Answer:
27,000 m
450 m/s
Explanation:
Assuming the initial velocity is 0 m/s:
v₀ = 0 m/s
a = 15 m/s²
t = 60 s
A) Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²
Δy = 27,000 m
B) Find: v_avg
v_avg = Δy / t
v_avg = 27,000 m / 60 s
v_avg = 450 m/s
Answer:
As pressure goes up, volume goes down.
Explanation:
Pressure and volume of a gas are inversely proportional. This means that as pressure goes up, volume goes down. And as volume goes up, pressure goes down.
Cheers.
Answer:
True.
Explanation:
A diode, which allows current to flow in one direction only, consists of two types of semiconductors joined together.
A semiconductor can be defined as a crystalline solid substance that has its conductivity lying between that of a metal and an insulator, due to the effects of temperature or an addition of an impurity. Semiconductors are classified into two main categories;
1. Extrinsic semiconductor.
2. Intrinsic semiconductor.
An intrinsic semiconductor is a crystalline solid substance that is in its purest form and having no impurities added to it. Examples of intrinsic semiconductor are Germanium and Silicon.
In an intrinsic semiconductor, the number of free electrons is equal to the number of holes. Also, in an intrinsic semiconductor the number of holes and free electrons is directly proportional to the temperature; as the temperature increases, the number of holes and free electrons increases and vice-versa.
In an intrinsic semiconductor, each free electrons (valence electrons) produces a covalent bond.
Answer:
(a)
(b)
Explanation:
<u>Electric Circuits</u>
Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:
(a) The electromagnetic force of the battery is and its internal resistance is . Knowing the equivalent resistance of the headlights is , we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:
Solving for i
i=2.28\ A
The potential difference across the headlight bulbs is
(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is
Answer:
a) > x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
b)
And if we compare this with the general model
We see that the slope is m= 0.98 and the intercept b = 1.10
Explanation:
Part a
For this case we have the following data:
x: 1,2,3,4,5
y: 1.9,3.5,3.7,5.1, 6
For this case we can use the following R code:
> x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
Part b
For this case we have the following trend equation given:
And if we compare this with the general model
We see that the slope is m= 0.98 and the intercept b = 1.10