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vazorg [7]
2 years ago
11

You go to another forest and measure the shadow of a tree as being 6 meters long. The shadow of your meter

Physics
1 answer:
krok68 [10]2 years ago
6 0
Well, 6/n=3/1, so n=2.
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An inductance L, resistance R, and ideal battery of emf are wired in series. A switch in the circuit is closed at time t = 0, at
Kay [80]

Explanation:

After some time t the current does not passing through the circuit

=>so the back emf is zero

=>here the inductor opposes decay of the circuit

- Ldi/dt = Ri

di/dt = - R/Li

di/i = - R/Ldt

now we applying the integration on both sides

log i=-R/Lt+C

here t=0=>i=io

Log io=C

=>Log i=-R/L*t + Log io

logi-Log io=-R/L*t

Log[i/io]=-R/L*t

i/io=e^-Rt/L

i=ioe^-Rt/L

the option D is correct

3 0
3 years ago
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How can wasted energy be made useful?
murzikaleks [220]

Answer:

by using it in a such place or thing which needs it or which can work with it

7 0
2 years ago
In a lab, the mass of object a is 52 kg. object a weighs
aksik [14]
If you are asking for the weight then the formula is F=mg where f is weight m is mass and g is acceleration due to gravity.m=52kg and g=9.8m/s2(the gravity of earth)
F=52*9.8=509.6
therefore the weight of the object is 509.6N
5 0
3 years ago
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What does Pascal's Law state? what are the implications of Pascal's law?
dybincka [34]

Answer:

Pascal's Law states that the pressure applied to a fluid in a closed container is transmitted equally to all points in the fluid and act in all directions of the container. ... Therefore, it can rightly be said that since the liquid does not flow, it definitely has equal pressure acting on it at all the points.

Explanation:

plzzzzzzz Mark my answer in brainlist

4 0
3 years ago
An 92-kg football player traveling 5.0m/s in stopped in 10s by a tackler. What is the original kinetic energy of the player? Exp
Artemon [7]

Explanation:

It is given that,

Mass of the football player, m = 92 kg

Velocity of player, v = 5 m/s

Time taken, t = 10 s

(1) We need to find the original kinetic energy of the player. It is given by :

k=\dfrac{1}{2}mv^2

k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2

k = 1150  J

In two significant figure, k=1.2\times 10^3\ J

(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0

i.e. P=\dfrac{W}{t}=\dfrac{\Delta K}{t}

P=\dfrac{\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2}{10\ s}

P = 115 watts

In two significant figures, P=1.2\times 10^2\ Watts

Hence, this is the required solution.  

6 0
2 years ago
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