Answer:
Explanation:
Initial moment of inertia of the carousel + girl
I₁ = 110 + 44 x 1.7²
= 110 + 127.16
= 237.16 kgm².
final moment of inertia of carousel + girl
I₂ = 110 + 44 x d²
applying law of conservation of angular momentum
I₁ ω₁ = I₂ω₂
ω₁ and ω₂ are angular velocities of the carousel before and after .
237.16 x 3.4 = (110 + 44 x d²)x 5.4
806.34 = 594 + 237.6 d²
237.6 d² = 212.34
d²= .8936
d = .9453 m
An iron railroad rail is 800 ft long when the temperature is 31°C. What is its length (in ft) when the temperature is −17°C?
Answer:
799.54 ft
Explanation:
Linear thermal expansion is:
ΔL = α L₀ ΔT
where ΔL is the change in length,
α is the linear thermal expansion coefficient,
L₀ is the original length,
and ΔT is the change in temperature.
Given:
α = 1.2×10⁻⁵ / °C
L₀ = 800 ft
ΔT = -17°C − 31°C = -48°C
Find: ΔL
ΔL = (1.2×10⁻⁵ / °C) (800 ft) (-48°C)
ΔL = -0.4608
Rounded to two significant figures, the change in length is -0.46 ft.
Therefore, the final length is approximately 800 ft − 0.46 ft = 799.54 ft.
Two circular coils are concentric and lie in the same plane. The inner coil contains 110 turns of wire, has a radius of 0.014 m, and carries a current of 9.0 A. The outer coil contains 160 turns and has a radius of 0.022 m. What must be the magnitude of the current in the outer coil, such that the net magnetic field at the common center of the two coils is zero?
Answer:
The current flowing through the outer coils is
Explanation:
From the question we are told that
The number of turn of inner coil is [tex]N _i = 110 \ turns[/tex]
The radius of inner coil is [tex]r_i = 0.014 \ m[/tex]
The current flowing through the inner coil is [tex]I_i = 9.0 \ A[/tex]
The number of turn of outer coil is [tex]N_o = 160 \ turns[/tex]
The radius of outer coil is [tex]r_o = 0.022\ m[/tex]
For net magnetic field at the common center of the two coils to be zero the current flowing in the outer coil must be opposite to current flowing inner coil
The magnetic field due to inner coils is mathematically represented as
[tex]B_i = \frac{N_i \mu I}{2 r_i}[/tex]
The magnetic field due to inner coils is mathematically represented as
[tex]B_o = \frac{N_o \mu I_o}{2 r_o}[/tex]
Now for magnetic field at center to be zero
[tex]B_o = B_i[/tex]
So
[tex]\frac{N_i \mu I_i}{2 r_i} = \frac{N_o \mu I_o}{2 r_o}[/tex]
=> [tex]\frac{110 * 9}{2 * 0.014} = \frac{160 *I_o}{2 0.022}[/tex]
[tex]I_o = 9.72 \ A[/tex]
Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.30 kg and is initially moving to the right at 2.4 m/s toward a second block of mass 0.80 kg that is initially at rest. When the blocks collide, a cocked spring releases 1.2 J of energy into the system. (For velocities, use to mean to the right, - to mean to the left).A) What is the velocity of the first block after the collision?
B) What is the velocity of the second block after the collision?
Answer:
a) 3.632 m/s
b) 0.462 m/s
Explanation:
Using the law of conservation of momentum:
[tex]m_{1} u_{1} + m_{2} u_{2}= m_{1} V_{1} + m_{2} V_{2}[/tex]..........(1)
[tex]m_{1} = 0.30 kg\\u_{1} = 2.4 m/s\\m_{2} = 0.80 kg\\u_{2} = 0 m/s[/tex]
Substituting the above values into equation (1) and make V2 the subject of the formula:
[tex]0.3(2.4) + 0.80(0)= 0.3 V_{1} + 0.8 V_{2}\\[/tex]
[tex]V_{2} = \frac{0.72 - 0.3 V_{1}}{0.8}[/tex]..................(2)
Using the law of conservation of kinetic energy:
[tex]0.5m_{1} u_{1} ^{2} + 1.2 = 0.5m_{1} V_{1} ^{2} + 0.5m_{2} V_{2} ^{2}\\0.5(0.3) (2.4) ^{2} + 1.2 = 0.5(0.3) V_{1} ^{2} + 0.5(0.8)V_{2} ^{2}\\[/tex]
[tex]2.064 = 0.15 V_{1} ^{2} + 0.4V_{2} ^{2}[/tex].......(3)
Substitute equation (2) into equation (3)
[tex]2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.72 - 0.3V_{1} }{0.8}) ^{2}\\2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.5184 - 0.432V_{1} + 0.09V_{1} ^{2} }{0.64}) \\1.32096 = 0.096 V_{1} ^{2} + 0.20736 - 0.1728V_{1} + 0.036V_{1} ^{2} \\0.132 V_{1} ^{2} - 0.1728V_{1} - 1.1136 = 0\\V_{1} = 3.632 m/s[/tex]
Substituting [tex]V_{1}[/tex] into equation(2)
[tex]V_{2} = \frac{0.72 - 0.3 *3.632}{0.8}\\V_{2} = \frac{0.72 - 0.3 *(3.632)}{0.8}\\V_{2} = 0.462 m/s[/tex]
A cube of ice is taken from the freezer at -5.5 ∘C and placed in a 75-g aluminum calorimeter filled with 300 g of water at room temperature of 20.0 ∘C. The final situation is observed to be all water at 17.0 ∘C. The specific heat of ice is 2100 J/kg⋅C∘, the specific heat of aluminum is 900 J/kg⋅C∘, the specific heat of water is is 4186 J/kg⋅C∘, the heat of fusion of water is 333 kJ/Kg.
What was the mass of the ice cube?Express your answer to two significant figures and include the appropriate units.
Answer:
Explanation:
Let mass of ice cube taken out be m kg .
ice will gain heat to raise its temperature from - 5.5° to 0° and then from 0° to 17° .
Total heat gained = m x 2.1 x 5.5 + m x 333 + m x 4.186 x 17
= (11.55 + 333 + 71.162 )m
= 415.712 m kJ
Heat lost by aluminium calorimeter
= .075 x .9 x 3
= .2025 kJ
Heat lost by water
= .3 x 4.186 x 3
= 3.7674
Total heat lost
= 3.9699 kJ
Heat lost = heat gained
415.712 m = 3.9699
m = .0095 kg
9.5 gm .
Answer:
0.00954g or 9.5x[tex]10^{-3}[/tex] kg
Explanation:
The only conversion that is needed is changing the heat of fusion of water from 333 kJ/kg to 333000 J/kg.
This is the condensed version of the equation needed for this problem: mcΔT + mL + mcΔT = mcΔT + mcΔT
This is the expanded version of the equation needed for this problem:
[tex]m_{ice}[/tex]([tex]c_{ice}[/tex])(temperature of ice from -5.5°C to 0°C) + [tex]m_{ice}[/tex](L) + [tex]m_{ice}[/tex]([tex]c_{water}[/tex])(temperature of water from 0°C to 17°C) = [tex]m_{water}[/tex]([tex]c_{water}[/tex])(ΔT) + [tex]m_{aluminum}[/tex]([tex]c_{aluminum}[/tex])(ΔT)
Use the equation to solve for the mass of ice:
m(2100)(5.5) + m(333000) + m(4186)(17) = 0.3(4186)(20-17) + 0.075(900)(20-17)
m [(2100x5.5) + 333000 + (4186x17)] = 3767.4 + 202.5
m(415712) = 3969.9
m = 0.00954g or 9.5x[tex]10^{-3}[/tex] kg
A student uses the right-hand rule as shown.
What is the direction of the magnetic field in front of the wire
closest to the student?
up
right
down
left
Answer:
right is the correct answer to the given question .
Explanation:
In this question figure is missing
The main objective right-hand rule to decide the position of the magnetic force on the positive force acting, either the position of the thumb of a right hand with in position of v, the fingers throughout the position of B1, and a right angles throughout the position of F1 to the hand positions.
So [tex]F1 \ =\ q\ v \ B1\ Sin\alpha[/tex]
So from the magnetic right hand rule the direction of the magnetic field in front of a wire is right .All the others options are incorrect because they do not give the direction of the magnetic field in front of a wire is right .Answer:
Right
Explanation:
Just did this on edgen, in the diagram, the fingers are pointing to the right, which indicates that the magnetic field is acting to the right.
Two fans are watching a baseball game from different positions. One fan is located directly behind home plate, 18.3 mfrom the batter. The other fan is located in the centerfield bleachers, 127 m from the batter. Both fans observe the batterstrike the ball at the same time(because the speed of light is about a million times faster than that of sound), but the fan behind home plate hears the sound first. What is the time difference between hearing the sound at the two locations? Use 345 m/s as the speed of sound.
Answer:
Δt = 0.315s
Explanation:
To calculate the time difference, in which both fans hear the batterstrike, you first calculate the time which takes the sound to travel the distances to both fans:
[tex]t_1=\frac{d_1}{v_s}[/tex]
[tex]t_2=\frac{d_2}{v_s}[/tex]
d1: distance to the first fan = 18.3 m
d2: distance to the second fan = 127 m
vs: speed of sound = 345 m/s
You replace the values of the parameters to calculate t1 and t2:
[tex]t_1=\frac{18.3m}{345m/s}=0.053s\\\\t_2=\frac{127m}{345m/s}=0.368s[/tex]
The difference in time will be:
[tex]\Delta t =t_2-t_2=0.368s-0.053s=0.315s[/tex]
Hence, the time difference between hearing the sound at the location s of both fans is 0.315s
R1=3 ohms
R2=6 ohms
R4=18 ohms
R5= 15 ohms
R5=9 ohms
90 volts
What is the current running through the entire circuit?
Answer: current I = 1.875A
Explanation:
If the resistors are connected in series,
Then the equivalent resistance will be
R = 6 + 18 + 15 + 9
R = 48 ohms
Using ohms law
V = IR
Make current I the subject of formula
I = V/R
I = 90/48
I = 1.875A
And if the resistors are connected in parallel, the equivalent resistance will be
1/R = 1/6 + 1/18 + 1/15 + 1/9
1/R = 0.166 + 0.055 + 0.066 + 0.111
R = 1/0.3999
R = 2.5 ohms
Using ohms law
V = IR
I = 90/2.5
Current I = 35.99A
You're carrying a 4.0-m-long, 21 kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 cm from its tip. How much force must you exert to keep the pole motionless in a horizontal position?
Answer:
The weight of the pole can be assumed to act at the pole's midpoint, which is 2m from the fence / pivot point, giving us a moment of 490Nm (245N x 2m). We have to counteract this moment by holding the end of the pole. So, we have a lever arm of 3.65m (4.0m - 0.35m), so we would need to exert a force of 134.4N (490N / 3.65m) at a point 35cm from the end of the pole.
Explanation:
happy to help:)
Use Hooke's Law to determine the variable force in the spring problem. A force of 450 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 40 centimeters to 70 centimeters?
Answer:
Work Done = 67.5 J
Explanation:
First we find the value of spring constant (k) using Hooke's Law. Hooke's is formulated as:
F = kx
where,
F = Force Applied = 450 N
k = Spring Constant = ?
x = Stretched Length = 30 cm = 0.3 m
Therefore,
450 N = k(0.3 m)
k = 450 N/0.3 m
k = 1500 N/m
Now, the formula for the work done in stretching the spring is given as:
W = (1/2)kx²
Where,
W = Work done = ?
k = 1500 N/m
x = 70 cm - 40 cm = 0.3 m
Therefore,
W = (1/2)(1500 N/m)(0.3 m)²
W = 67.5 J
A planetary nebula is a(n) ____. Group of answer choices a shell of gas ejected by and expanding away from an extremely hot dying low-mass star a shell of gas ejected by and expanding away from an extremely hot dying high-mass star an expanding atmosphere of a low-mass star as it becomes a red giant a contracting shell of ionized interstellar medium absorbed by a dying low-mass star a contracting shell of dusty material from planets destroyed by a dying low-mass star
Answer: Planetary nebular is
A shell of gas ejected by and expanding away from an extremely hot dying high-mass star
An expanding atmosphere of a low-mass star as it becomes a red Giants.
Explanation:
Planetary nebular is a form of nebula emission that comprises an expanding and glowing shell of gas that is ejected from red giant stars.
Planetary nebulae play an important role in the chemical evolution of the Milky Way by expelling elements into the interstellar medium from stars where those elements were produced.
A car 4m long moving at a velocity of 25m/s was beside a lorry 20m long with a velocity 19mls. At t =
0,
the distance between them was 10m. How long will it take the car to overtake the lorry (a) 9s (b) 5s (c) 3s
(d) 2s
Answer:
???????
Explanation:
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You are designing a hydraulic lift for a machine shop. The average mass of a car it needs to lift is about 1500 kg. You wish to exert a force on a smaller piston of not more than 550N .a) What should be the specifications on the dimensions of the pistons?Asmall piston/Alarge piston = ???b) How far down will you need to push the piston in order to lift the car 50cm ?h = ???
Answer:
(a) Area(small piston)/Area(large piston) = 0.037
(b) h = 1336.36 cm = 13.36 m
Explanation:
(a)
The stress on the smaller piston is equally transmitted to the larger piston, in a hydraulic lift. Therefore,
Stress (small piston) = Stress (large piston)
Force (small piston)/Area (small piston) = Force (Large Piston)/Area (Large Piston)
Area(small piston)/Area(large piston) = Force (small piston)/Force(Large piston)
Area(small piston)/Area(large piston) = 550 N/(1500 kg)(9.8 m/s²)
Area(small piston)/Area(large piston) = 0.037
(b)
The work is also transmitted equally to the large piston. So,
Work(small piston) = Work(Large Piston)
Force(small piston).Displacement(small piston) = Force(large piston).Displacement(small piston)
(550 N)(h) = (1500 kg)(9.8 m/s²)(50 cm)
h = 735000 N.cm/550 N
h = 1336.36 cm = 13.36 m
(a) The ratio of area smaller piston to area of larger piston is 0.037.
(b) The distance the smaller piston will be pushed down to lift the car at the given height is 13.36 m.
The given parameters;
mass of the car, m = 1500 kgforce on the smaller piston, F₁ = 550 Nlet the area of the small piston = A₁
let the area of the large piston = A₂
Apply constant pressure principle as shown below;
[tex]P = \frac{F_1}{A_1} = \frac{F_2}{A_2} \\\\\frac{A_1}{A_2} = \frac{A_{small}}{A_{large}} = \frac{F_1}{F_2} = \frac{550}{mg} \\\\ \frac{A_{small}}{A_{large}} = \frac{550}{1500 \times 9.8} \\\\ \frac{A_{small}}{A_{large}} = 0.037[/tex]
The height the car was raised = 50 cm = 0.5 mThe distance the effort will be applied is calculated as follows;
[tex]550 d = mgh\\\\550d = (1500 \times 9.8 \times 0.5)\\\\550 d = 7350\\\\d = \frac{7350}{550} \\\\d = 13.36 \ m[/tex]
Thus, the distance the smaller piston will be pushed down to lift the car at the given height is 13.36 m.
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What is meant by a charge carrier hole in a semiconductor? Can it be created in a conductor?
Answer:
The materials used to make electronic components like transistor and integrated, circuit behave as if effective particles known as electron through them, causing electrical properties
Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2.25 kg and is on an incline of 1=43.5∘ with coefficient of kinetic friction 1=0.205 . 2 has a mass of 5.45 kg and is on an incline of 2=32.5∘ with coefficient of kinetic friction 2=0.105 . The two‑block system is in
Answer:
The acceleration of [tex]M_2[/tex] is [tex]a = 0.7156 m/s^2[/tex]
Explanation:
From the question we are told that
The mass of first block is [tex]M_1 = 2.25 \ kg[/tex]
The angle of inclination of first block is [tex]\theta _1 = 43.5^o[/tex]
The coefficient of kinetic friction of the first block is [tex]\mu_1 = 0.205[/tex]
The mass of the second block is [tex]M_2 = 5.45 \ kg[/tex]
The angle of inclination of the second block is [tex]\theta _2 = 32.5^o[/tex]
The coefficient of kinetic friction of the second block is [tex]\mu _2 = 0.105[/tex]
The acceleration of [tex]M_1 \ and\ M_2[/tex] are same
The force acting on the mass [tex]M_1[/tex] is mathematically represented as
[tex]F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1[/tex]
=> [tex]M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1[/tex]
Where T is the tension on the rope
The force acting on the mass [tex]M_2[/tex] is mathematically represented as
[tex]F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]
[tex]M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]
At equilibrium
[tex]F_1 = F_2[/tex]
So
[tex]T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]
making a the subject of the formula
[tex]a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}[/tex]
substituting values [tex]a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}[/tex]
=> [tex]a = 0.7156 m/s^2[/tex]
The acceleration of the second block to the right is 2.21 m/s².
The normal force on block1 is calculated as follows;
[tex]F_n_1 = m_1g cos(\theta_1)[/tex]
The parallel force on block 1 is calculated as;
[tex]F_x_1 = m_1gsin(\theta)[/tex]
The frictional force on block 1 is calculated as;
[tex]F_k_1 = \mu_k F_n = \mu_k m_1gcos\theta_1[/tex]
The net force on block 1 is calculated as;
[tex]\Sigma F_x_1 = m_1gsin(\theta_1) - \mu_k_1m_1gcos(\theta_1)[/tex]
The normal force on block 2 is calculated as follows;
[tex]F_n_2 = m_2gcos\theta _2[/tex]
The frictional force on block 2 is calculated as;
[tex]F_k_2 = \mu k_2 m_2g cos\theta _2[/tex]
The net force on block 2 is calculated as follows;
[tex]\Sigma F_x_2 = m_2a_2\\\\ m_2g_2 sin(\theta _2) - F_k_2 - \Sigma F_x_1 = m_2a_2 \\\\m_2gsin(\theta) - F_k_2 - (m_1gsin(\theta) - \mu_k _1 m_1g cos(\theta)) = m_2a_2\\\\m_2gsin(\theta) -\mu_k_2 m_2gcos(\theta) + \mu_k _1 m_1g cos(\theta) - m_1gsin(\theta) = m_2a_2\\\\5.45( 9.8) sin(32.5) -(0.105)(5.45)(9.8)cos(32.5) + \\\\0.205( 2.25) ( 9.8)cos(43.5) - 2.25( 9.8) sin (43.5) = 5.45a_2\\\\ 12.07 = 5.45a_2\\\\a_2= \frac{12.07}{5.45} \\\\a_2 = 2.21 \ m/s^2[/tex]
Thus, the acceleration of the second block to the right is 2.21 m/s².
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Daniel Levinson found that men and women
A. Go through the same stages of development
B. Differ in terms of their social roles and identities
C. Deal with the development tasks in each stage differently
D. All of the above
Answer:
Option D. is correct
Explanation:
Daniel Levinson, a psychologist developed an adult development theory. This theory was referred to as the Seasons of Life theory. This theory describes the stage in which a person leaves adolescence and start making choices about adult life.
Daniel Levinson found that men and women go through the same stages of development, differ in terms of their social roles and identities, and deal with the development tasks in each stage differently.
Option D. is correct
Answer:
Daniel Levinson found that men and women
d. all of the above
Explanation:
cause i got a 100 on edg;)
Fill in the blanks for the following:
Arigid container of volume 100.0 Liters contains Oxygen gas. It is at room temperature (293 Kelvin), and is at atmospheric pressure (absolute pressure, meaning the gauge pressure is zero). Therefore, the number of moles of Oxygen molecules . Also, the rms-velocity inside is ________of these Oxygen molecules is most nearly ______, which is______ the rms-speed of the Nitrogen molecules just outside the container (the rigid container and its surroundings are in thermal equilibrium).
Answer:
a. 4.21 moles
b. 478.6 m/s
c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank
Explanation:
Volume of container = 100.0 L
Temperature = 293 K
pressure = 1 atm = 1.01325 bar
number of moles n = ?
using the gas equation PV = nRT
n = PV/RT
R = 0.08206 L-atm-[tex]mol^{-1}[/tex][tex]K^{-1}[/tex]
Therefore,
n = (1.01325 x 100)/(0.08206 x 293)
n = 101.325/24.04 = 4.21 moles
The equation for root mean square velocity is
Vrms = [tex]\sqrt{\frac{3RT}{M} }[/tex]
R = 8.314 J/mol-K
where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol
Vrms = [tex]\sqrt{\frac{3*8.314*293}{0.0319} }[/tex]= 478.6 m/s
For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship
[tex]\frac{Voxy}{Vnit}[/tex] = [tex]\sqrt{\frac{Mnit}{Moxy} }[/tex]
where
Voxy = root mean square velocity of oxygen = 478.6 m/s
Vnit = root mean square velocity of nitrogen = ?
Moxy = Molar mass of oxygen = 31.9 g/mol
Mnit = Molar mass of nitrogen = 14.00 g/mol
[tex]\frac{478.6}{Vnit}[/tex] = [tex]\sqrt{\frac{14.0}{31.9} }[/tex]
[tex]\frac{478.6}{Vnit}[/tex] = 0.66
Vnit = 0.66 x 478.6 = 315.876 m/s
the root mean square velocity of the oxygen gas is
478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank
A nuclear power plant generates 3000 MW of heat energy from nuclear reactions in the reactor's core. This energy is used to boil water and produce high-pressure steam at 280∘C. The steam spins a turbine, which produces 1100 MW of electric power, then the steam is condensed and the water is cooled to 25∘C before starting the cycle again.
a. What is the maximum possible thermal efficiency of the power plant?
b. What is the plant's actual efficiency?
Answer:
a) [tex]\eta_{th} = 46.1\,\%[/tex], b) [tex]\eta_{th,real} = 36.667\,\%[/tex]
Explanation:
a) The maximum possible thermal efficiency of the power plant is given by the Carnot's Cycle thermal efficiency, which consider a reversible power cycle according to the Second Law of Thermodynamics, whose formula is:
[tex]\eta_{th} = \left(1-\frac{T_{L}}{T_{H}} \right)\times 100\,\%[/tex]
Where:
[tex]T_{L}[/tex] - Temperature of the cold reservoir (Condenser), measured in K.
[tex]T_{H}[/tex] - Temperature of the hot reservoir (Evaporator), measured in K.
The maximum possible thermal efficiency is:
[tex]\eta_{th} = \left(1-\frac{298.15\,K}{553.15\,K} \right)\times 100\,\%[/tex]
[tex]\eta_{th} = 46.1\,\%[/tex]
b) The actual efficiency of the plant is the ratio of net power to input heat rate expressed in percentage:
[tex]\eta_{th, real} = \frac{\dot W}{\dot Q_{in}} \times 100\,\%[/tex]
[tex]\eta_{th, real} = \frac{1100\,MW}{3000\,MW}\times 100\,\%[/tex]
[tex]\eta_{th,real} = 36.667\,\%[/tex]
As per the question the nuclear power plant generates about 3000 MW of energy form the reactors and is due to the core of the reactor. The high pressure stream is 280 a C is used for the turbine to make it spin.
a) The maximum possible thermal efficiency of the power plant is given by the Carnot's Cycle thermal efficiency, b) The actual efficiency of the plant is the ratio of net power to input heat rate expressed in percentage.Learn more about the plant generates 3000 MW of heat energy.
brainly.com/question/18959084.
An 80.0 kg man sits on a scale in his car. The car is driving at a speed of 11.0 m/s right as it passes over the top of a semicircular hill of radius 17.0 m. What does the scale read right when he is at the top
Answer:
F / g = 138 kg
Explanation:
For this exercise let's use Newton's second law
F- W = m a
the force is equal to the back of the balance
in this case the acceleration is centripetal
a = v² / r
we substitute
F - m g = m v² / r
F = m (g + v²/ r)
calculus
F = 80 (9.8 + 11²/17)
F = 1353 N
the balance reading is this value between gravity
F / g = 1353 / 9.8
F / g = 138 kg
A distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. A) Decide whether this galaxy is approaching or receding from the earth. Give your reasoning. B) Find the speed of the galaxy relative to the earth.
Answer:
A) receding from the earth
B) [tex]3.078x10^6m/s[/tex]
Explanation:
A) receding from the earthThe wavelength went from 434.1nm to 438.6nm, there was an increase in wavelength (also knowecn as redshift due to the doppler efft), this increase is due to the fact that the source that emits the radiation (the distant galaxy) is moving away and therefore the light waves it emits are "stretched", causing us to see a wavelength greater than the original.
B) [tex]3.078x10^6m/s[/tex]to calculate the relative speed we use the following formula:
[tex]v_{rel}=c(1-\frac{\lambda_{1}}{\lambda_{2}} )[/tex]
where [tex]c[/tex] is the speed of light: [tex]c=3x10^8m/s[/tex]
[tex]\lambda_{1}[/tex] is the wavelength emited by the source, and
[tex]\lambda_{2}[/tex] is the wavelength measured on earth.
we substitute all the values and do the calculations:
[tex]v_{rel}=(3x10^8m/s)(1-\frac{434.1nm}{438.6nm} )\\\\v_{rel}=(3x10^8m/s)(1-0.98974)\\\\v_{rel}=(3x10^8m/s)(0.01026)\\\\v_{rel}=3.078x10^6m/s[/tex]
the relative speed is: [tex]3.078x10^6m/s[/tex]
What piece of equipment should be used to handle radioactive sources?
A stunt cyclist needs to make a calculation for an upcoming cycle jump. The cyclist is traveling 100 ft/sec toward an inclined ramp which ends 10 feet above a level landing zone. Assume the cyclist maintains a constant speed up the ramp and the ramp is inclined Ao (degrees) above horizontal. With the pictured imposed coordinate system, the parametric equations of the cyclist will be: x(t) = 100t cos(A) y(t) = –16t2 + 100t sin(A) + 10.
Calculate the horizontal velocity of the cyclist at time t; this is the function x'(t) = _______
What is the horizontal velocity if A = 20 degrees?
What is the horizontal velocity if A = 45 degrees? (Four decimal places.)
Calculate the vertical velocity of the cyclist at time t; this is the function y(t) =________.
What is the vertical velocity if A = 20 degrees? (Four decimal places.)
What is the vertical velocity if A = 45 degrees? (Four decimal places.)
The vertical velocity of the cyclist is zero at time ________ seconds.
If the cyclist wants to have a maximum height of 35 feet above the landing zone, then the required launch angle is A = _______ degrees. (Accurate to four decimal places.)
Answer:
Explanation:
The parametric equations of the cyclist are:
[tex]x(t)=100tcos(A)\\\\y(t)=-16t^2+100tsin(A)+10[/tex] (1)
A) The horizontal velocity is the derivative of x(t), in time:
[tex]x'(t)=100cos(A)[/tex]
B) For A=20° the horizontal velocity is:
[tex]v_x=x'(t)=100cos(20\°)=93.9692ft/s[/tex]
For A=45°:
[tex]v_x=100cos(45\°)=70.7106ft/s[/tex]
C) To find the time in which the vertical velocity is zero you first obtain the derivative of, in time:
[tex]v_y=y'(t)=-32t+100sin(A)+10[/tex]
Next, you equal the vertical velocity to zero and solve for time t:
[tex]-32t+100sin(A)+10=0\\\\t=\frac{100sin(A)+10}{32}[/tex]
D) The maximum height is reached when the derivative of y (height) is zero. You use the previous value of t in the equation (1), equals y to 35. Next, you solve for t:
[tex]y=35\\\\-16(\frac{100sin(A)+10}{32})^2+100(\frac{10sin(A)+10}{32})sinA+10=35\\\\-\frac{16}{1024}(10000sin^2A+2000sinA+100)+31.25sin^2A+31.25sinA+10=35\\\\-156.25sin^2A-31.25sinA-1.5625+31.25sin^2A+31.25sinA+10=35\\\\-125sin^2A-26.5625=0[/tex]
When a voltage difference is applied to a piece of metal wire, a 8-mA current flows through it. If this metal wire is now replaced with a silver wire having twice the diameter of the original wire, how much current will flow through the silver wire
Answer:
Explanation:
The resistance of a wire can be given by the following expression
[tex]R=\frac{\rho\times L}{A }[/tex]
where R is resistance , ρ is specific resistance , L is length of wire and A is cross sectional area
specific resistance of metals are almost the same . So in the present case ρ and l are same . Hence the formula becomes
R = k / A where k is a constant .
The diameter of wire becomes two times hence area of cross section becomes 4 times or 4A .
Resistance becomes 1/4 times . Hence if resistance of metal wire is R , resistance of silver wire will be R / 4 .
current = voltage / resistance
In case of metal wire
8 x 10⁻³ = V / R
In case of silver wire
I = V / (R / 4 ) , I is current , V is potential difference .
I = 4 x V/R
= 4 x 8 x 10⁻³ A
= 32 mA.
Displacement is the slope of a velocity vs. time graph.
True or false
Answer: false
Explanation:
The slope of a velocity–time graph is the acceleration.
A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target that is 30 meters away at a height of 1.8 meters?b) What is the time for the ball to reach the target?
Answer:
a.18.5 m/s
b.1.98 s
Explanation:
We are given that
[tex]\theta=35^{\circ}[/tex]
a.Let [tex]v_0[/tex] be the initial velocity of the ball.
Distance,x=30 m
Height,h=1.8 m
[tex]v_x=v_0cos\theta=v_0cos35[/tex]
[tex]v_y=v_0sin\theta=v_0sin35[/tex]
[tex]x=v_0cos\theta\times t=v_0cos35\times t[/tex]
[tex]t=\frac{30}{v_0cos35}[/tex]
[tex]h=v_yt-\frac{1}{2}gt^2[/tex]
Substitute the values
[tex]1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2[/tex]
[tex]1.8=30tan35-\frac{6574.6}{v^2_0}[/tex]
[tex]\frac{6574.6}{v^2_0}=21-1.8=19.2[/tex]
[tex]v^2_0=\frac{6574.6}{19.2}[/tex]
[tex]v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s[/tex]
Initial velocity of the ball=18.5 m/s
b.Substitute the value then we get
[tex]t=\frac{30}{18.5cos35}[/tex]
t=1.98 s
Hence, the time for the ball to reach the target=1.98 s
Frequency refers to the number of wavelengths that pass a fixed point in a minute. true or false
A car is traveling with a constant speed of 30.0 m/s when the driver suddenly applies the brakes, causing the car to slow down with a constant acceleration. The car comes to a stop in a distance of 120 m. What was the acceleration of the car as it slowed down?
Answer:
a = - 3.75 m/s²
negative sign indicates deceleration here.
Explanation:
In order to find the constant deceleration of the car, as it stops, we will use the 3rd equation of motion. The 3rd equation of motion is as follows:
2as = Vf² - Vi²
a = (Vf² - Vi²)/2s
where,
a = deceleration of the car = ?
Vf = Final Velocity = 0 m/s (Since, the car finally stops)
Vi = Initial Velocity = 30 m/s
s = distance covered by the car = 120 m
Therefore,
a = [(0 m/s)² - (30 m/s)²]/(2)(120 m)
a = - 3.75 m/s²
negative sign indicates deceleration here.
Select the correct answer.
Which person is vulnerable to identity theft?
A.
Beverley opens a line of credit to purchase a household appliance.
B.
Deborah fills out her income tax form and includes her Social Security number.
C. Josiah misses three monthly car loan payments in a row.
D.
Randell uses a computer at a public library to view his bank account online.
Reset
Next
Answer:
d
Explanation:
Randell uses a computer at a public library to view his bank account online represents one of the cases when a when person is vulnerable to identity theft, therefore the correct answer is option D.
What is identity theft?Identity theft occurs when criminals steal your confidential info and use it to create fresh accounts, rent or purchase property, file false financial records, or carry out other illegal activities.
This implies that a thief may steal sensitive data such as names, birthdates, Social Security numbers, information from driver's licenses, residences, and credit card or bank account numbers.
Once they have this information, they may use it to buy goods, apply for credit and debit cards, or even utilize it to pay for medical care with the assistance of health coverage.
Thus,the correct answer is option D.
To learn more about identity theft from here, refer to the link;
https://brainly.com/question/1531239
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Three balls, with masses of 3m,2m and m, are fastened to a massless rod of length L. The rotational inertias about the ledt
I = MR^2
The Attempt at a Solution:::
I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2
I total = 3ML^2/2
It says the answer is 3ML^2/4 though.
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The rotational inertia about the left is [tex]3ML^{2} /2[/tex].
What is meant by inertia?Inertia, property of a body by virtue of which it opposes any agency that attempts to put it in motion or, if it is moving, to change the magnitude or direction of its velocity. Inertia is a passive property and does not enable a body to do anything except oppose such active agents as forces and torques.To calculate the rotational inertia about the left[tex]I = I1 + I2 + I3\\ I= 3M(0^{2}) + 2M(L/2 )^{2} + M(L)^{2} \\I = 3ML^{2} /2[/tex]
The rotational inertia about the left is [tex]3ML^{2} /2[/tex]
To learn more about Inertia refer:https://brainly.com/question/18113232
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help me i cant solve it
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mark it as brainliest.... ✌✌A machinist needs to remove a tight-fitting pin of material A from a hole in a block made of material B. The machinist heats both the pin and the block to the same high temperature and removes the pin easily. What statement relates the coefficient of thermal expansion of material A to that of material B?
a. The situation is not possible, heating block B will shrink the hole in the material as the material expands with increasing temperature
b. Material B has the same coefficient of expansion as does material A
c. Material B has a negative coefficient of expansion and A has a positive coefficient of expansion
d. Material A has a greater coefficient of expansion than does material B
e. Material B has a greater coefficient of expansion than does material A
Answer:
C. Material B has a negative coefficient of expansion and A has a positive coefficient of expansion
Explanation:
If both material A and material B have the same coefficient of thermal expansion and they were heated to same temperature, both will expand making it impossible to remove the pin from the hole.
Also, if the coefficient of thermal expansion of any of the materials is higher than the other and they were subjected to the same temperature, the material with lower coefficient of thermal expansion will expand, making it impossible to remove the pin.
However, materials with negative coefficient of thermal expansion will contract on heating instead of expanding, while materials with positive coefficient of expansion will expand on heating. This makes it possible to remove the pin from the hole in the block.