Answer & Explanation:
1 N-m = 1 Joule
So 82 kJ of energy put into the system during (i).
45 kJ of heat leaves the system, so 82 kJ - 45 kJ = 37 kJ is remaining.
(ii) requires 100 kJ of energy but only 37 kJ is available, so 100 kJ - 37 kJ = 63 kJ of heat energy must be added to the system.
The heat given would be equal to the heat emitted from the system and by providing some external source of energy the volume or temperature of the system may increase.
The amount of heat added to the system is 63kJ.
The energy can be estimated as:
Given,
Work done by piston = 82000 NmHeat rejected in surrounding = 45 kJWork done during expansion stroke = 100000 NmQuantity of added heat = ?During compression stroke:
Work done by the piston [tex]\rm (W_{1-2})[/tex] = - 82000Nm or - 82kJHeat rejected to the system [tex]\rm (Q_{1-2})[/tex] = - 45kJWe know that,
[tex]\rm Q_{1-2} = (U_{2} - U_{1}) + W[/tex]
Therefore,
[tex]\begin{aligned}-45 &= \rm (U_{2} - \rm U_{1}) + (-82)\\\\\rm (U_{2}-U_{1}) &= 37\rm (U_{2} - U_{1}) = 37\; kJ \end{aligned}[/tex] (equation 1 )
During Expansion system:
Work done by the piston [tex]\rm (W_{2-1})[/tex] = 100000 Nm or 100 kJ
Now putting values in the equation:
[tex]\begin{aligned}\rm Q_{2-1} &= \rm U_{1} - U_{2} + W\\\\&=\rm (U_{1} - U_{2}) + W\end{aligned}[/tex]
Substituting value from equation 1:
[tex]\begin{aligned}\rm Q_{2-1} &= - 37+100\rm \;kJ\\&= 63\rm \; kJ\end{aligned}[/tex]
Therefore, 63kJ of energy is added to the system.
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Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 12.0 mg piece of tape held 0.55 cm above another. (The magnitude of this charge is consistent with what is typical of static electricity.)
Answer:
q = 2 10⁻⁸ C
Explanation:
For this exercise we use the translational equilibrium equation
F_e -A =
F_e = W
the electric force is given by Coulomb's law
F_e = [tex]k \frac{q_1q_2}{r^2}[/tex]
in this case they indicate that the loads on the tapes are equal
F_e = k q² / r²
we substitute
k q² / r² = m g
q = [tex]\sqrt{ \frac{mg r^2}{k} }[/tex]
calculate
q = [tex]\sqrt { \frac{ 12 \ 10^{-3} \ 9.8 (0.55 \ 10^{-2})^2 }{9 \ 10^9} }[/tex]
q = [tex]\sqrt{ 3.9526 \ 10^{-16}[/tex]
q = 1,999 10⁻⁸ C
q = 2 10⁻⁸ C
A diesel engine lifts the 225 kg hammer of a pile driver 20 m in 5 seconds. How much work is done on
the hammer? What is the power?
Answer:
a. Workdone = 44100 Joules
b. Power = 8820 Watts.
Explanation:
Given the following data:
Mass = 225kg
Distance = 20m
Time = 5 seconds
To find the workdone;
Workdone = force * distance
But force = mg
We know that acceleration due to gravity is equal to 9.8m/s²
Force = 225*9.8 = 2205N
Substituting the values into the equation, we have;
Workdone = 2205 * 20
Workdone = 44100 Joules
b. To find the power;
Power = workdone/time
Power = 44100/5
Power = 8820 Watts.
VP 3.12.1 Part APart complete A cyclist going around a circular track at 10.0 m/s has a centripetal acceleration of 5.00 m/s2. What is the radius of the curve? Express your answer with the appropriate units. R = 20.0 m Previous Answers Correct VP 3.12.2 Part B A race car is moving at 40.0 m/s around a circular racetrack of radius 265 m. Calculate the period of the motion. Express your answer in seconds. T = nothing s Request Answer Part C Calculate the car’s centripetal acceleration.
Answer:
A) r = 20.0 m
B) T = 41.6 s
C) = 6.1 m/s²
Explanation:
A)
The centripetal acceleration is the one that explains that even though the cyclist is moving at a constant speed, his velocity is changing the direction all the time, keeping him around a circle.This acceleration can be expressed as follows:[tex]a_{c} =\frac{v^{2}}{r} = \frac{(10.0m/s)^{2}}{r} = 5.00 m/s2 (1)[/tex]
Solving for r:[tex]r = \frac{v^{2}}{a_{c} } = \frac{(10.0m/s)^{2}}{5.00m/s2} = 20.0 m (2)[/tex]
B)
We can apply the definition of linear velocity, remembering that the period is the time needed to complete an entire circle (T).The arc around a circumference (the distance traveled) , is just 2*π*r, so applying the definition of linear velocity, we can write the following expression:[tex]v = \frac{\Delta s}{\Delta t} = \frac{2*\pi*r}{T} (3)[/tex]
Solving for T:[tex]T = \frac{\Delta s}{v} = \frac{2*\pi*r}{v} = \frac{2*\pi*265m}{40.0m/s} =41.6 s (4)[/tex]
C)
The centripetal acceleration of the car from B) can be found as follows:[tex]a_{c} =\frac{v^{2}}{r} = \frac{(40.0m/s)^{2}}{265m} = 6.1 m/s2 (5)[/tex]
A model shows a machine that works using electrical fields. What would this machine need for the electrical field to function properly?
at least two charged interacting parts
A car with mass m travels over a hill with a radius of curvature of r at a speed of 15 m/s. What is the normal force on the car when the car is at the top of the hill?
Answer:
zero
Explanation:
The computation of the normal force is shown below:
As we know that
F_c = mg - N
F_c = mv^2 ÷ r
N = mg - mv^2 ÷ r
N = m(g - v^2 ÷ r)
Assume that
The mass of the car is 1200 kg
And, r = 10 m
So,
= 1200 (9.8 - 15^2 ÷ 10)
= -15240 N
Since it comes in negative so the normal force would be zero
Galileo used marbles rolling down inclined planes to deduce some basic properties of constant accelerated motion. In particular, he measured the distance a marble rolled during specific time periods. For example, suppose a marble starts from rest and begins rolling down an inclined plane with constant acceleration a. After 1 s, you find that it moved a distance .
a. In terms of x, how far does it move in the next 1 s time period—that is, in the time between 1 s and 2 s?
b. How far does it move in the next second of the motion?
c. How far does it move in the nth second of the motion?
Answer:
a) y₁ = ½ a, b) y₂ = 4 y₁, c) y₃ = 9 y₁
Explanation:
For this exercise we can use the accelerated motion relationships.
Let's set a reference system where the x axis is parallel to the plane and its positive side is going down the plane.
y = y₀ + v₀ t + ½ a t²
in that case where we throw the marble is the zero point, y₀ = 0, as part of rest its initial velocity is zero v₀ = 0 and a is the acceleration along the inclined plane
y = ½ a t²
a) in the first second t = 1
y₁ = ½ a
b) in the next second of movement
t = 2 s
y₂ = ½ a 2²
y₂ = 4 ½ a
y₂ = 4 y₁
c) for the next second
t = 3 s
y₃ = ½ a 3²
y₃ = 9 ½ a
y₃ = 9 y₁
A wave in the ocean has a wavelength of 2 m and a frequency of 0’5 Hz. What is the speed of this wave?
Answer:
the speed of the wave =1m/s
Two objects are electrically charged. The net charge on one object is doubled.
Therefore, the electric force _____.
reverses
doubles
quadruples
divides
help please i will mark brainlist!!!
Answer:
.50 M
Explanation:
5*.50=2.5 + 2*.25=.5 = 3n
6*.50= 3N
Final answer is .50M
PLEASE HELP ASAP WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST!!!!
Because of the forces acting on the cart, it will
A. not accelerate
B. accelerate upwards
C. accelerate to the right
D. accelerate to the left
Answer:
D.
Explanation:
The First Law of Thermodynamics is the same as ______ with heat and work taken into consideration.
A. The First Law of Robotics
B. The Law of Conservation of Energy
C. Newton's First Law of Motion
D. The Law of Conservation of Momentum
Answer:
the law of conservation of energy
what is the average velocity of a van that moves from 0 to 60 m east and 20 seconds
Explanation:
I have a lot to say it was very nice to meet my parents are u doing well I dont want too its been so much I love you so I was like u know I am not a man but you are the auditions I have been in a long long long life is a triangle and a chair for me and my parents think about the way I
A 120 W lightbulb and a 90 W lightbulb each operate at a voltage of 120 V. Part A Which bulb carries more current? Which bulb carries more current? 120 W lightbulb 90 W lightbulb The currents are equal. It is impossible to determine.
Answer:
120 W lightbulb
Explanation:
Let the two lightbulb be A and B respectively.
Given the following data;
Power A = 120W
Power B = 90W
Voltage = 120V
To find the current flowing through each lightbulb;
a. For lightbulb A
Power = current * voltage
120 = current * 120
Current = 120/120
Current = 1 Ampere.
b. For lightbulb B
Current = power/voltage
Current = 90/120
Current = 0.75 Amperes
Therefore, the lightbulb that carries more current is A with 1 Ampere.
The bulb that carries more current is :
- A with 1 Ampere.
Let the two lightbulb be A and B respectively.
Given :Power A = 120WPower B = 90WVoltage = 120VTo find the current flowing through each lightbulb;
a. For lightbulb APower = current * voltage120 = current * 120Current = 120/120Current = 1 Ampere.b. For lightbulb BCurrent = power/voltageCurrent = 90/120Current = 0.75 AmperesThus, the lightbulb that carries more current is A with 1 Ampere.
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What power rating of resistors would you use in the application required it to handle
0.6W?
I would use a resistor rated for 1 W or more. Not less.
The power rating of a resistor that would be used in application that requires 0.6 W must be greater than 0.6 W.
Electrical powerThe electrical power of an appliance shows the rating of the appliance, in terms of energy consumed at a given period of time.
Electrical power is calculated as follows;
P = IV
where;
V is the voltage I is the current[tex]P = (\frac{V}{R} )V\\\\P = \frac{V^2}{R}[/tex]
Thus, the power rating of a resistor that would be used in application that requires 0.6 W must be greater than 0.6 W.
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You have 2 resistors of unknown values you label Ra and Rb. You have an old battery and a multimeter you bought years ago for 7$ at Harbor Freight. Using the meter in voltage mode, you measure 10 V across the battery. You then connect the 2 resistors in series across the battery and use the meter in current mode to find the current flowing through the circuit. It reads 0.111A. You then connect the 2 resistors in parallel across the battery and use the meter again to measure the current now coming from the battery to be 0.5A. With this information you have gathered, you find the value of the 2 resistors.
Value of smallest resistance in ohms.
a. 60
b. 90
c. 20
d. 30
Answer:
the answers, the correct one is D, Rb₂ = 29.97 ohm
Explanation:
For this exercise we use ohm's law and the equivalent resistance ratio for series and parallel circuits.
Serial circuit
(Ra + Rb) is = V
(Ra + Rb) 0.111 = 10
(Ra + Rb) = 10 / 0.111 = 90.09
parallel circuit
[tex]\frac{1}{R} = \frac{1}{Ra} + \frac{1}{Rb}[/tex]
R = [tex]\frac{Ra \ Rb}{Ra + Rb}[/tex]
\frac{Ra \ Rb}{Ra + Rb} i_p = V
\frac{Ra \ Rb}{Ra + Rb} 0.5 = 10
\frac{Ra \ Rb}{Ra + Rb} = 10 / 0.5 = 20
we write and solve our system of equations
Ra + Rb = 90.09
\frac{Ra \ Rb}{Ra + Rb} = 20
we solve for Ra in the first equation
Ra = 90.09 - Rb
RaRb = 20 (Ra + Rb)
we substitute Ra in the second equation
(90.09-Rb) Rb = 20 [(90.09-Rb) + Rb]
90.09 Rb - Rb² = 20 90.09
Rb² - 90.09 Rb + 1801.8 = 0
we solve the quadratic equation
Rb = [90.09 ±[tex]\sqrt{90.09^2 - 4 \ 1801.8}[/tex] ] / 2
Rb = [90.09 ± 30.15] / 2
Rb₁ = 60.12 ohm
Rb₂ = 29.97 ohm
the smallest value is Rb = 30 ohm
When checking the answers, the correct one is D
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 4.20 m/sm/s. Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 52.9 mm above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 4.50 ss after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.
Required:
a. With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground?
b. Where is Henrietta when she catches the bagels?
Answer:
a) v₀ₓ = 9.9 m / s, b) x_woman = 32.7 m
Explanation:
A) In this exercise, the movement of the bagels is parabolic, we find the time it takes to reach the floor.
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + 0 - ½ gt²
t = [tex]\sqrt{2y_o/g}[/tex]
let's calculate
t = [tex]\sqrt{2 \ 52.9/9.8}[/tex]
t = 3,286 s
Now we can analyze how long Henrieta has walked, she has a walking time before the bagel movement begins (t₀ = 4.50 s)
t_woman = t₀ + t
t_woman = 4.50 + 3.286
t_woman = 7.786 s
The distance traveled in this time is
x_{woman} = v_woman t_woman
x_{woman} = 4.20 7.786
x_{woman} = 32.7 m
For her to grab the bagel, the two of them must be at this point
x_bagel = x_woman
x_bael = vox t
v₀ₓ = x_bagel / t
v₀ₓ = 32.7 / 3,286
v₀ₓ = 9.9 m / s
b) when catching the bagels this point x_woman = 32.7 m
A baseball is thrown horizontally from a cliff at 30 m/s and lands 7 seconds after the baseball was thrown. Calculate the horizontal AND vertical distance.
Answer:
The horizontal and vertical distances are x = 210 m and y = -240.35 m, respectively.
Explanation:
Using the equation of the displacement in the x-direction, we have:
(let's recall we have a constant velocity in this direction)
[tex]x=v_{ix}t[/tex]
Where:
v(ix) is the initil velocity in the x direction (v(ix) = 30 m/s)t is the time (t = 7 s)[tex]x=30(7)[/tex]
[tex]x=210\: m[/tex]
Now, we need to use the equation of the displacement in the y-direction to find the vertical distance. Here we have an acceleration (g)
[tex]y=v_{iy}t-\frac{1}{2}gt^2[/tex]
Where:
v(iy) is the initial velocity at the y-direction. In this case, it will be 0t is the timeg is the acceleration of gravity (g=9.81 m/s²)Then, the vertical position at 7 s is:
[tex]y=-\frac{1}{2}(9.81)(7)^2[/tex]
[tex]y=-240.35\: m[/tex]
Therefore, the horizontal and vertical distances are x = 210 m and y = -240.35 m, respectively. The minus sign means the negative value in the y-direction.
I hope it helps you!
Help plsssssssssss I write it 100 time no one answer
Answer:
1.93×10²⁸ s
Explanation:
From the question given above, the following data were obtained:
Number of electron (e) = 2×10²⁴
Current (I) = 10 A
Time (t) =?
Next, we shall determine the quantity of electricity flowing through pasing through the point. This can be obtained as follow:
1 e = 96500 C
Therefore,
2×10²⁴ e = 2×10²⁴ e × 96500 / 1 e
2×10²⁴ e = 1.93×10²⁹ C
Thus, 1.93×10²⁹ C of electricity is passing through the point.
Finally, we shall determine the time. This can be obtained as follow:
Current (I) = 10 A
Quantity of electricity = 1.93×10²⁹ C
Time (t) =?
Q = it
1.93×10²⁹ = 10 × t
Divide both side by 10
t = 1.93×10²⁹ / 10
t = 1.93×10²⁸ s
Thus, it took 1.93×10²⁸ s for 2×10²⁴ electrons to pass through the point
The relationship between frequency and period is...
[tex] \\ [/tex]
Frequency, f, is how many cycles of an oscillation occur per second and is measured in cycles per second or hertz (Hz). The period of a wave, T, is the amount of time it takes a wave to vibrate one full cycle. These two terms are inversely proportional to each other: f = 1/T and T = 1/f.
[tex] \\ [/tex]
Hope It Helps!
Answer:
Inverse
Explanation:
Frequency is the number of cycles in a second. Frequency is the inverse of a period
frequency = 1 / period
A fuel tank for a rocket in space under a zero-g environment is rotated to keep the fuel in one end of the tank. The system is rotated at 3 rev/min. The end of the tank (point A) is 1.5 m from the axis of rotation, and the fuel level is 1 m from the rotation axis. The pressure in the nonliquid end of the tank is 0.1 kPa, and the density of the fuel is 800 kg/m3 . What is the pressure at the exit (point A)
Answer:
P₂ = 4098.96 Pa
Explanation:
For this exercise let's use Bernoulli's equation
Let's use the subscript 1 for the point of the liquid surface and the subscript 2 for the ends (point A)
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
the velocity at the end of the tank
v₂ = w r₂
the velocity at the surface of the liquid is
v₁ - w r₁
where r₂ = 1.5 m and r₁ = 1 m
the tank pressure is P₁ = P₀ = 0.1 10³ Pa
P₂ = P₁ + ½ ρ [w² (r₁² - r₂²)] + ρ g (y₁ -y₂)
We must remember that the pressure measurements the distances are measured from the lowest part to the surface that has zero height
let's reduce the magnitudes to the SI system
w = 3 rev / min (2π rad / 1rev) (1 min / 60 s) = 0.314159 rad / s
let's calculate
P₂ = 0.1 10³ + ½ 800 0.314159² |(1² -1.5²)| + 800 9.8 |(1-1.5)|
P₂ = 0.1 103 +78.96 + 3920
P₂ = 4098.96 Pa
current must flow if 0.56 coulombs is to be transferred 35ms
Answer:
the current is 16 amphere
Explanation:
The computation of Current is shown below:
As we know that
1 ms = 0.001s
So for 35 ms = 0.035
Now the current is
= 0.56 ÷ 0.035
= 16 AMphere
Hence, the current is 16 amphere
4.
How does the United Nations Development Program use its resources?
It provides health programs for mothers and children.
O It develops natural resources.
It works to eliminate poverty through development.
O It invests funds in industrialized nations.
Economic
Answer:
It develops natural resources.
It works to eliminate poverty through development.
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25 cm and the point-like object with charge q2 = −2.14 µC is located at x2 = −1.80 cm.
A) Determine the total electric potential (in V) at the origin.
B) Determine the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
Answer:
a) the total electric potential is 2282000 V
b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
Explanation:
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
so
Electric potential at p in the diagram 1 below is;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we know that; Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)
the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
r1² = 0.015² + 0.0125²
r1 = √[ 0.015² + 0.0125² ]
r1 = √0.00038125
r1 = 0.0195
Also
r2² = 0.015² + 0.018²
r2 = √[ 0.015² + 0.018² ]
r2 = √0.000549
r2 = 0.0234
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
a) The total electric potential is 2282000 V
b) The total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
What is electric potential?The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to a specific point in an electric field.
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
Electric potential at p in diagram 1 below is;
[tex]V_P=V_1+V_2[/tex]
[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]
we know that; the Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)The total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
[tex]r_1^2=0.015^2+0.0125^2[/tex]
[tex]r_1 = \sqrt{[ 0.015^2 + 0.0125^2 ][/tex]
[tex]r_1 = \sqrt{0.00038125}[/tex]
[tex]r_1 = 0.0195[/tex]
Also
[tex]r_2^2 = 0.015^2 + 0.018^2[/tex]
[tex]r_2 = \sqrt{0.015^2 + 0.018^2}[/tex]
[tex]r_2 = \sqrt{0.000549[/tex]
[tex]r_2 = 0.0234[/tex]
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
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The monkey experiment is an example of what?
A. top down processing
B. bottom up processing
C. inattentional blindness
D. sensory adaption
Answer:
D.) Sensory adaptation
Explanation:
Assuming you are talking about the cloth and metal monkey experiment performed in the field of psychology (not physics), the monkey formed an attachment to the cloth mother because it felt closer to it, as it was more appealing to its senses.
what's the dimension symbol for thermodynamic temperature
Answer: Throughout the scientific world where measurements are nearly always made in SI units, thermodynamic temperature is measured in kelvins (symbol: K). The Rankine scale uses the degree Rankine (symbol: °R) as its unit, which is the same magnitude as the degree Fahrenheit (symbol: °F).
Explanation:
Please mark me as the Brainiest if I got it right
what's the dimension symbol for thermodynamic temperature
Answer:
°R
Arun runs 9 meters across Mr. Scharff's classroom in 7.1 seconds. How fast did Arun run
Answer:
The answer would be 180 meters.
Explanation:
You are designing a thin transparent reflective coating for the front surface of a sheet of glass. The index of refraction of the glass is 1.52, and when it is in use the coated glass has air on both sides. Because the coating is expensive, you want to use a layer that has the minimum thickness possible, which you determine to be 104 nmnm. Part A What should the index of refraction of the coating be if it must cancel 500-nmnm light that hits the coated surface at normal incidence
Answer:
1.32
Explanation:
Index of refraction of the glass = 1.52
Thickness = 104 nm
Length = 550 nm
Using formula of index
n = L/4t
Where, L = length
t = thickness
Substituting the values into the formula we get
n = 500/(4×104)
n= 1.32
Hence, The index of refraction of the coating is 1.32.
To understand the nature of electric fields and how to draw field lines. Electric field lines are a tool used to visualize electric fields.
a. True
b. False
Answer:
a. True
Explanation:
Electric field is a region of space where the effect of electric field lines or lines of forces are felt.
The electric field lines creates electric field and these field lines help to visualize the electric field.
Therefore, electric field lines are tool used to visualize electric fields.
a. True
Which statement is correct?
A. If the electric field is zero everywhere inside a region of space, the potential must also be zero in that region.
B. When the electric field is zero at a point, the potential must also be zero there.
C. If the electrical potential in a region is constant, the electric field must be zero everywhere in that region.
D. If the electric potential at a point in space is zero, then the electric field at that point must also be zero.
Answer:
The answer is "Choice C ".
Explanation:
The relationship between the E and V can be defined as follows:
[tex]\to E= -\Delta V[/tex]
Let,
[tex]\to E= \frac{\delta V}{\delta x}[/tex]
When E=0
[tex]\to \frac{\delta V}{\delta x}=0[/tex]
v is a constant value
Therefore, In the electric potential in a region is a constant value then the electric-field must be into zero that is everywhere in the given region, that's why in this question the "choice c" is correct.
Which cell part controls all the other parts of a cell?
Answer:
Nucleus
Explanation:
Answer: Nucleolus
Explanation: The nucleolus is like the cells brain. It controls all the other organelles (cell parts).
Hope I helped!
