Am assuming the B-field has an rms strength of 1.10×10−10T which was omitted
Answer: E=1.686 x10⁻⁸J/hr
Explanation:
Intensity for electromagnetic wave is given as
I= cε₀(Emax)²....... equation 1
Also
B=Emax/c ....... equation 2
puting the value of B in equation 2 in equation 1 becomes
I= cε₀(Emax)²
I= c³ε₀B²
where
c = speed of light = 3 X 10⁸m/s
ε₀=permittivity of free space = 8.85x 10-12 F m-1
B=1.10×10−10T
I= (3 X 10⁸m/s)³ X (8.85 X 10⁻12F/m) x(1.10×10−10T)²
I=2.891 X 10⁻⁶ W/m²
Energy per unit hour
E= IAt
= 2.891 X 10⁻⁶ W/m² x πr²x 3600
2.891 X 10⁻⁶ W/m² x 3.142 x ( 1.80 x 10⁻³/2) x(3600/1hr)
E=1.686 x10⁻⁸J/hr
The total energy transported per hour will be 6.62 × 10⁻⁹J
Calculation of energy transported:The Intensity for an electromagnetic wave is given by:
I = cε₀E₀²
where c is the speed of light, and
E₀ is the maximum value of the electric vector
Also
B = E/c
where B is the magnetic field
Therefore:
I= cε₀E₀²
I= c³ε₀B²
Given that the magnetic field is:
B = 1.10×10⁻¹⁰ T
I= (3 X 10⁸m/s)³ X (8.85 X 10⁻12F/m) x(1.10×10−10T)²
I=2.891 X 10⁻⁶ W/m²
Energy transported per unit hour
E= IAt
where A is the area
and t is the time = 1hr = 3600s
E = 2.891 × 10⁻⁶ × πr² × 3600
E = 2.891 X 10⁻⁶ × 3.14 × ( 1.80 x 10⁻³/4)² x (3600)
E = 6.62 × 10⁻⁹J
Learn more about electromagnetic waves:
https://brainly.com/question/13803241?referrer=searchResults
5.3C The nozzle on a fire hose is connected to the hose via a coupling. When the fire hose is in use with water flowing through it and the hose is stationary, the coupling is: a) in equilibrium, so there is no force on the coupling. b) in tension. c) in compression
Answer:
Tension ( B )
Explanation:
When the fire hose is in use with water flowing through it and the hose is stationary, the coupling is in TENSION , this is because the force of flow in a stationary object will be in a compressive nature that will in turn cause the force on the coupling to be in a form of a Tension ( i.e. Tensile )
Which of the following statements is correct about the energy
of water stored in a tank?
1. Water stored in the tank has kinetic energy and potential
energy
2. Water stored in the tank has kinetic energy only.
3. Water stored in the tank has no energy.
4. Water stored in the tank has potential energy only.
Answer:
4. Water stored in the tank has potential energy only.
Explanation:
when water begins to move the potential energy is converted into kinetic but until it is being used it has potential energy only.
The village blacksmith dunks a red-hot horseshoe into a large bucket of 22.0°C water. How much heat was lost by the horseshoe in vaporizing 0.0100 kg of water?
Answer:
25776J
Explanation:
Recall that heat lost/gained is obtained from
H=mcθ
H= heat lost/gained
m= mass of water = 0.0100 kg
c= 4200 J/kg°C
θ = temperature
Heat gained by the water in rising to 100°
H= 0.0100 * 4200 * (100-22)
H= 3276 J
Latent heat of vaporization of water = mL
Latent heat of vapourization = 0.0100*(22.5*10^5)=22500J
Heat lost by iron = heat gained by water
total heat gained by water =
22500+3276=25776J
A bus is moving to the right at an absolute velocity U. You throw a water balloon at the bus at an absolute velocity V (greater than U) after the bus passes you. A person on the bus sees the balloon hit the bus at a speed of:
Answer:
The person on the bus will see the balloon hit the bus at a velocity = V - U
Explanation:
There is relative motion between the water balloon and the person sitting in the bus.
Relative motion is the motion of a body with respect to another body (eg. an observer) from the point of reference of the second body (observer).
The person on the bus has a velocity which is the same as that of the moving bus = U
The velocity of the thrown water balloon (greater than U) = V
The relative motion between the person and the water balloon is the difference between their velocities (since they are both moving in the same direction) = V - U
Therefore, the person on the bus will see the balloon hit the bus at a velocity = V - U
Two ropes are tied to a 50 kg boat. One pulls with a force of 120 N due West and the other rope pulls with a force of 140 N due South. Find the net force of the boat (magnitude and direction) and the acceleration.
The net force on the boat is the vector
F = (-120 N) i + (-140 N) j
where we take West and South to be the negative directions going left-and-right and up-and-down, respectively.
The magnitude of the net force is
F = √((-120 N)² + (-140 N)²) = 20 √(85) N ≈ 180 N
and its direction is
tan(θ) = (-140 N) / (-120 N) → θ ≈ -130°
relative to due East, or approximately 50° South of West.
In which disorder does a person find it difficult to fall asleep or stay asleep for a long time?
A.
somnambulism
B.
narcolepsy
C.
insomnia
D.
sleep apnea
Answer:c
Explanation:
I have it
Answer:
C.
insomnia
Explanation:
plato answer
What conditions are required so that the mechanical energy of a system is conserved? Select all that apply. View Available Hint(s) Select all that apply. No internal forces act on objects within the system. No net external force acts on the system No non-conservative forces, such as friction, act between objects within the system
Answer:
No non-conservative forces, such as friction, act between objects within the system
Explanation:
The law of conservation of mechanical energy in a closed system states that;'The total amount of mechanical energy, in a closed system in the absence of non-conservative forces, such as friction, remains constant.' Non-conservative forces cause energy to be lost from the system, this lost energy can't be gotten back
This implication of this is that energy can not disappear. Kinetic energy can be converted into potential energy and vice-versa in accordance with this principle.
3. What would be the mass of a truck if it is accelerating at a rate of 5 m/s and hits a parked car with a
force of 14,000 N?
I
Answer:
We are given both the Force applied by the truck and the acceleration of the truck:
Force applied (F) = 14,000 N
Acceleration (a)= 5 m/s²
Solving for the mass of the truck:
From Newton's second law of motion,
F = ma (where F is the force applied , m is the mass of the object and a is the acceleration of the object)
replacing the variables:
14000 = (m)* 5
m = 14000/5
m = 2800
Therefore, the mass of the truck is 2800 kg
Answer:
m = 2800 N
Explanation:
14000 = (m)* 5
m = 14000/5
F= 14,000 N
a= 5 m/s^2
A running back weighing 100 kg runs headlong into a defender weighing 120 kg. The two players are moving at 7 m/s and 5 m/s, respectively. Calculate the total momentum and the total kinetic energy of the system. [Calculations including velocity should reflect direction of travel.]
Answer:
Explanation:
we know that momentum is given as
P=mv
given data
P1=?
mass m1=100kg
velocity v1=7m/s
P2=?
mass m2=120kg
velocity v2=5m/s
the total momentum =P1+P2
total=m1v1+m2v2
total= 100*7+120*5
total=700+600
total= 1300kgm/s
total kinetic energy
KE total= 1/2m1v1^2+1/2m2v2^2
KE total= 1/2*100*(7)^2+1/2*120*(5)^2
KE total= 1/2*100*49+1/2*120*25
KE total= 1/2*4900+1/2*3000
KE total=2450+1500
KEtotal=3950 J
Given that the tensile strength of aluminum foil is 311 megapascals, its thickness is approximately 15.0 micrometers, and a roll of household aluminum foil is 30.0 centimeters wide, how much force F is needed to pull off a sheet to use
Answer:
Explanation:
cross sectional area = width x thickness
A = 30 x 10⁻² x 15 x 10⁻⁶ = 450 x 10⁻⁸ m
tensile strength F / A = 311 x 10⁶ N / m²
F = A x 311 x 10⁶
= 450 x 10⁻⁸ x 311 x 10⁶
= 139950 x 10⁻²
= 1399.5 N .
Find the resistance that must be placed in series with a 25.0Ω25.0 Ω galvanometer having a 50.0µA50.0 μA sensitivity (the same as the one discussed in the text) to allow it to be used as a voltmeter with a 0.100-V full-scale reading
Answer:
1975 ohm
Explanation:
The maximum current that can pass through the galvanometer is 50.0µA
p.d across the galvanometer at a full scale deflection is given by
V = IR
V = (50.0×10^-6)*25
V1 = 1/800 or 0.00125volts
Since the p.d across the multiplier and galvanometer at full scale deflection is 0.100volts,the p.d across the multiplier alone must be given by
V2 = 0.100-0.00125
= 0.09875volts
Current passing through the multiplier is (50.0×10^-6)A
Hence if R is the resistance of the multiplier, we have that
V2 = (50.0×10^-6)×R
0.09875volts = (50.0×10^-6)×R
Divide both sides by 0.09875volts
R = 0.09875/(50.0×10^-6)
R =( 0.09875) ÷ 1/20000
R = 0.09875×20000
R = 1975 ohms
Physics Freefall Name 1. You decide to drop a penny from the top of the hotel in New York. It hits a tourist in the head at 32 m/s. How far did the penny fall
Answer:
h = 52.24 m
Explanation:
Given that,
A penny is dropped from the top of the hotel in New York. It hits a tourist in the head at 32 m/s.
Let it falls at a height of h.
We can find the value of h using the conservation of energy such that,
[tex]\dfrac{1}{2}mv^2=mgh\\\\h=\dfrac{v^2}{2g}\\\\h=\dfrac{(32)^2}{2\times 9.8}\\\\h=52.24\ m[/tex]
So, it will fall at a distance of 52.24 m.
A spacecraft is traveling in space at a velocity of 8200 m's. Turning a dial elicits the engines to accelerate the spacecraft at 2 m/s. How many seconds will it take
before the spacecraft reaches a velocity of 8400 m's?
How can humans detect infrared waves?
А
They see them as color.
B
They feel them as heat.
с
They hear them as sound.
D
They feel them as vibrations.
What is the net work, in joules, required to stop a bullet of mass 7.73 g that is moving at a speed of 409 m/s
Answer:
KE=1.58J
Explanation:
In this problem we are required to solve for the total energy
"The kinetic energy is the energy possessed by a body by virtue of motion"
KE=1/2mv²
Given
Mass m=7.73g
Velocity v=409m/s
In kg= 0.00773kg
KE=1/2*0.00773*409
KE=3.16157/2
KE=1.58J
2. Which is a non-example of matter?
Answer: there are multiple
Explanation:
light from a torch, the heat from a fire, and the sound of a police siren. You cannot hold, taste, or smell these things. They are not types of matter, but forms of energy.
A very bouncy ball is dropped from a height of 2.47 m to an asphalt playground surface and the height of its 5th bounce is measured to be 1.51 m. Find the coefficient of restitution of the ball for a collision with asphalt
Answer: the coefficient of restitution of the ball for a collision with asphalt=0.95
Explanation:
Initial height of the bounce from the ball, h =2.47m
height of the 5th bounce of the ball , h₁= 1.51m
Making e the coefficient of the restitution, then the height covered by the ball after the 5th bounce is
h₁= e²ⁿh
Where n = 5
h₁= e²⁽⁵⁾h
e¹⁰=h₁/h
e¹⁰=1.51/2.47
e= (1.51/2.47)¹/¹⁰
e=0.95198≈ 0.95
Therefore, the coefficient of restitution of the ball for a collision with asphalt is 0.95
A Carnot engine receives 250 kJ/s of heat from a heat source reservoir at 525 o C and rejects heat to a heat-sink reservoir at 50 o C. What are the power developed and the heat rejected
Answer:
Explanation:
efficiency of carnot engine = (Q₁ - Q₂) / Q₁ = (T₁ - T₂) / T₁
Q₁ = heat absorbed , Q₂ = heat rejected
T₁ = Temperature of source = 525 + 273 = 798 K
T₂ = Temperature of sink = 50 + 273 = 323 K
efficiency = (798 - 323) / 798
= 475 / 798 = .595
(Q₁ - Q₂) / Q₁ = .595
(250 - Q₂) / 250 = .595
250 - Q₂ = 148.75
Q₂ = 101.25 kJ
Heat rejected = 101.25 kJ .
output work = Q₁ - Q₂ = 250 - 101.25 = 148.75 kJ / s
power = 148.75 kJ /s
PLEASE HELPP WITH QUESTIONS :(
Answer:no
Explanation: lol
How long will it take an object that falls from rest to attain a velocity of 147 m/s?
Answer:
15 seconds
Explanation:
We can use this formula to solve this problem:
Velocity = Initial Velocity + Acceleration * Time
We know that initial velocity is 0, the velocity is 147, and acceleration is 9.8(because of earth's gravitational pull), so we can plug those numbers in and solve for time:
147 = 0 + 9.8*t
t = 15
Therefore, it will take 15 seconds for an object that falls from rest to reach 147 m/s.
Determine the required height h of the crest of the roller coaster to thebottom so that when it is essentially at rest at the crest of the hill it will reacha speed of 28 m/s when it comes to the bottom.
Answer:
h = 40 m
Explanation:
Assuming no friction present, total mechanical energy must be conserved, so the following expression stands:ΔK + ΔU = 0 (1)Now, if the car is at rest at the crest of the hill, the change in kinetic energy is just as follows:[tex]\Delta K = \frac{1}{2} * m* v_{b} ^{2}[/tex] (2)
where vb = speed at the bottom = 28 m/s
If we define the bottom as our zero reference level for the gravitational potential energy, we can write the following equation:[tex]\Delta U = U_{f} - U_{i} = 0- m*g*h = -m*g*h[/tex] (3)
From (1) we get:ΔK = -ΔUReplacing by (2) and (3), we get:[tex]\frac{1}{2} * m* v^{2} = m*g*h[/tex]
Simplifying and rearranging terms, we can solve for h (height required) as follows:[tex]h = \frac{v_{b} ^{2} }{2*g} = \frac{(28m/s)^{2}}{2*9.8m/s2} = 40 m[/tex]
A speeding car is traveling at a constant 30.0 m/s when it passes a sationary police car. If the police car delays for 1.00s before starting, what must be the magnitude of the constant accleration of the police car. pdf
Answer:
30.0m/s²Explanation:
Given
final speed v = 30.0m/s
initial velocity u = 0m/s
Time taken t = 1.00s
Required
constant acceleration of the police car a
Substitute the given parameters into this equation of motion
v = u + at
30 = 0 +1a
30 = a
a = 30.0m/s²
Hence the magnitude of the constant acceleration of the police car is 30.0m/s²
If somebody asks me if i need help on my work and i say " no ty " , is that rude???
Answer:
no
Explanation:
A cart with a mass of 55 kg is pushed with a force of 220 N. What is the acceleration?
Answer:
4m/s^2
Explanation:
According to newton's second law of motion:
F= ma
a= F/m
F =220N
m =55kg
a= 220/55
= 4m/s^2
Problem 6 (20 points) A ballistic pendulum is a device often used for measuring the velocity of a projectile. Assume to projectile fully embeds itself in the pendulum. If the projectile mass is 25 grams, the pendulum mass is 2.5 kg, and the vertical displacement of the pendulum after the collision is 12 cm, find the velocity of the projectile just prior to the collision.
Answer:
The velocity of the projectile just prior to the collision is 154.934 meters per second.
Explanation:
Let consider the projectile-pendulum system as our system of study, from statement we know that projectile gives kinetic energy to the pendulum and turns into gravitational potential energy. By Principle of Energy Conservation we construct the following expression:
[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}[/tex] (Eq. 1)
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energies, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies, measured in joules.
By applying definitions of gravitational potential and kinetic energies, we expand and simplify the equation above:
[tex](m+M)\cdot g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (m+M)\cdot (v_{1}^{2}-v_{2}^{2})[/tex]
[tex]g\cdot (y_{2}-y_{1})=\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})[/tex]
And we clear the initial velocity of the pendulum-projectile system:
[tex]v_{1}=\sqrt{v_{2}^{2}+2\cdot g\cdot (y_{2}-y_{1})}[/tex] (Eq. 2)
Where:
[tex]m[/tex] - Mass of the projectile, measured in kilograms.
[tex]M[/tex] - Mass of the ballistic pendulum, measured in kilograms.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the pendulum-projectile system, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Initial and vertical position of the pendulum-projectile system, measured in meters.
Under the consideration of inellastic collision, we find the velocity of the projectile prior to the collision by Principle of Linear Momentum Conservation:
[tex]m\cdot v_{o, P} + M\cdot v_{o,B} = (m+M)\cdot v_{1}[/tex]
[tex]v_{o,P} = \frac{(m+M)\cdot v_{1}-M\cdot v_{o,B}}{m}[/tex] (Eq. 2)
Where:
[tex]v_{o, P}[/tex] - Velocity of the project prior to the collision, measured in meters per second.
[tex]v_{o,B}[/tex] - Velocity of the ballistic pendulum prior to the collision, measured in meters per second.
If we know that [tex]v_{2} = 0\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{2}-y_{1} = 0.12\,m[/tex], [tex]m = 0.025\,kg[/tex], [tex]M = 2.5\,kg[/tex] and [tex]v_{o,B} = 0\,\frac{m}{s}[/tex], then the velocity just prior to the collision is:
By (Eq. 1):
[tex]v_{1} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.12\,m)}[/tex]
[tex]v_{1} \approx 1.534\,\frac{m}{s}[/tex]
By (Eq. 2):
[tex]v_{o,P} = \frac{(0.025\,kg+2.5\,kg)\cdot \left(1.534\,\frac{m}{s} \right)-(2.5\,kg)\cdot \left(0\,\frac{m}{s} \right)}{0.025\,kg}[/tex]
[tex]v_{o,P} = 154.934\,\frac{m}{s}[/tex]
The velocity of the projectile just prior to the collision is 154.934 meters per second.
What is the angle between a wire carrying an 8.00-A current and the 1.20-T field it is in if 50.0 cm of the wire experiences a magnetic force of 2.40 N? (b) What is the force on the wire if it is rotated to make an angle of 90 ∘ 90∘ with the field?
Answer:
(a) the angle between the wire and field is 30⁰
(b) the force is 4.8 N
Explanation:
Given;
current in the wire, I = 8 A
magnetic field, B = 1.2 T
length of the wire, L = 50.0 cm = 0.5 m
magnetic force, F = 2.4 N
The magnetic force on the wire is given by;
F = BILsinθ
where;
θ is the angle between the wire and field
sinθ = F / BIL
sinθ = 2.4 / (1.2 x 8 x 0.5)
sinθ = 0.5
θ = sin ⁻¹ (0.5)
θ = 30⁰
(b) the force on the wire if it is rotated to make an angle of 90⁰
F = BIL(sin90⁰)
F = BIL(1)
F = BIL
F = 1.2 x 8 x 0.5
F = 4.8 N
A piece of pencil lead is connected in series with an ammeter and a power supply. The power supply is turned on. After a few minutes, although the potential difference across the pencil lead does not change, the current in the circuit increases significantly. Explain why the current increases.
Answer:
More electrons become free due to the potential which are charge carriers. More the charge carriers, more the current.
Current causes a heating effect. Resistance of lead decreases/ n increases [I=nAve]. Current increase
Explanation:
What is one way radio waves are used?
А
to carry signals for televisions
B
to kill germs in hospitals
с
to treat cancer with radiotherapy
D
to take images of bones
Answer:
A
Explanation:
because it is used in tvs all the time
A water balloon is thrown horizontally with a
speed of 8.31 m/s from the roof of a building of
height H=23.0 m. How far does the balloon travel
horizontally before striking the ground? (Use the
vertical formula Av=v0v+1 2aw/2 to solve for time
and plug your components into the equation for
the horizontal direction Ar-vxt)
Your answer
Answer:
18mExplanation:
Given
speed of water balloon = 8.31m/s
Height of building = 23.0m
Required
Horizontal distance
First we need to get the time using the equation of motion
S = ut + 1/2at²
23 = 0+1/2(9.81)t²
23 = 4.905t²
t² = 23/4.905
t² = 4.689
t = √4.689
t = 2.17secs
Next is to get the horizontal direction expressed as:
Sh = Vxt
Sh = 8.31 * 2.17
Sh = 17.99m
Sh ≈ 18m
Hence the horizontal distance of the object before striking the ground is approximately 18m
In riot control, the riot squad uses a water hose that shoots water at a rate of 5 m/s and volume of 30 L/s. What is the average force exerted on a person assuming that the water splashes sideways in all directions? The density of water is 1000 kg/m3.
Answer:
The average force exerted on a person assuming that the water splashes sideways in all directions is 150 newtons.
Explanation:
From Fluid Mechanics and Newton's Laws we know that water stream is an example of a variable mass system at constant speed and the magnitude of the net force done by the water splash ([tex]F[/tex]), measured in newtons, is given by the following expression:
[tex]F = \rho\cdot \frac{dV}{dt}\cdot v[/tex] (Eq. 1)
Where:
[tex]\rho[/tex] - Density of water, measured in kilograms per cubic meter.
[tex]\frac{dV}{dt}[/tex] - Volume flow rate, measured in cubic meters per second.
[tex]v[/tex] - Flow velocity, measured in meters per second.
If we know that [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]\frac{dV}{dt} = 0.03\,\frac{m^{3}}{s}[/tex] and [tex]v = 5\,\frac{m}{s}[/tex], the average force exerted on a person is:
[tex]F = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(0.03\,\frac{m^{3}}{s} \right)\cdot \left(5\,\frac{m}{s} \right)[/tex]
[tex]F = 150\,N[/tex]
The average force exerted on a person assuming that the water splashes sideways in all directions is 150 newtons.