Answer:
(b)
Explanation:
The voltage always lags the current by 90°, regardless of the frequency.
Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces an amount of electric energy with the hot reservoir at 500 K during Day One and then produces the same amount of electric energy with the hot reservoir at 600 K during Day Two. The thermal pollution was:
Answer: hello your question lacks some vital information below is the complete question
Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces 1 × 106 J of electricity with the hot reservoir at 500 K during Day One and then produces 1 × 106 J of electricity with the hot reservoir at 600 K during Day Two. The thermal pollution was
answer:
Total thermal pollution = 2.5 * 10^6 J
Explanation:
Low temperature reservoir = 300 K
hot reservoir temperature = 500 K
Electrical energy produced by plant ( W ) = 1 * 10^6 J
lets assume ; Q1 = energy absorbed , Q2 = energy emitted
W = Q1 - Q2 or Q2 = Q1 - W ( we will apply this as the formula for determining thermal pollution )
For day 1
T1 = 500k , T2 = 300k
applying Carnot engine formula
W / Q1 = 1 - T2/T1
∴ Q1 = 10^6 / ( 1 - (300/500)) = 2.5 * 10^6 J
thermal pollution ; Q2 = Q1 - W = ( 2.5 * 10^6 - 1 * 10^6 ) = 1.5 * 10^6 J
for Day 2
T1 = 600k, T2 = 300k
Q1 = 10^6 / ( 1 - (300/600)) = 2 * 10^6 J
Thermal pollution; Q2 = Q1 - W = 1 * 10^6 J
Therefore the Total thermal pollution = 1 * 10^6 + 1.5 * 10^6 = 2.5 * 10^6 J
a. What do you mean by chromatic aberration in lenses?
Three 15-Ω and two 25-Ω light bulbs and a 24 V battery are connected in a series circuit. What is the current that passes through each bulb?
1) 0.18 A
2) 0.25 A
3) 0.51 A
4) 0.74 A
5) The current will be 1.6 A in the 15-Ω bulbs and 0.96 A in the 25-Ω bulbs.
Answer:
I = 0.25 A
Explanation:
Given that,
Three 15 ohms and two 25 ohms light bulbs and a 24 V battery are connected in a series circuit.
In series combination, the equivalent resistance is given by :
[tex]R=R_1+R_2+R_3+....[/tex]
So,
[tex]R=15+15+15+25+25\\\\=95\ \Omega[/tex]
The current each resistor remains the same in series combination. It can be calculated using Ohm's law i.e.
V = IR
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{24}{95}\\\\I=0.25\ A[/tex]
So, the current of 0.25 A passes through each bulb.
An ice skater with a mass of 50 kg is gliding acrossthe ice at a speed of 8 m/s when herfriend comes up from behind and gives her a push,causing her speed to increase to 12m/s. How much work did the friend do on the skater
Answer:
[tex]W=2KJ[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=50kg[/tex]
Initial Velocity [tex]v_1=8m/s[/tex]
Final Velocity [tex]v_2=12m/s[/tex]
Generally the equation for Work-done is mathematically given by
W=\triangle K.E
Therefore
[tex]W=0.5M(v_2^2-v_1^2)[/tex]
[tex]W=0.5*50(12^2-8^2)[/tex]
[tex]W=2KJ[/tex]
if a body covers 100m in 5 second from rest find the acceleration produced by a body in 10 second
Answer:
a=10m/s^2
Explanation:
acceleration= final velocity+ initial velocity/time taken
v-u/t=a
100-0/5=a
100/5=a
a=20m/s^2
case2
100-0/10=a
100/10=a
a=10m/s^2
Don't forget to write the units.
Hope this helps
please mark me as brainliest.
Question 7 of 10
A railroad freight car with a mass of 32,000 kg is moving at 2.0 m/s when it
runs into an at-rest freight car with a mass of 28,000 kg. The cars lock
together. What is their final velocity?
A.1.1 m/s
B. 2.2 m/s
C. 60,000 kg•m/s
D. 0.5 m/s
Answer:
a
Explanation:
you take 32,000kg ÷2.0m
A wheel accelerates so that it's angular speed increases uniformly from 150 rads/s to 580 rads/s in 16 revolutions.Cakcjlate its angular acceleration.
Answer:
A = 26.875 rad/s²
Explanation:
Given the following data;
Initial angular speed, Uw = 150 rads/s.
Final angular speed, Vw = 580 rads/s.
Time = 16 seconds.
To calculate the angular acceleration;
From kinematics equation;
At = Vw - Uw
Where;
A is the angular acceleration.t is the timeVw is the final angular speed.Uw is the initial angular speed.Substituting into the formula, we have;
A*16 = 580 - 150
16A = 430
A = 430/16
A = 26.875 rad/s²
Question 8 a-e plz
Answer:
(a) t = 0 s
(b) t = 0 s, 30 s, 55 s
(c) t = 40 s to t = 60 s
(d) t = 10 s to t = 15 s
(e) a = 6 m/s^2
Explanation:
(a) The car is at starting position at t = 0 s and v = 0 m/s.
(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.
(c) from t = 40 s to 60 s the car is moving in the negative direction.
(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.
(e) The slope of the velocity time graph gives acceleration.
a = (60 - 0) / (10 - 0) = 6 m/s^2
Which number has four significant figures?
A. 4000
B. 3.008
C. 86.012
D. 0.0001
a. 4000
This has 4-digits.
Answer:
in my opinion letter d.
Explanation:
Sana pi tama
A 1.0 ball moving at 2.0 / perpendicular to a wall rebounds from the wall at 1.5 /. If the ball was in contact with the wall for 0.1 , what force did the wall impart onto the ball?
Answer:
5N
Explanation:
We have a simple problem of momentum here.
ΔMomentum= mΔv= FΔt
Solve for F
mΔv/Δt=F
Plug in givens
1*(2-1.5)/0.1=F
F=5N
The amount of force that the wall imparts on the ball is 5.0N
According to Newton's second law, the formula for calculating the force applied is expressed as:
[tex]F=ma[/tex]
m is the mass of the object
a is the acceleration of the object
Since acceleration is the change in velocity of an object, hence [tex]a=\frac{\triangle v}{t}[/tex]
The applied force formula becomes [tex]F=\frac{m\triangle v}{t}[/tex]
Given the following parameters
m = 1.0kg
[tex]\triangle v=2.0-1.5\\\triangle v=0.5m/s[/tex]
t = 0.1sec
Substitute the given parameter into the formula
[tex]F=\frac{1.0\times 0.5}{0.1}\\F=\frac{0.5}{0.1}\\F=5N[/tex]
Hence the amount of force that the wall imparts on the ball is 5.0N
Learn more here: https://brainly.com/question/17811936
convert 2.4 milimetre into metre
Answer is 0.0024
Explanation
divide the length value by 1000.
Typhoon signal number 2 is raised. What is the speed of the expected typhoon?
the simple answer is from 61kmph to 120kmph
Explanation:
no explanation is needed
A student measure the length of a laboratory bench with a meter ruler. Which of the following values is the most approbriate way to record the result ? a.4.022m b.4.02m c.4.0m d.4m
Answer:
Well a meter stick has increments of a centimeter, and since 1 cm=0.01m he should record it as 4.02m(b)
Explanation:
What is the rate of the entropy change of the universe as heat leaks out a window, consisting of a single pane of glass that is 0.5 cm thick and 1.0 m2 in area, where the indoor temperature is 25°C and the outdoor temperature is -10°C?
Answer:
The change in entropy is 1.6 W/K.
Explanation:
Thickness, d = 0.5 cm
Area, A = 1 m^2
T = 25°C
T' = - 10°C
Coefficient of thermal conductivity of glass, K = 0.8 W/mK
The change in entropy is given by
S = Q/T
Here,
[tex]S =\frac{Q}{T}\\\\S = \frac{K A (T - T')}{d(T - T')}\\\\S = \frac{0.8\times 1}{0.5} = 1.6 W/K[/tex]
A heavy truck moving with 20 km/hr hits a car at rest. A physics student argued that
the maximum velocity the car suddenly gains is 40 km/hr. Do you agree with it?
Explain with necessary theory
Answer:
Yes
Explanation:
speed of truck = 20 km/h
Initially the car at rest.
maximum velocity of car = 40 km/h
When the truck and the car collide, the momentum of the truck transferred to car.
So, the car can attain the speed of 40 km/h.
Of the following, which have the highest frequency in the electromagnetic
spectrum?
A. Visible light
B. Infrared waves
C. Ultraviolet rays
D. X-rays
The equations for calculating both the electric force and the gravitational force are above. Their equations are very similar. What is an important difference between these two forces?
A The electrical force is measured in coulombs; the gravitational force is measured in newtons.
B The electrical force between two charged objects will always be weaker than the gravitational force between them.
C The gravitational force decreases with the square of the distance between the objects; the electrical force increases with the square of the distance between the objects.
D Electrical forces can be attractions or repulsions; gravitational forces can only be attractions.
A, B, and C are hilarious. D is correct.
Charges can be positive or negative, so a pair of charges can be alike or opposite. But so far, we've never seen a negative mass.
93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures
Answer:
828 kg/m³ or 0.828 g/cm³
Explanation:
Applying,
D = m/V............. Equation 1
Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.
From the question,
Given: m = 77 g , V = 93 cm³
Substitute these values into equation 1
D = 77/93
D = 0.828 g/cm³
Converting to kg/m³
D = 828 kg/m³
A fan spins at 6.0 rev/s. You turn it off, and it slows at 1.0 rev/s2. What is the angular displacement before it stops
Answer:
Angular displacement before it stops = 18 rev
Explanation:
Given:
Speed of fan w(i) = 6 rev/s
Speed of fan (Slow) ∝ = 1 rev/s
Final speed of fan w(f) = 0 rev/s
Find:
Angular displacement before it stops
Computation:
w(f)² = w(i) + 2∝θ
0² = 6² + 2(1)θ
0 = 36 + 2θ
2θ = -36
Angular displacement before it stops = -36 / 2
θ = -18
Angular displacement before it stops = 18 rev
Turning a corner at a typical large intersection in a city means driving your car through a circular arc with a radius of about 25 m. if the maximum advisable acceleration of your vehicle through a turn on wet pavement is 0.40 times the free-fall acceleration, what is the maximum speed at which you should drive through this turn?
Answer:
9.89 m/s.
Explanation:
Given that,
The radius of the circular arc, r = 25 m
The acceleration of the vehicle is 0.40 times the free-fall acceleration i.e.,a = 0.4(9.8) = 3.92 m/s²
Let v is the maximum speed at which you should drive through this turn. It can be solved as follows :
[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{3.92\times 25} \\\\=9.89 m/s[/tex]
So, the maximum speed of the car should be 9.89 m/s.
The unit of kinetic energy is the _______. The unit of kinetic energy is the _______. hertz meter watt joule radian
Answer:
joule
Explanation:
Una cuerda horizontal tiene una longitud de 5 m y masa de 0,00145 kg. Si sobre esta cuerda se da un pulso generando una longitud de onda de 0,6 m y una frecuencia de 120 Hz. La tensión a la cual está sometida la cuerda es:
a. 1,5 N
b. 15,0 N
c. 3,1 N
d. 5,2 N
Answer:
Option (A) is correct.
Explanation:
A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is
mass of string, m = 0.00145 kg
Frequency, f = 120 Hz
wavelength = 0.6 m
Speed = frequency x wavelength
speed = 120 x 0.6 = 72 m/s
Let the tension is T.
Use the formula
[tex]v =\sqrt\frac{T L}{m}\\\\72 = \sqrt\frac{T\times 5}{0.00145}\\\\T = 1.5 N[/tex]
Option (A) is correct.
The force an ideal spring exerts on an object is given by , where measures the displacement of the object from its equilibrium position. If , how much work is done by this force as the object moves from to
Answer:
The correct answer is "1.2 J".
Explanation:
Seems that the given question is incomplete. Find the attachment of the complete query.
According to the question,
x₁ = -0.20 mx₂ = 0 mk = 60 N/mNow,
⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]
⇒ [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]
⇒ [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]
By putting the values, we get
⇒ [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]
⇒ [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]
⇒ [tex]=1.2 \ J[/tex]
Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 40.0 mph and half the distance at 60.0 mph . On her return trip, she drives half the time at 40.0 mph and half the time at 60.0 mph.
Required:
a. What is Julie's average speed on the way to grandmother's house?
b. What is her average speed in the return trip?
Answer:
a. The average speed on her way to Grandmother's house is 48.08 mph
b. The average speed in the return trip is 50 mph.
Explanation:
The average speed (S) can be calculated as follows:
[tex] S = \frac{D}{T} [/tex]
Where:
D: is the total distance
T: is the total time
a. To find the total distance in her way to Grandmother's house, we need to find the total time:
[tex]T_{i} = t_{1_{i}} + t_{2_{i}} = \frac{d_{1_{i}}}{v_{1_{i}}} + \frac{d_{2_{i}}}{v_{2_{i}}}[/tex]
Where v is for velocity
[tex] T = \frac{d_{1_{i}}}{v_{1_{i}}} + \frac{d_{2_{i}}}{v_{2_{i}}} = \frac{(100/2) mi}{40.0 mph} + \frac{(100/2) mi}{60.0 mph} = 1.25 h + 0.83 h = 2.08 [/tex]
Hence, the average speed on her way to Grandmother's house is:
[tex]S_{i} = \frac{D}{T_{i}} = \frac{100 mi}{2.08 h} = 48.08 mph[/tex]
b. Now, to calculate the average speed of the return trip we need to calculate the total time:
[tex]D = v_{1_{f}}\frac{T_{f}}{2} + v_{2_{f}}\frac{T_{f}}{2} = \frac{T_{f}}{2}(v_{1_{f}} + v_{2_{f}})[/tex]
[tex]100 mi = \frac{T_{f}}{2}(40 mph + 60 mph)[/tex]
[tex] T_{f} = \frac{200 mi}{40 mph + 60 mph} = 2 h [/tex]
Therefore, the average speed of the return trip is:
[tex]S_{f} = \frac{D}{T_{f}} = \frac{100 mi}{2 h} = 50 mph[/tex]
I hope it helps you!
what is time taken by radio wave to go and return back from communication satellite to earth??
Answer:
Radio waves are used to carry satellite signals. These waves travel at 300,000 km/s (the speed of light). This means that a signal sent to a satellite 38,000 km away takes 0.13 s to reach the satellite and another 0.13 s for the return signal to be received back on Earth.
Explanation:
hope it help
Please assist with solving this problem and showing the steps
Answer:
2.21 N
Explanation:
The force in this case is the total mass multiplied by the acceleration due to gravity. You are not asked for the solution to be in terms of the torque which is the usual way to solve these problems. That's why you are not given where the fulcrum is.
The fulcrum feels F1 + F2 + 34 * 980
F2 = 141.7 * 980 = 138866
F1 = 50.3 * 980 = 49294
Ruler = 34 * 980= 33320
Total Force = 221480 The units here are dynes
I just saw in the middle of the question that g = 9.80
So the answer becomes 221480 / 1000 = 221.48 because we needed kg
And that answer becomes 221.48/100 2.21 because the force of gravity should be 9.8 not 980
The total force exerted on the fulcrum is
A current of 5.50 A flows in a conductor for 7.5 s. How much charge passes a given point in the conductor during this time?
Hi, so i have to find T1, can some1 help?
30.1 N
Explanation:
Given:
[tex]W_1 = 16\:\text{N}[/tex]
[tex]W_2 = 8\:\text{N}[/tex]
Let's write the components of the net forces at the intersections. Note that the system is equilibrium so all the net forces are zero.
Forces involving W1:
[tex]x:\:\:\:-T_1 + T_3\cos \alpha = 0\:\: \\ \text{or}\:\:T_2 = T_3\cos \alpha\:\:\:\:\:(1)[/tex]
[tex]y:\:\:\:T_3\sin \alpha - W_1 = 0\:\:\: \\ \text{or}\:\:\:T_3\sin \alpha = W_1\:\:\:\:\:\:(2)[/tex]
Forces involving W2:
[tex]x:\:\:\:T_1\sin 53 - T_3\sin \alpha = W_2\:\:\:\:\:\:\:(3)[/tex]
[tex]y:\:\:\:T_4 - T_1\cos 53 - T_3\cos \alpha = 0\:\:\:\;(4)[/tex]
Substitute (2) into (3) and we get
[tex]T_1\sin 53 - W_1 = W_2[/tex]
Solving for [tex]T_1[/tex],
[tex]T_1 = \dfrac{W_1 + W_2}{\sin 53} = 30.1\:\text{N}[/tex]
A body starts from rest and accelerates uniformly at 5m/s. Calculate the time taken by the body to cover a distance of 1km
Answer:
20 seconds
Explanation:
We are given 2 givens in the first statement
v0=0 and a=5
And we are trying to find time needed to cover 1km or 1000m.
So we use
x-x0=v0t+1/2at²
Plug in givens
1000=0+2.5t²
solve for t
t²=400
t=20s
two point charges with charge q are initially separated by a distance d. if you double the charge on both charges, how far should the charges be separated for the potential energy between them to remain the same
Answer:
r ’= 4 r
Explanation:
Electric potential energy is
U = [tex]k \frac{q_1q_2}{r_{12}}[/tex]k q1q2 / r12
in this exercise
q₁ = q₂ = q
U = k q² / r
for when the charge change
U ’= k q’² / r’
indicate that
q ’= 2q
U ’= U
we substitute
U = k (2q) ² / r ’
U = 4 k q² / r ’
we substitute
[tex]k \ \frac{q^2}{r} = 4 k \ \frac{q^2}{r'}[/tex]k q² / r = 4 k q² / r ’
r ’= 4 r