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jeyben [28]
3 years ago
14

An electric heater is rated at 1400 W, a toaster is rated at 1150 W, and an electric grill is rated at 1560 W. The three applian

ces are connected in parallel across a 112 V emf source. Find the current in the heater.
Physics
1 answer:
gayaneshka [121]3 years ago
8 0

Answer:

The current in the heater is 12.5 A

Explanation:

It is given that,

Power of electric heater, P₁ = 1400 W

Power of toaster, P₂ = 1150 W

Power of electric grill, P₃ = 1560 W

All three appliances are connected in parallel across a 112 V emf source. We need to find the current in the heater. We know that in parallel combination of resistors the current flowing in every branch of resistor divides while the voltage is same.

Electric power, P_1=V\times I_1

I_1=\dfrac{P_1}{V}

I_1=\dfrac{1400\ W}{112\ V}

I_1=12.5\ A

So, the current in the heater is 12.5 A. Hence, this is the required solution.  

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If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

3 0
3 years ago
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