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inn [45]
3 years ago
7

One heater uses 490 W of power when connected by itself to a battery. Another heater uses 250 W of power when connected by itsel

f to the same battery. How much total power do the heaters use when they are both connected in series across the battery?
Physics
1 answer:
insens350 [35]3 years ago
6 0

Answer:

Power will be 166.66 watt when the heater are connected in series

Explanation:

We have given power of the first heater P_1 = 490 watt

And power of the second heater P_2 = 250 watt

We know that in series connection potential difference will be same

So let the potential difference is v

Power is given by P=\frac{V^2}{R}

So 490=\frac{V^2}{R_1}

R_1=\frac{V^2}{490}=0.0020V^2

Similarly R_2=\frac{V^2}{250}=0.04V^2

As resistances are connected in series so equivalent resistance R=R_1+R_2=0.0020V^2+0.04V^2=0.006V^2

So power P=\frac{V^2}{R}=\frac{V^2}{0.006V^2}=166.66Watt

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How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n
grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

dSin\theta = m\lambda

Where

m= Any integer which represent the number of repetition of spectrum

\lambda= Wavelength

d = Distance between the slits.

\theta= Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

d = 2.63*10^{-6}m

Therefore the number of lines per millimeter would be given as

\frac{1}{d} = \frac{1}{2.63*10^{-6} }

\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})

\frac{1}{d} = 379.4 lines/mm

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0
2 years ago
The position of a ball rolling in a straight line is given by x=2−3.6t+4.4t2, where x is in meters and t in seconds.
kobusy [5.1K]
I'm assuming the question is time it will take for ball to reach ground, if it is then set equation to zero then use the quadratic formula, the possible t value is your answer then
8 0
3 years ago
What do you think that astronomers mean when they use the term observable universe?
iren2701 [21]
The observable universe consists of galaxies and other matter that can, principally, be seen from Earth because the light signals have had time to reach us. Not everything in the sky is the way it is when we see it, because of the distance the light travels to reach us. 

Hope this helps :)
6 0
3 years ago
Question 15 of 25
vichka [17]
Answer: conduction :it transfers heat between objects that are in direct contact with eachother
7 0
2 years ago
A rock is launched at angle theta=53.2∘ above the horizontal from an altitude of ℎ=182 km with an initial speed ????0=1.61 km/s.
Mariulka [41]

Answer:

The rock's final speed at the required altitude will be 42.24 m/s.

Explanation:

Let's start by finding the initial vertical speed.

Vertical Speed = 1.61 * Sin (53.2°)

Vertical Speed = 0.8 m/s

We want to know the speed of the rock when it is at an altitude of 91 km.

The total displacement of the rock from its starting position will thus be equal to -91 km

We can use this in the following equation:

s=u*t+\frac{1}{2} (a*t^2)

-91=0.8*t+\frac{1}{2} (-9.8*t^2)

t = 4.3918 seconds

Thus it takes 4.3918 seconds to reach the required altitude. We can now find the speed as follows:

V=U+at

V=0.8+(-9.8)*(4.3918)

V = -42.24

Thus the rock's final speed at the required altitude will be 42.24 m/s.

8 0
3 years ago
Read 2 more answers
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