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3 years ago

How to rearrange E = VIT for V?

Physics

1 answer:

Ksivusya [100]3 years ago

7 0

The equation

E=VIT

On the left hand side there is only one variable i.e. E. On the right hand side there are three variables that are all multiplying with each other i.e. VIT. Now to make V subject of the formula use the basic mathematics techniques.

E/IT=V

V=E/IT

This is the final answer where we have made V to be the subject of the formula. Now if we have values for E,I and T we can get the value for V just by inserting the values in the above formula

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You are holding the axle of a bicycle wheel with radius

erastova [34]

Answer:

The angular acceleration of the wheel is -6.54 rad/s²

Explanation:

We'll use the equations of motion for this.

w = 2πf

f = 75 rpm = 1.25 rps = 1.25 rev/s

w₀ = initial angular velocity = 2π × 1.25 = 7.85 rad/s

w = final angular velocity = 0 rad/s

t = 1.2 s

α = ?

w = w₀ + αt

0 = 7.85 + 1.2α

α = 7.85/1.2 = - 6.54 rad/s²

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4 years ago

1. What coefficients would balance the following equation?

Stolb23 [73]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

Q7. A 2kg ball is released from a height of 20m.

Vesnalui [34]

Answer:

Given,

Mass, m = 2kg

Height or distance, s = 20m

Initial velocity, u = 0m/s

Acceleration due to gravity, g = 10m/s (It is 9.8m/s but can be assumed 10m/s for simplifying calculations)

(Because the object is is just dropped)

Final velocity, v = ?

Solution

v^2 = u^2 + 2as (u^2 means u raised to power 2)

=> v^2 = 0^2 + 2*10*20

=> v^2 = 400

=> v = 20

Thus the velocity just before the body strikes the ground will be 20m/s.

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3 years ago

Rank in order, from largest to smallest, the magnitudes of the electric field at the black dot. A. 2, 1, 3, 4 B. 1, 4, 2, 3 C. 3

sweet [91]

Given that,

Rank in order from largest to smallest the magnitude of the electric field at block dot.

Electric field :

Electric field is proportional to the charge divided by square of distance.

In mathematically,

$E\propto\dfrac{q}{r^2}$

Where, q = charge

r = distance

If the charge is greater then electric field will be greater.

If the distance is greater then electric field will be smaller.

We need to find the electric field at black dot

According to figure,

(I). The electric field at black dot due to positive charge point q to left. the distance is r.

The electric field will be

$E=\dfrac{kq}{r^2}$

The electric field will be largest.

(II). The electric field at black dot due to positive charge point 2q to left. The distance is 2r.

Then, the electric field will be

$E=\dfrac{k2q}{(2r)^2}$

$E=\dfrac{kq}{2r^2}$

The electric field will be smallest.

(III). The electric field at black dot due to positive charge point 2q to left. The distance is r.

Then, the electric field will be

$E=\dfrac{k2q}{(r)^2}$

The electric field will be very largest.

(IV). The electric field at black dot due to positive charge point q to left. The distance is 2r.

Then, the electric field will be

$E=\dfrac{kq}{(2r)^2}$

$E=\dfrac{kq}{4r^2}$

The electric field will be very smallest.

So, The electric field from largest to smallest will be

$E_{3}>E_{1}>E_{2}>E_{4}$

Hence, The ranking will be 3, 1, 2, 4.

(D) is correct option.

4 0

3 years ago

A force that varies with time is applied to an object. The force increases linearly from zero to twenty-four newtons during the

harkovskaia [24]

Answer:

144Ns

Explanation:

Impulse is the product of force and change in time or a measure of the change in momentum. Mathematically, Impulse is given as

J = FΔt

Where

J = Impulse,

F = Force,

Δt = change in time.

Due to the varying force, we know that the impulse varies at different points in the motion. Hence, the Net Impulse will be the addition of the individual impulses, i.e.

J = ΣJᵢ

J = J₁ + J₂ + J₃

J = F₁Δt₁ + F₂Δt₂ + F₃Δt₃

=> J = (24*2) + (24*4) + (0*2)

J = 48 + 96 + 0

J = 144 Ns

4 0

3 years ago