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Gala2k [10]
3 years ago
8

What happens to the acceleration when the velocity is zero

Physics
1 answer:
Ulleksa [173]3 years ago
7 0
<h2>Answer:</h2>

It's easy to fall into the temptation to say that when the velocity is zero, then the acceleration is also zero. But wait! To answer this question we need to bring out the concept of instantaneous velocity. This type of velocity stands for a specific moment, a specific instant of time, that is, t=1, \ t=2, \ t=3, \ t=3.2 \ t=4.5. If so, then acceleration may not be zero when velocity is zero. For example, suppose you throw an object upward, when it is at the top of the travel the instantaneous velocity is zero because it changes from positive to negative value and there is a moment when it must be zero, but yet there is a constant acceleration by the Earth's gravity at that moment. Even though the velocity at that stationary moment is zero, it doesn't imply the acceleration must be zero, so it has a value and in this case is -9.8m/s^{2}

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Ksivusya [100]

B is the correct answer

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  5. 2.25 N∙s
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8 0
3 years ago
5. The analytical method of adding vectors expressed in terms of their components may be applied to vectors in three dimensions,
Gwar [14]

Answer:

C = 17 i^ - 7 j^ + 16 k^ ,   | C| = 24.37

Explanation:

To work the vactor component method, we add the sum in each axis

C = A + B = (Aₓ + Bₓ) i ^ + (A_{y} + B_{y}) i ^ + (A_{z} + B_{z}) k ^

Cₓ = 12+ 5 = 17

C_{y} = -37 +30 = -7

C_{z} = 58 -42 = 16

Resulting vector

C = 17 i ^ - 7j ^ + 16k ^

The mangitude of the vector is

| C | = √ c²

| C | = √( 17² + 7² + 16²)

| C| = 24.37

6 0
3 years ago
Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el
lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}

The electrostatic force between them is

F_{e}=\frac{Kq^{2}}{d^{2}}

According to the question

1 % of Fg = Fe

0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}

2.927 \times 10^{35}=9 \times10^{9}q^{2}

3.25 \times 10^{25}=q^{2}

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0
3 years ago
As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of
kykrilka [37]

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.

Explanation:

Depression in freezing point :

\Delta T_f=K_f\times m

where,

\Delta T_f =depression in freezing point =  

K_f = freezing point constant  

m = molality  ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

  1. If molality of the solution in high the depression in freezing point of the solution will be more.
  2. If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

\frac{p_o-p_s}{p_o}=\chi_{solute}

p_o = Vapor pressure of pure solvent

p_s  = Vapor pressure of solution

\chi_{solute} = Mole fraction of solute

p_s\propto \frac{1}{\chi_{solute}}

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

  1. Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
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8 0
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