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vivado [14]
4 years ago
7

Define power and describe how to determine power. (4 points)

Physics
1 answer:
Allushta [10]4 years ago
7 0

Answer: The power is a measure of the rate at which work is done (or similarly, at which energy is transferred). The standard unit of power is Watt (1W). The power is determined by the change in energy, delta E (number of joules) and delta t, which is the time taken in seconds then: P= delta E/ delta t.

Mark me as Brainliest

Explanation:

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When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees
choli [55]

Answer:

a)  F = 1.26 10⁵ N, b)  F = 2.44 10³ N, c)   F_net = 1.82 10³ N  directed vertically upwards

Explanation:

For this exercise we must use the relationship between momentum and momentum

         I = Δp

         F t = p_f -p₀

a) It asks to find the force

as the man stops the final velocity is zero

         F = 0 - p₀ / t

the speed is directed downwards which is why it is negative, therefore the result is positive

         F = m v₀ / t

         F = 63.5 7.89 / 3.99 10⁻³

         F = 1.26 10⁵ N

b) in this case flex the knees giving a time of t = 0.205 s

          F = 63.5 7.89 / 0.205

          F = 2.44 10³ N

c) The net force is

         F_net = Sum F

         F_net = F - W

         F_net = F - mg

let's calculate

         F_net = 2.44 10³ - 63.5 9.8

         F_net = 1.82 10³ N

since it is positive it is directed vertically upwards

6 0
3 years ago
2 questions that I am just curious about!!! WILL NAME YOU BRAINLIEST!!! Hope you can figure out!! Worth 11 points!!
tensa zangetsu [6.8K]

Answer:

<em>#1:</em><em> Most likely it does so, and calls other paramedics to help the person it was on its way to go help. But usually people will stay out of its way when they hear one coming...</em>

<em>#2:</em><em> Probably, because they usually just don't want meat, and don't have problems with things such as crackers. I doubt that they can live solely from vegetables!! It seems impossible!!</em>

<em />

I have to say they are some quite interesting questions. Interesting to think over... :)  Anyway, I hope my answers help answer yours!!

5 0
3 years ago
We wish to obtain a equal sized inverted image of a candel flame on a screen kept at a distance of 4m from the candle flame a) n
soldier1979 [14.2K]

A) To obtain an inverted image of an object of same size a convex lens is used.

B)

In this question we have given

Distance between candle flame and screen=4

it means distance between candle flame and lens +distance between lens and screen =4m

u+v= 4m

u= v

therefore, 2u= 4m

or, u=-2m              ( u is taken negative because it lies on left side of the lens)

and v=2m

we have to find the focal length of convex lens

\frac{1}{f}  =  \frac{1}{v}   -  \frac{1}{u}\frac{1}{f}  =  \frac{1}{2} - ( -  \frac{1}{2} ) \\  \frac{1}{f}  = \frac{1}{1}\\ f = 1m

Hence, Focal length of the lens is 1m

and lens should be placed at a distance of 2m from the flame

4 0
3 years ago
What is the magnitude of the force a 25 mC charge exerts on a +2.5 mC charge 16 cm away?
kakasveta [241]
Keep up the good work too
5 0
3 years ago
A charged capacitor is connected to an ideal inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz. At time
IrinaK [193]

Answer:

8.0\mu C

Explanation:

We are given that

f=1.6 Hz

q=3.0\mu C=3.0\times 10^{-6} C

1\mu C=10^{-6} C

Current,I=75\mu A=75\times 10^{-6} A

1\mu A=10^{-6} A

We have to find the maximum charge of the capacitor.

Charge on the capacitor,q=q_0cos\omega t

\omega=2\pi f=2\pi\times 1.6=3.2\pi rad/s

3\times 10^{-6}=q_0cos3.2\pi t....(1)

I=\frac{dq}{dt}=-q_0\omega sin\omega t

75\times 10^{-6}=-q_0(3.2\pi)sin3.2\pi t....(2)

Equation (2) divided by equation (1)

-3.2\pi tan3.2\pi t=\frac{75\times 10^{-6}}{3\times 10^{-6}}=25

tan3.2\pi t=-\frac{25}{3.2\pi}=-2.488

3.2\pi t=tan^{-1}(-2.488)=-1.188rad

q_0=\frac{q}{cos\omega t}=\frac{3\times 10^{-6}}{cos(-1.188)}=8.0\times 10^{-6}=8\mu C

Hence, the maximum charge of the capacitor=8.0\mu C

4 0
4 years ago
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