MELAINE complete a long distance run at the average speed of 60 mph if it takes her 3 hours how far did she Run

Suppose a car is traveling in the negative x-direction and comes to a stop. What is the sign of that car’s acceleration? How do

Oduvanchick [21]

We have equation of motion v = u + at

Where v = final velocity, u = initial velocity, a = acceleration, and t = time

In this case car is travelling in -x direction

Velocity of car = displacement / time

Since displacement value is increasing in negative x axis it's initial velocity is negative

And it's final velocity is zero since it comes to rest, and time is also positive

So, v= u+ at => 0 = -u + at

So, a = u/t

Which is positive and along positive X - direction

7 0

4 years ago

Can you anser the two questions fast

Rasek [7]

Answer:

yes! what are the questions??

Explanation:

4 0

3 years ago

Why are lights in house wired in house wired in a parallel or nothing series

navik [9.2K]

Parallel circuits are mostly used in houses because if you turn one light off the rest stay on. While with a series circuit, if you turn one light off, all the lights turn off as well because they are all connected to one circuit (series). Parallel is usually 2 or more circuits which is why it lets you turn one thing off and everything stays on.

4 0

4 years ago

Why doesn't the moon move away from the earth

Aleks04 [339]

The earths gravitational pull keeps the moon orbiting around and from straying away from it and into the vast expanses of outer space.

4 0

4 years ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

4 0

4 years ago