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shusha [124]
3 years ago
9

Two objects separated by a distance r are each carrying a charge q The magnitude of the force exerted on the second object by th

e first is F If the first object is removed and replaced with an identical object that carries a charge 4q what is the magnitude of the electric force on the second object
Physics
1 answer:
Sindrei [870]3 years ago
4 0

Answer:F=4F

Explanation: Columbs law states that The force between the two point charges is directly proportional to the product of charges and inversely proportional to the square of distance between them

Force between the two charges is given by

F=K*q1*q2/r^2

if one charge become 4 times, new force is,

F=4(K*q1*q2)/r^2

F=4F

Where q1 and q2 are the point charges

r is the distance between the two charges

K is a constant of proportion called electrostatic force

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3 0
3 years ago
.Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric
maw [93]

Answer:

a) The uncertainty in calculated V, ΔV = 25.3

b) The uncertainty in calculated v, Δv = 0.41 m/s

c) The uncertainty in calculated V, ΔV = 22.2 V

Explanation:

We'll use Upper-Lower Bounds method of uncertainty to estimate the uncertainties.

a) I = 5.1 A, ΔI = 0.3 A

I = (5.1 ± 0.3) A

R = 77.5 ohms, ΔR = 0.4 ohms

R = (77.5 ± 0.4) ohms

V = IR = 5.1 × 77.5 = 395.25 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 5.1 - 0.3 = 4.8 A

Rₗ = 77.5 - 0.4 = 77.1 ohms

Vₗ = 4.8 × 77.1 = 370.08 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 5.1 + 0.3 = 5.4 A

Rᵤ = 77.5 + 0.4 = 77.9 ohms

Vᵤ = 5.4 × 77.9 = 420.66 V

The average of the differences from the mean voltage/true value is 25.3 V

V = 395.25 V, Δ = 25.3V

V = (395.25 ± 25.3) V

b) x = 2.9 m, Δx = 0.3 m

x = (2.9 ± 0.3) m

t = 4.4 s, Δt = 1.8 s

t = (4.4 ± 1.8) ohms

v = x/t = 2.9/4.4 = 0.659 m/s

The lower bound for average speed will be calculated using the lower bounds for distance and upper bounds for time.

xₗ = 2.9 - 0.3 = 2.6 m

tᵤ = 4.4 + 1.8 = 6.2 s

vₗ = 2.6/6.2 = 0.419 m/s

The upper bound for the average speed will be calculated using the upper bound for the distance and lower bound for time

xᵤ = 2.9 + 0.3 = 3.2 m

tₗ = 4.4 - 1.8 = 2.6 s

vᵤ = 3.2/2.6 = 1.231 m/s

The average of the differences from the mean average speed/true value is 0.41 m/s

v = 0.659 m/s, Δv = 0.41 m/s

v = (0.659 ± 0.41) m/s

c) ) I = 9.8 A, ΔI = 0.5 A

I = (9.8 ± 0.5) A

R = 40.5 ohms, ΔR = 0.2 ohms

R = (40.5 ± 0.2) ohms

V = IR = 9.8 × 40.5 = 396.9 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 9.8 - 0.5 = 9.3 A

Rₗ = 40.5 - 0.2 = 40.3 ohms

Vₗ = 9.3 × 40.3 = 374.79 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 9.8 + 0.5 = 10.3 A

Rᵤ = 40.5 + 0.2 = 40.7 ohms

Vᵤ = 10.3 × 40.7 = 419.21 V

The average of the differences from the mean voltage/true value is 22.2 V

V = 396.9 V, Δ = 22.2 V

V = (396.9 ± 22.2) V

7 0
3 years ago
What is the acceleration of a 10 kg mass pushed 5 n forceushed by a 5n(kg-m/s2) (use gresa method)
Pepsi [2]

In this problem,

Applied force(F) = 10 N

The object’s mass (m) is 5 kg.

Having said that,

An object’s force is equal to the product of its mass and the acceleration it experiences as a result of the applied force.

i.e., Mass + Acceleration = Force (a)

F= m×a

Therefore,

A= F÷m

A= (10÷5) m/sec²

A= 2 m/sec²

Consequently, the object’s acceleration,

A=2 m/sec²

Concept of force and acceleration:

This states that the rate of velocity change of an object is directly proportional to the applied force and moves in the direction of the applied force.

It can be expressed mathematically as force (N) = mass (kg) x acceleration (m/s2). Therefore, an object with constant mass will accelerate in direct proportion to the applied force.

To know more about such problems, visit:

brainly.com/question/16743612

#SPJ4

6 0
1 year ago
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