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Margarita [4]
2 years ago
8

Both Newton's Law of Universal Gravitation and Coulomb's Law follow the ...

Physics
1 answer:
Romashka [77]2 years ago
8 0

Answer:

inverse square relationship

Explanation:

Both the Newton's law of universal gravitation and coulomb's law have their force inversely proportion to the square of the distance between the bodies.

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What do you call a group of sea turtles?
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a bale

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a bale is a group of turtles

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A glass windowpane in a home is 0.62 cm thick and has dimen- sions of 1.0 m 3 2.0 m. On a certain day, the indoor temper- ature
DIA [1.3K]

Answer:

Explanation:

Thermal conductivity of glass pane = .8 W / m K

For conduction of heat , the formula is as follows

Q = \frac{kA(T_1 - T_2 )t}{d }   ; Q is heat conducted in time t by a medium of thickness d , area of cross section A and ( T₁ - T₂ ) is the temperature difference .

Putting the values in the equation

Q = \frac{.8 \times32\times(25 - 0 )\times1}{.0062 }

= 103225.8 W

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= 103225.8 x 24 x 60 x 60

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6 0
3 years ago
Two cars A and B, travel in a straight line. The distance of A from the starting point is given as a function of time by x????(?
Norma-Jean [14]

Answer:

a) They are in the same point

b) t = 0 s, t = 2.27 s, t = 5.73 s

c) t = 1 s, t = 4.33 s

d) t = 2.67 s

Explanation:

Given equations are:

x_{a}(t) = at+bt^2

x_{b}(t) = ct^2-dt^3

Constants are:

a = 2.60 m/s, b = 1.20 m/s^2, c= 2.80 m/s^2, d = 0.20 m/s^3

a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both x_a(t) and x_b(t) depend on t and don't have constant terms.

So both cars A and B are in the same point.

b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).

at+bt^2 = ct^2-dt^3

2.6t + 1.2t^2 = 2.8t^2 - 0.2t^3\\0.2t^2 - 1.6t + 2.6 = 0\\t^2 - 8t + 13 = 0

t_1 = 4 - \sqrt{3} = 2.27 s, t_1 = 4 + \sqrt{3} = 5.73 s

c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.

v_a(t) = \frac{d}{d(t)}x_a(t) = a + 2bt \\v_b(t) = \frac{d}{d(t)}x_b(t) = 2ct - 3dt^2\\a+2bt = 2ct - 3dt^2\\3dt^2+2(b-c)t+a = 0\\0.6t^2-3.2t+2.6 = 0

t_1 = 1 s, t_2 = 4.33 s

d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.

a_a(t) = \frac{d}{d(t)}v_a(t) = 2b \\a_b(t) = \frac{d}{d(t)}v_b(t) = 2c - 6dt\\2b = 2c - 6dt\\3dt = c - b\\t = (c - b)/3d = (2.8 - 1.2)/(3*0.2) = 2.67 s

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The answer is B.

The planet cannot be too hot or too cold it has to be the right distance from its sun to maintain life. 
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