Answers
Answer:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
Explanation:
also the answer is hit my dm on ig
Related Questions
15. A parallel-plate capacitor, with air between the plates, is connected across a voltage source. This source establishes a potential difference between the plates by placing charge of magnitude 4.15 x 10-6 C on each plate. The space between the plates is then filled with a dielectric material, with a dielectric constant of 7.74. What must the magnitude of the charge on each capacitor plate now be, to produce the same potential difference between the plates as before? (Input your answer in 3 significant figures without unit. For example, if the answer is 1.356 x 10-6 C; then just input 1.37)
Answers
Answer:
Q' = 3.21*10^{-5}C
Explanation:
To find the new magnitude of the charge you take into account that the voltage of the this capacitor is given by:
[tex]\frac{Q}{\epsilon_o\epsilon_r A}=\frac{V}{d}\\\\V=\frac{Qd}{\epsilon_o\epsilon_r A}[/tex]
Q: total charge
d: distance between parallel plates
A: area of the plates
εr: dielectric constant
εo = dielectric permittivity of vacuum
for the case of the air εr = 1, then,
[tex]V=\frac{Qd}{\epsilon_o A}[/tex] (1)
When a dielectric material is placed in between the plates, you have, for the same voltage, and for a different charge:
[tex]V=\frac{Q'd}{\epsilon_o\epsilon_rA}[/tex] (2)
you equal the equation (1) and (2) and obtain:
[tex]\frac{Qd}{\epsilon_o A}=\frac{Q'd}{\epsilon_o \epsilon_r A}\\\\Q'=\epsilon_r Q[/tex]
by replacing you obtain:
[tex]Q'=(7.74)(4.15*10^{-6}C)=3.21*10^{-5}C[/tex]
Which two factors affect the force between two masses, according to the universal law of gravitation? the masses of the objects the radii of the two objects the speed of the two objects the distance between the objects the weight of the two objects
Answers
Answer:1. The masses Of The objects 2.the distance between the objects
Explanation:
newton's law of universal gravitation states that the force of attraction between two masses is directly proportional to the product Of The masses, and inversely proportional to thermal square Of The distance between them.therefore we can conclude that the two factors affecting the force between masses are:(1)the masses of the objects (2)the distance between the objects
The two factors that affect the force between two masses, according to the universal law of gravitation are
the masses of the objectsthe distance between the objectsWhat is newton's Universal law of gravitation?Newton's Universal law of gravitation states that the force of attraction that exists between particle or objects is directly proportion to the product of their masses and inversely proportional to the distances between them .
Therefore, The two factors that affect the force between two masses, according to the universal law of gravitation are
the masses of the objects.the distance between the objects.Learn more about Newton's universal law of gravitation below.
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Which has the deer’s image? Why?
Answers
Answer: what ( younger brothers of hers )
Explanation:
K bye
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
Answers
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 1200 kg trailer is hitched to a 1400 kg car. the car and trailer are traveling at 72 km.h when the driver applies the brakes on both the car and the trailer. knowing that the braking forces exerted on the car and the trailer are 5000 N and 4000 N respectively, determine (a) the distance traveled by the car and trailer before they come to a stop and (b) the horizontal component of the force exerted by the trailer hitch
Answers
Answer:
a) 8.67m
b) 1000N
Explanation:
(a) To find the distance you use the second Newton Law for both car and trailer, in order to calculate the dis-acceleration of the system:
[tex]F=ma\\\\a_=\frac{F}{m}=\frac{5000N+4000N}{1400kg+1200kg}=3.46\frac{m}{s^2}[/tex]
once you have this value, you use the the following kinematic equation to calculate the distance traveled by both car and trailer:
[tex]v^2=v_o^2-2ax\\\\x=\frac{-v^2+v_o^2}{2a}[/tex]
v: final velocity=0
vo: initial velocity = 72km/h = 60 m/s
by replacing the values of these parameters you obtain for x:
[tex]x=\frac{-0m/s+60m/s}{2(3.46m/s^2)}\\\\x=8.67m[/tex]
(b) The horizontal component of the force exerted by the trailer hitch is given by:
[tex]F_T=5000N-4000N=1000N[/tex]
What is the resistance of a resistor attached to a 1.5 V battery if the current in the circuit is 0.15 A?
Answers
Answer:10 ohms
Explanation:
Current=0.15A
Voltage=1.5v
Resistance=voltage ➗ current
Resistance=1.5 ➗ 0.15
Resistance=10 ohms
Two atomic particles approach each other in a head-on collision. Each particle has a mass of 2.97 × 10-25 kg. The speed of each particle is 2.19 × 108 m/s when measured by an observer standing in the laboratory. (a) What is the speed of one particle as seen by the other particle? (b) Determine the relativistic momentum of one particle, as it would be observed by the other.
Answers
Answer:
a) [tex]\vec{v}_{12}=2.86*10^{8} m/s[/tex]
b) [tex]p=2.81*10^{-16} kg*m/s[/tex]
Explanation:
a) When we have two particles traveling in parallel directions, the formula for relative velocity is:
[tex]\vec{v}_{12}=\frac{\vec{v}_{1}-\vec{v}_{2}}{1-\frac{\vec{v}_{1}\vec{v}_{2}}{c^{2}}}[/tex]
Here we have that v(1) = -v(2), the speed of the of the second particle is the negative of the first one.
If we use these equivalence we have:
[tex]\vec{v}_{12}=\frac{2\vec{v}_{1}}{1+\frac{\vec{v}_{1}^{2}}{c^{2}}}[/tex]
[tex]\vec{v}_{12}=\frac{2*2.19*10^{8}}{1+\frac{2.19*10^{16}}{3*10^{16}}}[/tex]
[tex]\vec{v}_{12}=2.86*10^{8} m/s[/tex]
And, [tex]\vec{v}_{21}=-2.86*10^{8} m/s[/tex]
b) The relativistic momentum equation to one particle observed by the other particle, is:
[tex]p=\gamma mv[/tex]
Where gamma is:
[tex]\gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]
v is the speed of the first particle relative to the second particle ([tex]2.86*10^{8}[/tex])m is the mass of the particle [tex]2.97*10^{-25} kg[/tex]Then gamma will be:
[tex]\gamma=\frac{1}{\sqrt{1-\frac{(2.86*10^{8})^{2}}{(3*10^{8})^{2}}}}[/tex]
[tex]\gamma=3.31[/tex]
Finally, the value of the momentum will be:
[tex]p=3.31*2.97*10^{-25}*2.86*10^{8}[/tex]
[tex]p=2.81*10^{-16} kg*m/s[/tex]
I hope it helps you!
A mutation causes a dog to be born with a tail that is shorter than normal.
Which best describes this mutation?
Answers
Answer:
A mutation causes a dog to be born with a tail that is shorter than normal. Which best describes this mutation? It is harmful because it obviously affects the dog’s survival. It is harmful because it affects the dog’s physical appearance. It is neutral because it does not obviously affect the dog’s survival. It is beneficial because it affects the dog’s physical appearance.
Explanation:
Answer:
C
Explanation:
:)))
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?
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Exercise should challenge your body and be at a greater intensity than your usual bif daily activity. Discuss
Answers
Answer:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
Explanation:
25. You lift a book from the floor to a bookshelf 1.0 m above the ground. How much power is used if the
upward force is 15.0 N and you do the work in 2.0 ?
Answers
Explanation:
P=E/T
E=15N
T=2s
P=15/2
P=7.5
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertically up from the ground with an initial velocity of 0=40.0 m/s . At what height from the ground will the two objects first meet?
Answers
Answer:
33.2 m
Explanation:
For the first object:
y₀ = 81.5 m
v₀ = 0 m/s
a = -9.8 m/s²
t₀ = 0 s
y = y₀ + v₀ t + ½ at²
y = 81.5 − 4.9t²
For the second object:
y₀ = 0 m
v₀ = 40.0 m/s
a = -9.8 m/s²
t₀ = 2.20 s
y = y₀ + v₀ t + ½ at²
y = 40(t−2.2) − 4.9(t−2.2)²
When they meet:
81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²
81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)
81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716
81.5 = 61.56t − 111.716
193.216 = 61.56t
t = 3.139
The position at that time is:
y = 81.5 − 4.9(3.139)²
y = 33.2
A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed uniformly throughout the volume of the shell. What is the magnitude of the electric field at a distance r = 11.2 cm from the center of the shell? (ε0 = 8.85 × 10-12 C2/N ∙ m2) (Give your answer to the nearest 0.01 MN/C)
Answers
Answer:
E = 1580594.95 N/C
Explanation:
To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:
[tex]\int EdS=\frac{Q_{in}}{\epsilon_o}[/tex] (1)
dS: differential of the Gaussian surface
Qin: charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2
The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:
[tex]\int EdS=ES=E(4\pi r^2)[/tex] (2)
Qin is calculate by using the charge density:
[tex]Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho[/tex] (3)
Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.
The charge density is given by:
[tex]\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}[/tex]
Next, you use the results of (3), (2) and (1):
[tex]E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})[/tex]
Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:
[tex]E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}[/tex]
hence, the electric field is 1580594.95 N/C
A convex mirror of focal length 33 cm forms an image of a soda bottle at a distance of 19 cm behind the mirror.If the height of the image is 7.0 cm,where is the object located,and how tall is it? What is the magnification of the image? Is the image virtual or real? Is the image inverted or upright? Draw a ray diagram to confirm your results.
Answers
Answer:
Image distance = 44.8cm, Image height = 16.5cm, Magnification = 0.42
The image is a virtual and upright image.
Explanation:
The nature of image formed by an object placed in front of a convex mirror is always diminished, virtual and erect.
The focal length f and the image distance are always NEGATIVE beacause the image is formed behind the mirror.
Given f = -33.0cm, v = -19.0cm
using thr mirror formula to get the object distance u, we have;
[tex]\frac{1}{f}=\frac{1}{u} + \frac{1}{v}\\ \frac{1}{u}=\frac{1}{f} - \frac{1}{v}\\\frac{1}{u}=\frac{1}{-33} - \frac{1}{-19}\\\frac{1}{u}=\frac{-19+33}{627} \\\frac{1}{u}=\frac{14}{627} \\u=\frac{627}{14} \\u = 44.8cm[/tex]
To calculate the image height, we will use the magnification formula
M = [tex]\frac{image\ height}{object\ height}=\frac{image\ distance}{object\ distance} \\[/tex]
M = [tex]\frac{Hi}{HI}=\frac{v}{u}[/tex]
Given Hi = 7.0cm
v = 19.0cm
u = 44.8cm
HI = 7*44.8/19
HI = 16.5cm
The object height is 16.5cm
Magnification = v/u = 19.0/44.8 = 0.42
SInce the image is formed behind the mirror, the image is a VIRTUAL and UPRIGHT image
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.
Answers
Answer:
a and b are the correct answers
Explanation:
Answer:
A) physical property; mass.
Explanation:
took the test
Which element is malleable and ductile
Answers
Answer:
Gold, silver, platinum. Gold is the most malleable and ductile.
Explanation:
The elements which are malleable and ductile include the following:
CopperIronCobalt etc.What is Malleability and Ductility?Malleability is the ability of a substance to be hammered into thin sheets
while ductility involves the deformation of a substance without any
breakage occurring in it.
Transition metals are the group of elements which have both
characteristics and examples are listed above.
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An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 30.8 N, the spring is stretched by 17.7 cm from its equilibrium length. Calculate the additional work required by F to stretch the spring by an additional 12.4 cm from that position.
Answers
Answer:
[tex]W=5.16 J[/tex]
Explanation:
Using the Hooke's law we can find the elasticity constant:
[tex]F=-k\Delta x[/tex]
[tex]30.8=-k*0.177[/tex]
[tex]k=|-\frac{30.8}{0.177}|[/tex]
[tex]k=174 N/m[/tex]
Now, we know that the work done is equal to the elastic energy, so we will have:
[tex]W=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})[/tex]
x2 is the final distance (x2 = 0.177+0.124 = 0.301 m)
x1 is the initial distance (x1 = 0.177 m)
[tex]W=\frac{1}{2}*174(0.301^{2}-0.177^{2})[/tex]
[tex]W=5.16 J[/tex]
I hope it helps you!
What is an independent variable?
A. A variable that is intentionally changed during an experiment
B. A variable that depends on the experimental variable
C. A variable that is not used in an experiment
D. A variable that is unknown during the experiment
Answers
Answer:
The answer is A
Explanation:
Independent variables don't have to depend on other factors of the experiment because they're independent
Answer:
A.
Explanation:
Independent variables don't have to depend on other factors of the experiment because they're independent.
The strength of the force of friction depends on which two factors?
Answers
Answer:
coefficient of friction (μ) and normal force (N)
Answer: How hard the surfaces push together and the types of surfaces involved
Explanation:
What is believed to be at the center of the Milky Way galaxy or virtually all large galaxies?
Answers
Answer:
A black hole
Explanation:
I'm not sure, i just know that's the right answer
If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s=1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with a velocity of 200 m/s?
Answers
Answer:
c = 4,444.44
Explanation:
You have the following expression for the acceleration of the projectile:
[tex]a=cs[/tex] (1)
s: distance to the ground of the projectile
To find the value of the constant c you use the following formula:
[tex]v^2=v_o^2+2a \Delta s[/tex] (2)
vo: initial velocity = 0 m/s
v: final speed = 200 m/s
Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m
You replace the expression (1) into the expression (2):
[tex]v^2=2(cs)\Delta s[/tex]
You do the constant c in the last equation, then you replace the values of v, s and Δs:
[tex]c=\frac{v^2}{2s\Delta s}=\frac{(200m/s)^2}{2(3m/s^2)(1.5m)}=4444.44[/tex]
A 25kg box in released on a 27° incline and accelerates down the incline at 0.3 m/s2. Find the friction force impending its motion? What is the coefficient of kinetic friction?
A block is given an initial speed of 3m/s up a 25° incline. Coefficient of friction
Answers
Answer:
a) μ = 0.475 , b) μ = 0.433
Explanation:
a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it
X axis
Wₓ - fr = m a
the friction force has the expression
fr = μ N
y Axis
N - [tex]W_{y}[/tex] = 0
let's use trigonometry for the components the weight
sin 27 = Wₓ / W
Wₓ = W sin 27
cos 27 = W_{y} / W
W_{y} = W cos 27
N = W cos 27
W sin 27 - μ W cos 27 = m a
mg sin 27 - μ mg cos 27 = m a
μ = (g sin 27 - a) / (g cos 27)
very = tan 27 - a / g sec 27
μ = 0.510 - 0.0344
μ = 0.475
b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result
μ = tan 25 - 0.3 / 9.8 sec 25
μ = 0.466 -0.03378
μ = 0.433
An astronaut visiting Jupiter's satellite Europa leaves a canister of 1.20 mol of nitrogen gas (28.0 g/mol) at 50.0 ∘C on the satellite's surface. Europa has no significant atmosphere, and the acceleration due to gravity at its surface is 1.30 m/s2. The canister springs a leak, allowing molecules to escape from a small hole. Neglect the interaction with surrounding atmosphere. (a) What is the maximum height (in km) above Europa's surface that is reached by a nitrogen molecule whose speed equals the rms speed? Assume that the molecule is shot straight up out of the hole in the canister, and ignore the variation in g with altitude. (b) The escape speed from Europa is 2025 m/s. Can any of the nitrogen molecules escape from Europa and into space?
Answers
Answer:
the answer is a
Explanation:
The voltage in a circuit is given by the equation V= IR.in this equation v is the voltage Iis correct and R is the resistance which answer shows this equation solved for current?
Answers
Answer:I=V/R
Explanation:
V=IR
Divide both sides by R
V/R=IR/R
V/R=I
I=V/R
The voltage in a circuit is given by the equation V= IR, in this equation v is the voltage Iis correct and R is the resistance the solution for the current is given as follows,
V= IR
I = V/R
What is resistance?Resistance is the obstruction of electrons in an electrically conducting material. The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.
V=IR
As for the given problem if we have to solve for the current from the equation V=IR
V=IR
I = V/R
Let us suppose a 60-volt battery connected in a closed circuit with a resistor of 15 ohms then we have o find out the amount of current flowing in the circuit,
Voltage = 60V
Resistance = 15 Ohm
Current =?
By using Ohm,s Law,
V=IR
I = V/R
By substituting the respective values,
I = 60/15
I = 4 Ampere
Hence, we solved for the current from the equation V=IR.
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Cannon
Mass - M
Marble
Mass m
Note: Figure not drawn to scale.
Students launch identical marbles of mass mm horizontally from a toy cannon of mass MC,
where Mc > mm, as shown above. The cannon can be adjusted to change the launch speed v of
the marble relative to the ground. Each time a marble is launched, the cannon slides backward
before coming to rest. The coefficient of kinetic friction between the cannon and the ground is u
(mu). For each launch, the students vary the launch speed v and record the distance x the cannon
slides backward for each launch speed.
Answers
Answer:
the distance x increases as the student increase the mass of the cannon.
Explanation:
a) From the question ; we get to understand that for each launch, the students use a different mass which is launch at speed v relative to the ground.
This changes in the mass used brought about a change in the momentum at the same speed v ; perhaps an increase in momentum. However; since the conservation of the momentum is considered at each launching.
The momentum of the marble = momentum of the cannon
But since the momentum of the cannon increase ; therefore the same equivalent changes takes place in its kinetic energy . Therefore , the kinetic energy will increase and the distance will also increase in the bid to quench the amount of energy generated. Thus, the distance x increases as the student increase the mass of the cannon.
b)We all know that conservation of the momentum will definitely takes place after launching of the cannon.
Let assume that [tex]\rho[/tex] is the momentum of the cannon with mass [tex]M_C[/tex]
The kinetic energy of the canon will be:
[tex]\frac{\rho ^2 }{2 M_C}[/tex]
Also the frictional force acting on the cannon is :
[tex]f = \mu mg[/tex]
If the cannon move at an additional distance x; the frictional force acting at this area quench the amount of the energy generated and consume the kinetic energy of the cannon;
So;
[tex]fx = K.E[/tex]
[tex]fx = \frac{\rho ^2 }{2 M_C}[/tex]
[tex]\mu mg x = \frac{\rho ^2 }{2 M_C}[/tex]
[tex]x = \frac{\rho ^2 }{2 \mu mg M_C}[/tex]
[tex]x = \frac{m_m^2 V^2 }{2 \mu mg M_C}[/tex]
Thus; it is consistent with the answer in (a) as increase in the mass of the marble will bring about an increase in distance x
The increase in the mass of the marble increase the distance of the cannon.
Momentum:
Momentum of an object is equal to the product of the mass and the velocity of the object.
The kinetic energy of the cannon
[tex]K_C = \dfrac {\rho^2}{2 M_C}[/tex]
Where,
[tex]\bold{\rho}[/tex] - momentum of the cannon
Mc - mass
The frictional force on cannon
[tex]\bold{F_f = \mu mg}[/tex]
Cannon move a distance x,
Hence,
[tex]\bold {F_f \times x = K_C}\\\\\bold {\mu m g \times x = \dfrac {\rho^2}{2 M_C }}\\\\\bold {x = \dfrac {\rho^2}{2\mu m g M_C }} }\\\\\bold {x = \dfrac {m^2 V^2}{2\mu m g M_C } }[/tex]
Therefore, The increase in the mass of the marble increase the distance of the cannon.
To know more about kinetic energy,
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Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a great amount of angular momentum when rotating. A wind turbine has a total of 3 blades. Each blade has a mass of m = 5500 kg distributed uniformly along its length and extends a distance r = 46 m from the center of rotation. The turbine rotates with a frequency of f = 11 rpm.
Required:
a. Calculate the total moment of inertia of the wind turbine about its axis, in units of kilogram meters squared.
b. Calculate the angular momentum of the wind turbine, in units of kilogram meters squared per second.
Answers
Answer:
Explanation:
moment of inertia of each blade which is similar to rod rotating about its one end
= 1/3 ml²
moment of inertia of 3 blades = ml²
= 5500 x 46²
I = 11638 x 10³ kg m²
angular velocity = 2πn where n is rotation per second
n = 11 / 60
angular velocity = 2π x 11/60
= 1.1513 rad /s
angular momentum
= moment of inertia x angular velocity
= 11638 x 10³ x 1.1513
= 13399 x 10³ kg m² per second.
A cobalt-60 source with activity 2.60×10-4 Ci is embedded in a tumor that has
mas 0.20 kg. The source emits gamma photons with average energy 1.25 MeV.
Half the photons are absorbed in the tumor, and half escape.
i. What energy is delivered to the tumor per second? [4 marks]
ii. What absorbed dose, in rad, is delivered per second? [2 marks]
iii. What equivalent dose, in rem, is delivered per second if the RBE for
these gamma rays is 0.70? [2 marks]
Page 6 of 7
iv. What exposure time is required for an equivalent dose of 200 rem? [2
marks]
B. A laser with power output of 2.0 mW at a wavelength of 400 nm is projected
onto a Calcium metal. The binding energy is 2.31 eV.
i. How many electrons per second are ejected? [6 marks]
ii. What power is carried away by the electrons? [4 marks]
C. A hypodermic needle of diameter 1.19 mm and length 50 mm is used to
withdraw blood from a patient? How long would it take for 500 ml of blood to be
taken? Assume a blood viscosity of 0.0027 Pa.s and a pressure in the vein of
1,900 Pa. [10 marks]
D. A person with lymphoma receives a dose of 35 gray in the form of gamma
radiation during a course of radiotherapy. Most of this dose is absorbed in 18
grams of cancerous lymphatic tissue.
i. How much energy is absorbed by the cancerous tissue? [2 marks]
ii. If this treatment consists of five 15-minute sessions per week over the
course of 5 weeks and just one percent of the gamma photons in the
gamma ray beam are absorbed, what is the power of the gamma ray
beam? [4 marks]
iii. If the gamma ray beam consists of just 0.5 percent of the photons
emitted by the gamma source, each of which has an energy of 0.03
MeV, what is the activity, in Curies, of the gamma ray source? [4 marks]
E. A water heater that is connected across the terminals of a 15.0 V power supply
is able to heat 250 ml of water from room temperature of 25°C to boiling point
in 45.0 secs. What is the resistance of the heater? The density of water is 1,000
kg/m2 and the specific heat capacity of water is 4,200 J/kg/°C. [10 marks]
Answers
Answer:
A i. E = 9.62 × 10⁻⁷ J/s
ii. The absorbed dose is 4.81 × 10⁻⁶ Gy
iii. The equivalent dose is 3.37 × 10⁻⁴ rem/s
iv. t = 593471.81 seconds
B. i. 4.025 × 10¹⁵/s
ii. 0.512 mW
C. 7218092.2 seconds
D. i. 6.3 × 10⁻¹ J
ii. 1.4 × 10⁻² W
iii. 1.57 × 10³ Curie
E. 0.129 Ω
Explanation:
The given parameters are;
Mass of tumor = 0.20 kg
Activity of Cobalt-60 = 2.60 × 10⁻⁴ Ci
Photon energy = 1.25 MeV
(i) The energy, E, delivered to the tumor is given by the relation;
[tex]E = \frac{1}{2}\left (Number \, of \, decay / seconds \right )\times \left (Energy \, of \, photon \right )[/tex]
[tex]E = \frac{1}{2}\left (2.6\times 10^{-4}Ci )\times \left (\frac{3.70\times 10^{10}decays/s}{1 Ci} \right )\times 1.25\times 10^{6}eV\times \frac{1.6\times 10^{-19}J}{1eV}[/tex]
E = 9.62 × 10⁻⁷ J/s
(ii) The equation for absorbed dose is given as follows;
Absorbed dose, D, in Grays Gy = (Energy Absorbed Joules J)/Mass kg
Therefore, absorbed dose = (9.62 × 10⁻⁷ J/s)/( kg) = 4.81 × 10⁻⁶ Gy
1 Gray = 100 rad
4.81 × 10⁻⁷ Gy = 100 × 4.81 × 10⁻⁶ = 4.81 × 10⁻⁴ rad/s
(iii) Equivalent dose, H, is given by the relation;
H = D × Radiation factor, [tex]w_R[/tex]
∴ H = 0.7 × 4.81 × 10⁻⁴ rad/s = 3.37 × 10⁻⁴ Sv = 3.37 × 10⁻⁴ rem/s
(iv) The exposure time required for an equivalent dose of 200 rem is given as follows;
[tex]\dot{H} = \dfrac{H}{t}[/tex]
Therefore;
[tex]t= \dfrac{200}{{3.37 \times 10^{-4}} } = 593471.81 \, s[/tex]
∴ t = 6.9 days
B. The number of electrons ejected is given by the relation;
[tex]N = \frac{P}{E} = \frac{P \times \lambda}{hc}[/tex]
[tex]N = \dfrac{2.0 \times 10^{-3} \times 400 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} = 4.025 \times 10^{15}/s[/tex]
(ii) The power carried by the electron
The energy carried away by the electrons is given by the relation;
[tex]KE_e = hv - \Phi[/tex]
[tex]KE_e = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} - 2.31 \times \frac{1.6 \times 10 ^{-19} }{1}[/tex]
[tex]KE_e = 4.9695 \times 10^{-19} - 3.696 \times 10 ^{-19} = 1.2735 \times 10^{-19} J[/tex]
Power, P[tex]_e[/tex], carried away by the electron = 4.025 × 10¹⁵ × 1.2735 × 10⁻¹⁹ = 0.512 mW
C. The given parameters are;
d = 1.19 mm, ∴ r = 1.19/2 = 0.595 × 10⁻³ m
l = 50 mm = 5 × 10⁻³ m
V = 500 ml = 5 × 10⁻⁴ m³
η = 0.0027 Pa
p = 1,900 Pa.
[tex]\dfrac{V}{t} = \dfrac{\pi }{8} \times \dfrac{P/l}{\eta } \times r^4[/tex]
[tex]t = \dfrac{8\times \eta\times V\times l }{\pi \times P \times r^4}[/tex]
[tex]t = \dfrac{8\times 0.0027 \times 5 \times 10^{-4} \times 5 \times 10^{-2} }{\pi \times 1900 \times (0.595 \times 10^{-4} )^4}[/tex]
t = 7218092.2 seconds
D) i. Energy absorbed is given by the relation;
E = m×D
Where:
D = 35 Gray = 35 J/kg
m = 18 g = 18 × 10⁻³ kg
∴ E = 35 × 18 × 10⁻³ = 6.3 × 10⁻¹ J
ii. Total time for treatment = 15 × 5 = 75 minutes
Energy absorbed = 6.3 × 10⁻¹ × 100 = 63 J
Power = Energy(in Joules)/Time (in seconds)
∴ Power = 63/(75×60) = 1.4 × 10⁻² W
iii. Whereby the power is provided by 0.5% of the photons emitted by the source, we have;
[tex]P_{source}= \frac{P_{beam}}{0.005} =\frac{0.0014}{0.005} =0.28 \, W[/tex]
1 MeV = 1.60218 × 10⁻¹³ J
0.03 MeV = 0.03 × 1.60218 × 10⁻¹³ J = 4.80654 × 10⁻¹⁵ J/photon
Therefore, the number of disintegration per second = 0.28 J/s ÷ 4.80654 × 10⁻¹⁵ J/photon = 5.83 × 10¹³ disintegrations per second
1 Curie = 3.7 × 10¹⁰ disintegrations per second
Hence, 5.83 × 10¹³ disintegrations per second = (5.83 × 10¹³)/(3.7 × 10¹⁰) Curie
= 1.57 × 10³ Curie
E. The parameters given are;
Density of water = 1000 kg/m³
Volume of water = 250 ml = 0.00025 m³
Initial temperature, T₁, = 25°C
Final temperature, T₂, = 100°C
Change in temperature, ΔT = 100 - 25 = 75°
Specific heat capacity of the water = 4200 J/kg/°C
Mass of water = Density × Volume = 1000 × 0.00025 = 0.25 kg
∴ Heat supplied = 4200 × 0.25 × 75 = 78,750 J
Time to heat the water = 45.0 sec
Therefore, power = Energy/time = 78750/45 = 1750 W
The formula for electrical power = I²R =VI = V²/R
Therefore, where V = 15.0 V, we have;
15²/R = 1750
R = 15²/1750 = 0.129 Ω.
The resistance of the heater = 0.129 Ω.
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to a rest uniformly in a distance of 160 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?
Answers
Answer:
Force applied to stop the car = 1,250 N
Explanation:
Given:
Mass of car (M) = 1,000 kg
Initial velocity (U) = 20 m/s
Final velocity (V) = 0 m/s
Distance (S) = 160 m
Find:
Force applied to stop the car.
Computation:
[tex]v^2 = u^2 + 2as\\\\0^2=20^2+2(a)(160)\\\\0=400+320(a)\\\\Acceleration = a = -1.25m/s^2\\\\Force = ma \\\\Force= 1,000(1.25)\\\\Force = 1,250 N[/tex]
Force applied to stop the car = 1,250 N
How does a concave mirror form an image
Answers
Answer:
a magnified and erect virtual is found to be formed.
Four unequal resistors are connected in a series circuit. Which one of the following statements is correct about this circuit? The total resistance is equal to average of the four resistors. The total resistance is less than the smallest resistor. The total resistance is less than the largest resistor. The total resistance is equal to any one of the resistors. The total resistance is more than the largest resistor.
Answers
Answer:
The total resistance is more than the largest resistor.
Explanation:
In series connection of resistors, the total resistance of the connection is the sum of all the resistance of the resistors connected.
For example; let assume three resistors with resistance 3,4 and 5 ohms, the total resistance is;
Rt = 3+4+5 = 12 ohms
So, the total resistance is greater than the largest resistor since 12 ohms is greater than 5 ohms(largest resistor).
