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Daniel [21]
3 years ago
8

A 1,200-watt water heater is plugged into a 120-volt outlet and used for 1.5 hours. How much current runs through the water heat

er?
Physics
2 answers:
Ray Of Light [21]3 years ago
7 0

Answer:

Current (i) = 10 Ampere

Explanation:

As we know that power consumed by the electrical heater is given by the rate of electrical energy consumed by it.

So as we know that electrostatic potential energy of charge is given by

U = qV

now for electrical power we know that

P = \frac{dU}{dt}

P = \frac{d}{dt}(qV)

P = V\frac{dq}{dt}

P = Vi

so now by plug in data in this above equation we have

1200 = 120 \times i

i = 10 A

so current through the water heater will be i = 10 A

stiv31 [10]3 years ago
4 0
Current = power/voltage = 1200w/120v = 10 amperes. Constantly. Whenever the heater is turned on and heating.
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Answer:

Today, to benefit humankind, radiation is used in medicine, academics, and industry, as well as for generating electricity. In addition, radiation has useful applications in such areas as agriculture, archaeology (carbon dating), space exploration, law enforcement, geology (including mining), and many others.

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A car parked on level pavement exerts a force of 10,000 newtons on the ground. What force does the pavement exert back on the ca
natka813 [3]

Answer:

Normal force of 10,000N

Explanation:

From the question, the weight the car exerts on the pavement is 10,000N.

The pavement exerts upward and perpendicular contact force called normal force on the car to support its weight. Also, the normal force is equal and opposite to the weigh of the car.

Hence the pavement exerts normal force of 10,000N back on the car to prevent it from passing through it.

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3 years ago
The ratio of the lift force L to the drag force D for the simple airfoil is L/D = 10. If the lift force on a short section of th
Yuri [45]

Answer:

R=64.32\ lb\\\\\theta=84.3\°

Explanation:

Given:

Ratio of lift force to drag force is, \frac{L}{D}=10

Lift force on a short section is, L=64\ lb

Magnitude of resultant, R= ?

The angle of 'R' with the horizontal is, \theta=?

We know that, lift force and drag are at right angles to each other. So, the resultant can be computed using Pythagoras theorem.

For calculating 'R', we first compute drag force 'D'.

As per question:

\frac{L}{D}=10\\\\D=\frac{L}{10}=\frac{64\ lb}{10}=6.4\ lb

Now, the magnitude of resultant 'R' is given as:

R=\sqrt{L^2+D^2}

Plug in the given values and solve for 'R'. This gives,

R=\sqrt{64^2+6.4^2}\\\\R=\sqrt{4096+40.96}\\\\R=\sqrt{4136.96}=64.32\ lb

Therefore, the magnitude of the resultant force 'R' is 64.32 lb.

Now, the angle \theta is given as the arctan of the ratio of the lift and drag force.

Therefore,

\theta=\tan^{-1}(L/D)\\\\\theta=\tan^{-1}(10)\\\\\theta=84.3\°

Therefore, the angle made with the horizontal is 84.3°.

6 0
3 years ago
s=1/2(u+v)t check the dimensional consistency for this? where s is displacement, t is time, u is initial velocity, v is final ve
Anettt [7]

Answer:

See the explanation below.

Explanation:

The units of work are consistent since if we work in the international system of measures we have the following dimensional quantities for velocity, distance and time.

s = displacement [m]

v and u = velocity [m/s]

t = time [s]

Now using these units in the given equation.

s = 0.5*([m/s]+[m/s])*[s]\\s=0.5*[m/s]*[s]\\s = 0.5*[m]

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8 0
3 years ago
High-speed stroboscopic photographs show that the head of a 180 g golf club is traveling at 47 m/s just before it strikes a 46 g
inn [45]

Answer:

47 m/s

Explanation:

golf club mass, mc = 180 g

golf ball mass, mb = 46 g

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final golf club speed, vc2 = 35 m/s

initial golf ball speed, vb1 = 0 m/s

final golf ball speed, vb2 = ? m/s

The total momentum is conserved, then:

mc*vc1 + mb*vb1 = mc*vc2 + mb*vb2

Replacing with data and solving (dimension are omitted):

180*47 + 46*0 = 180*35 + 46*vb2

vb2 = (180*47 - 180*35)/46

vb2 = 47 m/s

7 0
3 years ago
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