Answer:
72 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 30 × 2.4
We have the final answer as
72 NHope this helps you
A bowling ball is 21.6 cm in diameter. What is the angular speed of these ball whenit is moving at 3.0 m/s?
Answer:
Angular speed = 27.78 rad/s (Approx)
Explanation:
Given:
Diameter = 21.6 cm
Speed = 3 m/s
Find:
Angular speed
Computation:
Radius = 21.6 / 2 = 10.8 cm = 0.108 m
Angular speed = v / r
Angular speed = 3 / 0.108
Angular speed = 27.78 rad/s (Approx)
A book is sitting on a table. Which of the following is true about the table?
Answer:
Its pulling down on the book
Explanation: if it was pushing up the book would be floating and the other choices don't make sense
A school bus has a mass of 18,200 kg. The bus moves at 13.5 m/s. How fast must a 0.142-kg baseball move in order to have the same momentum as the bus?
Answer:
bus momentum
p_bus= m_bus x v_bus
=18,200 x 16.5
basball momentum
pball=mball x vball
=0.142 x v
p_bus = pball
18200 x 16.5 = 0.142 x v
v=(18200 x 16.5)/0.142
v is the answer for baseball
Explanation:
⚠️not my answer tryna be honest here⚠️
The momentum of the bus of 18200 kg and velocity of 13.5 m/s is 245700 Kg m/s. To have equal momentum the base ball with 145 g have to throw in a speed of 1.7 × 10 ⁶ m/s.
What is momentum?Momentum of a moving body is the product of mass and velocity. Thus it have the unit of g m/s or Kg m/s. Momentum is a vector quantity and thus having magnitude and direction.
Given that one bus is having a mass of 18200 Kg and 13.2 m/s speed. The momentum is:
p = mv
=18200 kg × 13.5 m/s
= 245700 Kg m/s
To have a momentum of 245700 Kg m/s the base ball with 0. 142 g have to have a velocity = 245700 Kg m/s / 0.142 g
=1.7 × 10 ⁶ m/s
Hence, the baseball weighs0. 142 g have to move in 1.7 × 10 ⁶ m/s
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What happens to the molecules of water when it moves from a liquid to a gas?
A. Water molecules condense and move slower.
B. Water molecules spread out and move slower.
C. Water molecules spread out and move faster.
D. Water molecules condense and move faster.
its A or D but im not sure which one ik it moves fast
A stone dropped from a bridge strikes the water 5.6 seconds later. What is the final velocity in meters/s?
A) 179.78 meters/s
B) 5.71 meters/s
C) 1.75 meters/s
D) 54.88 meters/s
Answer: 54.88 meters/s
Explanation:
The final velocity will be calculated by using the formula:
v = u + at
where,
v = final velocity
u = initial velocity = 0
a = 9.8
t = 5.6
Therefore, we slot the value back into the formula. This will be:
v = u + at
v = 0 + (9.8 × 5.6)
v = 0 + 54.88
v = 54.88 meters per second
Therefore, the final velocity is 54.88m/s
During a hockey game, a puck is given an initial speed of 10 m/s. It slides 50 m on the horizontal ice before it stops due to friction. What is the coefficient of kinetic friction between the pick and the ice.A) 0.12B) 0.10C) 0.11D) 0.090
Answer:
The value is [tex]\mu_k = 0.102[/tex]
Explanation:
From the question we are told that
The initial speed of the pluck is [tex]u = 10 \ m/s[/tex]
The distance it slides on the horizontal ice is [tex]s = 50 \ m[/tex]
Generally from kinematic equation we have that
[tex]v^2 = u^2 + 2as[/tex]
Here v is is the final velocity and the value is 0 m/s given that the pluck came to rest, so
[tex]0^2 = 10 ^2 + 2* a * 50[/tex]
=> [tex]a = - 1 \ m/s^2[/tex]
Here the negative sign show that the pluck is decelerating
Generally the force applied on the pluck is equal to the frictional force experienced by the pluck
So
[tex]F = F_f[/tex]
=> [tex]m * a = m* g * \mu_k[/tex]
=> [tex]1 = 9.8 * \mu_k[/tex]
=> [tex]\mu_k = 0.102[/tex]
Brandon hits a golf ball with an initial velocity of 30 m/s at an angle of 30 above the horizontal. How long is it in the air?
Given :
Brandon hits a golf ball with an initial velocity of 30 m/s at an angle of 30 above the horizontal.
To Find :
How long is it in the air.
Solution :
We know, the formula of time of flight is :
[tex]T = \dfrac{2usin\ \theta}{g}\\\\T = \dfrac{2\times 30\times sin\ 30^o}{9.8}\\\\T = 3.06\ seconds[/tex]
Therefore, the ball is in air for 3.06 seconds.
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion
[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\ [/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}[/tex]
[tex]\\[/tex]
☯ As we know that,
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }[/tex]
[tex]\\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\ [/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = 0.04 \: m }[/tex]
[tex]\\[/tex]
☯ For left penetration (s₂)
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}[/tex]
[tex]\\[/tex]
[tex]\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\ [/tex]
A truck pushes a pile of dirt horizontally on a frictionless road with a net force of
20
N
20N20, start text, N, end text for
15.0
m
15.0m15, point, 0, start text, m, end text.
How much kinetic energy does the dirt gain?
Answer:
300 JoulesExplanation:
This is a common question on Khan Academy's "Calculating change in kinetic energy from a force" practice exercises. (AP Physics 1)
The simplest method to use is the following: [tex]W = F * d * cos(theta)[/tex], where W represents work (joules), F represents force (newtons), d represents distance (meters), and theta represents the angle of the force that's being applied.In this scenario, the force being applied is horizontal, so we can remove the [tex]cos(theta)[/tex] from our equation.So, our equation is now: [tex]W = F * d[/tex]. This would mean that [tex]W = 20 * 15[/tex], which is equal to [tex]300[/tex]. Our answer is 300 joules. (this value is positive and not negative because kinetic energy is being GAINED, not LOST)Here's the real question without all the formatting:
A truck pushes a pile of dirt horizontally on a frictionless road with a net force of 20 N for 15.0 m. How much kinetic energy does the dirt gain?A truck pushes a pile of dirt in the horizontal direction with a force of 20 N to a distance of 15 m then the kinetic energy of the dirt is 300 J.
What is work?In physics, the term "work" refers to the measurement of energy transfer that takes place when an item is moved over a distance by an externally applied, at least a portion of which is applied inside the direction of the displacement.
The duration of the path is multiplied by the component of a force acting along the path to calculate work if the forces are constant. The work W is theoretically equivalent to the pressure f times the range d, or W = fd, to represent this idea. Work is done when a force is applied at an angle of to a displacement, or W = fd cos.
The work done, [tex]W = Force * Displacement[/tex]
W = 15× 20
W = 300 J
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The image below shows four boxes that each contain a different sample of gas. The atoms of each gas are represented by dots, 1 2 3 4 Which box contains the gas with the greatest density?
A. 1
B. 2
C. 3
D. 4
What differentiates galaxy groups from clusters?
A.
Clusters are bigger than groups.
B.
Clusters are more massive than groups.
C.
Clusters contain a hot intracluster medium, whereas groups do not.
D.
Clusters are collections of galaxy groups, whereas groups are collections of galaxies.
E.
Clusters don't gravitationally bind galaxies together, while groups bind galaxies gravitationally.
Answer:
A
Explanation:
Galaxy clusters are basically very large (>50 galaxies) groups
Answer:
The correct answer would be:
C.
Clusters contain a hot intracluster medium, whereas groups do not.
#PLATOFAM
Have a nice day!
A pmdc has a stall torque of 10 and maximum mechanical power of 200. What is the maximum angular velocity?
Answer:
The maximum angular velocity is 20 rad/s
Explanation:
Given;
torque, τ = 10 N
maximum mechanical power, P = 200 J/s
The output power of the pmdc is given as;
P = τω
where;
P is the maximum mechanical power
ω is the maximum angular velocity
ω = P / τ
ω = (200) / (10)
ω = 20 rad/s
Therefore, the maximum angular velocity is 20 rad/s
A 500-eV electron and a 300-eV electron trapped in a uniform magnetic field move in circular paths in a plane perpendicular to the magnetic field. What is the ratio of the radii of their orbits?
a. 2.8
b. 1.7.
c. 1.3.
d. 4.0.
e. 1.0.
Answer:ratio of the radii of their orbits = 1.3 --- C
Explanation:
1- eV = to the kinetic energy of the electrons
and kinetic energy is given as
K.E= 1/2mv2
v = √(2E/m)----- equation 1
The force on the particles relating to the magnetic and circular motion ( centripetal force is given as
F = magnetic force = centripetal force
F= qvB = mv2/r
qvB = mv2/r
r = mv/qB ------ equation 2
We know from equation 1 that v = √(2E/m)
Therefore,
r = √(2mE)/qB------ equation 3
We can now say that the ratio of the two radii of their orbits can be calculated as
r1/r2 =(√(2mE1)/qB) /(√(2mE2)/qB
Where E1 = 500-eV and E2 = 300-eV (1- eV = to the kinetic energy of the electrons)
r1/r2 = (√(2m x500)/qB) /(√(2mx 300)/qB
Cancelling out common variables, we are left with
r1/r2 =[tex]\sqrt{500/300}[/tex]
r1/r2= 1.29 ≈ 1.3
A 1 200-kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s2. How far does the car travel before it stops?
Answer:
Δx = 39.1 m
Explanation:
Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:[tex]v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)[/tex]
where vf is the final velocity (0 in our case), v₀ is the initial velocity
(25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance
traveled since the brakes are applied.
Solving (1) for Δx, we have:[tex]\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)[/tex]
The car will travel a distance of 39.1 m before its stops.
To solve the problem above, use the equations of motion below.
Equation:
v² = u²+2as................... Equation 1Where:
v = final velocity of the automobileu = initial velocity of the automobilea = accelerations = distance coveredFrom the question,
Given:
v = 0 m/s (before its stops)u = 25 m/sa = -8 m/s² (decelerating)Substitute these values into equation 1
⇒ 0² = 25²+2(-8)(s)Solve for s
⇒ 0²-25² = -16s⇒ -16s = -625⇒ s = -625/16⇒ s = 39.1 mHence, The car will travel a distance of 39.1 m before its stops.
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A motorcycle skids for a distance of 2.0 m with the icy road pushing on its tires with force of 120 N as its
brakes are applied
What is the change in kinetic energy for the motorcycle?
Round the answer to two significant digits.
Answer:
-240
Explanation:
A motorcycle skids for a distance of 2.0 m on an icy road, then the change in kinetic energy for the motorcycle will be equal to -240 J.
What is kinetic energy?The force which a moving object has is referred to as kinetic energy in physics. It is defined as the number of effort required to propel a person of a specific mass from still to a specific velocity.
Aside from slight fluctuations in speed, your body holds onto the kinetic energy it obtains during acceleration.
When the body slows down from its present level to a condition of rest, the same quantity of energy is used.
Formally, kinetic energy is any quantity that has a gradient concerning time in the Lagrangian of a system.
As per the given information in the question,
Distance, d = 2.0 m
Friction, f = 120 N
The angle between displacement and friction force, θ = 180°
Now, the change in kinetic energy for the motorcycle = Work done by the friction.
K.E = f × d(cos θ)
= 120 (2.0 m)(cos 180°)
Δ K.E = -240 J
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The speed of revolution of particle going around a circlr is doubled and its angular speed is havled. What happen to the centripetal acceleration?
a) unchanged
b) doubles
c) halves
d) becomes four times
Answer: The correct answer is C
Explanation:
Lab - Wave Properties in a Spring
11-05
The wave characteristics you will observe in this lab are common to all waves (water, light, sound,
etc.). Use your prior knowledge and the book to fill in the following blanks, then go in the hall and
perform the lab.
A wave is a disturbance that moves through (propagates) through empty space or through a
_____________. There are two types of waves. A _____________________ wave requires
matter to travel. List some examples of this type:
A _____________________ wave does not require a medium. Examples include:
In order to start and transmit a mechanical wave, a source of _____________ and an
_______________ medium are required. A single disturbance is referred to as a
_______________, and a series of disturbances is a wave __________.
The questions in bold are those you should observe directly. Others will be answered using the book.
A. TYPES OF MECHANICAL WAVES: In the hall, stretch the slinky on the floor until it is
stretched (but still loose). Practice sending single pulses down the slinky by popping your wrist
from the center to the side and back to the center. Then send a continuous wave train along as
your partner holds the other end still. A piece of ribbon should be tied to one coil. Watch the
motion of this ribbon (representing a particle) as the wave travels through the spring.
In this type of wave, the particles move (perpendicular, parallel)
to the direction the wave travels. This type of wave is called a __________________ wave.
Its pulses are called ________________ and ________________.
Now send a pulse by quickly pushing the spring forward and pulling
it back, as shown. This type of wave is called _______________. Watch the motion of the ribbon.
In this type, the particles move _____________ to the direction the wave travels. Its pulses
are called _____________ and _____________. Label each.
Note that all waves transfer _____________ without transferring _______________. In
mechanical waves, particles of the medium vibrate back and forth in simple harmonic motion while
the disturbance (or _____________) moves from one place to another.
B. WAVE SPEED
Send a large pulse, followed by a small one. Does one pulse catch up to the other? ______
(Hint: The person who sends these waves should watch how the waves look when they return. Make
sure that both pulses are large enough initially to make it back to the sender!) The size of the
pulse is called the __________________ of the wave. Did the size affect the speed? ______
Generate a single transverse pulse in the slinky, keeping the stretch constant. Using a stopwatch,
time the journey of the pulse from one end to the other and back again. Take the average of
several trials. _________
Without changing your positions on the floor (therefore keeping the _____________ the pulse
travels the same), pull the slinky tighter using only about 3/4 of the coils. This makes a completely
different medium through which the pulse will travel. Time the journey as before. ___________
Does the kind of medium affect the speed of the pulse? ___________
Lab – Wave Properties in a Spring ____________________
PHYSICSFundamentals
© 2004, GPB
11-06
C. WAVELENGTH AND FREQUENCY
Shake the slinky back and forth steadily to send a
transverse wave train while your partner holds the other end still. On the diagram, label wavelength
(- Greek letter lambda). The frequency of the wave depends on how fast you shake the slinky.
Shake it regularly but slowly, then regularly but rapidly.
Higher frequency waves are generated by shaking the spring (slowly, rapidly). High frequency
waves have (short, long) wavelengths, and low frequency waves have __________.
The speed of a wave in any medium is equal to the _______________ of the wave X
________________. This wave equation ___________________ shows that f and are
______________ proportional. Write the units for each of the variables in this equation.
The exercise involves filling in the gaps with the possible wave
properties that can be obtained from a spring.
How is the Wave Properties in a Spring Lab exercise correctly completed?The correctly completed exercise is presented as follows;
A wave is a disturbance that moves through a medium. There are two
types of waves. A mechanical wave requires matter to travel. List some
examples of this type: sound wave, water wave, spring waves.
A electromagnetic wave does not require a medium. Examples include: Light waves
In order to start and transmit a mechanical wave, a source of
disturbance and a physical medium are required. A single disturbance is
referred to as a pulse, and a series of disturbance is a wave train.
This type of wave is called transverse wave. Its pulses are called crest
and troughs.
Now send a pulse by quickly pushing the spring forward and pulling it
back, as shown. This type of wave is called longitudinal wave. Watch the
motion of the ribbon. In this type, the particles move parallel to the
direction the wave travels. Its pulses are called compression and
rarefactions. Note that all waves transfer energy without transferring
matter. In mechanical waves, particle of the medium vibrate back and
forth in simple harmonic motion while the disturbance (or energy)
moves from one place to another.
B. Wave speed
Does the pulse catch up to the other? yes. The size of the pulse is called
the amplitude of the wave.
Did the size of the pulse affect the speed? No.
The average time wave it takes the wave to travel
Without changing your positions therefore keeping the distance the
pulse travels the same), pull the slinky tighter using only about 3/4 of
coils. This makes a completely different medium through which the
pulse will travel. Time the journey as before time record. Does the kind
of medium affect the speed of the pulse? Yes
C. Wavelength and Frequency
High frequency waves have short wavelengths and low frequency waves
have long wavelengths.
The speed of a wave in any medium is equal to the frequency of the wave × the wavelength. This wave equation [tex]\underline{f = \dfrac{v}{\lambda } }[/tex] shows that f and λ are
inversely proportional. The units of the variables are;
Units of the frequency, f is hertz unit HzUnits of the velocity, v, is m/sUnits of the wavelength, λ, is meters (m)Learn more about waves here:
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A balloon contains 0.075 m^3 of
gas. The pressure is reduced to
100kPa and fills a box of 0.45 m^3.
What is the initial pressure inside the
balloon if the temperature remains
constant?
Answer:
600 KPa.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V1) = 0.075 m³
Final volume (V2) = 0.45 m³
Final pressure (P2) = 100 KPa
Initial pressure (P1) =?
Temperature = constant
The initial pressure can be obtained by using the Boyle's law equation as shown below:
P1V1 = P2V2
P1 × 0.075 = 100 × 0.45
P1 × 0.075 = 45
Divide both side by 0.075
P1 = 45 / 0.075
P1 = 600 KPa.
Thus, the initial pressure in the balloon is 600 KPa.
The decibel scale intensity for busy traffic is 80 dB. Two people having a loud conversation have a decibel intensity of 70 dB. What is the approximate combined sound intensity?
Given :
The decibel scale intensity for busy traffic is 80 dB.
Two people having a loud conversation have a decibel intensity of 70 dB.
To Find :
The approximate combined sound intensity.
Solution :
We know, intensity in decibel can be converted to W/m² by :
[tex]\beta(dB) = 10\ log_{10}( \dfrac{I}{I_o})[/tex]
Putting intensity in decibel scale, we get :
[tex]I(80\ dB ) =10^{-4}\ W/m^2\\\\I(70\ dB ) = 10^{-5}\ W/m^2[/tex]
Let, combine intensity is I .
I = I(80 dB) + I(70 dB)
[tex]I = 10^{-4} + 10^{-5} \ W/m^2\\\\I = 1.1 \times 10^{-4} \ W/m^2[/tex]
Therefore, the combined sound intensity is [tex]1.1 \times 10^{-4} \ W/m^2[/tex] .
Calculate the mechanical advantage of a ramp if the box you are trying to move has a mass of 10 kilograms, the
board is 15 feet long and the height of the ramp is 5 feet .
3
300
150
45
Answer:
[tex]MA = 3[/tex]
Explanation:
Given
[tex]Box = 10kg[/tex]
[tex]Ramp\ Height = 5ft[/tex]
[tex]Ramp\ Length = 15ft[/tex]
Required
Determine the mechanical advantage
This is calculated as follows:
[tex]MA = \frac{Ramp\ Length}{Ramp\ Height}[/tex]
[tex]MA = \frac{15ft}{5ft}[/tex]
[tex]MA = 3[/tex]
Hence, the mechanical advantage is 3
g You heard the sound of a distant explosion (3.50 A/10) seconds after you saw it happen. If the temperature of the air is (15.0 B) oC, how far were you from the site of the explosion
Answer:
The answer is "1557 meters".
Explanation:
speed of sound in ([tex]\frac{m}{s}[/tex]) [tex]= 331.5 + 0.60 \ T^{\circ}\ C\\\\[/tex]
[tex]\to V = 331.5 + 0.6 \times 24 = 346 \frac{m}{s}\\\\\to t = 4.5 \ seconds \\\\\to S = vt = 346 \times 4.5 = 1557 \ meters[/tex]
why do feet smell and noses run?
Answer:
Nose has mucous glands with hairs which helps the body in trapping pollutants and infectants from entering inside the body. On the other hand,our feet is composed of millions of sweat pores when dirt and other things accumulate,it smells because of sweat mixed with the dirt and other dirty things of the ground.
Explanation:
Hope this helps
Can I get help on this question please
it would be the 3rd one. so C
Part A:
The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size-AA battery that supplies a constant voltage of 1.5 volts. What voltage would be measured across the secondary coil?
Part B:
A transformer is intended to decrease the value of the alternating voltage from 500 volts to 25 volts. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondary coil.
Part C:
A transformer is intended to decrease the value of the alternating current from 500 amperes to 25 amperes. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondary coil.
Part D:
In a transformer, the primary coil contains 400 turns, and the secondary coil contains 80 turns. If the primary current is 2.5 amperes, what is the secondary current I2?
Part E:
The primary coil of a transformer has 200 turns and the secondary coil has 800 turns. The power supplied to the primary coil is 400 watts. What is the power generated in the secondary coil if it is terminated by a 20-ohm resistor?
Part F:
A transformer supplies 60 watts of power to a device that is rated at 20 volts. The primary coil is connected to a 120-volt ac source. What is the current I1 in the primary coil?
Part G:
The voltage and the current in the primary coil of a nonideal transformer are 120 volts and 2.0 amperes. The voltage and the current in the secondary coil are 19.4 volts and 11.8 amperes. What is the efficiency e of the transformer? The efficiency of a transformer is defined as the ratio of the output power to the input power, expressed as a percentage: e=100Pout/Pin.
Answer:
a) 0 V
b) 10 turns
c) 4000 turns
d) 12.5 A
e) 400 W
f) 0.5 A
g) 95.4%
Explanation:
A
0
B
To solve this, we would be using the simple relationship between voltage and number of turns
V1/V2 = N1/N2
500/25 = 200/N2
20 = 200/N2
N2 = 200/20
N2 = 10 turns
C
Here also, we would be using the relationship between current and the number of turns
I1/I2 = N2/N1
500/25 = N2/20
20 = N2/20
N2 = 20 * 20
N2 = 4000 turns
D
Like in the previous question, current and the number of turn relationship is used
N1/N2 = I2/I1
400/80 = I2/2.5
5 = I2/2.5
I2 = 5 * 2.5
I2 = 12.5 A
E
The power remains unchanged at 400 W
F
Power = Voltage * Current
P = VI
I = P/V
I = 60/120
I = 0.5 A
G
95.4%
The transformer is a device used to step up or step down voltage.
Part A;
Given that;
Es/Ep = Ns/Np
Es = voltage in the secondary coil
Ep = voltage in primary coil
Ns = Number of turns in secondary coil
Np = Number of coils in primary coil
Es = Ns/Np × Ep
Es = 200/100 × 1.5 V
Es = 3 V
Part B
Ns = Es/Ep × Np
Ns = 25/500 × 200
Ns = 10 turns
Part C
Ns/Np = Ip/Is
Ns = Ip/Is × Np
Ns = 500/25 × 200
Ns = 4000 turns
Part D
Ns/Np = Ip/Is
NsIs = NpIp
Is = NpIp/Ns
Is = 400 × 2.5/80
Is =12.5 A
Part E
The power in the primary coil is the same as the power in the secondary coil. The power in the secondary coil is 400 watts.
Part F
Power supplied = 60 watts
Voltage of primary coil = 120 V
Since;
P = IV
I = P/V = 60/120 = 0.5 A
Part G
Since;
E = 100Pout/Pin
Pin = 120 V × 2 A = 240 W
Pout = 19.4 V × 11.8 A = 228.92 W
E = 100(228.92/240)
E = 95.4%
Learn more: https://brainly.com/question/8646601
WHAT DOES DENSITY HAVE TO DO WITH PLATE TECTONICS?
Explain
Answer: The reason for the differences in density is the composition of rock in the plates. When two plates come in contact with each other through plate tectonics, scientists can use the density of the plates to predict what will happen. Whichever plate is more dense will sink, and the less dense plate will float over it.
Explanation:
Hope this helps ( not copied and pasted, this answer was done by me so I don't know if it's good or not)
A ball is throw at an angle of 30 degrees off the horizontal, with an initial velocity of 28 m/s. what is the maximum height the ball will reach?
[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Angle \ of \ projection = 30^{\circ} }[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ projectile = 28 \: m/s^{-1} }[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Height_{\:(max)}\: reached\: by \:the \:projectile }[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]
☯ As we know that,
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{u^2\;sin^2\theta}{2\;g} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{H = \dfrac{(28)^2\;sin^2 30^{\circ}}{2\;(9.8)} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{H = \dfrac{784 \times \;sin^230^{\circ}}{19.6} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times sin^2 30^{\circ}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \bigg(\dfrac{1}{2}\bigg)^2 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \dfrac{1}{4} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: {\boxed{\sf{H=10\:m }}}[/tex]
which factor does not affect the strength of an electromagnet
Answer:
the placement of the ammeter in the circuit
Explanation:
A 5 kg block rests on an inclined plane with a coefficient of static friction equal to 0.30. What is the minimum angle at which the block will begin to slide
Answer:
[tex]\theta = 16.70 ^{\circ}[/tex]
Explanation:
The coefficient of static friction is equal to the tangent of the minimum angle at which an object will begin to start sliding down a ramp.
[tex]\displaystyle u_s=\frac{F_f}{F_N} = \frac{F_g\ \text{sin}\theta}{F_g\ \text{cos} \theta} = \text{tan} \theta[/tex]Since we are given the coefficient of static friction we can solve for the minimum angle that the block will begin to slide.
Let's solve for the force of gravity that is acting on the block. The force of gravity is also known as the weight force, which can be calculated by using w = mg.
[tex]w=mg[/tex]We are given the mass of the block (kg) and we know that g = 9.8 m/s².
[tex]w=(5)(9.8) = 49 \ \text{N}[/tex]Now we can use this force in the equation:
[tex]\displaystyle u_s = \frac{F_g \ \text{sin} \theta }{F_g \ \text{cos} \theta}[/tex]Plug [tex]\displaystyle u_s = 0.30[/tex] and 49 N into the equation.
[tex]\displaystyle 0.30 = \frac{(49) \ \text{sin} \theta }{(49) \ \text{cos} \theta}[/tex] [tex]0.30=\text{tan} \theta[/tex]Notice that the gravitational force cancels out in the end, so we can actually start with [tex]0.30=\text{tan} \theta[/tex].
Evaluate this equation by taking the inverse tangent of both sides of the equation.
[tex]\text{tan}^-^1 (0.30) = \text{tan}^-^1 (\text{tan}\theta)[/tex] [tex]\text{tan}^-^1 (0.30) =\theta[/tex] [tex]\theta = 16.69924423[/tex]The minimum angle at which the block will begin to slide is about 16.70 degrees.
Two clear but non-mixing liquids each of depth 15 cm are placed together in a glass container. The liquids have refractive indices of 1.75 and 1.33. What is the apparent depth of the combination when viewed from above?
Answer:
The apparent depth d = 19.8495 cm
Explanation:
The equation for apparent depth can be expressed as:
[tex]d = \dfrac{d_1} {\mu_1}+\dfrac {d_2}{\mu_2}[/tex]
here;
[tex]d_1 = d_2 = 15 \ cm[/tex]
[tex]\mu_1[/tex] = refractive index in the first liquid = 1.75
[tex]\mu_2[/tex] = refractive index in the second liqquid= 1.33
∴
[tex]d = \dfrac{15}{1.75}+\dfrac{15}{1.33}[/tex]
[tex]d = 15( \dfrac{1}{1.75}+\dfrac{1}{1.33})[/tex]
[tex]d = 15( 0.5714 +0.7519)[/tex]
d = 15(1.3233 ) cm
d = 19.8495 cm
A particular engine has a power output of 2 kW and an efficiency of 27%. If the engine expels 9085 J of thermal energy in each cycle, find the heat absorbed in each cycle. Answer in units of J.
Answer:
12445 J
Explanation:
Given that
Power output, P = 5 kW
efficiency of the engine, e = 27% = 0.27
Thermal energy expelled, Q(c) = 9085 J
Heat absorbed, Q(h) = ?
Using the formula
e = W/Q(h)
e = [Q(h) - Q(c)] / Q(h)
e = 1 - Q(c)/Q(h)
Now, substituting the values into the formula, we have
0.27 = 1 - 9085/Q(h)
9085/Q(h) = 1 - 0.27
9085/Q(h) = 0.73
Q(h) = 9085 / 0.73
Q(h) = 12445 J
Thus, the heat absorbed is 12445 J