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adell [148]
3 years ago
12

4.) The diagram to the right is the orbit of a comet:

Physics
1 answer:
Brilliant_brown [7]3 years ago
6 0

Answer: This is what I found hope it helps

Explanation:

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What element atomic number is in Group 17 and Period 6?
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The halogen elements are the six elements in Group 17 of the periodic table. Group 17 is the second column from the right in the periodic table and contains six elements: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (As), and tennessine (Ts).

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3 years ago
In a series circuit with three resistors connected to a power source, what would happen to the equivalent resistance and current
igor_vitrenko [27]

<u>Answer:</u>

<h2><em></em></h2>

<em>The equivalent resistance would decrease and the current would increase.</em>

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<u><em>Mark me as the brainlyest</em></u>

3 0
2 years ago
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Which term describes the upward force that acts on an object submerged in a fluid?
MaRussiya [10]
The answer would be: buoyancy


6 0
4 years ago
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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

3 0
4 years ago
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ANSWER FAST FOR BRAINLIEST!
Ne4ueva [31]

Answer:

the last one

Explanation:

7 0
3 years ago
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