Answer:
The minimum running time is 319.47 s.
Explanation:
First we find the distance covered and time taken by the train to reach its maximum speed:
We have:
Initial Speed = Vi = 0 m/s (Since, train is initially at rest)
Final Speed = Vf = 29.17 m/s
Acceleration = a = 0.25 m/s²
Distance Covered to reach maximum speed = s₁
Time taken to reach maximum speed = t₁
Using 1st equation of motion:
Vf = Vi + at₁
t₁ = (Vf - Vi)/a
t₁ = (29.17 m/s - 0 m/s)/(0.25 m/s²)
t₁ = 116.68 s
Using 2nd equation of motion:
s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²
s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²
s₁ = 1701.78 m = 1.7 km
Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.
We have:
Final Speed = Vf = 0 m/s (Since, train is finally stops)
Initial Speed = Vi = 29.17 m/s (The train must maintain max. speed for min time)
Deceleration = a = - 0.7 m/s²
Distance Covered to stop = s₂
Time taken to stop = t₂
Using 1st equation of motion:
Vf = Vi + at₂
t₂ = (Vf - Vi)/a
t₂ = (0 m/s - 29.17 m/s)/(- 0.7 m/s²)
t₂ = 41.67 s
Using 2nd equation of motion:
s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²
s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²
s₂ = 607.78 m = 0.6 km
Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.
The remaining distance is:
s₃ = 7 km - s₂ - s₁
s₃ = 7 km - 0.6 km - 1.7 km
s₃ = 4.7 km
Now, for uniform speed we use the relation:
s₃ = vt₃
t₃ = s₃/v
t₃ = (4700 m)/(29.17 m/s)
t₃ = 161.12 s
So, the minimum running time will be:
t = t₁ + t₂ + t₃
t = 116.68 s + 41.67 s + 161.12 s
t = 319.47 s
A steam engine takes in superheated steam at 270 °C and discharges condensed steam from its cylinder at 50 °C. The engine has an efficiency of 30%, and taken in 50 kJ from the hot steam per cycle. If a Carnot engine takes in the same amount of heat per cycle and operates at these temperatures, the work it can turn into is most likely to be:a) 15 kJ. b) 20 kJ. c) 10 kJ. d) 50 kJ.
Answer:
b) 20 kJ
Explanation:
Efficiency of carnot engine = (T₁ - T₂ ) / T₁ Where T₁ is temperature of hot source and T₂ is temperature of sink .
T₁ = 270 + 273 = 543K
T₂ = 50 + 273 = 323 K
Putting the given values of temperatures
efficiency = (543 - 323) / 543
= .405
heat input = 50 KJ
efficiency = output work / input heat energy
.405 = output work / 50
output work = 20.25 KJ.
= 20 KJ .
What must x be so that the handle end of the bat remains at rest as the bat begins to move? (Hint: Consider the motion of the center of mass and the rotation about the center of mass. Find x so that these two motions combine to give v=0 for the end of the bat just after the collision. Also, note that integration of equation ∑τ⃗ =dL⃗ dt gives ΔL=∫t1t2(∑τ)dt. )
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told that
The mass of the bat is [tex]m_b = 0.800 \ kg[/tex]
The bat length is [tex]L_b = 0.900 \ m[/tex]
The distance of the bat's center of mass to the handle end is [tex]z_c = 0.600 \ m[/tex]
The moment of inertia of the bat is [tex]I = 0.0530 \ kg \cdot m^2[/tex]
The objective of the solution is to find x which is the distance from the handle of the bat to the point where the baseball hit the bat
Generally the velocity change at the end of the bat is mathematically represented as
[tex]\Delta v_e = \Delta v_c - \Delta w* z_c[/tex]
Where [tex]\Delta v_c[/tex] is the velocity change at the center of the bat which is mathematically represented as
[tex]\Delta v_c = \frac{Impulse}{m_b }[/tex]
We are told that the impulse is J so
[tex]\Delta v_c = \frac{J}{m_b }[/tex]
And [tex]\Delta w[/tex] is the change in angular velocity which is mathematically represented as
[tex]\Delta w = \frac{J (z -z_c)}{I}[/tex]
Now we have that
[tex]\Delta v_e = \frac{J}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]
Before a swing when the bat is at rest the velocity change a the end of the bat handle is zero and the impulse will be 1
So
[tex]0 = \frac{1}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]
=> [tex]x = \frac{I}{m_b z_c} + m_b[/tex]
substituting values
[tex]x = \frac{0.530}{0.800 * 0.600} + 0.600[/tex]
[tex]x = 0.710 \ m[/tex]
Part A - At what angle does it leave?
Part B - At what distance x does it exit the field?
Answer:
Total internal reflection (TIR) is the phenomenon that involves the reflection of all the incident light off the boundary. TIR only takes place when both of the following two conditions are met: the light is in the more dense medium and approaching the less dense medium.
Explanation: Hope i helped!!!
Block A, with a mass of 4 kg, is moving with a speed of 2 m/s while Block B, with a mass of 8.4 kg, is moving in the opposite direction with a speed of 6.1 m/s. The center of mass of the two block system is moving with a velocity of ____ m/s. Round your answer to the nearest tenth. Assume Block A is moving in the positive direction.
Answer:
The center of mass move with the velocity of -3.487 m/s.
Explanation:
Given values of block A.
Mass of block A, (M1) = 4 kg
Speed of block A, (V1) = 2 m/s
Given values of block B.
Mass of block B, (M2) = 8.4 kg
Speed of block B, (V2) = -6.1 m/s
Below is the formula to find the velocity of center of mass.
[tex]Velocity = \frac{M1V1 + M2V2}{M1 + M2} \\[/tex]
[tex]= \frac{4 \times 2 + 8.4 \times (-6.1) }{4 + 8.4} \\[/tex]
[tex]= \frac{- 43.24}{12.4}\\[/tex]
[tex]= - 3.487 m/s[/tex]
A 0.009 kg bullet fired through a door enters at 803 m/s and leaves at 617 m/s. If the door material is known to exert an average resistive force of 5620 N on bullets of this type at usual speeds, find the thickness of the door.
Answer:
The thickness of the door is 0.4230 m
Explanation:
Given;
mass of bullet, m = 0.009 kg
initial velocity of the bullet, u = 803 m/s
final velocity of the bullet, v = 617 m/s
average resistive force of the door on the bullet, F = 5620 N
Apply Newton's second law of motion;
Force exerted by the door on the bullet = Force of the moving bullet
F = ma
where;
F is applied force
m is mass
a is acceleration
Also, Force exerted by the door on the bullet = Force of the moving bullet
[tex]F =ma, \ But \ a =\frac{dv}{dt} = \frac{u-v}{t} \\\\F = \frac{m(u-v)}{t}[/tex]
where;
v is the final velocity of the bullet
u is initial velocity of the bullet
t is time
We need to calculate the time spent by the bullet before it passes through the door.
[tex]t = \frac{m(u-v)}{F} \\\\t = \frac{0.009(803-617)}{5620} = 0.0002979 \ s[/tex]
Distance traveled by the bullet within this time period = thickness of the door
This distance is equivalent to the product of average velocity and time
[tex]S = (\frac{u+v}{2}) t[/tex]
where;
s is the distance traveled
[tex]S = (\frac{u+v}{2}) t\\\\S = (\frac{803+617}{2}) 0.0002979\\\\S = 0.4230 \ m[/tex]
Therefore, the thickness of the door is 0.4230 m
What types of mediums are involved in the energy transfer
Answer:
In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles.
6. The two ends of an iron rod are maintained at different temperatures. The amount of heat thatflows through the rod by conduction during a given time interval does notdepend uponA) the length of the iron rod.B) the thermal conductivity of iron.C) the temperature difference between the ends of the rod.D) the mass of the iron rod.E) the duration of the time interval.Ans: DDifficulty: MediumSectionDef: Section 13-27. The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twiceas long and twice the diameter conduct heat between the same two temperatures
Answer:
20cal/s
Explanation:
Question:
There are two questions. The first one has been answered:
From the formular, Power = Q/t = (kA∆T)/l
the amount heat depends on the duration of time interval, length of the iron rod, the thermal conductivity of iron and the temperature difference between the ends of the rod.
The amount of heat that flows through the rod by conduction during a given time interval does not depend upon the mass of the iron rod (D).
Second question:
The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?
Solution:
Power = 10cal/s
Power = energy per unit time = Q/t
Where Q = energy
Power = (kA∆T)/l
k = thermal conductivity of iron
A = area
Area = πr^2
r = radius
Diameter = d = 2r
r = d/2
Area = (πd^2)/4
Length = l
∆T = change in temperature
10 = (kA∆T)/l
For a steel rod with length doubled and diameter doubled:
Let Length (L) = 2l
Diameter (D)= 2d
Area = π [(2d)^2]/4 = (π4d^2)/4
Area = 4(πd^2)/4
Using the formula Power = (kA∆T)/l, insert the new values for A and l
Power = [k × 4(πd^2)/4 × ∆T]/2l
Power = [4k((πd^2)/4) ∆T]/2l
Power = [(4/2)×k((πd^2)/4) ∆T]/l
Power = [2k(A) ×∆T]/l = 2(kA∆T)/l
Power of a steel that has its length doubled and diameter doubled = 2(kA∆T)/l
Recall initial Power = (kA∆T)/l = 10cal/s
And ∆T is the same
2[(kA∆T)/l] = 2 × 10
Power of a steel that has its length doubled and diameter doubled = 20cal/s
The strength of the force of friction depends on which two factors?
Answer:
coefficient of friction (μ) and normal force (N)
Answer: How hard the surfaces push together and the types of surfaces involved
Explanation:
A long solid conducting cylinder with radius a = 12 cm carries current I1 = 5 A going into the page. This current is distributed uniformly over the cross section of the cylinder. A cylindrical shell with radius b = 21 cm is concentric with the solid cylinder and carries a current I2 = 3 A coming out of the page. 1)Calculate the y component of the magnetic field By at point P, which lies on the x axis a distance r = 41 cm from the center of the cylinders.
Answer:
Explanation:
We shall use Ampere's circuital law to find magnetic field at required point.
The point is outside the circumference of two given wires so whole current will be accounted for .
Ampere's circuital law
B = ∫ Bdl = μ₀ I
line integral will be over circular path of radius r = 41 cm .
Total current I = 5A -3A = 2A .
∫ Bdl = μ₀ I
2π r B = μ₀ I
2π x .41 B = 4π x 10⁻⁷ x 2
B = 2 x 10⁻⁷ x 2 / .41
= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.
Consider a circular vertical loop-the-loop on a roller coaster. A car coasts without power around the loop. Determine the difference between the normal force exerted by the car on a passenger with a mass of mm at the top of the loop and the normal force exerted by the car on her at the bottom of the loop. Express your answer in terms of mmm and the acceleration due to gravity ggg.
Answer:
Explanation:
Let v₁ and v₂ be velocities at lowest and topmost position . Let r be the radius of the circle .
Let N₁ and N₂ be the normal reaction force .
At the top position
centripetal force = N₂ + mg ; so
N₂ + mg = m v₂² / r
At the bottom position
centripetal force = N₁ - mg ; so
N₁ - mg = m v₁² / r
subtracting these two equations
N₁ - mg - N₂ - mg = m v₁² / r - m v₂² / r
N₁ - N₂ - 2mg = 1/r (m v₁² - m v₂² )
N₁ - N₂ - 2mg = 1/r x mg x 2r ( loss of potential energy = gain of kinetic energy )
N₁ - N₂ = 2mg + 2mg
= 4 mg .
The greater the distance between two objects in space, the _______ their gravitational
Answer is Weaker. If it is talking about the objects' gravitational forces.
Electric fields are MOST associated with ________.
If Jim could drive a Jetson's flying car at a constant speed of 440 km/hr across oceans and space, approximately how long (in millions of years, in 106 years) would he take to drive to a nearby star that is 12.0 light-years away? Use 9.461 × 1012 km/light-year and 8766 hours per year (365.25 days).
Answer:
t = 2.94 x 10⁶ years
Explanation:
The equation used in the case of constant speed is:
s = vt
t = s/v
where,
s = distance = 12 light years
s = (12 light years)(9.461 x 10¹² km/light year) = 113.532 x 10¹² km
v = speed = 440 km/hr
t = time passed = ?
Therefore,
t = (113.532 x 10¹² km)/(440 km/hr)
t = 2.58 x 10¹¹ hr
Now, converting it to years:
t = (2.58 x 10¹¹ hr)(1 year/8766 hr)
t = 2.94 x 10⁶ years
You could use an analytical or triple beam balance to determine a ___ called ____
A)
physical property; mass.
B)
chemical property, mass.
C)
physical property; weight.
D)
physical property; density.
Answer:
a and b are the correct answers
Explanation:
Answer:
A) physical property; mass.
Explanation:
took the test
Refracted light rays...
A- are bent as they pass into a different medium
B- are absorbed by an object
C- are reflected from an object at a variety of angles
D- bounce off a medium
Answer:
A.
Explanation:
Refraction light rays are bent as they pass into a different medium where its speed is different. As refracted light rays pass from a fast medium to a slow medium, the light ray bends toward the normal to the boundary between the two medium. Light refracts as it travels at an angle into a medium with a different refractive index.
A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 3.7 mm will experience only elastic deformation when a tensile load of 1890 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.
Answer:
L= 276.4 mm
Explanation:
Given that
E= 180 GPa
d= 3.7 mm
F= 1890 N
ΔL= 0.45 mm
We know that ,elongation due to load F in a cylindrical bar is given as follows
[tex]\Delta L =\dfrac{FL}{AE}[/tex]
[tex]L=\dfrac{\Delta L\times AE}{F}[/tex]
Now by putting the values in the above equation we get
[tex]L=\dfrac{0.45\times 10^{-3}\times \dfrac{\pi}{4}\times (3.7\times 10^{-3})^2\times 108\times 10^9}{1890}\ m[/tex]
L=0.2764 m
L= 276.4 mm
Therefore the length of the specimen will be 276.4 mm
A Texas cockroach of mass 0.157 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has a radius 14.9 cm, rotational inertia 5.92 x 10-3 kg·m2, and frictionless bearings. The cockroach's speed (relative to the ground) is 2.92 m/s, and the lazy Susan turns clockwise with angular velocity ω0 = 3.89 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
Answer:
-7.23 rad/s
Explanation:
Given that
Mass of the cockroach, m = 0.157 kg
Radius of the disk, r = 14.9 cm = 0.149 m
Rotational Inertia, I = 5.92*10^-3 kgm²
Speed of the cockroach, v = 2.92 m/s
Angular velocity of the rim, w = 3.89 rad/s
The initial angular momentum of rim is
Iw = 5.92*10^-3 * 3.89
Iw = 2.3*10^-2 kgm²/s
The initial angular momentum of cockroach about the axle of the disk is
L = -mvr
L = -0.157 * 2.92 * 0.149
L = -0.068 kgm²/s
This means that we can get the initial angular momentum of the system by summing both together
2.3*10^-2 + -0.068
L' = -0.045 kgm²/s
After the cockroach stops, the total inertia of the spinning disk is
I(f) = I + mr²
I(f) = 5.92*10^-3 + 0.157 * 0.149²
I(f) = 5.92*10^-3 + 3.49*10^-3
I(f) = 9.41*10^-3 kgm²
Final angular momentum of the disk is
L'' = I(f).w(f)
L''= 9.41*10^-3w(f)
Using the conservation of total angular momentum, we have
-0.068 = 9.41*10^-3w(f) + 0
w(f) = -0.068 / 9.41*10^-3
w(f) = -7.23 rad/s
Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed
b)
The mechanical energy of the cockroach is not converted as it stops
To understand thermal linear expansion in solid materials. Most materials expand when their temperatures increase. Such thermal expansion, which is explained by the increase in the average distance between the constituent molecules, plays an important role in engineering. In fact, as the temperature increases or decreases, the changes in the dimensions of various parts of bridges, machines, etc., may be significant enough to cause trouble if not taken into account. That is why power lines are always sagging and parts of metal bridges fit loosely together, allowing for some movement. It turns out that for relatively small changes in temperature, the linear dimensions change in direct proportion to the temperature.
For instance, if a rod has length L0 at a certain temperature T0 and length L at a higher temperature T, then the change in length of the rod is proportional to the change in temperature and to the initial length of the rod: L - L0 = αL0(T - T0),
or
ΔL = αL0ΔT.
Here, α is a constant called the coefficient of linear expansion; its value depends on the material. A large value of α means that the material expands substantially as the temperature increases; smaller values of α indicate that the material tends to retain its dimensions. For instance, quartz does not expand much; aluminum expands a lot. The value of α for aluminum is about 60 times that of quartz!
Questions:
A) Compared to its length in the spring, by what amount ΔLwinter does the length of the bridge decrease during the Teharian winter when the temperature hovers around -150°C?
B) Compared to its length in the spring, by what amount ΔLsummer does the length of the bridge increase during the Teharian summer when the temperature hovers around 700°C?
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. If the coffee has a temperature of 210 degrees Fahrenheit when freshly poured, and 2.5 minutes later has cooled to 191 degrees in a room at 64 degrees, determine when the coffee reaches a temperature of 156 degrees.
Answer:
Explanation:
The problem is based on Newton's law of cooling .
According to Newton's law
dQ / dt = k ( T - T₀ ) ,
dT / dt = k' ( T - T₀ ) ; dT / dt is rate of fall of temperature.
T is average temperature of hot body , T₀ is temperature of surrounding .
In the first case rate of fall of temperature = (210 - 191) / 2.5
= 7.6 degree / s
average temperature T = (210 + 191) /2
= 200.5
Putting in the equation
7.6 = k' ( 200.5 - 64 )
k' = 7.6 / 136.5
= .055677
In the second case :---
In the second case, rate of fall of temperature = (191 - 156) / t
= 35 / t , t is time required.
average temperature T = (156 + 191) /2
= 173.5
Putting in the equation
35 / t = .05567 ( 173.5 - 64 )
t = 5.74 minute .
3. The current in a flashlight powered by 4.5 Volts is 0.5 A. What is the power delivered to the flashlight?
4.If the flashlight in the previous problem is left on for 3 minutes, how much electric energy is delivered to the bulb?
Answer:
Question 3: 2.25 watts
Question 4: 405 joules
Explanation:
question 3:
Current =0.5 amps
Voltage =4.5 volts
Power= current x voltage
Power=0.5 x 4.5
power=2.25 watts
Question 4
Current =0.5 amps
Voltage =4.5v
Time=3 minutes
Time =3x60
Time =180 seconds
Energy=current x voltage x time
Energy =0.5 x 4.5 x 180
Energy =405 joules
PIUDICITIS CONSECulvely and Circle your aliswers. Lilyo
proper significant digits.
53. When you turn on your CD player, the turntable accelerates from zero to 41.8 rad/s in
3.0 s. What is the angular acceleration?
or
Answer:
The angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].
Explanation:
Initial angular speed of a CD player is 0 and final angular speed is 41.8 rad/s. Time to change the angular speed is 3 s.
It is required to find the angular acceleration. The change in angular speed of the CD player divided by time taken is called its angular acceleration. It can be given by :
[tex]a=\dfrac{\omega_f-\omega_i}{t}\\\\a=\dfrac{41.8-0}{3}\\\\a=13.94\ rad/s^2[/tex]
So, the angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].
Calculate potential energy of a 5 kg object sitting on 3 meter ledge
Answer:147 joules
Explanation:
Mass=m=5kg
Acceleration due to gravity=g=9.8m/s^2
Height=h=3 meter
Potential energy=m x g x h
Potential energy=5 x 9.8 x 3
Potential energy=147 joules
Which term BEST describes the movement of air from the ocean toward the land in the daytime? (AKS 4b DOK 1) *
1 point
Sea breeze
Land Breeze
Valley Breeze
Current Breeze
Answer:
Option A, Sea Breeze
Explanation:
Ssea breeze is a wind that blows from the ocean or any water body to the nearby land mass. This breeze is cold as compared to the air on land. The water in water bodies has high specific heat capacity and hence takes longer time to cool as compared to the surrounding objects. The warmer air over the land rises upward thereby reducing the pressure on land and hence the sea breeze starts flowing from region of high pressure (i.e above the water body) towards the low pressure region that is the land.
Hence, option A is correct
The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters per second, the time t is measured in seconds, and the magnitude of the constant a is measured in meters per second squared. What is its maximum speed, expressed as a multiple of a? (Do not include units in your answer.)
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
[tex]v(t)=ate^{-6t}[/tex] (1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
[tex]\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))][/tex] (2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
[tex]a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}[/tex]
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
[tex]v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a[/tex]
hence, the maximum speed is v_max = ((1/6)e^-1)a
A turntable has a moment of inertia of 3.00 x 10-2 kgm2 and spins freely on a frictionless bearing at 25.0 rev/min. A 0.300 kg ball of putty is dropped vertically on the turntable and sticks at a point 0.10m from the center. The total moment of inertia of the system increases, and the turntable slows down. But by what factor does the angular momentum of the system change after the putty is dropped onto the turntable
Answer:
There will be no change in the angular momentum of the system.
Explanation:
Total angular momentum of the system will remain unchanged . We can apply law of conservation of momentum because no external torque is acting on the system . There is increase in the momentum of inertia due to dropping of ball of putty . In order to conserve angular momentum , the system decreases its angular velocity . Hence the final angular momentum remains unchanged .
A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.
a) What is the current flowing through this light bulb?
b) What is the resistance of the light bulb?
Given Information:
Power = P = 100 Watts
Voltage = V = 220 Volts
Required Information:
a) Current = I = ?
b) Resistance = R = ?
Answer:
a) Current = I = 0.4545 A
b) Resistance = R = 484 Ω
Explanation:
According to the Ohm’s law, the power dissipated in the light bulb is given by
[tex]P = VI[/tex]
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.
Re-arranging the above equation for current I yields,
[tex]I = \frac{P}{V} \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\[/tex]
Therefore, 0.4545 A current is flowing through the light bulb.
According to the Ohm’s law, the voltage across the light bulb is given by
[tex]V = IR[/tex]
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.
Re-arranging the above equation for resistance R yields,
[tex]R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega[/tex]
Therefore, the resistance of the bulb is 484 Ω
Answer:
bulb will burn out!
Explanation:
what do hydroelectric plants use to generate electrical energy?
Answer:
A. falling water
Explanation:
I got it right on Edgenuity. Good luck on your quiz.
In hydroelectric plants, water falls on turbine and makes it rotate. In generator, this mechanical energy transforms into electrical energy.
What is hydroelectric power?Hydroelectric power is generated by turbines that turn the potential energy of falling or swiftly flowing water into mechanical energy, which is then used to power generators. The most popular renewable energy source in the early 21st century was hydroelectricity, which in 2019 accounted for more than 18% of the world's total power producing capacity.
Water is gathered or stored at a higher elevation during the production of hydroelectric power and then transported through substantial pipes or tunnels (penstocks) to a lower elevation; the difference between these two elevations is referred to as the head. The falling water turns turbines as it nears the bottom of the pipelines. In turn, the turbines power generators, which transfer the mechanical energy of the turbines into electricity.
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A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500.0 N/m (assume that the spring is massless). The block is held in position such that the spring is compressed 4.00 cm shorter than its undisturbed length. The block is suddenly released and allowed to slide away on the frictionless surface. Find the speed the block will be traveling when it leaves the spring.
Answer:
6 m/s
Explanation:
Given that :
mass of the block m = 200.0 g = 200 × 10⁻³ kg
the horizontal spring constant k = 4500.0 N/m
position of the block (distance x) = 4.00 cm = 0.04 m
To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.
i.e [tex]\frac{1}{2} kx^2 = \frac{1}{2} mv^2[/tex]
[tex]kx^2 = mv^2[/tex]
[tex]4500* 0.04^2 = 200*10^{-3} *v^2[/tex]
[tex]7.2 =200*10^{-3}*v^{2}[/tex]
[tex]v^{2} =\frac{7.2}{200*10^{-3}}[/tex]
[tex]v =\sqrt{\frac{7.2}{200*10^{-3}}}[/tex]
v = 6 m/s
Hence,the speed the block will be traveling when it leaves the spring is 6 m/s
A particle is projected at an angle 60 degrees to the horizontal with a speed of 20m/s. (i) calculate total time of flight of the particle. (i) speed of the particle at its maximum height
Answer:
Time of flight=3.5 seconds
Speed at maximum height is 0
Explanation:
Φ=60°
initial velocity=u=20m/s
Acceleration due to gravity=g=9.8 m/s^2
Total time of flight=T
Final speed=v
question 1:
T=(2 x u x sinΦ)/g
T=(2 x 20 x sin60)/9.8
T=(2 x 20 x 0.8660)/9.8
T=34.64/9.8
T=3.5 seconds
Question 2
Speed at maximum height is 0
A 46.0-kg box is being pushed a distance of 8.80 m across the floor by a force P whose magnitude is 171 N. The force P is parallel to the displacement of the box. The coefficient of kinetic friction is 0.250. Determine the work done on the box by (a) the applied force, (b) the friction force, (c) the normal force, and (d) by the force of gravity. Be sure to include the proper plus or minus sign for the work done by each force.
Answer:
a) 1504.8 J
b) 991.76 J
c) 0J
d) 0J
Explanation:
(a) The work done by the force P on the box is given by the following formula:
[tex]W_P=Px[/tex]
P: applied force = 171N
x: distance in which the for P is applied = 8.80m
you replace the values of P and x and obtain:
[tex]W_P=(171N)(8.80m)=1504.8J[/tex]
(b) The work don by the friction force is:
[tex]W_f=F_fx=\mu N x=\mu Mg x[/tex]
μ = coefficient of kinetic friction = 0.250
M: mass of the box = 46.0kg
g: gravitational constant = 9.8 m/s^2
[tex]W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J[/tex]
(c) The Normal force is
[tex]N=Mg=(46.0kg)(9,8m/s^2)=450.8N[/tex]
but this force does not do work on the box because the direction is perpendicular to the direction of the force P.
[tex]W_N=0J[/tex]
(d) the same as before:
[tex]W_g=0J[/tex]
