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jeka57 [31]
3 years ago
9

Please help, RIGHT NOWW!!! PLEASE

Physics
1 answer:
Alex73 [517]3 years ago
7 0

If a substance could reach absolute zero, then


-- the heat in it would be zero;

(if it had any heat in it, then the temperature would not be zero)


-- the temperature would be zero

(you just said so in the question)


-- the motion of its particles would be zero

(any particle motion would show up as temperature, which it doesn't have)


-- all of these

___________________________________


When somebody hands you a °Celsius temperature, it's not hard

to figure out the equivalent °Fahrenheit temperature.


°Fahrenheit = (1.8 · °Celsius) + 32° .


So it turns out that 100°C is equivalent to 212°F.

That's equal to A and less than B .


Also, it's not hard to figure out the equivalent absolute (Kelvin) temperature.

Just take the °Celsius temperature and add 273.15 .


So 100°C is equivalent to 373.15 K. (No degrees. Just K .)

That's equal to D.


The only choice that 100°C is greater than is choice-C .

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A ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec. Assuming that the air resistance can
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Answer:

Maximum height of the ball, h(t) = 27.56 m

Explanation:

It is given that, a ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec.      

The height of the ball as a function of time t is given by :

h(t)=h_o+v_ot-16t^2

h₀ is initial height, h₀ = 0

So, h(t)=42t-16t^2 .........(1)

For maximum/minimum height,  \dfrac{dh(t)}{dt}=0

42-32t=0...(2)

t = 1.31 s

Differentiating equation (2) wrt t

h''(t) = -32 < 0

So, at t = 1.31 seconds we will get the maximum height.

Put the value of t in equation (1)

h(t)=42\times 1.31-16\times (1.31)^2

h(t) = 27.56 m

Hence, this is the required solution.

7 0
4 years ago
At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.855 m/s2 if the acceleration du
natulia [17]
Given: Normal pull of gravity g = 9.8 m/s²; 

 g = 0.855 m/s²  (at a certain distance)

Universal gravitational constant G = 6.67 x 10⁻¹¹ N.m²/Kg²

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Radius r = ?

g = GMe/r²

r = √GMe/g

r = √(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)/(0.855 m/s²)

r = 2.16 x 10⁷ m or 

r =  21,610 Km





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5 0
3 years ago
Which of the following is an example of charging by friction?
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Answer: where is the examples?

Explanation:

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3 years ago
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Katie and Mark sit next to one another in class. She has a mass of 40 kg and his mass is 65 kg.
OlgaM077 [116]
A :-) for this question , we should apply
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( For making the calculation easy , first remove the decimals )
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M = 65 kg
m = 40 kg
d = 0.5 m
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F = GMm by d^2
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Answer:

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Explanation:

Simple Physics question!!!!!

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