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lawyer [7]
3 years ago
10

What are the formulas for Work, KE, GPE and Power?

Physics
1 answer:
Katyanochek1 [597]3 years ago
6 0

Answer:

work=f×d

GPE =m×g×h

kinetic energy =  \frac{1}{2} mv {}^{2}  \\ power =  \frac{w}{t}

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Which quantity must equal zero if a car is moving at a constant rate along a straight line?
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B the energy of the car. Sorry if this is not right
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3 years ago
A plant dying after being exposed to poison represents a physical change.
Sphinxa [80]

Answer:

False is the correct answer.

It is considered more of a chemical change because all of the cells inside the plant decompose (change chemicaly)

5 0
3 years ago
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A concrete piling of 50 kg is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. How much will the wire stretch?
pentagon [3]

Explanation:

It is given that,

Mass of concrete pilling, m = 50 kg

Diameter of wire, d = 1 mm

Radius of wire, r = 0.0005 m

Length of wire, L = 11.2

Young modulus of steel, Y=20\times 10^{10}\ N/m^2

The young modulus of a wire is given by :

Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}

Y=\dfrac{F.L}{A\Delta L}

\Delta L=\dfrac{F.L}{A.Y}

\Delta L=\dfrac{50\ kg\times 9.8\ m/s^2\times 11.2\ m}{\pi (0.0005\ m)^2\times 20\times 10^{10}\ N/m^2}

\Delta L=0.034\ m

So, the wire will stretch 0.034 meters. Hence, this is the required solution.

8 0
4 years ago
A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho
Crank

Answer:

a)N = 3.125 * 10¹¹

b) I(avg)  = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

         = (500)(0.5)(0.1 × 10⁻⁶)

         = 2.5 × 10⁻⁵A

C)  If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}

= 1250W

d) Final peak=

P= Ik/e

= = P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W

P = 2.5 × 10⁷W

5 0
3 years ago
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Which of the following tools is used to observe objects that are very far away? A. a anemometer B. a hand lens C. a microscope D
inn [45]

Answer:

D. a telescope

Explanation:

3 0
3 years ago
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