How many aluminum atoms are in the following compound?<br> 4 Al(OH)3

The security alarm on a parked car goes off and produces a frequency of 960 Hz. The speed of sound is 343 m/s. As you drive towa

Greeley [361]

Answer: $13.4\ m/s$

Explanation:

Given

The frequency of the source is $f_o=960\ Hz$

Change in frequency is $75\ Hz$

Speed of sound $c=343\ m/s$

Suppose $v$ is the velocity of the observer

Doppler frequency is given by

$f'=f_o\left(\dfrac{c\pm v_o}{c\pm v_s}\right)$

Here, the source is at rest

While approaching source, frequency is

$f_1=f_o\left(\dfrac{c+v}{c}\right)\quad \ldots(i)$

While leaving, frequency is

$f_2=f_o\left(\dfrac{c-v}{c}\right)\quad \ldots(ii)$

The difference in the frequency is

$\Rightarrow f_1-f_2=75\\\\\Rightarrow f_o\left(\dfrac{c+v}{c}\right)-f_o\left(\dfrac{c-v}{c}\right)=75\\\\\Rightarrow f_o\left(\dfrac{2v}{c}\right)=75\\\\\Rightarrow v=\dfrac{75\times 343}{2\times 960}\\\\\Rightarrow v=13.39\approx 13.4\ m/s$

7 0

3 years ago

please read the question and answer. it is quite complicated so someone clever please reply. many thanks :)

Lelechka [254]

Answer:

If he's 500 kg and 1 kg = 10 N then it would be 5,000 N

Explanation:

The surface area is unimportant and they'll use it to trick you.

5 0

4 years ago

Force = mass x acceleration

zloy xaker [14]

The answer is (B 60N) hope it helps and have a good day!

4 0

4 years ago

What is air resistance means explain it with free falling body

NikAS [45]

In freely falling body, there is no force acting on it other than the force of gravity (g).

8 0

3 years ago

Two charges, each 9 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field on the x axis at

bearhunter [10]

a) Electric field at x = -2 m: 21,060 N/C to the left

b) Electric field at x = 2 m: 18,000 N/C to the right

c) Electric field at x = 6 m: 18,000 N/C to the left

d) Electric field at x = 10 m: 21,060 N/C to the right

e) Electric field is zero at x = 4 m

Explanation:

a)

The electric field produced by a single-point charge is given by

$E=k\frac{q}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

Here we have two charges of

$q=9\mu C = 9\cdot 10^{-6} C$

each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.

We call the charge at x = 0 as $q_0$ , and the charge at x = 8 m as $q_8$ .

For a point located at x = -2 m, both the fields $E_0$ and $E_8$ produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:

$E=-\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=-kq(\frac{1}{(-2)^2}+\frac{1}{(8-(-2))^2})=-21060 N/C$

b)

In this case, we are analyzing a point located at

x = 2 m

The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(2)^2}-\frac{1}{(8-2)^2})=18000 N/C$

And since the sign is positive, the direction is to the right.

c)

In this case, we are considering a point located at

x = 6 m

The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(6)^2}-\frac{1}{(8-6)^2})=-18000 N/C$

And the negative sign indicates that the electric field in this case is towards the left.

d)

In this case, we are considering a point located at

x = 10 m

This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:

$E=\frac{kq_0}{(0-x)^2}+\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(10)^2}+\frac{1}{(8-(10))^2})=21060 N/C$

And the positive sign means the field is to the right.

e)

We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(-x)^2}-\frac{1}{(8-x)^2})=0$