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nata0808 [166]
3 years ago
14

Two objects experienced the same displacement in the same amount of time. The two objects must have the same -

Physics
1 answer:
nikdorinn [45]3 years ago
6 0

If two object experiences the same displacement at the same time, the two objects must have the same average velocity.

The average velocity of an object is defined as the change in displacement per change in time of motion. This can be written as follows;

avg. \ velocity = \frac{\Delta \ displacement }{\Delta \ time} \\\\V = \frac{\Delta \ x}{\Delta t}

If two object experiences the same displacement at the same time, it means that the change in displacement with time is constant.

\frac{\Delta x_1}{\Delta t_1} = \frac{\Delta x_2}{\Delta t_2} = V_{avg}

Thus, we can conclude that if two object experiences the same displacement at the same time, the two objects must have the same average velocity.

Learn more here: brainly.com/question/23856383

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A charge q1 of -5.00X10^-9 C and a charge q2 of -2.00X10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium posit
Alex73 [517]

Answer:

The equilibrium position for the third charge is 69.28 cm

Explanation:

Given;

q₁ = -5.00 x 10⁻⁹ C

q₂ = -2.00 x 10⁻⁹ C

q₃ = 15.00 x 10⁻⁹ C

distance between q₁  and q₂ = 40.0 cm = 0.4 m                                    

(-q₁)--------------------------------------(-q₂)---------------------------------(+q₃)

At equilibrium the repulsive force between q₁ and q₂ must be equal to attractive force between q₂ and q₃

According to Coulomb's law, repulsive or attractive force between charges is calculated as;

F = \frac{Kq_1q_2}{r_1^2} =  \frac{Kq_2q_3}{r_2^2}

where;

F is repulsive or attractive force between charges

K is Coulomb's constant = 8.99 x 10⁹ Nm²/c²

r₁ is the distance between q₁ and q₂

q₁, q₂ and q₃ are the charge

distance between q₂ and q₃, r₂ is calculated as;

\frac{Kq_1q_2}{r_1^2} = \frac{Kq_2q_3}{r_2^2}\\\\\frac{q_1q_2}{r_1^2} = \frac{q_2q_3}{r_2^2}\\\\r_2^2= \frac{r_1^2q_2q_3}{q_1q_2}\\\\r_2^2= \frac{r_1^2q_3}{q_1} = \frac{0.4^2*15*10^{-9}}{5*10^{-9}} = 0.48\\\\r_2 = \sqrt{0.48} = 0.6928 \ m

Therefore, the equilibrium position for the third charge is 69.28 cm

3 0
3 years ago
A block slides forward 4.73 m while being pulled back by a 39.7 N force at 142° from the direction of motion. How much work does
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Answer:

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Given:

displacement of the box, s = 4.73 m

angle between force and displacement, \theta = 142^{\circ}

Force, F = 39.7 N

Solution:

Work done by the force can be expressed as the dot product of Force, F and displacement 's' is given by:

W = Fscos\theta

W = 39.7\times 4.73\times cos142^{\circ}

W = -147.97 J

where negative sign indicates that work is done on the block

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