The energy flow per unit time per unit area (S) of an electromagnetic wave has an average value of 310 mW/m2. The maximum value of the magnetic field in the wave is closest to:
Answer:
5.096*10^-8
Explanation:
Given that
The average value of the electromagnetic wave is 310 mW/m²
To find the maximum value of the magnetic field the wave is closest to, we say
Emax = √Erms
Emax = √[(2 * 0.310 * 3*10^8 * 4π*10^-7)]
Emax = √233.7648
Emax = 15.289
Now, with our value of maximum electromagnetic wave gotten, we divide it by speed of light to get our final answer
15.289 / (3*10^8) = 5.096*10^-8 T
Suffice to say, The maximum value of the magnetic field in the wave is closest to 5.096*10^-8
why did the following change occur? the snow that covered the ground all winter began to melt, temperatures rose, and flowers started to bloom
A.) Days became shorter.
B.) The Earth tilted away from the sun
C.) The sun's rats became for concentrated.
D.) Earth completed one revolution
Answer:C
Explanation:
Answer:
B
Explanation:
Because the earth shifts axes points so it would make it cllser hotter etc
5. 3 women push a stalled car. Each woman pushes with a 400N force. What is the mass of the car if the car accelerates at 0.85 m/s??
Answer:
470.59 kgExplanation:
The the mass of the car can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{400}{0.85} \\ = 470.58823...[/tex]
We have the final answer as
470.59 kgHope this helps you
A television camera is positioned 4 km from the base of a rocket launching pad. In order to keep the rocket in focus as it takes off, the camera must be programmed with the distance to the rocket and how that distance is changing. If we assume that the rocket rises vertically at a speed of 200 km/hr, how fast is the distance from the camera to the rocket changing when the rocket has risen 3 km
Answer:
120 km/hr
Explanation:
Let D be the distance between the rocket and the camera as the rocket is moving upwards. Let d be the distance the rocket moves and L be the distance between the camera and the base of the rocket = 4 km.
Now, at any instant, D² = d² + L²
= d² + 4²
= d² + 16 since the three distances form a right-angled triangle with the distance between the rocket and the camera as the rocket is moving upwards as the hypotenuse side.
differentiating the expression to find the rate of change of D with respect to time, dD/dt ,we have
d(D²)/dt = d(d² + 16)/dt
2DdD/dt = 2d[d(d)/dt]
dD/dt = 2d[d(d)/dt] ÷ 2D
Now d(d)/dt = vertical speed of rocket = 200 km/hr
dD/dt = 200d/D [D = √(d² + 16)]
dD/dt = 200d/[√d² + 16]
Now substituting d = 3 km, the distance the rocket has risen into the equation, we have
dD/dt = 200(3)/[√(3² + 16)]
dD/dt = 600/[√(9 + 16)]
dD/dt = 600/√25
dD/dt = 600/5
dD/dt = 120 km/hr
So, the speed at which the distance from the camera to the rocket changing when the rocket has risen 3 km is 120 km/hr.
A jet pilot puts an aircraft with a constant speed into a vertical loop. (a) Which is greater, the normal force exerted on the seat by the pilot at the bottom of the loop or that at the top of the loop
Answer:
A jet pilot puts an aircraft with a constant speed into a vertical loop is explained below in complete details.
Explanation:
Well, the difficulty does not provide the pilot's mass (or weight in regular gravity), but the difficulty can be resolved and declared in courses of m (the pilot's mass).
When the jet is at the foundation of the circuit, a free-body chart displays the centripetal energy working upward approaching the middle of the loop, and the sound force of the chair and the pilot also upward. The pilot's weight (mg) is earthward. From Newton's second law:
?F(c) = ma(c) = n - mg
n = mg + ma(c)
= m[g + a(c)]
Since centripetal acceleration equals v² / r, the equalization enhances:
n = m[g + (v² / r)]
A solid sphere made of plastic density of 1350 kg/m3 has a radius of 35.0 cm. It is suspended by a massless cord. 75% of its volume is in water and 25% of it is in oil. The density of water is 1000 kg/m3 and the density of the oil is 850 kg/m3. What is the total buoyant foce on the sphere
Answer:
B_total = 1694 N , T = - 937.7 N
Explanation:
This is an exercise of the Archimedes principle that establishes that the thrust of a liquid is equal to the weight of the dislodged volume
B = ρ g V_body
For this case we will assume that the weight of the body is in equilibrium with the thrust of the liquids and the tension of the rope
B₁ + B₂ - W + T = 0
suppose liquid 1 is water and liquid 2 is oil
ρ₁ g V₁ + ρ₂ g V₂ = W
for body weight let's use the definition of density
ρ = m / V
m = ρ V
we substitute
T = ρ g V - (ρ₁ g V₁ + ρ₂ g V₂)
In the problem we are told that the volume in the water is 75% and the volume in the oil is 25% of the body's volume.
T = ρ g V - g (ρ₁ 0.75 V + ρ₂ 0.25 V)
let's calculate the volumes
Body
V = 4/3 π r³
liquid 1 (water)
V₁ = 0.75 V = ¾ V
liquid 2 (oil)
V₂ = 0.25 V = ¼ V
we substitute
T = ρ g [tex]\frac{4}{3}[/tex] π r³ - g (ρ₁ [tex]\frac{3}{4}[/tex] [tex]\frac{4}{3}[/tex] π r³ + ρ₂ [tex]\frac{1}{4}[/tex] [tex]\frac{4}{3}[/tex] π r³)
T = [tex]\frac{4}{3}[/tex] ρ g r³ - g π r³ (ρ₁ + [tex]\frac{1}{3}[/tex] ρ₂)
The term of the floating force is
B_total = g π r³ (ρ₁ + [tex]\frac{1}{3}[/tex] ρ₂)
let's calculate its value
B_total = 9.8 π 0.35³ (1000 + ⅓ 850)
B_total = 1694 N
therefore the tension in the string is
T = [tex]\frac{4}{3}[/tex] 1350 9.8 0.35³ - 1694
T = 756.3 - 1694
T = - 937.7 N
The negative sign indicates that the system tends to come out of the liquid, for which a downward force must be applied to keep it in position.
which statement about force is incorrect
Answer:
What are the options?
Explanation:
when is a bandage best suited to be used?
A horizontal force F is used to pull a 5 kg block across a floor at constant speed of 3 m/s. The frictional force between the block and the floor is 10 N. The net work done on the block in 1 minutes is most nearly:
Answer:
5× 3 over 10 = 1.5
Explanation:
F = 1m/s × kg over n
The net work done on the block in 1 minutes is most nearly 1800 Joule.
What is work?
A force must be applied in order for work to be completed, and there must also be motion or displacement in the force's direction. The amount of force multiplied by the distance moved in the force's direction is known as the work done by a force acting on an item.
Work has no direction and only magnitude. Work is a scalar quantity as a result.
Given that: A horizontal force F is used to pull a 5 kg block across a floor at constant speed of 3 m/s.
As the block is moving with uniform velocity; this force is equal to the frictional force in magnitude.
So, The net work done on the block in 1 minutes is =
force× displacement
= force × velocity × time
= 10 × 3 × 60 Joule
= 1800 Joule.
Learn more about work here:
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A force of 22.7 N stretches an elastic band at room temperature. The rate at which its entropy changes as it stretches is about _____ J/Km. Round your answer to 3 decimal places.
Answer:
The value is [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]
Explanation:
From the question we are told that
The force is [tex]F = 22.7 \ N[/tex]
The value of room temperature is [tex]T = 298 \ K[/tex]
Generally the rate at which its entropy changes as it stretches is mathematically represented as
[tex]\frac{\Delta S }{ L} = - \frac{F}{T}[/tex]
=> [tex]\frac{\Delta S }{ L} = - \frac{21.5}{ 298 }[/tex]
=> [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]
You want to move a spacecraft that is in 300 km circular orbit around Venus into another circular orbit around Venus of 600 km. Explain how you would accomplish this. Solve for the velocities that need to be applied.
Answer:
The final velocity will be half of the initial velocity of the spacecraft.
Explanation:
Angular momentum is conserved for the circular force motion and central force motion.
Considering
L = MVR = Constant
Where
M = Mass of the object
V = Velocity of the object
r = radius of circle
We know that
V = [tex]\frac{1}{R}[/tex]
So,
[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{R_{1} }{R_{2} }[/tex]
As per the given data
[tex]R_{1}[/tex] = Initial Radius = 300 km
[tex]R_{2}[/tex] = Final Radius = 600 km
[tex]V_{1}[/tex] = Initial Velocity =
[tex]V_{2}[/tex] = Final Velocity =
Placing values in the formula
[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{300 km}{600 km }[/tex]
[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{1}{2}[/tex]
[tex]{V_{2}[/tex] = [tex]\frac{1}{2} V_{1}[/tex]
Hence, The final velocity will half of the initial velocity of the spacecraft.
PLEASE HELP!!! Which process in living things evolved as oxygen levels increased in Earth’s atmosphere?
glycolysis
photosynthesis
fermentation
aerobic respiration
Answer: Areobic Resporation
Explanation; -Aerobic respiration is the process in living things that evolved as oxygen levels increased in earth's atmosphere.
A 2.00 kg copper pot is placed on a glass shelf and then one side of the shelf falls so that it becomes angled at 35.0° to the horizontal. The coefficient of static friction between copper and glass is 0.680 and the coefficient of kinetic friction is 0.530. Will the pot slide, and if so, what is its acceleration?
Answer:
Explanation: about 2 Mph when it falls and sorry if I’m wrong
______ is the total distance traveled divided by the total time of travel.
(5 Points)
speed
total speed
average speed
displacement
Do clouds have respiration
Answer:
yes it also does have but not in the exact form but it does 30 percent of respiration to produce rain
A ball is thrown horizontally at 30 m/s from a height of 45m. How long is it in the air?
(i) How fast is it moving horizontally when it hits the ground?
Answer:
1.25s
1) 90.9m
Explanation:
From the question, we are given the following;
Speed v = 30m/s
Maximum height H = 45m
Required;
Time
Using the equation of motion S = ut + 1/2gt²
Substitute;
45= 30t + 1/2(9.8)t²
45 = 30t + 4.9t²
4.9t² + 30t - 45 = 0
Factorize;
t = -30 ±√30²-4(-45)(4.9)/2(4.9)
t = -30 ±√900+882/9.8
t = -30 ±42.21/9.8
t = -30 + 42.21/9.8
t = 12.21/9.8
t = 1.25secs
Hence the ball spent 1.25secs in air.
i) To get the horizontal distance, we will use the formula;
R = U√2H/g
R = 30√2(45)/9.8
R = 30√90/9.8
R = 30√9.18
R = 30(3.03)
R = 90.9m
Hence the horizontal distance is 90.9m
The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?
Answer:
The second distance of the sound from the source is 431.78 m..
Explanation:
Given;
first distance of the sound from the source, r₁ = 1.48 m
first sound intensity level, I₁ = 120 dB
second sound intensity level, I₂ = 70.7 dB
second distance of the sound from the source, r₂ = ?
The intensity of sound in W/m² is given as;
[tex]dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2[/tex]
[tex]For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2[/tex]
The second distance, r₂, can be determined from sound intensity formula given as;
[tex]I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m[/tex]
Therefore, the second distance of the sound from the source is 431.78 m.
If vector A= aj^ and vector B = bj^, then vector A×B is equal to
Answer:
0 because j×j =0
Explanation:
so a×b is 0
Two spherical balls are placed so that their centers are 3.61 m apart. The
force between them is 1.65 x 10-7 N. If the mass of the smaller ball is 81 kg,
what is the mass in kilograms of the other ball?
Answer:
The mass of the other ball is 397.775 kg.
Explanation:
Gravitation is the force of mutual attraction that bodies experience due to the fact that they have a certain mass.
The universal law of gravitation is a classical physical law that describes the gravitational interaction between different bodies with mass.
The law was formulated by Newton, who deduced that the force with which two bodies of different masses are attracted only depends on the value of their masses and the square of the distance that separates them.
In other words, the Law of Universal Gravitation predicts that the force exerted between two bodies of masses M1 and M2 separated by a distance "d" is proportional to the product of their masses and inversely proportional to the square of the distance, that is:
[tex]F=G\frac{M1*M2}{d^{2} }[/tex]
where:
F = It is the module of the force exerted between both bodies, and its direction is found on the axis that joins both bodies. G = It is the constant of Universal Gravitation, whose value is 6.67384*10⁻¹¹ [tex]\frac{N*m^{2} }{kg^{2} }[/tex]In this case:
F= 1.65*10⁻⁷ NG= 6.67384*10⁻¹¹ [tex]\frac{N*m^{2} }{kg^{2} }[/tex]M1= 81 kgM2= ?d= 3.61 mReplacing:
[tex]1.65*10^{-7} N=6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }\frac{81 kg*M2}{(3.61 m)^{2} }[/tex]
Solving for M2:
[tex]M2=\frac{1.65*10^{-7} N*(3.61 m)^{2}}{6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }*81 kg}[/tex]
M2= 397.775 kg
The mass of the other ball is 397.775 kg.
a 60N at an angle of 30°from
horizontal
Explanation:
Force, F = 60 N
Angle, [tex]\theta=30^{\circ}[/tex]
We need to find horizontal and vertical component of the force.
Horizontal component,
[tex]F_x=F\cos\theta\\\\=60\times \cos(30)\\\\=51.96\ N[/tex]
Vertical component,
[tex]F_=F\sin\theta\\\\=60\times \sin(30)\\\\=30\ N[/tex]
So, the horizontal and vertical component are 51.96 N and 30 N respectively.
Q. At what point in a waterfall do the drops of water contain the most kinetic energy ?
1.When they are halfway to the bottom
2.When they reach the bottom of the fall
Answer:
2.When they reach the bottom of the fall
Explanation:
The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).
Thus, the correct option is "2" When they reach the bottom of the fall
1 What is the resistance of the inductive coil
takes 5A current across 240V, 50Hz supply
at 0.8 power factor?
A 48
B 4250
C 38.40
D 26.60
Answer:
38.40
Explanation:
This is the answer for this question
Question 1 of 25
Two asteroids with masses 3.71 x 10 kg and 1.88 x 104 kg are separated by
a distance of 1,300 m. What is the gravitational force between the asteroids?
Newton's law of gravitation is F gravity
Gm, 2 The gravitational
constant Gis 6.67 x 10-11 Nm²/kg?
A. 275 x 10"N
B. 4.13 x 10°N
C. 2.04 x 10°N
O D. 3.58 x 10-N
SUBMIT
Answer:
(A)
Explanation:
Answer:
275 x 10"N
Explanation:
A 52.0-kg woman wearing high-heeled shoes is invited into a home in which the kitchen has vinyl floor covering. The heel on each shoe is circular and has a radius of 0.600 cm. If the woman balances on one heel, what pressure does she exert on the floor?
Answer:
4.5 × 10^6 Nm
Explanation:
Given that a 52.0-kg woman wearing high-heeled shoes is invited into a home in which the kitchen has vinyl floor covering. The heel on each shoe is circular and has a radius of 0.600 cm. If the woman balances on one heel, what pressure does she exert on the floor?
Let's first calculate the area covered by the heel shoe by using the area of a circle.
Convert cm to m
0.6 / 100 = 0.006
A = πr^2
A = 22/7 × 0.006^2
A = 1.131 × 10^-4 m^2
The force exerted by the woman = mg
Force exerted = 52 × 9.8
Force exerted = 509.6 N
Pressure = force / Area
Pressure = 509.6 / 1.131 × 10^-4
Pressure = 4505853.3 N/m
Therefore, she exerts pressure of 4.5 × 10^6 Nm approximately on the floor.
The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.
Answer:
The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.
1. What is the total distance traveled?
A 3.0m
B 4.0m
C 5.0m
D 6.0m
Answer:
c
Explanation:
(c) The plates are moved farther apart with each plate maintaining the same net charge. In a coherent paragraph-length response, apply concepts of work and energy to explain how the electric potential difference between the plates changes, if at all, when the plates are moved farther apart.
Answer:
Energy is applied on the charge to do work.
Explanation:
Work is only be done when one charge moves against the electric field of anther charge that require huge amount of energy because both have same charge and the force of repulsion occurs between them. Work is defined as the product of force and displacement so from this equation we can conclude that work is done when a force is applied on an object and it moves in the direction in which force is applied. If the force or energy is removed from the charge which is present in the electric field of another charge so it moves away from that charge and the work is also be done..
Can we use a hydrometer to
measure the density of milk?
Answer:
yes
Explanation:
i hope this helps not sure im right
The intensity of sound is measured on the decibel scale, dB. The equation dB=10 log I represents the decibel level, where I is the ratio of the sound to the human hearing threshold. A noise is 150,000 times greater than the human hearing threshold. Which shows a valid step in the process of finding the decibel level of the noise?
a. 150,000 = 10 log I
b. 15,000 = log I
c. dB = 10 log 150,000
d. 10dB = log 150,000
e. 10/dB= log 150,000
Answer:
The correct option is c
Explanation:
From the question we are told that
The ratio of the noise to human hearing threshold is [tex]I = 150 000[/tex]
Generally from the equation given we have that
dB = 10 log I
So
dB = 10 log 150000
The expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000. Option C is correct
Given the equation for calculating the intensity of sound which is measured in decibel expressed as:
dB=10 log I
where;
I is the ratio of the sound to the human hearing threshold
Given that noise is 150,000 times greater than the human hearing threshold.
Substitute I = 150,000 into the expression above;
dB = 10 log 150,000
Hence the expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000
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A 7750 kg space probe, moving nose-first toward Jupiter at 179 m/s relative to the Sun, fires its rocket engine, ejecting 72.0 kg of exhaust at a speed of 230 m/s relative to the space probe. What is the final velocity of the probe?
Answer:
179.47m/s
Explanation:
Using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity
Substitute
7750(179)+72(230) = (7750+72)v
1,387,250+16560 = 7822v
1,403,810 = 7822v
v = 1,403,810/7822
v= 179.47m/s
Hence the final velocity of the probe is 179.47m/s