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The velocity of an object increases from -10m/s to -15m/s in 2.0s what is the average acceleration of the object?

-2.5m/s^2
+2.5m/s^2
+12.5m/s^2
-12.5m/s^2

Answers

Answer 1

To find the average acceleration of an object, subtract the initial velocity from the final velocity and divide it by the time taken. In this case, the average acceleration is -2.5 m/s².

The average acceleration of an object can be calculated by using the formula:

acceleration = (change in velocity) / (time taken)

In this case, the change in velocity is  -15 m/s - (-10 m/s) = -5 m/s and the time taken is 2.0 s. So, acceleration = (-5 m/s) / (2.0 s) = -2.5 m/s² which means the correct answer is -2.5 m/s².

Therefore, the first option is correct.

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the sun is in the middle of the milky way and the planets in our solar system rotate around it

The resistivity of a silver wire with a radius of 5.04 × 10–4 m is 1.59 × 10–8 ω · m. if the length of the wire is 3.00 m, what is the resistance of the wire?

Answers

5.98 x 10^-2 ohms. Resistance is defined as: R = rl/A where R = resistance in ohms r = resistivity (given as 1.59x10^-8) l = length of wire. A = Cross sectional area of wire. So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives: R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2) R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7) R = (4.77 x 10^-8) / (7.98015 x 10^-7) R = 5.98 x 10^-2 ohms So that wire has a resistance of 5.98 x 10^-2 ohms.

The resistivity of a silver wire is 5.98 x 10^-2 ohms.

How do find the resistivity of a silver wire?

Resistance is defined as: R = r*l/A where R = resistance in ohms r = resistivity (given as 1.59x10^-8) l = length of wire. A = Cross sectional area of wire.

Substituting into the formula, the known values, including the area of a circle being pi*r^2, gives:

R = 1.59x[tex]10^{-8}[/tex] * 3.00 / ([tex]\pi[/tex] * (5.04 x [tex]10^{-4}[/tex])²)

R = (4.77 x[tex]10^{-8}[/tex]) / (pi * 2.54016 x [tex]10^{-7}[/tex])

R = (4.77 x [tex]10^{-8}[/tex]) / (7.98015 x [tex]10^{-7}[/tex])

R = 5.98 x [tex]10^{-2}[/tex] ohms

The wire has a resistance of 5.98 x 10^-2 ohms.

Learn more about Resistance here: https://brainly.com/question/14883923

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What is the magnitude of the electric field due to the dipole at location a, a distance d?

Answers

Zero. Dipoles do not have an electric field as it is cancelled out by the dipole it's self. However, the general rule for the propagation of an electric field is the inverse square law.

k is simply a constant of the value 9.9 x[tex]10^{9} N.m^{2} / C^{2}[/tex]

[tex]E = {k .Q}/d^2[/tex]
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