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2 years ago

Tires can be recycled instead of thrown out. True False

Engineering

1 answer:

Arisa [49]2 years ago

8 0

Answer:

True :)

Explanation:

You can recycle it! Tire recycling is the most practical and environment-friendly way of disposing of old and worn-out tires. Due to their inherent durability, large volume and environment and health risks, tires are one of the most problematic sources of solid wastes.

Hope it helped have a nice day! :)

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If a signal is transmitted at a power of 250 mWatts (mW) and the noise in the channel is 10 uWatts (uW), if the signal BW is 20M

Bess [88]

Answer:

C = 292 Mbps

Explanation:

Given:

- Signal Transmitted Power P = 250mW

- The noise in channel N = 10 uW

- The signal bandwidth W = 20 MHz

Find:

what is the maximum capacity of the channel?

Solution:

-The capacity of the channel is given by Shannon's Formula:

C = W*log_2 ( 1 + P/N)

- Plug the values in:

C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)

C = (20*10^6)*log_2 (25001)

C = (20*10^6)*14.6096

C = 292 Mbps

3 0

3 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

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

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

For each of the resistors shown below, use Ohm's law to calculate the unknown quantity, Be sure to put your answer in proper eng

daser333 [38]

Answer:

the hurts my brain sorry bud cant help

Explanation:

6 0

2 years ago

According to the

zysi [14]

Answer:

The part of the system that is considered the resistance force is;

Explanation:

The simple machine is a system of pulley that has two pulleys

The effort, which is the input force at A gives the value of the tension at C and D which are used to lift the load B

Therefore, we have;

A = C = D

B = C + D = C + C = 2·C

∴ C = B/2

We have;

C = B/2 = A

Therefore, with the pulley only a force, A equivalent to half the weight, B, of the load is required to lift the load, B

The resistance force is the constant force in the system that that requires an input force to overcome in order for work to be done

It is the force acting to oppose the sum of the other forces system, such as a force acting in opposition to an input force

Therefore, the resistance force is the load force, B, for which the input force, A, is required in order for the load to be lifted.

3 0

3 years ago

An example of Ferrous alloy is Brass a)-True b)-False

djyliett [7]

Answer: False

Explanation: No, brass is not a ferrous alloy.

Ferrous alloys are those alloy which contain iron like cast iron, steel, strain-less steel, high carbon steel. Brass on the other hand does not contain any composition. of iron hence it can not be considered as a ferrous alloy. Brass comes under the category of non- ferrous made with a composition of copper and zinc, however their proportion is not strict and we can add other elements like aluminium or lead to alter its durability or corrosiveness.

5 0

3 years ago