Answer:
T = 3.14 hours
Explanation:
We need to find the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.
We know that the radius of Mars is 3,389.5 km.
So, r = 1,787 + 3,389.5 = 5176.5 km
Using Kepler's law,
[tex]T^2=\dfrac{4\pi ^2}{GM}r^3[/tex]
M is mass of Mars, [tex]M=6.39\times 10^{23}\ kg[/tex]
So,
[tex]T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 6.39\times 10^{23}}\times (5176.5 \times 10^3)^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times6.39\times10^{23}}\times(5176.5\times10^{3})^{3}}\\\\T=11334.98\ s[/tex]
or
T = 3.14 hours
So, the orbital period is 3.14 hours
A plane monochromatic light wave is incident on a double slit as illustrated in Figure 37.1.
(i) As the viewing screen is moved away from the double slit, what happens to the separation between the interference fringes on the screen?
(a) It increases,
(b) It decreases,
(c) It remains the same,
(d) It may increase or decrease, depending on the wavelength of the light.
(e) More information is required,
(ii) As the slit separation increases, what happens to the separation between the interference fringes on the screen? Select from the same choices.
Explanation:
The separation between the interference fringes on the screen increases because the distance of the screen from the slit is increased. Therefore option (a) is correct.
The separation between the interference fringes on the screen increases because the distance of the screen from the slit is increased, which is contradictory. Therefore option (b) is incorrect.
The separation between the interference fringes on the screen increases because the distance of the screen from the slit is increased, which is contradictory. Therefore option (c) is incorrect.
The separation between the interference fringes on the screen increases because the distance of the screen from the slit is increased, which is contradictory. Therefore option (d) is incorrect.
The separation between the interference fringes on the screen increases because the distance of the screen from the slit is increased, which is contradictory. Therefore option (e) is incorrect.
A plastic rod that has been charged to − 15 nC touches a metal sphere. Afterward, the rod's charge is − 5.0 nC.
1) What kind of charged particle was transferred between the rod and the sphere, and in which direction?
A) electrons transferred from rod to sphere.
B) electrons transferred from sphere to rod.
C) protons transferred from rod to sphere.
D) protons transferred from sphere to rod.
2) How many charged particles were transferred?
Answer:
B) electrons transferred from sphere to rod.
(2) 1.248 x 10¹¹ electrons were transferred
Explanation:
Given;
initial charge on the plastic rod, q₁ = 15nC
final charge on the plastic rod, q₂ = - 5nC
let the charge acquired by the plastic rod = q
q + 15nC = -5nC
q = -5nC - 15nC
q = -20 nC
Thus, the plastic rod acquired excess negative charge from the metal sphere.
Hence, electrons transferred from sphere to rod
B) electrons transferred from sphere to rod.
2) How many charged particles were transferred?
1.602 x 10⁻¹⁹ C = 1 electron
20 x 10⁻⁹ C = ?
= 1.248 x 10¹¹ electrons
Thus,1.248 x 10¹¹ electrons were transferred
1. Electrons transferred from sphere to rod.
Option B is correct.
2. There are [tex]6.24*10^{10}[/tex] electrons transferred from sphere to rod.
Given that initial charge on the plastic rod, q₁ = 15nC
final charge on the plastic rod, q₂ = - 5nC
let us consider that the charge absorbed by the plastic rod is q
[tex]q - 15nC = -5nC\\q = -5nC +15nC\\q = -10 nC[/tex]
Thus, the plastic rod acquired excess negative charge from the metal sphere.
Therefore, electrons transferred from sphere to rod
The charge on one electron is, 1.602 x 10⁻¹⁹ C .
Number of electrons, [tex]n=\frac{10*10^{-9} }{1.602*10^{-19} }= 6.24*10^{10}[/tex]
Thus,[tex]6.24*10^{10}[/tex] electrons were transferred.
Learn more:
https://brainly.com/question/13822373
A group of students conducted several trials of an experiment to study Newton’s second law of motion. They concluded that tripling the mass required tripling the net force applied. What quantity were the students holding constant?
Answer:
1) Mass and acceleration
2) 4.0
3)When the net force applied to an object changes, the acceleration changes by the same factor.
4)acceleration
5)The acceleration is half of its original value
Explanation:
A wise man once said if you cheat together you succeed together! lol i am jk
A dipole is oriented along the x axis. The dipole moment is p (= qs). (Assume the center of the dipole is located at the origin with positive charge to the right and negative charge to the left.)
Calculate exactly the potential V (relative to infinity) at a location x, 0, 0 on the x axis and at a location 0, y, 0 on the y axis, by superposition of the individual 1/r contributions to the potential. (Use the following as necessary: q, ε0, x, s and y.)
Answer:
Explanation:
dipole moment = qs = q x s
= charge x charge separation
charge = q
separation between charge = s
half separation l = s / 2
dipole has two charges + q and - q separated by distance s .
Potential at distance x along x axis due to + q
[tex]v_1=\frac{1}{4\pi \epsilon } \times\frac{q}{x-l}[/tex]
Potential at distance x along x axis due to - q
[tex]v_2=\frac{1}{4\pi \epsilon } \times\frac{-q}{x+l}[/tex]
Total potential
v = v₁ + v₂
[tex]v=\frac{1}{4\pi \epsilon } \times( \frac{q}{x-l}-\frac{q}{x+l})[/tex]
[tex]v=\frac{1}{4\pi \epsilon } \times\frac{2ql}{x^2-l^2}[/tex]
[tex]v=\frac{1}{4\pi \epsilon } \times\frac{qs}{x^2-(\frac{s}{2}) ^2}[/tex]
Potential at distance y along y axis due to + q
[tex]v_1=\frac{1}{4\pi \epsilon } \times\frac{qs}{(y^2+\frac{s^2}{4})^\frac{1}{2} }[/tex]
Potential at distance y along y axis due to - q
[tex]v_1=\frac{1}{4\pi \epsilon } \times\frac{-qs}{(y^2+\frac{s^2}{4})^\frac{1}{2} }[/tex]
Total potential
v = v₁ + v₂
[tex]v= 0[/tex]
A/An ____________________ is a small, flexible tube with a light and lens on the end that is used for examination. Question 96 options:
Answer:
"Endoscope" is the correct answer.
Explanation:
A surgical tool sometimes used visually to view the internal of either a body cavity or maybe even an empty organ like the lung, bladder, as well as stomach. There seems to be a solid or elastic tube filled with optics, a source of fiber-optic light, and sometimes even a sample, epidurals, suction tool, and perhaps other equipment for sample analysis or recovery.Write the equation in slope-intercept form
x-2y=4
Answer:
[tex]y=\frac{1}{2}x-2[/tex]
Explanation:
Slope-intercept form means we want the y to be by itself in the equation. Every thing we do will be about getting the y alone on the left side of the equation
To start we should move x to the left hand side. We can do this by subtracting x from both sides. That way, there is an x on the right, but not the left.
x-x-2y=4-x
this gives us
-2y=4-x
Great! So now what? Well, the y isn't by itself yet because it still is being multiplied by negative two (-2). In order to move it from the left side to the right side, we have to do the opposite of multiply; divide. So, we will divide both sides by -2
[tex]\frac{-2y}{-2} =\frac{4}{-2} -\frac{x}{-2}[/tex]
-2 divided by -2 is 1, 4 divided by -2 is -2, and -x divided by -2 is [tex]\frac{1}{2}x[/tex]
This gives us the answer: [tex]y=\frac{1}{2}x-2[/tex]
Tips:
A negative divided by a negative is a positive ex: -4 divided by -2 is positive 2
If you are subtracting by a negative number, you are actually adding by a positive ex: 2-(-2) is actually 2+2
Don't be afraid to have fractions in your equations
Whatever you do to the one side of the equation, you must do it to the other side as well. Multiply the left side by 2? You HAVE TO multiply the right side by two as well. Add 3 to the right side? You HAVE TO add 3 to the left.
For problems like this (and when you have access to the internet), where you need to rewrite an equation, double check your work on desmos, which is an online graphing calculator. Input both the original equation and the equation you rewrote, and if they don't create the same line, you did something wrong.
A bus carrying 10 people has over turned on a remote hillside during an intense thunderstorm. What three factors could contribute to creating a delay in advanced care
Answer:
The three factors that can contribute to creating a delay in advanced care for the passengers in the overturned bus include:
1. Lack of communication: Since the accident happened on the remote hillside, there is a possibility that, there would be no communication network which could have afforded them the opportunity to call medical or technical team.
2. Steep Nature of the Hill: This is another factor which will affect the care which they could have received. Steeply area tends to be difficult for climbing in or out from.
3. Thunderstorm: This factor is another reason which could contribute to delay in receiving advance care. Thunderstorm create barriers for location f the area where the bus overturned or in other situation complicate the rescue efforts of the team sent out to rescue.
Explanation:
At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.10 ✕ 10−5 T
Answer:
The speed of the proton is 4059.39 m/s
Explanation:
The centripetal force on the particle is given by;
[tex]F = \frac{mv^2}{r}[/tex]
The magnetic force on the particle is given by;
[tex]F = qvB[/tex]
The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.
[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}[/tex]
where;
r is the radius of the circular path moved by both electron and proton;
⇒For electron;
[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]
⇒For proton
The speed of the proton is given by;
[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]
Therefore, the speed of the proton is 4059.39 m/s
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to
Answer:
Ok, the question is incomplete buy ill try to answer this in a general way.
Suppose that you have no-polarized light.
When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.
Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:
I(θ) = I0*cos^2(θ)
For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°
and:
I(0°) = I0*cos^2(0°) = I0
So the intensity does not change.
Now, knowing the initial intensity, you can find the angle needed to get a given intensity.
For example, if the question was:
"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"
We should solve:
I(θ) = A = I0*cos^2(θ)
(A/i0) = cos^2(θ)
√(A/I0) = cos(θ)
Acos(√(A/I0)) = θ
Q9 A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor.
Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
Which is one criterion that materials of a technological design should meet? They must be imported. They must be affordable. They must be naturally made. They must be locally produced.
Answer:
they must be affordable because they have to pay for it or they wont get the stuff they are bying.
Explanation:
need a brainliest please.
Answer: B, they must be affordable.
Explanation:
A tank whose bottom is a mirror is filled with water to a depth of 19.6 cm. A small fish floats motionless a distance of 6.40 cm under the surface of the water.
A) What is the apparent depth of the fish when viewed at normal incidence?
B) What is the apparent depth of the image of the fish when viewed at normal incidence?
Answer:
A. 4.82 cm
B. 24.66 cm
Explanation:
The depth of water = 19.6 cm
Distance of fish = 6.40 cm
Index of refraction of water = 1.33
(A). Now use the below formula to compute the apparent depth.
[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\= \frac{1}{1.33} \times 6.40 \\= 4.82 cm.[/tex]
(B). the depth of the fish in the mirror.
[tex]d_{real} = 19.6 cm + (19.6 cm – 6.40 cm) = 32.8 cm[/tex]
Now find the depth of reflection of the fish in the bottom of the tank.
[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\d_{app} = \frac{1}{1.33} \times 32.8 = 24.66\\[/tex]
Inside the wall of a house, an L-shaped section of hot-water pipe consists of three parts: a straight, horizontal piece 28.0 cm long, an elbow, and a straight vertical piece ℓ = 159 cm long. A stud and a second-story floorboard hold the ends of this section of copper pipe stationary. Find the magnitude and direction of the displacement of the pipe elbow when the water flow is turned on, raising the temperature of the pipe from 18.0°C to 40.2°C. (The coefficient of linear expansion of copper is
Answer:
The magnitude and direction are
7.638×10-4m
80.01°
Explanation:
We know that the coefficient of linear expansion for copper = 16.6×10^-6 m/m-C
ΔT = 40.2 - 18.0 = 28.5 C°
The expansion of horizontal pipe length can be calculated as
= (0.28)(16.6×10^-6)(28.5) = 13247×10^-8
= 0.0001325 m
The expansion of vertical pipe length = (1.28)(16.6×10^-6)(28.5) = 60557×10^-8 = 0.000752229 m
horizontal displacement = 0.1325 mm
= 1.356×10^-4m
vertical displacement = 0.75223mm
=7.5223×10-4m
size of total displacement can be calculated as
√(x²+y²)
Where x and y are vertical and horizontal displacement respectively
= √(0.1325)²+(0.75223)² =
= 0.7638 mm
= 7.638×10-4m
Angle below horizontal = arctan Θ
= 0.75223/0.1325
=5.6772
= arctan (5.6772)
= 80.01°
Therefore, the the magnitude and direction of the displacement of the pipe elbow when the water flow is turned at (7.638×10-4m) 0.7638 mm and 80.01°
an electromagnetic wave propagates in a vacuum in the x-direction. In what direction does the electric field oscilate
Answer:
The electric field can either oscillates in the z-direction, or the y-direction, but must oscillate in a direction perpendicular to the direction of propagation, and the direction of oscillation of the magnetic field.
Explanation:
Electromagnetic waves are waves that have an oscillating magnetic and electric field, that oscillates perpendicularly to one another. Electromagnetic waves are propagated in a direction perpendicular to both the electric and the magnetic field. If the wave is propagated in the x-direction, then the electric field can either oscillate in the y-direction, or the z-direction but must oscillate perpendicularly to both the the direction of oscillation of the magnetic field, and the direction of propagation of the wave.
Rays of light coming from the sun (a very distant object) are near and parallel to the principal axis of a concave mirror. After reflecting from the mirror, where will the rays cross each other at a single point?
The rays __________
a. will not cross each other after reflecting from a concave mirror.
b. will cross at the center of curvature.
c. will cross at the point where the principal axis intersects the mirror.
d. will cross at the focal point. will cross at a point beyond the center of curvature.
A concave mirror is an example of curved mirrors. So that the appropriate answer to the given question is option D. The rays will cross at the focal point.
A concave mirror is a type of mirror in which its inner part is the reflecting surface, while its outer part is the back of the mirror. This mirror reflects all parallel rays close to the principal axis to a point of convergence. It can also be referred to as the converging mirror.
In this type of mirror, all rays of light parallel to the principal axis of the mirror after reflection will cross at the focal point.
Therefore, the required answer to the given question is option D. i.e The rays will cross at the focal point.
For reference: https://brainly.com/question/20380620
A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. The skin and structure of the balloon has a mass of 950 kg. Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift. Express your answer to two significant figures and include the appropriate units.
volume of balloon
= 4/3 T R3
= 4/3 x 3.14 x 6.953
= 1405.47 m3
uplift force
= volume of balloon x density of air x 9.8
= = 1405.47 x 1.29 x 9.8
= 1813.05 x 9.8 N
weight of helium gas
= volume of balloon x density of helium x
9.8
= 1405.47 x .179 x 9.8
= 251.58 x 9.8 N
Weight of other mass = 930 x 9.8 N Total weight acting downwards
= 251.58 x 9.8 +930 x 9.8
= 1181.58 x 9.8 N
If W be extra weight the uplift can balance
1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8
1181.58+W=1813.05
W= 631.47 kg
The difference between a DC and an AC generator is that
a. the DC generator has one unbroken slip ring.
b. the AC generator has one unbroken slip ring
c. the DC generator has one slip ring splitin two halves.
d. the AC generator has one slip ring split in two halves.
e The DC generator has twounbroken sip rings
Answer:
The AC generator has one unbroken slip ring
Explanation:
In physics, the application of electromagnetic induction can be seen in generators and dynamos. Electromagnetic induction is the process of generating electricity using magnets. It found applications in generators and the types of generator they found application is in AC and DC generator.
An AC generator is also called a Dynamo. A DC generator contains what is called a SPLIT RING fixed to the end of the coil which can be separated and coupled back according to the name "split". An AC generator also called a Dynamo makes use of a SLIP ring which cannot be divided into two. It comes as an entity. The presence of this rings is what differentiates a DC generator from an AC generator.
We can replace split rings with slip rings when converting a DC generator to an AC generator and vice versa.
It can therefore be concluded that the difference between a DC and an AC generator is that the AC generator has one unbroken slip ring.
A red laser beam goes from crown glass with refraction index n=1.3 to air (n=1) with an incident angle of 0.23 radians. What is the angle of refraction in degrees?
Answer:
θ = 10.28º
Explanation:
To find the angle of refraction use the equation of refraction
n₁ sin θ₁ = n₂ sin θ₂
where index 1 is for incident light and index 2 is for refracted light.
sin θ₂ = n₁ / n₂ sin θ
let's calculate
sin = 1 / 1.3 sin 0.23
sin = 0.175
θ= 0.17528 rad
let's reduce to degrees
θ = 0.17528 rad (180ª / pi rad)
θ = 10.28º
Identify 2 different ways that data can be displayed or represented.
Answer:
tables, charts and graphs
Explanation:
The microwaves in a microwave oven are produced in a special tube called a magnetron. The electrons orbit the magnetic field at 2.4 GHz, and as they do so they emit 2.4 GHz electromagnetic waves. What is the strength of the magnetic field?
Answer:
The magnetic field is 0.0857 T.
Explanation:
The electrons orbit the magnetic field with a centripetal force equal to
F = [tex]\frac{mv^{2} }{r}[/tex]
also, the force on an electron in a magnetic field is gotten as
F = Bqv
equating this two equations give
[tex]\frac{mv^{2} }{r}[/tex] = Bqv
mv/r = Bq
where m is the mass of the electron = 9.11 x 10^-31 kg
v is the the linear speed of the electron
B is the magnetic field on the electron
r is the radius of the orbital movement
q is the charge on an electron = 1.602 x 10^-19 C
but, the linear speed v = ωr
where ω is the angular speed of the electron
substituting into equation above, we have
mωr/r = Bq
which reduces to
mω = Bq
finally, w know that the angular speed is related to the frequency of the electron by
ω = 2πf
we then finally have
2mπf = Bq
where f is the frequency emitted by the electron = 2.4 GHz = 2.4 x 10^9 Hz
substituting values into the equation, we have
2 x 9.11 x 10^-31 x 3.142 x 2.4 x 10^9 = B x 1.602 x 10^-19
B = (1.3734 x 10^-20)/(1.602 x 10^-19) = 0.0857 T
= 85.7 mT
: A spaceship is traveling at the speed 2t 2 1 km/s (t is time in seconds). It is pointing directly away from earth and at time t 0 it is 1000 kilometers from earth. How far from earth is it at one minute from time t 0
Answer:
145060km
Explanation: Given that
speed = dx/dt = 2t^2 +1
integrate
x = 2/3t^3 + t + c (c is constant, x is in km, t is in second)
given that at t=0, x = 1000
so 1000 = 2/3 X (0)^3 + 0 + c
or c = 1000
So x = 2/3t^3 + t + 1000
for t = 1 min = 60s
x = 2/3 X 60^3 + 60 + 1000
x = 2/3×216000+ 1060
x = 144000+1060
= 145060km
At one minute, it will be 145060km far from the earth
Consider 2 converging lenses of focal lengths 5 mm (objective) and 50 mm.(eyepiece) An object 0.1 mm in size is placed a distance of 5.2 mm from the objective.
1. What is the size and location of the image from the objective? What is the linear magnification of this objective?
2. Treat the image from the objective as an object for the eyepiece. If the eyepiece creates an image at infinity, how far apart are the two lenses?
3. What is the angular magnification of the pair of lenses?
Answer:
1) q₁ = 12.987 cm , b) L = 17.987 cm , c) m = 179.87
Explanation:
We can solve the geometric optics exercises with the equation of the constructor
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distance to the object and the image respectively.
Let's apply this equation to our case
1) f = 5mm = 0.5 cm
p₁ = 5.2 mm = 0.52 cm
h = 0.1 mm = 0.01 cm
1 / q₁ = 1 / f- 1 / p
1 / q₁ = 1 / 0.5 - 1 / 0.52 = 2 - 1.923
1 / q₁ = 0.077
q₁ = 12.987 cm
2) in this part they tell us that the eyepiece creates an image at infinity, therefore the object that comes from being at the focal length of the eyepiece
p₂ = 5 cm
The absolute thing that goes through the two lenses is
L = q₁ + p₂
L = 12.987 +5
L = 17.987 cm
3) This lens configuration forms the so-called microscope, whose expression for the magnifications
m = -L / f_target 25 cm / f_ocular
m = - 17.987 / 0.5 25 / 5.0
m = 179.87
A car travels at 45 km/h. If the driver breaks 0.65 seconds after seeing the traffic light turn yellow, how far will the car continue to travel before it begins to slow?
Answer:
8.1 m
Explanation:
Convert km/h to m/s.
45 km/h × (1000 m/km) × (1 h / 3600 s) = 12.5 m/s
Distance = speed × time
d = (12.5 m/s) (0.65 s)
d = 8.125 m
Approximating the eye as a single thin lens 2.70 cmcm from the retina, find the focal length of the eye when it is focused on an object at a distance of 265 cmcm
Answer:
0.37 cm
Explanation:
The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.
The object is at a distant of 265 cm to the lens of the eye.
From lens formula,
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]
where: f is the focal length, u is the object distance and v is the image distance.
Thus, u = 265.00 cm and v = 2.70 cm.
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{265}[/tex] + [tex]\frac{27}{10}[/tex]
= [tex]\frac{10+7155}{2650}[/tex]
[tex]\frac{1}{f}[/tex] = [tex]\frac{7165}{2650}[/tex]
⇒ f = [tex]\frac{2650}{7165}[/tex]
= 0.37
The focal length of the eye is 0.37 cm.
front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the rear tire
Answer:
The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.
Explanation:
This question suggests that the car is accelerating forward. Thus, the easiest way for us to know what friction is doing is for us to know what happens when we turn friction off.
Now, if there is no friction and the car is stopped, if we push down on the accelerator, it will make the front wheels to spin in a clockwise manner. This spin occurs on the frictionless surface with the rear wheels doing nothing while the car doesn't move.
Now, if we apply friction to just the front wheels, the car will accelerate forward while the back wheels would be dragging along the road and not be spinning. Thus, friction opposes the motion and as such, it must act im a direction opposite to where the car is going. This must be static friction.
The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.
How wide is the central diffraction peak on a screen 2.30 m behind a 0.0368-mm-wide slit illuminated by 558-nm light
Answer:
The value [tex]y = 0.0349 \ m[/tex]
Explanation:
From the question we are told that
The distance of the screen is [tex]D = 2.30 \ m[/tex]
The width of the slit is [tex]d = 0.0368 \ nm = 0.0368 *10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 558 \ nm = 558 *10^{-9} \ m[/tex]
The width of the central diffraction peak is mathematically represented as
[tex]k = 2 * y[/tex]
Where y is the distance from the center to the high peak which is mathematically represented as
[tex]y = \frac{\lambda * D }{d }[/tex]
substituting values
[tex]y = \frac{ 558 *10^{-8} * 2.30 }{0.0368 *10^{-3} }[/tex]
[tex]y = 0.0349 \ m[/tex]
Approximately how many calories are in one gram of carbohydrates?
Answer:
4 caloriesExplanation:
Carbohydrates are known as energy giving foods and they are contained in our diets no matter how small. Other components of foods are protein, fat and oil, vitamin, water etc. A calorie is a unit of energy. It tells us the amount of energy the weight of any food contains. Knowing the amount of calories of food we are taking in helps us to monitor our weight. Each component of food have their own calories depending on the weight of the food component.
For carbohydrate as a component of food, 1gram of carbohydrates contains 4 calories. With this conversion, we can therefore calculate the calories of carbohydrate we are taking in by taking the weight of the food we want to eat.
Hence, approximately 4 calories are in one gram of carbohydrates
One gram of carbohydrates contains approximately 4 calories.
how many calories are in one gram of carbohydrates?The amount of energy in macronutrients, like carbohydrates, is measured in calories per gram. In terms of nutrition, a "calorie" is the amount of energy needed to warm up one gram of water by one degree Celsius.
Carbohydrates are one of the three main types of nutrients that our body needs, along with proteins and fats. They give the body fuel. When you eat carbohydrates, your body breaks them down into a type of sugar called glucose. Cells in your body use this glucose for energy.
Read more about carbohydrates here:
https://brainly.com/question/336775
#SPJ6
A guitar string is 90 cm long and has a mass of 3.5g . The distance from the bridge to the support post is L=62cm, and the string is under a tension of 540N . What are the frequencies of the fundamental and first two overtones? Express your answers using two significant figures. Enter your answers in ascending order separated by commas.
Answer:
[tex]v_1 = 301 Hz[/tex]
[tex]v_2 = 601 \ \ Hz[/tex]
[tex]v_3 = 901 \ Hz[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]l = 90 \ cm = 0.9 \ m[/tex]
The mass of the string is [tex]m_s = 3.5 \ g =0.0035 \ kg[/tex]
The distance from the bridge to the support post [tex]L = 62 \ c m = 0.62 \ m[/tex]
The tension is [tex]T = 540 \ N[/tex]
Generally the frequency is mathematically represented as
[tex]v = \frac{n}{2 * L } [\sqrt{ \frac{T}{\mu} } ][/tex]
Where n is and integer that defines that overtones
i.e n = 1 is for fundamental frequency
n = 2 first overtone
n =3 second overtone
Also [tex]\mu[/tex] is the linear density of the string which is mathematically represented as
[tex]\mu = \frac{m_s}{l}[/tex]
=> [tex]\mu = \frac{0.0035 }{ 0.9 }[/tex]
=> [tex]\mu = 0.003889 \ kg/m[/tex]
So for n = 1
[tex]v_1 = \frac{1}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v_1 = 301 \ Hz[/tex]
So for n = 2
[tex]v_2 = \frac{2}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v_2 = 601 \ Hz[/tex]
So for n = 3
[tex]v_3 = \frac{3}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v =901 \ Hz[/tex]
A 1000 kg car experiences a net force of 9500 N while slowing down from 30 m/s to 16 m/s. How far does it travel while slowing down?
Answer:
33.89 m
Explanation:
We must first obtain the acceleration of the car from;
F=ma
Where
F= force= 9500 N
m= mass of the car= 1000kg
a= acceleration
a= F/m= 9500/1000
a= 9.5 m/s^2
From;
V^2=u^2 + 2as
Where;
V= final velocity
u= initial velocity
s= distance covered
a= acceleration
s= v^2 -u^2/2a
s= (30)^2 -(16)^2/2×9.5
s= 900 - 256/19
s= 644/19
s= 33.89 m
The distance is 33.89 m
The first step is to calculate the acceleration
F= ma
force= 9500N
mass= 1000 kg
9500= 1000 × a
a= 9500/1000
= 9.5 m/s
v²= u² + 2as
30²= 16² + 2(9.5)(s)
900= 256 + 19s
900-256= 19s
644= 19s
s= 644/19
s= 33.89 m
Hence the distance traveled by the car is 33.89 m
Please see the link below for more information
https://brainly.com/question/18313681?referrer=searchResults
What are the systems of units? Explain each of them.
THERE ARE COMMONLY THREE SYSTEMS OF UNIT. THEY ARE:-
• CGS System- (Centimeter-Gram-Second system) A metric system of measurement that uses the centimeter, gram and second for length, mass and time.
• FPS System- (Foot–Pound–Second system).
The system of units in which length is measured in foot , mass in pound and time in second is called FPS system. It is also known as British system of units.
• MKS System- (Meter-Kilogram-Second system) A metric system of measurement that uses the meter, kilogram, gram and second for length, mass and time. The units of force and energy are the "newton" and "joule."