2
v2 = [53.08 - 5.25 ] / 96.2
v2 = 47.83 / 96.2
v2 = 0.497 ~= 0.50 m/s
Answer:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second
The number of protons in an atom is represented by the atomic number.
hope this helps
Answer:
Explanation:
We can use Newton's Universal Law of Gravitation to solve this problem:
., where 
is acceleration due to gravity at the planet's surface, 
is gravitational constant 
, 
is the mass of the planet, and 
is the radius of the planet.
Since acceleration due to gravity is given as 
, our radius should be meters. Therefore, convert 
kilometers to meters:
.
Now plugging in our values, we get:
,
Solving for :
.
