Answer:
The voltage across a semiconductor bar is 0.068 V.
Explanation:
Given that,
Current = 0.17 A
Electron concentration 
Electron mobility 
Length = 0.1 mm
Area = 500 μm²
We need to calculate the resistivity
Using formula of resistivity


Put the value into the formula


We need to calculate the resistance
Using formula of resistance



We need to calculate the voltage
Using formula of voltage

Put the value into the formula


Hence, The voltage across a semiconductor bar is 0.068 V.
Answer:
It traveled 4 centimeters.
Explanation:
In a speed versus time graph, the distance travelled is given by the area under the graph.
In this graph we have the following:
- The speed of the object is v = 1 cm/s between time t = 0 s and t = 4 s
- The speed of the object is v = 0 cm/s between time t = 4 s and t = 8 s
Since the speed in the second part is zero, the distance travelled in the second part is zero. So, the only distance travelled by the object is the distance travelled during the first part, which is equal to the area of the first rectangle:

Here we can use coulomb's law to find the force between two charges
As per coulombs law
]tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we have




now by using the above equation we have


so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.
When two surfaces slide against each other, a force called friction makes them stick very slightly together. Smooth surfaces, like ice and glass, are easy to slide over. They create very little friction. Rough surfaces like rock and sand create much more friction, and are easy to grip on to.
hope it helps...!!!
<span>The answers are as follows:
(a) how many meters are there in 11.0 light-years?
11.0 light years ( 365 days / 1 year ) ( 24 h / 1 day ) ( 60 min / 1 h ) ( 60 s / 1 min ) ( 2.998x10^8 m/s ) = 1.04x10^17 m
(b) an astronomical unit (au) is the average distance from the sun to earth, 1.50 × 108 km. how many au are there in 11.0 light-years?
1.04x10^17 m ( 1 au / </span>1.50 × 10^8 km <span>) ( 1 km / 1000 m) = 693329.472 au
(c) what is the speed of light in au/h? au/h
</span>2.998 × 10^8 m/s ( 1 au / 1.50 × 10^8 km ) ( 1 km / 1000 m) ( 3600 s / 1 h ) = 7.1952 au/h