Can you please answer this question?

Which two statements about composite materials are true?

Lostsunrise [7]

Answer:

Here down below is 3 things true about composite materials

Explanation:

-They’re made up of more than one substance.

-They have the same or similar properties as the materials used to make them.

-They’re always made of metal.

-They’re readily available in nature.

4 0

2 years ago

A man is walking while riding a train. He says he is moving at 2 mph. A woman standing on a platform at a train station says the

Semenov [28]

The woman on the platform is correct because it is the pace of the man moving on the train not walking.

6 0

2 years ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

3 years ago

An 8.00- W resistor is dissipating 100 watts. What are the current through it and the difference of potential across it?

hjlf

Answer:

I= 3.5 amps

Explanation:

Step one:

given data

rating of resistor R= 8 ohms

power P= 100W

Required

The current I

Step two

Yet this power is also given by

$P = I^2R$

make I subject of the formula we have

$I= \sqrt{\frac{P}{R} }$

substitute

$I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps$

8 0

2 years ago

For a photoelectric tube, calculate the voltage which will be just sufficient to stop electrons emitted by the sodium photo-plat

Talja [164]

Answer:

1.11 V

Explanation:

Given that the Einstein photoelectric equation states that;

KE = E - Wo

E = energy of incident photon

Wo= work function of the metal

E = hf = 6.64 x 10-34 * 6 x 1014

E = 39.84 * 10^-20 J or 3.98 * 10^-19 J

KE = 3.98 * 10^-19 J - 2.2 x 10-19J

KE = 1.78 * 10^-19J

We convert this value of KE to electron volts

KE = 1.78 * 10^-19J/1.6 x 10-19C

KE = 1.11 eV

Hence; 1.11 V will be just sufficient to stop electrons emitted by the sodium photo-plate reaching the collector plate.

8 0

2 years ago