Answer:
Average speed of Olympic runner will be 23.08 km/hr
Explanation:
We have given distance d = 27.3 miles
We know that 1 mile = 1.6 km
So 27.3 mile =
Given time = 2 hour 45 minutes and 35 sec
We know that 1 hour = 60 minutes
So 45 minutes
We also know that 1 hour = 3600 sec
So 1 sec
So total time t =
We know that speed
A. 0.77 A
Using the relationship:
where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.
For the first light bulb, P = 60 W and V = 120 V, so the resistance is
For the second light bulb, P = 200 W and V = 120 V, so the resistance is
The two light bulbs are connected in series, so their equivalent resistance is
The two light bulbs are connected to a voltage of
V = 240 V
So we can find the current through the two bulbs by using Ohm's law:
B. 142.3 W
The power dissipated in the first bulb is given by:
where
I = 0.77 A is the current
is the resistance of the bulb
Substituting numbers, we get
C. 42.7 W
The power dissipated in the second bulb is given by:
where
I = 0.77 A is the current
is the resistance of the bulb
Substituting numbers, we get
D. The 60-W bulb burns out very quickly
The power dissipated by the resistance of each light bulb is equal to:
where
E is the amount of energy dissipated
t is the time interval
From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.
Answer:
Explanation:
Slope-intercept form means we want the y to be by itself in the equation. Every thing we do will be about getting the y alone on the left side of the equation
To start we should move x to the left hand side. We can do this by subtracting x from both sides. That way, there is an x on the right, but not the left.
x-x-2y=4-x
this gives us
-2y=4-x
Great! So now what? Well, the y isn't by itself yet because it still is being multiplied by negative two (-2). In order to move it from the left side to the right side, we have to do the opposite of multiply; divide. So, we will divide both sides by -2
-2 divided by -2 is 1, 4 divided by -2 is -2, and -x divided by -2 is
This gives us the answer:
Tips:
A negative divided by a negative is a positive ex: -4 divided by -2 is positive 2
If you are subtracting by a negative number, you are actually adding by a positive ex: 2-(-2) is actually 2+2
Don't be afraid to have fractions in your equations
Whatever you do to the one side of the equation, you must do it to the other side as well. Multiply the left side by 2? You HAVE TO multiply the right side by two as well. Add 3 to the right side? You HAVE TO add 3 to the left.
For problems like this (and when you have access to the internet), where you need to rewrite an equation, double check your work on desmos, which is an online graphing calculator. Input both the original equation and the equation you rewrote, and if they don't create the same line, you did something wrong.
Answer:
0.06 A
Explanation:
We have given mass =0.954 kg
velocity =1.27 m/sec
efficiency =0.361 the output kinetic energy of the motor
Efficiency =0.361
so input to the motor = output/efficiency
so input to the motor =
we know that where E is energy and T is time so
We know that power P=VI we have given V=4.5 VOLT
So current
Answer:
a) T_1 = 938 K , T_2 = 931 K
b) To prevent softening of the materials, which would occur below their melting points, the reactor should not be operated much above:
q = 3*10^8 W/m^3
Explanation:
Given:
- See the attachment for the figure for this question.
- Melting point of Thorium T_th = 2000 K
- Melting point of Thorium T_g = 2300 K
Find:
a) If the thermal energy is uniformly generated in the fuel element at a rate q = 10^8 W/m^3 then what are the temperatures T_1 and T_2 at the inner and outer surfaces, respectively, of the fuel element?
b) Compute and plot the temperature distribution in the composite wall for selected values of q. What is the maximum allowable value of q.
Solution:
part a)
- The outer surface temperature of the fuel, T_2, may be determined from the rate equation:
q*A_th = T_2 - T_inf / R'_total
Where,
A_th: Area of the thorium section
T_inf: The temperature of coolant = 600 K
R'_total: The resistance per unit length.
- Calculate the resistance per unit length R' from thorium surface to coolant:
R'_total = Ln(r_3/r_2) / 2*pi*k_g + 1 / 2*pi*r_3*h
Plug in values:
R'_total = Ln(14/11) / 2*pi*3 + 1 / 2*pi*0.014*2000
R'_total = 0.0185 mK / W
- And the heat rate per unit length may be determined by applying an energy balance to a control surface about the fuel element. Since the interior surface of the element is essentially adiabatic, it follows that:
q' = q*A_th = q*pi*(r_2^2 - r_1^2)
q' = 10^8*pi*(0.011^2 - 0.008^2) = 17,907 W / m
Hence,
T_2 = q' * R'_total + T_inf
T_2 = 17,907*0.0185 + 600
T_2 = 931 K
- With zero heat flux at the inner surface of the fuel element, We will apply the derived results for boundary conditions as follows:
T_1 = T_2 + (q*r_2^2/4*k_th)*( 1 - (r_1/r_2)^2) - (q*r_1^2/2*k_th)*Ln(r_2/r_1)
Plug values in:
T_1 = 931+(10^8*0.011^2/4*57)*( 1 - (.8/1.1)^2) - (10^8*0.008^2/2*57)*Ln(1.1/.8)
T_1 = 931 + 25 - 18 = 938 K
part b)
The temperature distributions may be obtained by using the IHT model for one-dimensional, steady state conduction in a hollow tube. For the fuel element (q > 0), an adiabatic surface condition is prescribed at r_1 while heat transfer from the outer surface at r_2 to the coolant is governed by the thermal resistance:
R"_total = 2*pi*r_2*R'_total
R"_total = 2*pi*0.011*0.0185 = 0.00128 m^2K/W
- For the graphite ( q = 0 ), the value of T_2 obtained from the foregoing solution is prescribed as an inner boundary condition at r_2, while a convection condition is prescribed at the outer surface (r_3).
- For 5*10^8 < q and q > 5*10^8, the distributions are given in attachment.
The graphs obtained:
- The comparatively large value of k_t yields small temperature variations across the fuel element, while the small value of k_g results in large temperature variations across the graphite.
Operation at q = 5*10^8 W/^3 is clearly unacceptable, since the melting points of thorium and graphite are exceeded and approached, respectively. To prevent softening of the materials, which would occur below their melting points, the reactor should not be operated much above:
q = 3*10^8 W/m^3
