I need help with question 6 and 7!!! Also, how to do the back???

1. Stops: Using the information learned in this course, explain three things you will not do when driving.

vazorg [7]

Don't text while driving
don't get your eyes off the road
don't get distracted

4 0

2 years ago

A 7.00 kg bowling ball is held 2.00 m above the ground. Using g- 9.8 m/s^2, how much energy does the bowling ball have due to it

Alexeev081 [22]

It should be B.137 just took the test.

3 0

2 years ago

Two trumpet players are trying to tune their instruments. When they are in tune, they will both be playing the same note and no

marusya05 [52]

Answer:

The second trumpeter will be playing at frequency = 515 Hz

Explanation: Given that the note sounds lower and they can hear 20 beats in 4.0 s.

Beat frequency = 20/4 = 5 Hz

Beat frequency = F2 - F1

5 = 520 - F1

F1 = 520 - 5

F1 = 515 Hz

Since the note sound lower, the second trumpeter will be playing at 515 Hz frequency

8 0

2 years ago

The intensity of the Sun's light in the vicinity of the Earth is about 1000W/m^2. Imagine a spacecraft with a mirrored square sa

NeX [460]

Explanation:

Formula to represent thrust is as follows.

F = $\frac{dP}{dt}$

= $\frac{2p}{dt}$

or, p = $\frac{E}{c}$

$\frac{p}{dt} = \frac{W}{c}$

F = $\frac{2IA}{c}$

= $\frac{2 (1000 W/m^{2})(5.5 \times 10^{3} m)^{2}}{3 \times 10^{8} m/s}$

= 201.67 N

Thus, we can conclude that the thrust is 201.67 N.

8 0

2 years ago

A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.

bija089 [108]

a)

i) Potential for r < a: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

ii) Potential for a < r < b: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

iii) Potential for r > b: $V(r)=0$

b) Potential difference between the two cylinders: $V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c) Electric field between the two cylinders: $E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

Explanation:

a)

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

$E=\frac{\lambda}{2\pi \epsilon_0 r}$

where

$\lambda$ is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

$V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)$

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

$E=0$

So the potential where the electric field is zero is constant:

$V=const.$

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density $+\lambda$ and an equal negative charge density $-\lambda$ . Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

$\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)$

However, we know that the potential at b is zero, so

$V(r)=V(b)=0$

ii) The electric field in the region a < r < b instead it is given only by the positive charge $+\lambda$ distributed over the surface of the inner cylinder of radius a, therefore it is

$E=\frac{\lambda}{2\pi r \epsilon_0}$

And so the potential in this region is given by:

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$ (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$