When the speed and direction of an object remain the same, what happens to its velocity?

1 answer:

4 0

Answer: Velocity is the vector, which means it has both direction and magnitude . Velocity is defined as the speed of something in a given direction .

Explanation:

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In the 1887 experiment by Michelson and Morley, the length of each interferometer arm was 11m. The experimental limit on the mea

Vikentia [17]

For the answer to the question above,
we can get the number of fringes by dividing (delta t) by the period of the light (Which is λ/c).

fringe = (delta t) / (λ/c)

We can find (delta t) with the equation:

delta t = [v^2(L1+L2)]/c^3

Derivation of this formula can be found in your physics text book. From here we find (delta t):

600,000^2 x (11+11) / [(3x10^8)^3] = 2.93x10^-13

2.93x10^-13/ (589x10^-9 / 3x10^8) = 149 fringes

This answer is correct but may seem large. That is because of your point of reference with the ether which is usually at rest with respect to the sun, making v = 3km/s.

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3 years ago

A car goes from rest to a velocity of 108 km/h north in 10s what is the car's acceleration in m/s2

fomenos

initial velocity of the car given as

$v_i = 0$

final velocity is given as

$v_f = 108 km/h$

as we know that

$1 km/h = 0.277 m/s$

now we can convert final speed into m/s

$v_f = 108 * 0.277 = 30 m/s$

now acceleration is rate of change in velocity

$a = \frac{v_f - v_i}{t}$

$a = \frac{30 - 0}{10}$

$a = 3 m/s^2$

so the acceleration of the car is 3 m/s^2

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3 years ago

a baseball player is dashing toward home plate with a speed of 200 feet per hour. When he decides to hit the dirt. he slides for

Serggg [28]

To solve this problem we will use the kinematic formula for the final velocity.

$V_f_x = V_0_x + a_xt$

The final speed is 0 at the moment the player stops.

The time until it stops is 1.3 s

The initial speed is 200 feet / s Note (check the speed units in the problem statement, 200ft / s is very much and 200ft / h is very small)

Then, we clear the formula.

$a_x = \frac{(Vfx-V0x)}{t}\\ a_x = \frac{(0-200)}{1.3}\\ a_x = -153.5 ft / s ^ 2$

Because the player is slowing down, the acceleration goes in the opposite direction to the player's movement, and that is why it is negative.

To answer part b) we use the following formula.

$Vf ^ 2 = Vo ^ 2 + 2a * (x_2 - x_1)\\\\ (x_2 - x_1)= \frac{(V_f ^ 2-V_0 ^ 2)}{2a}\\\\ (x_2 - x_1)= \frac{(0-200 ^ 2)}{- 2 * 153.5}\\\\ (x_2 - x_1)= 130.29 feet$