Answer:
the rms value of the electric field component transmitted is 3.295 V/m
Explanation:
Given;
intensity of the unpolarized light, I = 0.0288 W/m²
For unpolarized light, the relationship between the amplitude electric field and intensity is given as;
[tex]E_{max} = \sqrt{2\mu_0cI} \\\\E_{max} = \sqrt{2(4\pi \times 10^{-7})(3\times 10^8)(0.0288)} \\\\E_{max} = 4.66 \ V/m[/tex]
The relationship between the rms value of the electric field and the amplitude electric field is given as;
[tex]E_{rms} = \frac{E_0}{\sqrt{2} } =\frac{E_{max}}{\sqrt{2} } \\\\E_{rms} = \frac{4.66}{\sqrt{2} }\\\\E_{rms} = 3.295 \ V/m[/tex]
Therefore, the rms value of the electric field component transmitted is 3.295 V/m
A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level. Assuming the pressure in both locations are the same and the density of water is 1000 kg/m3. How fast will the water flow into the plant?
Answer:
v₂ = 70 m / s
Explanation:
For this exercise let's use Bernoulli's equation
where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂
indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute
ρ g y₁ = ½ ρ v₂²
v₂ = [tex]\sqrt {2g \ y_1}[/tex]
let's calculate
v₂ = √( 2 9.8 250)
v₂ = 70 m / s
A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the following distances from the axis of the rod, where distances are measured perpendicular to the rod's axis.
Answer:
Explanation:
From the question;
We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.
We are to calculate the following task, i.e. to determine the electric field at the distances:
a) at 4.75 cm
b) at 20.5 cm
c) at 125.0 cm
Given that:
the charge (q) = 33.3 nC/m
= 33.3 × 10⁻⁹ c/m
radius of rod = 5.75 cm
a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.
Then, the electric field will be zero.
b) The electric field formula [tex]E = \dfrac{kq }{d}[/tex]
[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}[/tex]
E = 1461.95 N/C
c) The electric field E is calculated as:
[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}[/tex]
E = 239.76 N/C
A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that
the acceleration due to gravity is 9.8 m/s2, what is the crate's potential energy
at this point?
Answer:
[tex]\boxed {\boxed {\sf 29,400 \ Joules}}[/tex]
Explanation:
Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.
[tex]E_P= m \times g \times h[/tex]
The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.
m= 150 kg g= 9.8 m/s²h= 20 mSubstitute the values into the formula.
[tex]E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m[/tex]
Multiply the three numbers and their units together.
[tex]E_p=1470 \ kg*m/s^2 \times 20 m[/tex]
[tex]E_p=29400 \ kg*m^2/s^2[/tex]
Convert the units.
1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.
[tex]E_p= 29,400 \ J[/tex]
The crate has 29,400 Joules of potential energy.
Answer:
29,400 J
Explanation:
did the quiz <3
A vertical wall (8.7 m x 3.2 m) in a house faces due east. A uniform electric field has a magnitude of 210 N/C. This field is parallel to the ground and points 42o north of east. What is the electric flux through the wall
Answer:
[tex]\phi=4344.72Nm^2/c[/tex]
Explanation:
From the question we are told that:
Dimension of Wall:
[tex](L*B)=(8.7 m * 3.2 m)[/tex]
Electric field [tex]B=210 N/C[/tex]
Angle [tex]\theta =42 \textdegree North[/tex]
Generally the equation for electric Flux is mathematically given by
[tex]\phi=EAcos\theta[/tex]
[tex]\phi=210*(8.7*3.2)*cos 42[/tex]
[tex]\phi=4344.72Nm^2/c[/tex]
A flat (unbanked) curve on a highway has a radius of 260 mm . A car successfully rounds the curve at a speed of 32 m/sm/s but is on the verge of skidding out.
Required:
a. If the coefficient of static friction between the car’s tires and the road surface were reduced by a factor of 2, with what maximum speed could the car round the curve?
b. Suppose the coefficient of friction were increased by a factor of 2; what would be the maximum speed?
I suppose you meant to say the radius of the curve is 260 m, not mm?
There are 3 forces acting on the car as it makes the turn,
• its weight mg pulling it downward;
• the normal force exerted by the road pointing upward, also with magnitude mg since the car is in equilibrium in the vertical direction; and
• static friction keeping the car from skidding with magnitude µmg (since it's proportional to the normal force), pointing horizontally toward the center of the curve.
By Newton's second law, the net force on the car acting in the horizontal direction is
F = ma => µmg = ma => a = µg
where a is the car's radial acceleration given by
a = v ^2 / R
with v = the car's tangential speed and R = radius of the curve. At the start, the car's radial acceleration is
a = (32 m/s)^2 / (260 m) ≈ 3.94 m/s^2
(a) If µ were reduced by a factor of 2, then the radial acceleration would also be halved:
1/2 a = 1/2 µg
Then the car can have a maximum speed v of
1/2 a = v ^2 / R => v = √(aR/2) = √((3.94 m/s^2) (260 m) / 2) ≈ 22.6 m/s
(b) If µ were increased by a factor of 2, then the acceleration would also get doubled. Then the maximum speed v would be
2a = v ^2 / R => v = √(2aR) = √(2 (3.94 m/s^2) (260 m)) ≈ 45.3 m/s
The food calorie, equal to 4186 J, is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit-and-cereal bar contains 160 food calories per bar.
Part A
If a 67.0 kg hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy?
Express your answer in meters.
Part B
If, as is typical, only 20.0 % of the food calories go into mechanical energy, what would be the answer to Part A? (Note: In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is actually not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.)
Express your answer in meters.
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?
A) 1.67 x 10^-4 s^-1
B) 5.43 x 10^-4 s^-1
C) 1.40 x 10^-4 s^-1
D) 2.22 x 10^-4 s^-1
OPTION C is the correct answer.
When a mass of 3.0-kg is hung on a vertical spring, it stretches by 0.085 m. Determine
the period of oscillation of a 4.0-kg object suspended from this spring.
Answer:
the period of oscillation of the given object is 0.14 s
Explanation:
Given;
mass of the object, m = 3 kg
extension of the spring, x = 0.085 m
The spring constant is calculated as follows;
[tex]F = mg = \frac{1}{2} ke^2\\\\2mg = ke^2\\\\k = \frac{2mg}{e^2} \\\\k = \frac{2\times 3 \times 9.8}{(0.085)^2} \\\\k = 8,138.41 \ N/m[/tex]
The angular speed of a 4 kg object is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\frac{2\pi }{T} = \sqrt{\frac{k}{m} } \\\\T= 2\pi \sqrt{\frac{m}{k} } \\\\T = 2\pi \sqrt{\frac{4}{8138.41} }\\\\T = 0.14 \ s[/tex]
Therefore, the period of oscillation of the given object is 0.14 s
You are using a constant force to speed up a toy car from an initial speed of 6.5 m/s
to a final speed of 22.9 m/s. If the toy car has a mass of 340 g, what is the work
needed to speed this car up?
By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:
W = ∆K
W = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²
W ≈ 82 J
A bug crawls 3.0 mm east, 4.0mm north, and then 5.0 mm at 45 north of east. Draw a diagram showing its displacements and determine its resultant displacement vector by use of the diagram.
Answer:
Explanation is given
Mark Me as Branliest
True or false quarterbacks should not expect to have bad passes
Answer:
false
Explanation:
As part of a safety investigation, two 1300 kg cars traveling at 17 m/s are crashed into different barriers. Find the average forces exerted on:
a. the car that hits a line of water barrels and takes 1.5 s to stop
b. the car that hits a concrete barrier and takes 0.10 s to stop.
Answer:
a. F = 14,733.33 N
b. F = 221,000 N
Explanation:
Given;
mass of the cars, m = 1300 kg
velocity of the cars, v = 17 m/s
time taken for the first car to stop after hitting a barrier, t = 1.5 s
time taken for the second car to stop after hitting a barrier, t = 0.1 s
The average forces exerted on each car is calculated as follows;
a. the car that hits a line of water barrels and takes 1.5 s to stop
[tex]F = ma = m\times \frac{v}{t} = 1300 \times \frac{17}{1.5} = 14,733.33 \ N\\\\F = 14,733.33 \ N[/tex]
b. the car that hits a concrete barrier and takes 0.10 s to stop
[tex]F = ma = m\times \frac{v}{t}= 1300 \times \frac{17}{0.1} = 221,000 \ N\\\\F = 221,000 \ N[/tex]
Calculate the current flowing when the voltage across is 35V and the resistance is 7ohms.
Explanation:
V= IR
35=I×7
I=35/7
I=5amperes
pls give brainliest
20 points and brainliest‼️‼️‼️‼️
A 4.88 x 10-6 C charge moves 265 m/s
parallel (at 0°) to a magnetic field of
0.0579 T. What is the magnetic force
on the charge?
Answer:
0 N
Explanation:
Applying,
F = qvBsin∅................. Equation 1
Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.
From the question,
Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°
Substitute these values into equation 1
F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)
Since sin0° = 0,
Therefore,
F = 0 N
50 POINTS/BRAINLIEST TO CORRECT!
A 1.25 x 10-4 C charge is moving
5200 m/s at 37.0° to a magnetic field
of 8.49 x 10-4 T. What is the magnetic
force on the charge?
Answer: 3.32x10^-4
Explanation: Works for Acellus
Magnitude of magnetic force F= qvB Sin0®
q is the magnitude of charge moving with speed v in magnetic field B. Theta is the angle between velocity and magnetic field.
F=1.25×10⁻⁴C×5200m/s×8.49×10⁻⁴T(sin37deg).
F=3.32×10⁻⁴N.
What is charge?An electric charge is the property of matter where it has more or fewer electrons than protons in its atoms. Electrons carry a negative charge and protons carry a positive charge.
Matter is positively charged if it contains more protons than electrons, and negatively charged if it contains more electrons than protons.
Thus, the magnetic force on the charge is 3.32×10⁻⁴N.
Learn more about charge here,
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does net force stay the same when a massless pulley is replaced by a pulley with mass
Hold a small piece of paper (e.g., an index card) flat in front of you. The paper can be thought of as a part of a larger plane surface. What single line could you use to specify the orientation of the plane of the paper (i.e., so that someone else could hold the paper in the same, or in a parallel, plane)?
Answer:
NORMAL
Explanation:
In mathematics to define a plane you need a line perpendicular to the plane called NORMAL and the point of application of this line at a point on the plane.
Based on this definition you only need to specify the normal plane is perpendicular to this line, it should be noted that this does not define a single plane but the whole family of plane is contains the normal.
Consequently only the NORMAL is needed
The magnetic field at the center of a 1.0-cm-diameter loop is 2.5 mT.
a. What is the current in the loop?
b. A long straight wire carries the same current from part a. At what distance from the wire is the magnetic field 2.5 mT?
Answer:
(a) The current in the wire is 19.89 A
(b) The distance from the wire is 0.159 cm
Explanation:
Given;
magnetic field, B = 2.5 mT
diameter of the wire, d = 1 cm
radius of the wire, r = 0.5 cm = 0.005 m
(a) The current in the wire is calculated as;
[tex]I = \frac{2Br}{\mu_0} \\\\I = \frac{2\times 2.5 \times 10^{-3} \times 0.005 }{4\pi \times 10^{-7} } \\\\I = 19.89 \ A[/tex]
(b) The distance from the wire where the magnetic field is 2.5 mT is calculated as;
[tex]B = \frac{\mu_0 I}{2\pi d} \\\\where;\\\\d \ is \ the \ distance \ from \ the \ wire\\\\d = \frac{\mu_0 I}{2\pi B} = \frac{4 \pi \times 10^{-7} \times 19.89}{2\pi \times 2.5 \times 10^{-3}} = 0.00159 \ m = 0.159 \ cm[/tex]
how do you convert micrometer to killometer
Answer:
1 x 10^-9 kilometers
Hope this helps
Have a good day :)
Explanation:
During hockey practice, two pucks are sliding across the ice in the same direction. At one instant, a 0.18-kg puck is moving at 16 m/s while the other puck has a mass of 0.14 kg and a speed of 3.8 m/s. What is the velocity of the center of mass of the two pucks?
Answer:
10.66 m/s
Explanation:
Applying,
The law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m').............. Equation 1
Where m = mass of the first puck, u = initial velocity of the first puck, m' = mass of the second puck, u' = initial velocity of the second puck, V = Velocity of the center of the two pucks
make V the subject of the equation
V = (mu+m'u')/(m+m').............. Equation 2
Given: m = 0.18 kg, u = 16 m/s, m' = 0.14 kg, u' = 3.8 m/s
Substitute these values into equation 2
V = [(0.18×16)+(0.14×3.8)]/(0.18+0.14)
V = (2.88+0.532)/(0.32)
V = 3.412/0.32
V = 10.66 m/s
A 12 kg hanging sculpture is suspended by a 80-cm-long, 6.0 g steel wire. When the wind blows hard, the wire hums at its fundamental frequency. What is the frequency of the hum
Answer:
[tex]F=78.3hz[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=12[/tex]
Length [tex]l=80cm=0.8m[/tex]
Linear density [tex]\mu= 6.0g[/tex]
Generally the equation for Frequency is mathematically given by
[tex]F=\frac{1}{2l}\sqrt{\frac{T}{K}}[/tex]
[tex]F=\frac{1}{2(0.8)}\sqrt{\frac{12*9.8*0.8}{6*10^{-3}}}[/tex]
[tex]F=78.3hz[/tex]
PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
A) 8.18 x 10^-14 J
B) 2.73 x 10^-22 J
C) 1.5053 x 10^-10 J
D) 1.5032 x 10^-10 J
Answer:
D) 1.5032 x 10^-10 J
Explanation:
The rest energy of a proton, E₀, follows the equation:
E₀ = mp*rate²
Where mass of proton, mp = 1.6726 x 10^-27kg
rate = 2.9979 x 10^8 m/s (2.9979 x 10^9 m/s is not the speed light)
E₀ = 1.6726 x 10^-27kg * (2.9979 x 10^8 m/s)²
E₀ =1.5032 x 10^-10 J
Right answer is:
D) 1.5032 x 10^-10 JCalculate the heat energy conducted per hour through the side walls of a cylindrical steel
boiler of 1.00 m diameter and 3.0 m long if the internal and external temperatures of the
walls are 140 °C and 40 °C respectively and the thickness of the walls is 6.0 mm. (Thermal
conductivity of steel, k = 42 Wm-4°C-4)
Explanation:
heat caoacity and heat is difference
The heat energy conducted per hour through the side walls of the cylindrical steel boiler is 27708847 kJ.
What is thermal conductivity?The rate at which heat is transported by conduction through a material's unit cross-section area when a temperature gradient exits perpendicular to the area is known as thermal conductivity.
In the International System of Units (SI), thermal conductivity is measured by Wm⁻¹K⁻¹.
Diameter of the cylindrical steel boiler: d = 1.00m.
Length of the cylindrical steel boiler: l = 3.00m.
thickness of the walls is = 6.0 mm = 0.006 m
Temperature gradient is = (140-40) °C/0.006 m = 1666.67 °C/m
Thermal conductivity of steel, = 42 W/m-°C.
Hence, the heat energy conducted per hour through the side walls of the cylindrical steel boiler = 42×3600×1666.67 ×2π×0.5(0.5+3.0) Joule
= 27708847 kJ
Learn more about thermal conductivity here:
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1. Calculate the density of a 5.00 g marble which, when placed in a
graduated cylinder containing 60 milliliters of water, raised the volume of
the water to 68 milliliters.
2. An elevator with a mass of 1000kg is lifted 20 meters. How much work
was done on the elevator?
3. Calculate the potential energy of a 12 kg cat at the top of a 42 meter-
high hill.
4. Calculate the kinetic energy in joules of a 1500 kg car that is moving at a
speed of 10 m/s.
Answer:
1. 0.625 g/mL
2. 2×10⁵ J
3. 5040 J
4. 75000 J
Explanation:
1. Determination of the density.
We'll begin by calculating the volume of the marble. This can be obtained as follow:
Volume of water = 60 mL
Volume of water + Marble = 68 mL
Volume of marble =?
Volume of marble = (Volume of water + Marble) – (Volume of water)
Volume of marble = 68 – 60
Volume of marble = 8 mL
Finally, we shall determine the density of the marble. This can be obtained as follow:
Mass of marble = 5 g
Volume of marble = 8 mL
Density of marble =?
Density = mass / volume
Density of marble = 5 / 8
Density of marble = 0.625 g/mL
2. Determination of the work done.
Mass (m) = 1000 Kg
Height (h) = 20 m
Acceleration due to gravity (g) = 10 m/s²
Workdone (Wd) =?
The work done can be obtained as follow:
Wd = mgh
Wd = 1000 × 10 × 20
Wd= 2×10⁵ J
3. Determination of the potential energy.
Mass (m) = 12 Kg
Height (h) = 42 m
Acceleration due to gravity (g) = 10 m/s²
Potential energy (PE) =?
The potential energy can be obtained as follow:
PE = mgh
PE = 12 × 10 × 42
PE = 5040 J
4. Determination of the kinetic energy.
Mass (m) = 1500 Kg
Velocity (v) = 10 m/s
Kinetic energy (KE) =?
The kinetic energy can be obtained as follow:
KE = ½mv²
KE = ½ × 1500 × 10²
KE = 750 × 100
KE = 75000 J
A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if the length of the runway is 2.00 km.ii) At this acceleration, how much time would the plane need from starting to takeoff. iii) What force must the engines exert to attain this acceleration
Answer:
i) the minimum acceleration to take off is 22500 km/h²
ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs
iii) required force that the engine must exert to attain acceleration is 625 kN
Explanation:
Given the data in the question;
mass of plane m = 360,000 kg
take of speed v = 300 km/hr = 83.33 m/s
i)
What should be the minimum acceleration to take off if the length of the runway is 2.00 km
from Newton's equation of motion;
v² = u² + 2as
we know that a plane starts from rest, so; u = 0
given that distance S = 2 km
we substitute
(300)² = 0² + ( 2 × a × 2 )
90000 = 4 × a
a = 90000 / 4
a = 22500 km/h²
Therefore, the minimum acceleration to take off is 22500 km/h²
ii) At this acceleration, how much time would the plane need from starting to takeoff.
from Newton's equation of motion;
v = u + at
we substitute
300 = 0 + 22500 × t
t = 300 / 22500
t = 0.0133 hrs
Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs
iii) What force must the engines exert to attain this acceleration
we know that;
F = ma
acceleration a = 22500 km/hr² = 1.736 m/s²
so we substitute
F = 360,000 kg × 1.736 m/s²
F = 624960 N
F = 625 kN
Therefore, required force that the engine must exert to attain acceleration is 625 kN
At 20 ◦C a copper wire has a resistance of 4×10−3 Ω and a temperature coefficient of resistivity of 3.9×10−3 (C◦)−1, its resistance at 100 ◦C is
A.
52.5 × 10-3 Ω
B.
5.25 × 10-3 Ω
C.
5.25 × 10-4 Ω
D.
5.25 × 10-2 Ω
E.
25.5 × 10-3 Ω
Answer:
25.5×10_3 1928 82i93874 89_/ 9299
a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficient of static friction between the car and the road is 0.78. What is the magnitude of the force of static friction acting on the car
Answer:
The magnitude of the friction force is 8197.60 N
Explanation:
Using the definition of the centripetal force we have:
[tex]\Sigma F=ma_{c}=m\frac{v^{2}}{R}[/tex]
Where:
m is the mass of the carv is the speed R is the radius of the curvatureNow, the force acting in the motion is just the friction force, so we have:
[tex]F_{f}=m\frac{v^{2}}{R}[/tex]
[tex]F_{f}=2100\frac{18^{2}}{83}[/tex]
[tex]F_{f}=8197.60 \: N[/tex]
Therefore the magnitude of the friction force is 8197.60 N
I hope it helps you!
FROM THE _____ WHOLE WATER CYCLE STARTS ALL OVER AGAIN
From the water whole water cycle starts again.
Most possibly water should be the answer.how many continents do have in africa
Answer:
There was 7 continents in africa
1. What different types of shots are taken on the basketball court?
Answer:
Here are a few commonly used types of shooting in basketball.
Jump Shot. A jump shot is most frequently used for a mid to long-range shots, including shooting beyond the arc. ...Hook Shot. A hook shot is when the shot is made while your body is not directly facing the basket. ...Bank Shot. ...Free Throw. ...Layup. ...Slam Dunk.