The average velocity of a car driving down the highway if it’s displacement is 123 m west during a time period of 14.0 s is 8.79 m/sec.
What is Velocity?The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity. Unit of velocity is m/sec.
Given in the question a car driving down the highway it’s displacement is 123 m west during a time period of 14.0 s then the average velocity is given as,
Average velocity = displacement / time
= 123/14
= 8.79 m/sec
The average velocity of a car driving down the highway if it’s displacement is 123 m west during a time period of 14.0 s is 8.79 m/sec.
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what is the difference between these two formulas [tex]fc = \frac{m(v {}^{2}) }{r} [/tex][tex]ac = \frac{v {}^{2} }{r} [/tex]and how would I apply them to a promblem?
One is centripetal acceleration (ac) and the other one is centripetal force (fc)
Since = F = m* a
F = force
m= mass
a= acceleration
For centripetal force (Fc), we use centripetal acceleration
Fc= m * ac
ac= centripetal acceleration = v^2 / r
r= radius
v= linear velocity
So, in the end, we have:
Fc = m * v^2 /r
You can apply the to find the Centripetal force of an object with a certain mass, that negotiates a certain radius curve, with a certain speed.
What is the specific heat c of a 0.500 kg metal sample that rises 5.40 C when 305J of heat is added to it?
ANSWER
[tex]\begin{equation*} 112.96\text{ }J\/kgK \end{equation*}[/tex]EXPLANATION
Parameters given:
Mass of sample, m = 0.5 kg
Temperature change, ΔT = 5.40 °C = 5.40 K
Heat energy, H = 305 J
To find the specific heat capacity of the sample, we have to apply the formula for heat energy:
[tex]H=mc\Delta T[/tex]Where c = specific heat capacity
Therefore, solving for c, the specific heat capacity of the metal sample is:
[tex]\begin{gathered} 305=0.5*c*5.4 \\ c=\frac{305}{0.5*5.4} \\ c=112.96\text{ }J\/kgK \end{gathered}[/tex]That is the answer.
An object is formed by attaching a uniform, thin rod with a mass of mr = 8kg and length L = 6 m to a uniform sphere with mass ms = 36.25 kg and radius R = 1.5m.1) What is the moment of inertia of the object about an axis at the left end of the rod?
ANSWER:
2167.68 kg*m^2
STEP-BY-STEP EXPLANATION:
Given:
mr = 8 kg
L = 6 m
ms = 36.25 kg
R = 1.5 m
Moment of inertia of sphere about its center is:
[tex]I_{CM}=\frac{2}{5}m_s\cdot R^2[/tex]Using paraller theorem, moment of inertia of shpere about end of rop is:
[tex]I_{\text{rod}}=m_s\cdot(R+L)^2+\frac{1}{3}m_r\cdot L^2[/tex]Therefore:
[tex]\begin{gathered} I=I_{cm}+I_{\text{rod}}_{} \\ I=\frac{2}{5}\cdot m_s\cdot R^2+m_s\cdot(R+L)^2+\frac{1}{3}\cdot m_r\cdot L^2 \end{gathered}[/tex]Replacing:
[tex]\begin{gathered} I=\frac{2}{5}\cdot36.25\cdot1.5^2+36.25\cdot(1.5+6)^2+\frac{1}{3}\cdot8\cdot6^2 \\ I=2167.69\text{ kg}\cdot m^2 \end{gathered}[/tex]The moment of inertia is 2167.68 kg*m^2
A flat coil of wire has an area of 0.020 m2 and contains 50 turns. Initially the coil is oriented so that the normal to its surface is parallel to and in the same direction as a constant magnetic field of 0.18 T. The coil is then rotated through an angle of 30o in a time of 0.10 s. What is the average induced emf? -0.44 V +0.44 V +0.24 V
ANSWER:
3rd option: +0.24 V
STEP-BY-STEP EXPLANATION:
Given:
N = 50
Area = 0.020 m^2
B = 0.18 T
θf = 30°
time = 0.10 s
We can calculate the average induced emf by the following formula
[tex]\epsilon=N\cdot B\cdot A\cdot\left(\frac{\cos\theta_i-\cos\theta_f}{t}\right)[/tex]We replacing:
[tex]\begin{gathered} \epsilon=\left(50\right)\left(0.18\right)\left(0.02\right)\left(\frac{\cos\:0\degree\:-\cos\:30\degree}{\:0.1}\right) \\ \epsilon=0.241\cong0.24\text{ V} \end{gathered}[/tex]The correct answer is 0.24V
The cross product of two vectors (X and Y) is a negative vector when the angle between them is:
A. 0
B. 90
C. 180
D. 270
Answer:
C. 180
Explanation:
Answer:
Explanation:
The cross product:
[ a × b ] = | a | · | b | · sin α
sin α < 0 if α ∈ (180°, 360°)
Answer:
D. 270°
what is velocity ratio
Answer:
Definition of velocity ratio
: the ratio of a distance through which any part of a machine moves to that which the driving part moves during the same time.
pls brainlies me me new pls
Explanation:
The cubic expansivity of mercury is 1.8×10^-4K^-1 and the linear expansivity of glass is 8.0×10^-6K^-1, calculate the apparent expansivity of mercury in a glass container?
Therefore,
[tex]Apparent\text{ cubic expansivity=}real\text{ cubic expansivity - cubic expansivity of the container}[/tex][tex]\begin{gathered} \text{ cubic expansivity of the container =3(}8.0\times10^{-6})=24\times10^{-6}=2.4\times10^{-5} \\ \text{note we converted linear expansivity to cubic expansivity} \end{gathered}[/tex]Finally,
[tex]\text{Apparent cubic expansivity=}1.8\times10^{-4}-2.4\times10^{-5}=1.56\times10^{-4}K^{-1}[/tex]Which of the following sources is most likely to be free of bias? OA. Textbook OB. Blog post c. Diary D. Video interview
Answer:
Textbook
Explanation:
Pretty obvious thi :v
Answer:
OA
Explanation:
Text books are often reviewed by experts hence, is likely to be biased.
The plow was an important invention for the Sumerian civilization because it:
A. made it easier for farmers to prepare land for farming.
B. simplified communication with other cultures.
C. allowed important information to be recorded.
D. protected city-states from foreign invaders.
I am in the middle of the test help pls
The plow was an important invention for the Sumerian civilization because it made it easier for farmers to prepare land for farming.
The plow made farming more efficient for the Sumerians, as people no longer had to waste time planting food by hand. The cart allowed people to carry goods back and forth from different civilizations in a much easier way as people did not have to carry the goods on their backs.
It is estimated that the ancient Sumerians invented the plow. The plow was designed to utilize domestic animals as a means of easing the workload on farmers. Oxen were harnessed to a wooden plow and pulled it through the fields to loosen the soil for easier planting of crops.
Plow, also spelled plough, most important agricultural implement since the beginning of history, used to turn and break up soil, to bury crop residues, and to help control weeds.
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The outer ear _____ sound waves
According to the research, the correct term to complete the statement is collect. The outer ear collects sound waves.
What is the outer ear?It is a structure that captures or collects sound or sound vibrations before channeling them to the middle ear.
In this sense, it is made up of the pinna that projects from the side of the head and the external auditory meatus that automatically amplifies sounds that pass inward to the tympanic membrane.
Therefore, we can conclude that the outer ear captures sound waves and conducts them to the middle ear, thus the correct term is collect.
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A boy walks 4 km east and then turns around and walks 1.5 km west. If east is taken as the positive direction and the west as the negative direction on a number line then what is the distance the boy covers and his displacement?
ANSWER
[tex]undefined[/tex]EXPLANATION
First, let us make a sketch of the question:
From the diagram, the blue circle represents his starting position while the black circle represents his final position.
The total distance the boy covers is the sum of his two journeys, to the east and then to the west.
That is:
[tex]\begin{gathered} 4+1.5 \\ 5.5\operatorname{km} \end{gathered}[/tex]The boy's displacement is the distance between his starting position and his final position.
Since the west is taken as negative direction and the east taken as the positive direction, it means that 4km east means +4km and 1.5
A roller coaster car is traveling through a loop at 13 m/s. If the loop has a 23m radius, what centripetal force will the 53kg rider feel?
Given,
The velocity of the car, v=13 m/s
The radius of the loop, r=23 m
The mass of the rider, m=53 kg
The centripetal force is the force that keeps an object in its circular path.
The centripetal force is given by,
[tex]F=\frac{mv^2}{r}[/tex]On substituting the known values,
[tex]\begin{gathered} F=\frac{53\times13^2}{23} \\ =389.43\text{ N} \end{gathered}[/tex]Therefore, the centripetal force on the rider is 389.43 N
A football blocking sled has a mass of roughly 100 kg. If a football player applies a force of 500 N to the sled, and there is a 350 N frictional force acting on the sled, what is the acceleration of the sled?
The the acceleration of the sled of mass 100 Kg will be 1.5 m/s².
What is Friction?A drag is force that opposes the motion of an object by acting in the direction opposite to its motion. It is of two types namely - Static friction and kinetic friction. The static friction is given by - F[S] = μ[s] x η and the kinetic friction is given by F[K] = μ[k] x η.
Given is a football blocking sled which has a mass of roughly 100 kg. A football player applies a force of 500 N to the sled, and there is a 350 N frictional force acting on the sled.
Assume that the force applied by the player is - F[P] = 500 N and the force of friction is - F = 350 N.
Now, for the motion of the sled with acceleration [a], we can write -
F[P] - F = ma
a = {F[P] - F}/m
Substituting the values, we get -
a = (500 - 350)/100
a = 150/100
a = 15/10
a = 1.5 m/s²
Therefore, the the acceleration of the sled of mass 100 Kg will be
1.5 m/s².
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A ball rolls off a table and falls 0.75m to the floor, landing with a speed of 4 m/s. (A) What is the acceleration of the ball just before it strikes the ground? (B) What was the initial speed of the ball? (C) What initial speed must the ball have if it is to land with a speed of 5 m/s?
Answer:
We had this question yesterday, let me check my book quickly
(A) The acceleration of the ball is 9.81 m/s² just before it strikes the ground.
(B) The initial speed of the ball is equal to 1.14 m/s.
(C) The initial speed must be 3.21 m/s if it is to land with a speed of 5 m/s.
What are the equations of motion?The equation of motion is the way to represent the relation between the time, acceleration, initial and final velocity, and distance covered by a moving object.
The three equations of motion are:
[tex]v= u+ at\\v^2 = u^2 +2aS\\S = ut +(1/2) at^2[/tex]
The acceleration of the ball just before it strikes the ground is equal to gravitational acceleration, g = 9.81 m/s².
Given, the final velocity of the ball, v = 4m/s
The height of the table, h = 0.75 m
The initial velocity of the ball, [tex]u = \sqrt{v^2-2gh}[/tex]
[tex]u = \sqrt{(4)^2-2\times 9.8\times 0.75}[/tex]
u = 1.14 m/s
When the final velocity of the ball, v = 5m/s
The initial velocity will be :[tex]u = \sqrt{(5)^2-2\times 9.8\times 0.75}[/tex]
u = 3.21 m/s.
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A turntable turning at rotational speed 78 rpm stops in 45 s when turned off. The turntable's rotational inertia is 1.2 ×10−2kg⋅m2.
Determine the magnitude of the resistive torque that slows the turntable.
The magnitude of the torque that slows the turntable is equal to 2.17 × 10⁻³ N.m.
What is torque?Torque can be described as the measure of the force that can cause the rotation of an object about an axis. The force causes an object to accelerate similarly, torque causes an angular acceleration. Therefore, torque is the rotational equivalent of linear force.
Given, the rotational speed of the turntable, ω = 78 rpm = 8.16 rad/s
The time turntable takes to stop, t = 45 s
The rotational inertia of the turntable, I = 1.2 × 10⁻² Kg.m²
From the rotational kinetics equation, the angular acceleration:
[tex]\omega_f = \omega +\alpha t[/tex]
The final speed of the turntable = 0
[tex]\alpha =\frac{\omega_f-\omega}{t}[/tex]
[tex]\alpha =\frac{0-8.16}{45}[/tex]
[tex]\alpha =-0.1814 \;rad/s^2[/tex]
Calculation of torque by using inertia and angular acceleration:
[tex]\tau = I\alpha[/tex]
[tex]\tau =1.2 \times 10^{-2} \times (-0.1814)[/tex]
[tex]\tau = 2.17\times 10^{-3} \; N.m[/tex]
Therefore, the magnitude of the torque that slows the turntable is 2.17 × 10⁻³ N.m.
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Exercise 1 :On a circuit, a pilot covers 600 m in 7.2 s.1. Calculate its speed in m / s.2. Convert this speed to km / h in two different ways.
Given data
*The distance covers by the pilot is d = 600 m
*The given time is t = 7.2 s
(1)
The formula for the speed is given as
[tex]s=\frac{d}{t}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} s=\frac{600}{7.2} \\ =83.3\text{ m/s} \end{gathered}[/tex]Hence, the speed is s = 83.3 m/s
(2)
The speed converted into km/h as
[tex]\begin{gathered} s=83.3\times(\frac{5}{18}) \\ =299.88\text{ km/h} \end{gathered}[/tex]The second way to convert the speed into kilometer per hour as,
[tex]undefined[/tex]If sound waves travel at 350 m/s on a warm humid day. And you are submerged underwater. Is the sound you hear while below the surface traveling faster, slower or at the same rate as the sound you hear above the water? Explain your answer
ANSWER:
Sound waves are faster in water
STEP-BY-STEP EXPLANATION:
Sound travels differently in water than it does in air.
Sound waves travel faster in denser substances because neighboring particles will more easily collide with each other.
Water is denser than air, which means that sound waves travel much faster in water than in air.
Although they travel faster in water, we must keep in mind that the human ear evolved to hear sounds in the air and is not as useful when submerged in water.
draw each of the following vectors, label an angle that specifies the vectors direction, then find its magnitude and direction.a. B= -4.0 I+ 4.0jb. r= (-2.0i-1.0j) cmc. v= (-10-100j) m/s d. a= (20i+10j) m/s^2the I's and j's have the hat and its the vector simble on the letters
A cyclist travels a distance S from A to B consisting of three segments: s = S1 + $2 + $3, Time and speed over the segments are t1, t2, t3 and v1= 20(km/h), v2 = 15(km/h), V3= 10(km/h). Calculate the average speed over the distance s. Consider two cases: a) s1=s2=s3 b) t1=t2=t3
The average speed over the distance s consisting of three segments s1, s2, and s3 in first case a) s1 = s2= s3 is 13.85 km/h and in the second case b) t1 = t2 = t3 is 15 km/h.
In the first case where the distance for three intervals is the same a) s₁ = s₂ = s₃, the average speed for three segments is calculated by the formula:
Average speed = 3 ÷ ( 1/v₁ + 1/v₂ +1/v₃)
Average speed = 3 ÷ ( ¹/₂₀ + ¹/₁₅ + ¹/₁₀)
Average speed = 13.85 km/h
In the second case where the time taken for all three intervals is the same b) t₁ = t₂ = t₃, the average speed for three segments is calculated by the formula:
Average speed = (v₁ + v₂ +v₃) ÷ 3
Average speed = (20 + 15 + 10) ÷ 3
Average speed = 15 km/h
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If an astronaut weighed 200 pounds on the surface of the earth how much would the astronaut weigh 4000 miles above the earths surface. Show working equation
Recall that the gravitational pull is given by
[tex]F=G\frac{m_{}\cdot M_E}{r^2}[/tex]Where m is the mass of the astronaut, ME is the mass of Earth, r is the distance between them, and G is the gravitational constant.
The above relation states that the force of gravitation is directly proportional to the mass of two objects and is inversely proportional to the square of the distance between them.
If the astronaut weighs 200 pounds and the radius of the Earth is 4000 miles then
[tex]\begin{gathered} 200=G\frac{m\cdot M_E}{4000^2} \\ G\cdot m\cdot M_E=200\cdot4000^2 \end{gathered}[/tex]When the astronaut moves 4000 miles above the earth's surface then the distance between them is
4000+4000 = 8000 miles
The distance is measured from the center of the earth so the radius must be included.
So, the new force of gravitation is
[tex]\begin{gathered} F=\frac{G\cdot m_{}\cdot M_E}{r^2} \\ F=\frac{200\cdot4000^2}{8000^2} \\ F=50\: lb \end{gathered}[/tex]Therefore, the astronaut weighs 50 pounds 4000 miles above the earth's surface.
This makes sense because as you go further away from the center of the Earth then your weight becomes less due to less force of gravity.
if an astronaut weighs 981 N on Earth and only 160 N on the moon, then what is the mass on the moon
If an astronaut weighs 981 N on Earth and only 160 N on the Moon, then his mass on the Moon will be 98.1 kg.
Let's calculate the mass of Earth as per the Earth's acceleration due to gravity.
Now, considering the acceleration due to gravity as 10m/s².
Mass = Weight/Acceleration due to gravity
Mass = 981/10
Mass = 98.1 kg
Now, as per the established fact, the mass is independent of acceleration due to the gravity of the planet i.e. The mass of the person on earth and on the moon is same.
Hence, his mass on the Moon will be 98.1 kg.
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How is the direction of the velocity of a satellite differ from the direction of its acceleration?
PLEASE HELP!!!!!
The direction of the velocity of a satellite differ from the direction of its acceleration by an angle of 90° as they act perpendicular to each other.
In an orbital motion, the acceleration of the object is always directed towards the center of the orbit. This acceleration is called as centripetal acceleration. It is denoted by [tex]a_{c}[/tex].
[tex]a_{c}[/tex] = v² / r
In an orbital motion, the velocity of the object is always tangential to the orbit. It can be calculated in two ways,
v = 2 π r / T
v = r ω
Therefore, the direction of the velocity of a satellite differ from the direction of its acceleration by an angle of 90°.
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An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid has a
mass of 5.8x 104 kg, and the force causes its speed to change from 6800 to 5600m/s. (a) What is the work done by the force? (b) If the
asteroid slows down over a distance of 2x 106 m determine the magnitude of the force.
The work done on the asteroid by the constant force is 4.06 × 10¹¹ joules and the force experienced by the asteroid is 1.89 × 10⁵ newtons
what is work energy theorem?
work energy theorem is an important phenomenon in physics which states that the change in kinetic energy is equal to the work done by the object.
This law is basic part of conservation of energy.
what is newtons law of motion?
Newton's laws of motion are three basic laws of classical mechanics that describe the relationship between the motion of an object and the forces acting on it. These laws can be paraphrased as follows: A body remains at rest, or in motion at a constant speed in a straight line, unless acted upon by a force.
given:
mass of the asteroid = 5.8 x 10⁴ kg
initial speed of the asteroid v₁ = 6800m/s
final speed of the asteroid v₂ = 5600 m/s
displacement of the asteroid = 2 × 10⁶ meters
a)
using work energy theorem for the asteroid we get,
work done = change in kinetic energy
work done = 1/2 × m × (v₂) ² - 1/2 × m × (v₁) ²
work done by the force on the asteroid is = 1/2 × 5.8 x 10⁴ (1.4 × 10⁷)
work done by the force on the asteroid is = 4.06 × 10¹¹ joules
b)
first, we will have to find the acceleration of the asteroid,
using newtons third law of motion we get,
a = 14/4
a = 3.5 m/s²
therefore, the force can be calculated using relation f = m × a
here,
m is mass
a is acceleration
therefore, the force on the asteroid is = 5.8 x 10⁴ × 3.5
the force on the asteroid is = 1.89 × 10⁵ newtons
therefore, the work done on the asteroid by the constant force is 4.06 × 10¹¹ joules and the force experienced by the asteroid is 1.89 × 10⁵ newtons
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The x-component of a force on a 46-g golf ball by a 7-iron versus time is plotted in the following figure: a. Find the x-component of the impulse during the intervals i. [0, 50 ms], and ii. [50 ms, 100 ms] b. Find the change in the x-component of the momentum during the intervals iii. [0, 50 ms], and iv. [50 ms, 100 ms]
The x-component of the impulse during the intervals [0, 50 ms] is 750 Nms.
The x-component of the impulse during the intervals [50, 100 ms] is 1,500 Nms.
The change in the x-component of the momentum during the intervals [0, 50 ms] is 0.75 kgm/s.
The change in the x-component of the momentum during the intervals [50, 100 ms] is 1.5 kgm/s.
What is the impulse experienced by the ball?
The impulse experienced by the ball is calculated from the product of force and time of motion of the ball.
J = Ft
where;
F is the applied forcet is the time of motionThe x-component of the impulse during the intervals i. [0, 50 ms] is calculated as follows;
From the diagram, the impulse between (0, 50 ms) is the area of the triangle.
Jₓ = ¹/₂(b)(h)
where;
b is the base of the triangle = 50 ms h is the height of the triangle = 30 NJₓ = ¹/₂(50 ms)(30 N) = 750 N.ms
The impulse during [50 ms, 100 ms] is the area of the rectangle,
Jₓ = Lb
where;
L is the length = 100 ms - 50 ms = 50 msb is the breadth = 30 NJₓ = 50 ms x 30 N
Jₓ = 1,500 Nms
Impulse is the change in momentum of an object.
The change in the x-component of the momentum during the intervals [0, 50 ms] is calculated as follows;
ΔP = Jₓ = ¹/₂(50 ms)(30 N) = 750 N.ms = 0.75 Ns = 0.75 kgm/s
For interval of [50 ms, 100 ms];
ΔP = 1,500 Nms = 1.5 Ns = 1.5 kgm/s
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A particle starts from rest of move with constant acceleration 4 m/s² along straight line then, Calculate velocity when it travel distance of 80 m step - by step explaination
Answer:
25.3 m/s
Explanation:
According to the laws of motion, the final velocity and the distance traveled by a particle moving at constant acceleration a m/s² is given by the equation
[tex]v^2 = v_0^2 + 2a(x - x_0)[/tex]
where
v= final velocity m/s
v₀ = initial velocity m/s
x = final position m
x₀ = initial position m
a = acceleration in m/s²
x - x₀ is called the displacement, d
We are given
v₀ = 0 since particle is starting at rest
d = 80 m
a = 4 m/s²
Plugging these values into the above equation.
v² = 0 + 2 x 4 x 80 m²/s²
v² = 640 m²/s²
v = √640 = 8√10 ≈ 25.3 m/s
When doing a chin-up, a student lifts her body with a force of 400N and a distance of 0.25 meters in 2 seconds. What is the power delivered by the students biceps?
Given:
The student lifts her body with a force F = 400 N
The distance is d = 0.25 N
The time is t = 2 s
To find: The power delivered by student's biceps.
Explanation:
The formula to calculate power is
[tex]P=\frac{F\times d}{t}[/tex]Substituting the values, the power will be
[tex]\begin{gathered} P=\frac{400\times0.25}{2} \\ =\text{ 50 W} \end{gathered}[/tex]Final Answer: The power delivered by student's biceps is 50 W.
assume that the brakes in your car create a constant deceleration if you double your driving speed how does this affect a0 the time required to come to a stop and b) the distance needed to stop?
For the moving car, if you double your driving speed;
a. the time required to come to a stop will increase by a factor of 2.
b. the distance needed to stop will increase by a factor of 2.
What is the formula relating acceleration, speed, time, and distance traveled?The formula relating to the acceleration, speed, time, and distance traveled of an object is a formula of the equations of motion.
The equations of motion are given below as follows:
v = u + at
s = ut + ¹/₂at²
v² = u² + 2as
where:
v = final speed
u = initial speed
t = time taken
s = distance traveled
a = acceleration
Considering the given scenario:
a. if the speed is doubled, the equation v = u + at applies for the time required to come to a stop;
t =2 * v - u / t
the time will be increase.
b. if the speed is doubled, the equation v² = u² + 2as applies for the distance needed to stop;
s = 2 * v² - u²
the distance will increase with an increase in speed.
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Four charges are arranged in a square formation. Take q to be 1 C of charge and a to be 2 cm in length. Four charges are arranged in a square formation. Find the net electric field at the center of the square.
We have to calculate the electric field in the center, so we need the distance to the center R and the interaction of each charge with that point
[tex]E=\sum_{n\mathop{=}0}^{\infty}E_i[/tex]To calculate the distance R we have a triangle
[tex]R=\sqrt{\frac{2a^2}{4}}=\frac{a}{\sqrt{2}}=0.014m=1.41cm[/tex][tex]\begin{gathered} \sum_{n\mathop{=}0}^{\infty}Ex=\frac{9\cdot10^9}{2\cdot10^{-4}}\cdot(cos45)\cdot(6C)=1.91\cdot10^{14}N/C \\ \sum_{n\mathop{=}0}^{\infty}Ey=4.5\cdot10^{13}\cdot(cos45\degree)(2C)=0.636\cdot10^{14}N/C \\ Etot=\sqrt{Ex^2+Ey^2}=2.01\cdot10^{14}N/C \end{gathered}[/tex]Is important to remember that E has direction so you have to calculate each axis, x, and y
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Given that hotter blackbodies produce more energy than cooler blackbodies, why do cooler red giants have much higher luminosities than much hotter white dwarfs?
Given that hotter blackbodies produce more energy than cooler blackbodies, cooler red giants have much higher luminosities than much hotter white dwarfs.
The hotter blackbodies emits more light as compared to the cooler blackbodies over all the wavelength. A blackbody emits light over all wavelength. the red giant is a star , hydrogen is fused to helium core to stat the fusion reactions. this hydrogen fusion caused the star to expand. the sun becomes red giant it will swell up to size of earth orbit. This gives a red color and make red giant much hotter. while the white dwarf comes after the red giant . white dwarf is also very hot but it is not hot enough to star fusion reaction in carbon and oxygen.
Thus, Given that hotter blackbodies produce more energy than cooler blackbodies, cooler red giants have much higher luminosities than much hotter white dwarfs.
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how fast is the angle of depression of the telescope changing when the boat is 190 meters from the shore
ANSWER:
- 0.01943 rad/sec
STEP-BY-STEP EXPLANATION:
The first thing is to make a drawing of what is mentioned in the statement, it would be the following:
Now, we have the following information:
[tex]\begin{gathered} \frac{dy}{dt}=15\text{ m/s} \\ x=50\text{ m} \\ y=190\text{ m} \end{gathered}[/tex]In this right angle triangle formed by telescope of the boat, e can apply the tangent trigonometric ratio, like this:
[tex]\begin{gathered} \tan \theta=\frac{x}{y} \\ \text{ replacing} \\ \theta=\tan ^{-1}\mleft(\frac{x}{y}\mright) \end{gathered}[/tex]Now, we implicitly derive with respect to t:
[tex]\begin{gathered} \frac{d}{dt}(\theta)=\frac{d}{dt}(\tan ^{-1}(\frac{x}{y})) \\ \frac{d}{dt}(\theta)=\frac{1}{1+(\frac{x}{y})^2}\cdot\frac{d}{dt}(\frac{x}{y}) \\ \frac{d}{dt}(\theta)=\frac{y^2}{x^2+y^2}\cdot x\cdot(-\frac{1}{y^2}\cdot\frac{dy}{dt}) \\ \frac{d}{dt}(\theta)=\frac{-x}{x^2+y^2}(\frac{dy}{dt}) \\ \text{ replacing} \\ \frac{d}{dt}(\theta)=\frac{-50}{50^2+190^2}\cdot(15) \\ \frac{d}{dt}(\theta)=-0.01943 \end{gathered}[/tex]The angle of depression is changing at a rate of -0.01943 rad/sec when the boat is 190 m from the shore