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4 years ago

What is the equation for calculating the heat energy released during heating effect of electric current?

2 answers:

5 0

The heat energy released from a piece of wire or any other section of a circuit is:

Energy = (voltage between its ends) x (current through it) x (time it's been going)

anygoal [31]4 years ago

3 0

The equation for calculating the heat energy released by the heating effect of electric current is given by the following equation : -

H = I²Rt

Where H stands for heat energy, I for electric current, R for resistance and t for time.

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Determine whether each of the following combinations of forces can be in equilibrium if their directions

lesya692 [45]

Answer:

I think it is a or c hope this helps

6 0

3 years ago

4. A 5.00-kg sled starts at rest and accelerates down a 45.0° incline. The coefficient of

Tresset [83]

Answer: $5.64\ m/s^2$

Explanation:

Given

mass of sled m=5 kg

Elevation $\theta =45^{\circ}$

$\mu _k=0.185$ (coefficient of kinetic friction)

Friction will oppose the motion of the sled while sled weight sin component helps it to move down the incline

$W\sin\theta-F_r=ma\\where\\W=weight(mg)\\F_r=Friction(\mu_kmg\cos\theta)\\a=acceleration$

Substituting values

$mg\sin \theta-\mu_kmgcos\theta=ma\\g\sin \theta-\mu_kg\cos \theta=a\\a=g(\sin 45^{\circ}-0.185\times \cos 45^{\circ})\\a=5.64\ m/s^2$

4 0

3 years ago

Give me three examples of fossil fuel-based energy sources pls

svet-max [94.6K]

Gasoline, kerosene, and coal.

4 0

4 years ago

A block of mass m = 2.20 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 6.80

Bogdan [553]

Answer:

(a) The speed of the lighter block is $v_{2x} = 3.7~m/s$ .

The speed of the heavier block is $v_{3x} = 3.55~m/s$ .

(b) The smaller block goes up to 0.69 m.

Explanation:

We will divide this question into three parts: Part A is for the smaller mass from the top of the incline to the collision. Part B is the collision. And Part C is for the smaller mass from the bottom to the highest point it can achieve.

In order to solve this question, I will assume that the smaller mass is initially at rest.

Part A:

We will use conservation of energy.

$K_1 + U_1 = K_2 + U_2\\0 + mgh = \frac{1}{2}mv_1^2 + 0\\(2.2)(9.8)(3.6) = \frac{1}{2}(2.2)v_1^2\\v_1 = 8.4 ~m/s$

This is the speed of the smaller mass just before the collision. The velocity of the mass is directed 30° above horizontal, since the mass is sliding down the incline.

Part B:

Momentum is a vector identity, so the x- and y-components of momentum are to be investigated separately. Since the collision occurs at the horizontal surface, only the x-component of momentum is conserved.

$P_1 = P_2\\mv_{1x} = mv_{2x} + Mv_{3x}\\(2.2)(8.4\cos(30^\circ)) = (2.2)v_{2x} + (6.8)v_{3x}\\16 = 2.2v_{2x} + 6.8v_{3x}$

During the collision kinetic energy is also conserved. Since kinetic energy is a scalar quantity, we don't have to separate its components.

$K_{initial} = K_{final}\\\frac{1}{2}mv_1^2 = \frac{1}{2}mv_{2}^2 + \frac{1}{2}Mv_{3x}^2\\(2.2)(8.4)^2 = (2.2)v_{2}^2 + (6.8)v_{3x}^2\\155.23 = 2.2v_{2}^2 + 6.8v_{3x}^2$