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hichkok12 [17]
2 years ago
5

Find out the name of metals which can be obtained from the following .a) argentite b)hematite c)chalcopryite d)bauxite e)calveri

te​
Physics
1 answer:
topjm [15]2 years ago
8 0

Answer:

a). <u>Silver</u><u>,</u><u> </u><u>Ag</u>

b). <u>Iron</u><u>,</u><u> </u><u>Fe</u>

c). <u>Copper</u><u>,</u><u> </u><u>Cu</u>

d). <u>Aluminium</u><u>,</u><u> </u><u>Al</u>

e). <u>Gold</u><u>,</u><u> </u><u>Au</u>

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Please Answer the question in the picture ASAP PLEASE
attashe74 [19]

Answer:

HERE IS YOUR ANSWER

Explanation:

PLEASE MARK MY ANSWER AS BRAINLIEST IF THE ANSWERS ARE CORRECT .

Beacuse of the loose connection of the wire .

Straight

5 0
2 years ago
Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr
slava [35]

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

W = F_{total} .d

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W  = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} .d =\frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}

F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

F_{sprinter} = F_{total} + F_{wind}  = 39.7 + 30 = 69.68 N

7 0
3 years ago
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The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward a
tino4ka555 [31]

Answer:

The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.

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2 years ago
In a clock, the average speed is maximum for the tip of&lt;br /&gt;a) Seconds hand b) Hours hand c) Minutes hand​
Simora [160]
A, hope this helped! I didn’t really get it but I think it’s correct?
3 0
3 years ago
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On the planet Xenophous a 1.00 m long pendulum on a clock has a period of 1.32 s. What is the free fall acceleration on Xenophou
Vilka [71]
The period of the pendulum is given by the following equation

T = 2π * sqrt (L/g)

Where g is the gravity (free fall acceleration)

L is the longitude of the pendulum

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g = L/(T /2π)^2...........> g = 22.657 m/s^2
3 0
3 years ago
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