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Degger [83]
3 years ago
6

An electric device, which heats water by immersing a resistance wire in the water, generates 50 cal of heat per second when an e

lectric potential difference of 12 V is placed across its leads. What is the resistance of the heater wire? (Note: 1 cal = 4.186 J)
Physics
1 answer:
Oksi-84 [34.3K]3 years ago
3 0

Answer:

0.69 ohm

Explanation:

Heat generated per second, H = 50 cal/s

Potential difference, V = 12 V

Let R is the resistance of coil.

The formula for the heat is given by

H = \frac{V^{2}}{R}t

50\times 4.186 = \frac{12^{2}}{R}\times 1

R = 0.69 ohm

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You lay a mirror flat on the floor with one edge against a wall and aim a laser at the mirror. The ray reflects from the mirror
joja [24]

Answer:

the angle of incidence θ is 45.56 º

Explanation:

Given data

strikes the mirror before wall x = 30.7 cm

reflected ray strikes the wall y =  30.1 cm

to find out

the angle of incidence θ

solution

let us consider ray is strike at angle  θ so after strike on surface ray strike to wall at angle 90 - θ

we will apply here right angle triangle rule that is

tan( 90 - θ) = y /x

tan( 90 - θ)  = 30.1 / 30.7

90 - θ = tan^-1 (30.1/30.7)

90 - θ = 44.4345

θ = 45.56 º

the angle of incidence θ is 45.56 º

4 0
4 years ago
A cyclist travels from point A to point B for 10 minutes. During the first 2.0 minutes of her trip, she maintains a uniform acce
bogdanovich [222]

Answer:

im not really good at explaining, but i found this website url:

https://www.numerade.com/questions/a-cyclist-travels-from-point-a-to-point-b-in-10-min-during-the-first-20-min-of-her-trip-she-maintain/

same question just with the explanation

6 0
2 years ago
The diagram shows a position-time graph What is the displacement of the object
algol13

The object is moving, so at different times, it has different displacement.  I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.

Displacement is the distance and the direction FROM (the position at the  beginning) TO (the position at the end).

At the beginning ... time=0 ... the position is 1 meter.

At the end ... time=5 ... the position is zero.

The distance FROM the beginning TO the end is (zero - 1m) .  That's  <em>-1m </em>.


5 0
3 years ago
When and where did the first nuclear reactor generate electricity
masha68 [24]
<h2>Answer: On December 20th, 1951 in Idaho, United States. </h2>

The world's first experimental nuclear power plant was the Experimental Breeder Reactor Number One (EBR-I), which was built in a desert in Idaho, United States.

This reactor made history when, on December 20th, 1951, four 200-watt light bulbs were illuminated by means of atomic energy, specifically by nuclear fission reaction.

5 0
3 years ago
What is the value of the composite constant (Gme,/r2e) to be multiplied by the mass of the object mo, in equation below:
Sedbober [7]

To solve this problem we will apply the definitions given in Newtonian theory about the Force of gravity, and the Force caused by weight. Both will be defined below, and in equal equilibrium condition to clear the variable concerning acceleration due to gravity. Finally, with the values provided in the statement, it will be replaced.

The equation for the gravitational force between the Earth and the object on the surface of the Earth is

F_g = \frac{Gm_em_o}{r^2_e}

Where,

G = Universal gravitational constant

m_e = Mass of Earth

r_e= Distance between object and center of earth

m_o= Mass of Object

The equation for the gravitational pulling force on the object due to gravitational acceleration is

F_g = m_o g

Equation the two expression we have

m_o g = \frac{Gm_em_o}{r_e^2}

g = \frac{Gm_e}{r_e^2}

This the acceleration due to gravity which is composite constant.

Replacing with our values we have then

g = \frac{(6.67*10^{-11}N\cdot m^2/kg^2)(5.98*10^{24}kg)}{6378km(\frac{10^3m}{1km})^2}

g = 9.8m/s^2

The value of composite constant is 9.8m/s^2. Here, the composite constant is nothing but the acceleration due to gravity which is constant always.

8 0
3 years ago
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