Answer:
22.8°
Explanation:
Given conditions:
When light beams to the east it is inclined at 3 degrees upwards
When the light beams points to the north it is inclined 10 degrees upwards
Determine the maximum angle of elevation of the light beam
let slope of the line ( m ) = tan ∅
In x-direction ( eastwards ) = df / dx = tan 3°
In y-direction ( north ) = df / dy = tan 10°
| νf | = | < tan 3° , tan 10° > |
therefore | νf | = [tex]\sqrt{(tan 3)^2 + (tan10)^2}[/tex]
| νf | = [tex]\sqrt{0.02032 +0.42037 }[/tex] = √ 0.44069 = 0.42037
hence the maximum angle of elevation = tan^-1 ( 0.42037 ) = 0.39794 rad
to degrees = 22.8°
The maximum angle of elevation of the light beam will be 22.8°. An elevation is its altitude beyond or below a specified reference point, It is calculated in degrees.
What is elevation?An elevation is its height above or below a fixed reference point, It is measured in degrees.
The given data in the problem is;
Angle inclined when light beams to the east = 3° upwards
Angle inclined light beams point to the north = 10° upwards
m is the slope of the line = tan ∅
The maximum angle of elevation of the light beam is found as;
The angle inclined in the x and y-direction;
In x-direction ( eastwards ) = df / dx = tan 3°
In y-direction ( north ) = df / dy = tan 10°
The resultant angle of both the above angle is;
[tex]\rm v_f = \sqrt{(tan3^0)^2 +(tan 10^0)^2 } \\\\ \rm v_f = \sqrt{(0.020320)^ +(0.42037) } \\\\ \rm v_f = \sqrt{0.44069 \\\\[/tex]
[tex]\rm v_f = 0.42037[/tex]
[tex]\rm \theta= tan^-1 (v_f) \\\\ \rm \theta= tan^-1 (0.42037) \\\\ \rm \theta= 22.8 ^0[/tex]
Hence the maximum angle of elevation of the light beam will be 22.8°.
To learn more about the elevation refer to the link;
https://brainly.com/question/481548
Does a convex lens always form a real image? Explain
Answer:
no
Explanation:
can form either real or virtual images. Convex lens converge parallel rays coming from object at infinity and highly diminished point sized, real and inverted image is formed.
Answer:
no it does not forms a real image
To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth.
The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were observed by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as the speed of the space shuttle and the height of geostationary satellites) is the simplest orbit--a circular one. This problem concerns the properties of circular orbits for a satellite orbiting a planet of mass M. For all parts of this problem, where appropriate, use G for the universal gravitational constant.
1. Find an expression for the square of the orbital period.
2. The quantities v, K, U, and L all represent physical quantities characterizing the orbit that depend on radius R. Indicate the exponent (power) of the radial dependence of the absolute value of each.
Answer:
a) T² = ([tex]\frac{4\pi ^2}{GM}[/tex]) r³
b) veloicity the dependency is the inverse of the root of the distance
kinetic energy depends on the inverse of the distance
potential energy dependency is the inverse of distance
angular momentum depends directly on the root of the distance
Explanation:
1) for this exercise we will use Newton's second law
F = ma
in this case the acceleration is centripetal
a = v² / r
the linear and angular variable are related
v = w r
we substitute
a = w² r
force is the universal force of attraction
F = [tex]G \frac{m M}{r^2}[/tex]
we substitute
[tex]G \frac{m M}{r^2} = m w^2 r[/tex]
w² = [tex]\frac{GM}{r^3}[/tex]
angular velocity is related to frequency and period
w = 2π f = 2π / T
we substitute
[tex]( \frac{2\pi }{T} ) = \frac{GM}{r^3}[/tex]
the final equation is
T² = () r³
b) the speed of the orbit can be found
v = w r
v = [tex]\sqrt{\frac{GM}{r^3} } \ r[/tex]
v = [tex]\sqrt{\frac{GM}{r} }[/tex]
in this case the dependency is the inverse of the root of the distance
Kinetic energy
K = ½ M v²
K = ½ M GM / r
K = ½ GM² 1 / r
the kinetic energy depends on the inverse of the distance
Potential energy
U =
U = -G mM / r
dependency is the inverse of distance
Angular momentum
L = r x p
for a circular orbit
L = r p = r Mv
L =
L =
The angular momentum depends directly on the root of the distance
Maurice is playing some of the arcades at the county fair. At one booth he throws a 1.2-kg ball forward with a velocity of 15 m/s and hits a 0.5-kg can
at rest. The ball continues forward at 9 m/s. What is the new velocity of the can, in m/s? (Round your answer to one decimal place if necessary.)
Answer:
9.6
Hope this helps.
ASSIGNMENT
1) A force of 2N stretches an elastic material by 30mm. what additional
force will stretch the material 35mm? Assume that the elastic limit is not
exceeded
Answer:
If 2N = 30mm
Then ? =35mm
2N×(35mm÷30mm)
2N×1.167
= 2.33N
To get the additional force required, subtract 2N from 2.33N
2.33N - 2N
= 0.33N
∴ The additional force needed is 0.33N
Explanation:
With this question, we establish the principles of proportion. That is, if more less divides and vice versa. If 2N is stretching the material to 30mm, then how more force will be needed to stretch the material to 35mm?
A water park has a slide that finishes with the rider flying horizontally of the bottom of the slide. The slide is designed to end 1.5 m above the water level, and the average rider is estimated to leave the bottom of the slide at 25 m/s. How far Horizontally will the rider travel from the bottom of the slide before hitting the water?
Answer:
25 m/s * √(1.5 m/g) ≈ 9.78 m
Explanation:
When effects of air resistance is neglected, the motion can be modeled as
x(t) = 25 m/s * t
y(t) = 1.5 m - g * t²
based on a coordinate system where the x axis is on the water level (modeled as a plane), y axis being parallel to the gravitational force and the point of origin sitting at the point on the water surface directly below the exit of the slide. The x value is the horizontal travel we’re interested in. Time t begins at the exit of the slide by this rider. g is the gravitational acceleration.
At the time when the rider hits water, we know y = 0.
0 = 1.5 m - g * t² ⇔ t = ±√(1.5 m/g)
Since we are only interested in the non-negative (not past) time range, it’s t = √(1.5 m/g)
The horizontal travel is thus:
x(√(1.5 m/g)) = 25 m/s * √(1.5 m/g) ≈ 25 * √(1.5/9.81) m ≈ 9.78 m
A child swings with a small amplitude on a playground with a 2.5m long chain.
a) what is the period of the child’s motion
b) what is the frequency of vibration
a. The period of the child's motion is 3.171 seconds.
b. The frequency of the vibration is 0.315 Hz.
Calculation of the period & frequency:Since A child swings with a small amplitude on a playground with a 2.5m long chain.
a. So here the time period should be
[tex]2\times \ pi(L/g)^{1/2}\\\\2\times 3.14(2.5/9.81)^{1/2}[/tex]
T=3.171s
b) Now the frequency is
[tex]T=1\div f\\\\f=1\div 3.171[/tex]
f=0.315Hz
learn more about frequency here: https://brainly.com/question/22971850
Which letter on the map represents the
Southern Ocean
Answer:
B cus is in the south but that pfp tho
most bones are connected at joints by
Answer:
Ligaments
Explanation:
I answered your question before it's just connective tissue
What color is a carrot?
Answer:
reddish-orrange
Explanation:
please mark me as brainliest
PLEASE HELP ME WITH ONE QUESTION
If 1000 J of heat energy is lost from a piece of hot metal which is placed into a cup of cool water, how much energy is gained by the water?
Answer:
1000J
Explanation:
According to the law of calorimetry;
Heat lost by cold substance = Heat gained by hot substance
If 1000 J of heat energy is lost from a piece of hot metal,
heat lost by hot substance = 1000J
Since we are to determine how much energy is gained by the water, using the law
Heat lost by hot metal = heat gained by the water
Hence heat gained by the water = 1000J
What is the increase in gravitational potential energy of an object of mass 9.00 kg when it is raised from the ground to a height of 2.41 m, in units of joules?
Answer:
23 asden
Explanation:
first clothes and shoes
please help me with the working too.
Answer:
1 : b 2: d
Explanation:
A marching band consists of rows of musicians walking in straight, even lines. When a marching band performs in an event, such as a parade, and must round a curve in the road, the musician on the outside of the curve must walk around the curve in the same amount of time as the musician on the inside of the curve. This motion can be approximated by a disk rotating at a constant rate about an axis perpendicular to its plane. In this case, the axis of rotation is at the inside of the curve. Consider two musicians, Alf and Beth. Beth is four times the distance from the inside of the curve as Alf. Knowing that If Beth travels a distance s during time Δt, how far does Alf travel during the same amount of time= (1/4)s
If Alf moves with speed v, what is Beth's speed?
a. 4v
b. v
c. v/4
Answer:
A) Total distance Alf travel during the same amount of time = [tex]\frac{1}{4}s[/tex]
B) Speed of Beth = 4v
Explanation:
Given - Consider two musicians, Alf and Beth. Beth is four times the
distance from the inside of the curve as Alf.
To find - A) Knowing that If Beth travels a distance s during time Δt, how
far does Alf travel during the same amount of time.
B) If Alf moves with speed v, what is Beth's speed?
Proof -
A)
As we know that
Speed = Distance / Time
⇒Distance = Speed ×Time
Now,
Given Speed of Alf = v
⇒Distance of Alf = vt
Also,
Distance of Beth = Speed of beth×time
Given that
Beth is four times the distance from the inside of the curve as Alf. Knowing that If Beth travels a distance s during time Δt, how far does Alf travel during the same amount of time=
⇒Distance of Alf = Speed of Alf×time
= [tex]\frac{1}{4}[/tex] Distance of Beth
⇒Distance of Alf = [tex]\frac{1}{4}s[/tex]
∴ we get
Total distance Alf travel during the same amount of time = [tex]\frac{1}{4}s[/tex]
B)
We know the conversion of angular velocity
ω(Alf) = ω(Beth)
⇒V(Alf)/ r = V(Beth)/4r
⇒V(Alf) = V(Beth) / 4
⇒V(Beth) = 4 V(Alf)
As given, Alf moves with speed v
⇒V(Beth) = 4v
So, the correct option is - a.4v
A substance is poured from one container to
another, but the shape does not change. What
state is it in?
Answer:
Its liquid because when you pour water into different cups of different sizes its still the same shape
6. If you lift an object, to find the work you did on the object, you must find
the__ first
a velocity
b. power
c. acceleration
d. weight
1. An astronaut is getting ready to explore. If his mass is 60 kg, what is his weight
on Earth?
Which group is an international organization in which the United States is involved?
A. The Foreign Relations Committee
B. The Senate Armed Services Committee
C. The Department of Defense
D. The World Trade Organization
Ccorrect Answer Is D: The World Trade Organization
Answer:
the answer is d the world trade organization
Explanation:
studyhero A violin string has a length, from the bridge to the end of the fingerboard, of 50 cm. That section of the string has a mass of only 2 g. When the violinist plays an open string (the full length) a 440 Hz A-note is heard. Determine the length of the string needed to play a 528 Hz note without adjusting the tension in the string. This is accomplished by pressing on the fingerboard at the appropriate location.
Answer:
= 0.517 m
Explanation:
This is a resonance exercise where the ends of the string are fixed, therefore it has a node of them, the fundamental (longest) wavelength created has the form
λ = 2L
wave speed is related to wavelength and frequency
v = λ f
v = 2L f
let's calculate
v = 2 0.50 440
v = 440 m / s
since they indicate that the tension of the string does not change and the linear density of the string is constant, the speed of the wave also remains constant
f =[tex]\frac{v}{2L}[/tex]
let's find the length for the new resonance frequency
L = [tex]\ \frac{v}{2f}[/tex]
let's calculate
L = [tex]\frac{440}{2 \ 528}[/tex]
L = 0.5166 m
the moon's mass is ____?
Answer:
i hope this is helpful
Explanation:
7.34767309 × 1022 kilograms
Answer:
Hellooo
7.35 x 1022 kg
The moon's mass is 7.35 x 1022 kg, about 1.2 percent of Earth's mass. Put another way, Earth weighs 81 times more than the moon.
Explanation:
thanks
g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosynchronous satellite has an orbital period of 24 hours. The radius of Earth is 6400 km. Use Kepler's 3rd Law to compare these two and find the orbital altitude of the geosynchronous satellite.
Answer:
r = 4.21 10⁷ m
Explanation:
Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining
T² = ([tex]\frac{4\pi }{G M_s}[/tex] ) r³ (1)
in this case the period of the season is
T₁ = 93 min (60 s / 1 min) = 5580 s
r₁ = 410 + 6370 = 6780 km
r₁ = 6.780 10⁶ m
for the satellite
T₂ = 24 h (3600 s / 1h) = 86 400 s
if we substitute in equation 1
T² = K r³
K = T₁²/r₁³
K = [tex]\frac{ 5580^2}{ (6.780 10^6)^2}[/tex]
K = 9.99 10⁻¹⁴ s² / m³
we can replace the satellite values
r³ = T² / K
r³ = 86400² / 9.99 10⁻¹⁴
r = ∛(7.4724 10²²)
r = 4.21 10⁷ m
this distance is from the center of the earth
The orbital altitude of the geosynchronous satellite is mathematically given as
r = 4.21 10⁷ m
What is the orbital altitude of the geosynchronous satellite?Question Parameters:
Generally, the equation for the Gravitational force is mathematically given as
T^2 =[tex]\frac{4\pi }{G M_s} * r^3[/tex]
Therefore
K = T₁^2/r₁^3
K = \frac{ 5580^2}{ (6.780 10^6)^2}
K = 9.99 10^{-14} s^2 / m^3
Since
r^3 = T^2 / K
r^3 = 86400² / 9.99 10^{-14}
r = 4.21 10⁷ m
For more information on distance
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In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal different information. You will use the both momentum principle and the energy principle in this problem.
A satellite of mass 3500 kg orbits the Earth in a circular orbit of radius of 7.3 106 m (this is above the Earth's atmosphere).The mass of the Earth is 6.0 1024 kg.
What is the magnitude of the gravitational force on the satellite due to the earth?
F= ________N
Using the momentum principle, find the speed of the satellite in orbit. (HINT: Think about the components of (dp^^\->)\/(dt) parallel and perpendicular to p^^\->.)
v = ________ m/s
Using the energy principle, find the minimum amount of work needed to move the satellite from this orbit to a location very far away from the Earth. (You can think of this energy as being supplied by work due to something outside of the system of the Earth and the satellite.)
work= ________J
Answer:
A) F_g = 26284.48 N
B) v = 7404.18 m/s
C) E = 19.19 × 10^(10) J
Explanation:
We are given;
Mass of satellite; m = 3500 kg
Mass of the earth; M = 6 x 10²⁴ Kg
Earth circular orbit radius; R = 7.3 x 10⁶ m
A) Formula for the gravitational force is;
F_g = GmM/r²
Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²
Plugging in the relevant values, we have;
F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²
F_g = 26284.48 N
B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.
Thus;
GmM/r² = mv²/r
Making v th subject, we have;
v = √(GM/r)
Plugging in the relevant values;
v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))
v = 7404.18 m/s
C) From the energy principle, the minimum amount of work is given by;
E = (GmM/r) - ½mv²
Plugging in the relevant values;
E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)
E = 19.19 × 10^(10) J
A cylindrical tank with radius 3 m is being filled with water at a rate of 2 m3/min. How fast is the height of the water increasing (in m/min)
Answer:
.0707 m / min
Explanation:
The computation is shown below:
= Flow of the volume ÷ crossectional area
A = π × r^2
= 9 π
= 28.27 m^2
so, the fast of the height of the water rise is
=(2 m^3 / min) ÷ 28.27 m^2
= .0707 m / min
I need help with both 3 and 4 please help?
Why do we use copper wiring?
Explanation:
The lower the level of resistivity the more electrical conductivity a metal has. Copper has low resistivity, and therefore is an excellent conductor. Copper is also less oxidative than other metals
[1 mark]
29. A ball rolls at a constant speed of 3.0 m/s for 8.0 seconds. How far does it roll in this time!
A. 2.7 m
B. 3.0 m
C. 8.0 m
D. 24.0 m
Please help me
Answer:
D. 24.0m
Explanation:
formula
S=d/t
s=speed
d=distance
t=time
s=3.0m/s
d=x
t=8.0s
s=d/t
d=s*t
d=3.0*8.0
d=24.0m
what is agrculture and it means
Agriculture is the science, the practice of cultivating the soil, art, producing crops, raising livestock, and in varying degrees the preparation and marketing of the resulting products cleared the land to use for agriculture.
Explanation:This is what I found during my research and I hope this helps. Correct me if I am incorrect! Have a good one.
≧◉◡◉≦2 points
13. The law of conservation of mass can be demonstrated by a chemical
reaction. Which of the following models of a chemical reaction best
represents the law of conservation of mass?
A
C
B
D.
8
8.
a.
Ο Ο Ο Ο
Answer:
A
Explanation:
the mass in the option a is conserved of two elements
who compose the Ghana national anthem
Answer:
Philip Gbeho
Explanation:
PLZ HELP PLZ ILL GIVE BRAINLIEST TO BEST ANSWERED
The Earth's gravity pulling on a skydiver: Contact or Non Contact
Friction from a car's brakes causing it to stop: Contact or Non Contact
The repulsive magnetic force between the north poles of two magnets: Contact or Non Contact
The attractive electrostatic force between an electrically charged balloon and a wall: Contact or Non Contact
Answer:
1.Contact because its a contact force
2 Contact because when the car is stopping it is coming in to contact with the ground.
3 non Contact because magnets don't need to have contact they pull
4 non Contact because its not a contact force
You hold your physics textbook in your hand. (Assume that no other objects are in contact with the book.)
a. Identify the forces acting on the book.
1. book on hand
2. hand on book
3. floor on book
4. earth on book
b. For each force you identified in part (a), indicate the direction. (If a force is not applicable, choose NA.)
1, book on hand __________
2. hand on book __________
3. floor on book __________
4. earth on book _________
Answer:
a) Please see below as the answer is self-explanatory.
b) 1. vertically downward
2. vertically upward
3 N/A
4 vertically downward.
Explanation:
a)
If the book is hold by the hand, this means (according to Newton's third law) that both exert a force on each other, equal and opposite.In this case this pair is formed by normal forces, the normal force (upward) that the hand does on the book (avoiding it to be accelerated downward) and the normal force that the book does on the hand (downward) avoiding it to be accelerated in the upward direction.The other force that acts on the book is the one due to gravity, and is always downward, which is numerically equal to the normal force that the hand does on the book, due to the book is at rest. There is no interaction between the book and the floor, so there is no force between them, neglecting the gravitational attraction between their masses.