Answer:
Explanation:
The earth-atmosphere energy balance is the balance between incoming energy from the Sun and outgoing energy from the Earth. Energy released from the Sun is emitted as shortwave light and ultraviolet energy. When it reaches the Earth, some is reflected back to space by clouds, some is absorbed by the atmosphere, and some is absorbed at the Earth's surface.
Hope this helped you!
100 yards from his house. What kind of antenna should he use on the house and barn to get the best signal
Complete question:
a farmer wants to use a wireless camera he has installed in his barn which is about 100 yards from his house. what kind of antenna should he use on the house and barn to get the best signal.
Answer: Unidirectional antenna
Explanation: Based on the description given in the scenario above, the farmer should use a unidirectional antenna in other to get the best signal taking advantage of the fact that he knows where he wants the Transmission to come from. The good transmission signal attributed to unidirectional antennas stems from the fact that, the area covered is relatively small as signal transmission and radiofrequecy energy is focused in a particular direction, meaning that less area is covered using a unidirectional antenna,hiwevwr, it leverages this limitation to provide very good signal within the limited area covered.
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm. The explorer finds that the pendulum completes 105 full swing cycles in a time of 125 s.
What is the value of the acceleration of gravity on this planet?Express your answer in meters per second per second.
Answer:
The value is [tex]g = 16.104 \ m/s[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 53.0 \ cm = 0.53 \ m[/tex]
The number of cycle is [tex]n = 105[/tex]
The time taken is [tex]t = 125 \ s[/tex]
The period is mathematically represented as
[tex]T = 2 \pi \sqrt{\frac{ l}{g} }[/tex]
Also the period is mathematically represented as
[tex]T = \frac{ t}{n}[/tex]
[tex]T = \frac{125}{110}[/tex]
[tex]T = 1.14 \ s/cycle[/tex]
So
[tex]1.14 = 2 * 3.142 \sqrt{\frac{ 0.53}{g} }[/tex]
=> [tex]g = \frac{ 4 * 3.142^2 * 0.53}{1.14^2}[/tex]
=> [tex]g = 16.104 \ m/s[/tex]
If "38 %" of the light passes through this combination of filters, what is the angle between the transmission axes of the filters
Answer:
52°
Explanation:
The initial intensity [tex]I_{0}[/tex] = [tex]I[/tex]
The final intensity [tex]I_{f}[/tex] = 38% of [tex]I[/tex] = 0.38
From the polarizing equation
[tex]I_{f} = I_{0} cos^{2}[/tex]θ
substituting values, we have
[tex]0.38I = I cos^{2}[/tex]θ
0.38 = [tex]cos^2[/tex]θ
cosθ = [tex]\sqrt{0.38}[/tex]
cosθ = 0.6164
θ = [tex]cos^{-1}[/tex] 0.6164
θ = 51.9 ≅ 52°
What is the electric field at a location vector b = <-0.2, -0.4, 0> m, due to a particle with charge +3 nC located at the origin?
Answer:
E = 134.85 N/C
Explanation:
We are given;
Electric field location; b = <-0.2, -0.4, 0> m
Charge: q = 3 nc = 3 × 10^(-9) C
From the location given, we have total distance;
r = √((-0.2²) + (-0.4)² + (0²))
r = √0.2
Formula for Electric field is;
E = kq/r²
where;
k is a constant with a value of 8.99 x 10^(9) N.m²/C²
q is charge on the proton particle with a costant value = 1.6 × 10^(-19) C
r is the distance
Plugging in the relevant values, we have;
E = (8.99 x 10^(9) × 3 × 10^(-9))/(√0.2)²
E = 134.85 N/C
Find out more information about <-0.2, -0.4, 0> here:
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A worker is pushing a crate of tools up a ramp into a truck. The crate has a mass of 120 kg and is accelerating at a rate of 1.05 m/s^2. Once he finishes pushing that crate up, he then takes another crate that is 3 times as massive and pushes it up the ramp with an acceleration of 0.71 m/s^2. What is the ratio of the force the worker used on the on the 120 kg crate and the more massive crate? a. 0.218 b. 0.493 c. 1.68 d. 1.07
Answer:
b. 0.493
Explanation:
A woman exerts a horizontal force of 5 pounds on a box as she pushes it up a ramp that is 6 feet long and inclined at an angle of 30 degrees above the horizontal. Find the work done on the box.
Answer:
26 lbft
Explanation:
Given that
Force exerted by the woman, F = 5 lb
Length of the ramp, d = 6 ft
angle of inclination, θ = 30° above the horizontal
Work done on the box, W = ?
This is very much a straightforward question..
Work done, W = F * d, where the force takes the factor of the Angie if inclination. So that,
W = Fcosθ * d
On substituting, we have
W = 5 * cos 30 * 6
W = 5 * 0.866 * 6
W = 30 * 0.866
W = 25.98 lbft
Therefore, the work done on the box is 25.98 or approximately 26 lbft
what was the average speed in km/h of a car that travels 490.0 km in 4.2 h?
Answer:
BelowExplanation:
[tex]Distance = 490 \:km\\Time = 4.2 \:hours\\\\Average\:speed = \frac{Distance}{Time} \\\\A.V = \frac{490}{4.2} \\\\A.V = 116.67km/h[/tex]
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A spherically spreading EM wave comes from an 1800-W source. At a distance of 5.0 m, what is the intensity, and what is the rms value of the electric field?
Explanation:
It is given that,
Power of EM waves, P = 1800 W
We need to find the intensity at a distance of 5 m. Also, the rms value of the electric field.
Intensity,
[tex]I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{1800}{4\pi\times (5)^2}\\\\I=5.72\ W/m^2[/tex]
The formula that is used to find the rms value of the electric field is as follows :
[tex]I=\epsilon_o cE^2_{rms}[/tex]
c is speed of light and [tex]\epsilon_o[/tex] is permittivity of free space
So,
[tex]E_{rms}=\sqrt{\dfrac{I}{\epsilon_o c}}\\\\E_{rms}=\sqrt{\dfrac{5.72}{8.85\times 10^{-12}\times 3\times 10^8}}\\\\E_{rms}=46.41\ V/m[/tex]
Hence, this is the required solution.
How large a net force is required to accelerate a 1600-kg SUV from rest to a speed of 25 m/s in a distance of 200 m
Answer:
F=2496 N
Explanation:
Given that,
Mass of SUV, m = 1600 kg
Initial speed, u = 0
Final speed, v = 25 m/s
Distance, d = 200 m
We need to find the net force. Firstly, let's find acceleration using equation of motion.
[tex]v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(25)^2-(0)^2}{2\times 200}\\\\a=1.56\ m/s^2[/tex]
Net force, F = ma
[tex]F=1600\times 1.56\\\\F=2496\ N[/tex]
So, the net force is 2496 N.
The amount of net force that will be required to accelerate a 1600kg SUV from rest to a speed of 25 m/s in a distance of 200m is 2500N
HOW TO CALCULATE NET FORCE:
The net force of a body can be calculated by multiplying its mass by acceleration. However, the acceleration of this SUV needs to be calculated using the following equation of motion:v² - u² = 2asa = v² - u²/2sWhere:
a = acceleration (m/s²)v = final velocity (m/s)u = initial velocity (m/s)s = distance (m)a = 25² - 0²/2(200)a = 625/400a = 1.563m/s²Since a = 1.563m/s²
F = 1600 × 1.563F = 2500NTherefore, the amount of net force that will be required to accelerate a 1600kg SUV from rest to a speed of 25 m/s in a distance of 200m is 2500N.Learn more at: https://brainly.com/question/18031807?referrer=searchResults
All Houston Methodist buildings system wide have an emergency power generator that turns on to supply emergency power after normal power shuts down within:__________A. 5 secondsB. 30 secondsC. 60 secondsD. 10 seconds
Answer:
D. 10 seconds
Explanation:
All Houston Methodist Hospital buildings employs the use of EPSS(Emergency Power Supply System). This system employs the use of an emergency power generator that turns on to supply emergency power after normal power shuts down within 10 seconds during power outage.
Due to it being used in a hospital the importance of this emergency power system cannot be overemphasized as it is used to provide power supply to hospital equipment such as life support machines etc.
A thin slab of Germanium is used as a Hall Effect probe. How would you orient a magnetic field to make the side facing out of the page be at a positive voltage with respect to the opposite side facing into the page
Answer:
the magnetic field must go in a direction parallel to the page perpendicular to the current.
Explanation:
The Hall effect is the voltage produced by the movement of electrons due to the effect of electric and magnetic fields in a material
F = eE + v x B
The electric field goes in the direction of the current that is opposite to the direction of the electrons, therefore the magnetic force must be perpendicular to it.
Therefore, if the current goes in a direction parallel to the page, in the x direction, the magnetic field must be perpendicular to it if we use the rule of the right wizard,
thumb points in the direction of E, x axis parallel page
The fingers extended should go parallel to the page in the direction and up
The palm is the direction of the Force, where the voltage will be produced points out the page, this is for positive charges, as in germanium the charges are negative, the real force goes into the page.
Therefore the electrons accumulate on the inside of the page and the voltage is negative in this part.
Therefore the voltage is positive on the outside of the sheet. In conclusion the magnetic field must go in a direction parallel to the page perpendicular to the current.
g a circular area with a radius of cm lies in the plane. What is the magnitude of 0.250 T in the
Complete Question
A circular area with a radius of 6.60 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field B = 0.250 T oriented in the following ways?
(a) in the +z-direction
Wb
(b) at an angle of 54
Answer:
a
[tex]\phi = 0.00342 \ Wb[/tex]
b
[tex]\phi = 0.00201 \ Wb[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 6.60 \ cm = 0.066 \ m[/tex]
The magnitude of the magnetic field is [tex]B = 0.250 \ T[/tex]
Generally the cross -sectional area is mathematically represented as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (0.066)^2[/tex]
[tex]A = 0.01369 \ m^2[/tex]
Generally when the magnetic field is oriented in the +z-direction the magnetic flux is mathematically represented as
[tex]\phi = B* A cos(0)[/tex]
=> [tex]\phi = 0.01369 * 0.250 * cos (0)[/tex]
=> [tex]\phi = 0.00342 \ Wb[/tex]
Now when the magnetic field is oriented at an angle of 54° the magnetic flux is mathematically represented as
[tex]\phi = B* A cos(54)[/tex]
[tex]\phi = 0.01369 * 0.250 * cos (54)[/tex]
[tex]\phi = 0.00201 \ Wb[/tex]
A certain automobile is 6.0 m long if at rest. If it is measured to be 4.8 m long while moving, its speed is:
Answer:
1.8*10^8m/s
Explanation:
Using
L= lo√1-v²/c²
So making v subject we have
V= c√1-4.8²/6²
V= 0.6*c
V= 0.6*3E8m/s
V= 1.8*10^8m/s
The path of an object projected at a 45 degree angle with initial velocity of 80 feet per second is given by the −32 2 function h(x) = (80)2 x + x where x is the horizontal distance traveled and h(x) is the height in feet. Use the TRACE feature of your calculator to determine the height of the object when it has traveled 100 feet away horizontally.
Answer:
50 feetExplanation:
Given the the path of an object at a 45° with initial velocity of 80 feet per second modeled by the equation [tex]h(x) = (-\dfrac{32}{80^2} )x^2 + x[/tex] where;
x is the horizontal distance traveled and h(x) is the height in feet, if x is given as 100 ft, the height of the object will be gotten by simply substituting x = 100 into the modeled function as shown;
[tex]h(x) = (-\dfrac{32}{80^2} )x^2 + x\\\\h(100) = (-\dfrac{32}{80^2} )100^2 + 100\\\\h(100) = (-0.005 )100^2 + 100\\\\h(100) = (-0.005 )10,000 + 100\\\\h(100) = -50 + 100\\\\h(100) =50 feet[/tex]
Hence height of the object when it has traveled 100 feet away horizontally is 50 feet.
Refer to a long, straight wire carrying constant current I. What can be concluded about the magnitude of the magnetic field at distance a from the wire?
Answer:
"the magnitude of the magnetic field at a point of distance a around a wire, carrying a constant current I, is inversely proportional to the distance a of the wire from that point"
Explanation:
The magnitude of the magnetic field from a long straight wire (A approximately a finite length of wire at least for close points around the wire.) decreases with distance from the wire. It does not follow the inverse square rule as is the electric field from a point charge. We can then say that "the magnitude of the magnetic field at a point of distance a around a wire, carrying a constant current I, is inversely proportional to the distance a of the wire from that point"
From the Biot-Savart rule,
B = μI/2πR
where B is the magnitude of the magnetic field
I is the current through the wire
μ is the permeability of free space or vacuum
R is the distance between the point and the wire, in this case is = a
define nuclear forces
The nuclear force is a force that acts between the protons and neutrons of atoms.
The nuclear force is a force that binds the protons and neutrons in a nucleus together.This force can exist between protons and protons,neutrons and protons and neutrons and neutrons.This force is what hold the nucleus together.
What is the threshold velocity vthreshold(water) (i.e., the minimum velocity) for creating Cherenkov light from a charged particle as it travels through water (which has an index of refraction of n
Answer:
2.3*10^8m/s
Explanation:
Using
V = c/n
Where c= speed of light
n = refractive index of water
By substituting
We have
V= 3*10^8m/s/1.33
= 2.3*10^8m/s
What is the average value of the magnitude of the Poynting vector S at 1 meter from a 100-watt lightbulb radiating in all directions
Answer:
The value is [tex]S = 7.96[/tex]
Explanation:
From the question we are told that
The power is [tex]P = 100 \ W[/tex]
The radius is [tex]r = 1 \ m[/tex]
Generally the average value of the magnitude of the Poynting vector is mathematically represented
[tex]S = \frac{P}{4 \pi r^2}[/tex]
=> [tex]S = \frac{ 100 }{ 4 *3.142 *1^2 }[/tex]
=> [tex]S = 7.96[/tex]
If the first-order maximum for monochromatic light falling on a double slit is at an angle of 10.0∘, at what angle is the second-order maximum?
Answer:
The value is [tex]\theta_2 = 20.322^o[/tex]
Explanation:
From the question we are told that
The angle of the first order maximum is [tex]\theta _1 = 10.0^o[/tex]
Generally the condition for constructive interference is
[tex]dsin\theta = n \lambda[/tex]
Here d is the separation between the slit ,
n is the order of maxima with values n = 1, 2 , 3 ... for first , second , third ... order of maxima
Now for first order of maximum
[tex]dsin\theta_1 = \lambda \ \ ... \ \ ( 1)[/tex]
=> [tex]dsin(10) = \lambda \ \ ... \ \ ( 1)[/tex]
Now for second order of maximum
[tex]dsin\theta = 2\lambda \ \ ... \ \ ( 2)[/tex]
dividing equation 1 by 2
[tex]\frac{d sin (10)}{d sin (\theta_2 )} = \frac{\lambda}{2\lambda}[/tex]
[tex]\frac{ sin (10)}{ sin (\theta_2 )} = \frac{1}{2}[/tex]
=> [tex]2sin(10) = sin (\theta_2 )[/tex]
=> [tex]0.3473 = sin(\theta_2)[/tex]
=> [tex]\theta_2 = sin^{-1} [0.3473][/tex]
=> [tex]\theta_2 = 20.322^o[/tex]
What is the potential energy of a 2 kg mass at a height of 40 meters?
Answer:
The potential energy of a 2 kg mass at a height of 40 meters is 784 J
Explanation:
Potential energy is that energy that a body possesses due to the height at which it is located and whose unit of measurement of the International System of Units is the joule (J).
The potential energy of a body is the result of multiplying its mass by its height and by gravity:
Ep=m*g*h
Potential energy Ep, is measured in joules (J), mass m is measured in kilograms (kg), gravity, g, in meters / second-squared ([tex]\frac{m}{s^{2} }[/tex]), and height, h , in meters (m).
In this case:
Ep=?m= 2 kgg= 9.8 [tex]\frac{m}{s^{2} }[/tex]h= 40 mReplacing:
Ep= 2 kg* 9.8 [tex]\frac{m}{s^{2} }[/tex] * 40 m
Solving:
Ep= 784 J
The potential energy of a 2 kg mass at a height of 40 meters is 784 J
Ondrea could drive a Jetson's flying car at a constant speed of 540.0 km/hr across oceans and space, approximately how long would she take to drive from the Sun to Pluto in years? (Assume Pluto is at its average distance of 5.9 × 109 km from the Sun)
Answer:
The time taken in years is [tex]x = 125 \ years[/tex]
Explanation:
From the question we are told that
The speed is [tex]v = 540.00 \ km /hr = \frac{540 *1000}{3600} = 150\ m/s[/tex]
The distance from the sun to Pluto is [tex]d = 5.9*10^{9} \ k m = 5.9*10^9 * 1000 = 5.9*10^{12} \ m[/tex]
Generally the time taken is mathematically represented as
[tex]t = \frac{d}{v}[/tex]
=> [tex]t = \frac{5.9*10^{11}}{150}[/tex]
=> [tex]t = 3.933*10^{9}[/tex]
Converting to years
[tex]1 year \to 3.154*10^7 \ s[/tex]
[tex]x \ years \to 3.933*10^{9}[/tex]
=> [tex]x = \frac{ 3.933*10^{9} * 1 }{ 3.154 *10^7}[/tex]
=> [tex]x = 125 \ years[/tex]
Help!!! PLEASEEEEEEE
Answer:
I think the answer is D.
1. Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fish really is
Answer: I believe farther. Correct me if I am wrong
Explanation:
Answer:
Closer.
Explanation:
Due to the refraction of light passing through water and air.
A laser technician measures the wavelength of light from a new semiconductor laser. The wavelength is . What is the wavelength in nanometers
Answer:
The wavelength in nanometers is 849.81 nanometers (nm)
Explanation:
Here is the complete question:
A laser technician measures the wavelength of light from a new semiconductor laser. The wavelength is 8.4981 × 10⁻⁷m . What is the wavelength in nanometers. Write your answer as a decimal.
Explanation:
The given wavelength is 8.4981 × 10⁻⁷m.
To determine the wavelength in nanometers, we will convert the given wavelength from meters (m) to nanometers (nm).
1 nanometer = 1 × 10⁻⁹ meters
∴ 10⁻⁹ meters = 1 nanometer
To convert 8.4981 × 10⁻⁷m to nanometer (nm):
If 10⁻⁹ meters (m) = 1 nanometer (nm)
Then, 8.4981 × 10⁻⁷m = (8.4981 × 10⁻⁷m × 1 nm) / 10⁻⁹ m
8.4981 × 10⁻⁷m = 8.4981 × 10⁻⁷ × 10⁹ nm
= 8.4981 × 10² nm
= 849.81 nm
Hence, the wavelength in nanometers is 849.81 nm
Objects fall near the surface of the earth with a constant downward acceleration of 10 m/s2 . Suppose a falling object is moving downward at 10 m/s at a certain instant. How fast is it falling 2 sec later
Answer:
The final velocity of the object after 2 seconds is 30 m/s
Explanation:
Given;
constant downward acceleration, a = 10 m/s²
initial velocity of the object falling down, v = 10 m/s
time of fall, t = 2 s
The final velocity of the object is given by;
v = u + at
where;
v is the final velocity
v = 10 + (10)(2)
v = 10 + 20
v = 30 m/s
Therefore, the final velocity of the object after 2 seconds is 30 m/s
If an object is rolling without slipping, how does its linear speed compare to its rotational speed?
Answer:
v = rw
Explanation:
When an object is rolling continuously without slipping, then every angle it rotates through, is equal to a distance the perimeter has rotated.
If the object completes 10 revolutions and takes a particular time, let's say t to complete it. The angular distance would then be 20 π rad, while its angular velocity will be 20 π/t
The circumference will somehow translate to the distance it covers, which is 20πr, this means that the speed is 20πr/t
So, like the question asked, the linear speed compared to angular speed is
v : w
20πr/t : 20πt, which can be simplified to
r : 1
In essence, v = rw
In an 8.00km race, one runner runs at a steady 11.8 km/hr and another runs at 15.0 km/hr. How far from the finish line is the slower runner when the faster runner finishes the race?
Answer:
The slower runner is 1.71 km from the finish line when the fastest runner finishes the race.
Explanation:
Given;
the speed of the slower runner, u₁ = 11.8 km/hr
the speed of the fastest runner, u₂ = 15 km/hr
distance, d = 8 km
The time when the fastest runner finishes the race is given by;
[tex]Time = \frac{Distance }{speed}\\\\Time = \frac{8}{15} \\\\Time = 0.533 \ hr[/tex]
The distance covered by the slower runner at this time is given by;
d₁ = u₁ x 0.533 hr
d₁ = 11.8 km/hr x 0.533 hr
d₁ = 6.29 km
Additional distance (x) the slower runner need to finish is given by;
6.29 km + x = 8km
x = 8 k m - 6.29 km
x = 1.71 km
Therefore, the slower runner is 1.71 km from the finish line when the fastest runner finishes the race.
if you walk at 0.7 m/s, how long would it take to walk a mile which is 1609 m
Answer:
[tex]\Huge \boxed{\mathrm{2298.57 \ seconds}}[/tex]
[tex]\rule[225]{225}{2}[/tex]
Explanation:
[tex]\displaystyle \sf Speed = \frac{Distance \ covered }{Time \ taken}[/tex]
[tex]\displaystyle \sf s = \frac{d }{t}[/tex]
The speed is 0.7 m/s.
The distance covered is 1609 m.
[tex]\displaystyle \sf 0.7 = \frac{1609 }{t}[/tex]
Multiplying both sides by t.
Then dividing both sides by 0.7.
[tex]\displaystyle \sf t = \frac{1609 }{0.7}[/tex]
[tex]\sf t= 2298.57[/tex]
It would take 2298.57 seconds.
[tex]\rule[225]{225}{2}[/tex]
Answer:
[tex]\huge\boxed{\sf t = 2298.57\ secs}[/tex]
Explanation:
Given:
Speed = v = 0.7 m/s
Distance = S = 1 mile = 1609 metre
Required:
Time = t = ?
Formula :
v = S/t
Solution:
v = S/t
t = S/v
t = 1609 / 0.7
t = 2298.57 secs
The level of toluene (a flammable hydrocarbon) in a storage tank may fluctuate between 10 and 400 cm from the top of the tank. Since it is impossible to see inside the tank, an open-end manometer with water or mercury as the manometer fluid is to be used to determine the toluene level. One leg of the manometer is attached to the tank 500 cm from the top. A nitrogen blanket at atmospheric pressure is maintained over the tank contents. To atmosphere 500 e Toluene Manoseer luld (H,0 or Hg)
a) When the toluene level in the tank is 150 cm below the top (h=150cm),the manometer fluid level in the open arm is at the height of the point where the manometer connects to the tank. What manometer reading, R (cm), would be observed if the manometer fluid is (i) mercury, (ii) water? Which manometer fluid would you use, and why?
b) Briefly describe how the system would work if the manometer were simply filled with toluene. Give several advantages of using the fluid you chose in Part (a) over using toluene.
(c) What is the purpose of the nitrogen blanket?
Answer:
A. 23.9
B.22.9
C. The levels will be equal
D. Obviously that will be to maintain atmospheric pressure
Explanation:
For mercury the pressure in both tubes at R is same so
P_left = P_right
Thus
=>>>Po + rho_t x g x (5 + R - 1.5) = Po + rho_ m x g x R
rho_t x g x (5 + R - 1.5) = rho_m x g xR
rho_t x (3.5 + R) = rho_m x R
3.5 + R = (rho_m/rho_t) x R
3.5 + R = (13560/867) x R
3.5 + R = 15.64 x R
R x (15.64 - 1) = 3.5
R = 3.5/14.64
= 0.239 m
= 23.9 cm this is for Mercury
ii)water
similarly,
3.5 + R = (rho_w/rho_t) x R
3.5 + R = (1000/867) x R
3.5 + R = 1.153 x R
R X (1.153 - 1) = 3.5
R = 3.5/0.153
= 22.9m for water