Answer:
E = 3495.96 J
Explanation:
From the law of conservation of energy:
Total Mechanical Energy at h1 = Total Mechanical Energy at h2
Kinetic energy at h1 + potential energy at h1 = Kinetic energy at h2 + potential energy at h2 + Mechanical Energy Lost due to Friction
[tex]K.E_{h1}+P.E_{h1} = K.E_{h2}+P.E_{h2} + E\\\\\frac{1}{2}mv_1^2\ J + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2 + E\\\\\frac{1}{2}(36\ kg)(0\ m/s)_1^2\ J + (36\ kg)(9.81\ m/s^2)(24\ m)_1 = \frac{1}{2}(36\ kg)(7.8\ m/s)_2^2 + (36\ kg)(9.81\ m/s^2)(11\ m)_2 + E\\\\0\ J + 8475.84\ J = 1095.12\ J + 3884.76\ J + E\\E = 8475.84\ J - 1095.12\ J - 3884.76\ J\\[/tex]
E = 3495.96 J
A submarine has a "crush depth" (that is, the depth at which
water pressure will crush the submarine) of 400 m. What is
the approximate pressure (water plus atmospheric) at this
depth? (Recall that the density of seawater is 1025 kg/m3, g=
9.81 m/s2, and 1 kg/(m-s2) = 1 Pa = 9.8692 x 10-6 atm.)
Answer:
P =40.69 atm
Explanation:
We need to find the approximate pressure at a depth of 400 m.
It can be calculated as follows :
P = Patm + ρgh
Put all the values,
[tex]P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm[/tex]
So, the approximate pressure is equal to 40.69 atm.
Question: A NEO distance from the Sun is 1.17 AU. What is the speed of the NEO (round your answer to 2 decimal places)
Answer:
v = 2.75 10⁴ m / s
Explanation:
For this exercise we must use Kepler's third law which is an application of Newton's second law to the solar system
F = ma
where force is the force of gravity
F = [tex]G \frac{m M}{r^2}[/tex]
acceleration is centripetal
a = [tex]\frac{v^2}{r}[/tex]
we substitute
G m M / r² = m v² / r
[tex]\frac{GM}{r}[/tex] = v²
v = [tex]\sqrt{GM/r}[/tex]
indicate that the radius of the orbit is r = 1.17 AU, let's reduce to the SI system
r = 1.17 AU (1.496 10¹¹ m / 1 AI) = 1.76 10¹¹ m
let's calculate
v = [tex]\sqrt{\frac{6.67 \ 10^{-11} 1.991 \ 10^{30} }{ 1.76 \ 10^{11}} }[/tex]Ra (6.67 10-11 1.991 10 30 / 1.76 10 11
v = [tex]\sqrt{7.5454 \ 10^8 }[/tex]ra 7.5454 10 8
v = 2.75 10⁴ m / s
A 0.20 mass on a horizontal spring is pulled back a certain distance and released. The maximum speed of the mass is measured to be 0.20 . If, instead, a 0.40 mass were used in this same experiment, choose the correct value for the maximum speed.
a. 0.40 m/s.
b. 0.20 m/s.
c. 0.28 m/s.
d. 0.14 m/s.
e. 0.10 m/s.
Answer:
d
Explanation:
Ya gon find the Kenitic Energy first
K=½mv²===> K=½×0.2×(0.2)²===> 0.1(0.04)===> 0.004
and now the replacement:
0.004=½×0.4V²====> v²=0.02===> V=0.14m/s
A typical incandescent light bulb consumes 75 W of power and has a mass of 20 g. You want to save electrical energy by dropping the bulb from a height great enough so that the kinetic energy of the bulb when it reaches the floor will be the same as the energy it took to keep the bulb on for 2.0 hours. From what height should you drop the bulb, assuming no air resistance and constant g?
Answer:
h = 2755102 m = 2755.102 km
Explanation:
According to the given condition:
Potential Energy = Energy Consumed by Bulb
[tex]mgh = Pt\\\\h = \frac{Pt}{mg}[/tex]
where,
h = height = ?
P = Power of bulb = 75 W
t = time = (2 h)(3600 s/1 h) = 7200 s
m = mass of bulb = 20 g = 0.02 kg
g = acceleration due to gravity = 9.8 m/s²
Therefore,
[tex]h = \frac{(75\ W)(7200\ s)}{(0.02\ kg)(9.8\ m/s^2)}[/tex]
h = 2755102 m = 2755.102 km
A car is moving at a speed of 60 mi/hr (88 ft/sec) on a straight road when the driver steps on the brake pedal and begins decelerating at a constant rate of 10ft/s2 for 3 seconds. How far did the car go during this 3 second interval?
Answer:
219 ft
Explanation:
Here we can define the value t = 0s as the moment when the car starts decelerating.
At this point, the acceleration of the car is given by the equation:
A(t) = -10 ft/s^2
Where the negative sign is because the car is decelerating.
To get the velocity equation of the car, we integrate over time, to get:
V(t) = (-10 ft/s^2)*t + V0
Where V0 is the initial velocity of the car, we know that this is 88 ft/s
Then the velocity equation is:
V(t) = (-10 ft/s^2)*t + 88ft/s
To get the position equation we need to integrate again, this time we get:
P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0
Where P0 is the initial position of the car, we do not know this, but it does not matter for now.
We want to find the total distance that the car traveled in a 3 seconds interval.
This will be equal to the difference in the position at t = 3s and the position at t = 0s
distance = P(3s) - P(0s)
= ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)
= ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)
= (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft
The car advanced a distance of 219 ft in the 3 seconds interval.
Question 3 of 10
Which statement describes the law of conservation of energy?
A. Air resistance has no effect on the energy of a system.
B. Energy cannot be created or destroyed.
C. The total energy in a system can only increase.
D. Energy cannot change forms.
هما
SUBMIT
Answer:
B . energy cannot be created or destroyed
Two children sit on a seesaw that is in rotational equilibrium. The first child has weight W and sits at distance d from the pivot. If the second child sits at a distance of 7*d from the pivot, what must be the weight of the second child
Answer:
W/7
Explanation:
By principle of moments,
Sum of clockwise moment = sum of anticlockwise moment
Weight × 7d = W × d
Weight = W/7
Since the two children are in rotational equilibrium, the weight of the second child is W/7.
How can the weight of the second child be determined?The weight of the second child can be determined from the principle of moments.
The principle of moments states that for a body in equilibrium, the sum of the clockwise moments and anticlockwise moments about a point is zero.
Let the weight of the second child be X
From the principle of moments:
W × d = 7×d × X
X = W/7
Therefore, the weight of the second child is W/7.
Learn more about principle of moments at: https://brainly.com/question/20298772
when blueshift occurs,the preceived frequency of the wave would be?
Answer:
When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.
Explanation:
As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.
Blueshifts happens when the source of the wave and the observer are moving closer towards one another.
Assume that the wave is of frequency [tex]f\; {\rm Hz}[/tex] at the source. In other words, the source of the wave sends out a peak after every [tex](1/f)\; {\text{seconds}}[/tex].
Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.
The source of this wave sends out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. It would appear to the observer that consecutive peaks arrive every [tex](1/f)\; {\text{seconds}}\![/tex]. That would correspond to a frequency of [tex]f\; {\rm Hz}[/tex].
On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of [tex]v \; {\rm m \cdot s^{-1}}[/tex].
Again, the source of this wave would send out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. However, by the time the source sends out the second peak, the source would have been [tex]v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}[/tex] closer to the observer then when the source sent out the first peak.
When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than [tex](1/f)\; {\text{seconds}}[/tex]. The observed frequency of this wave would be larger than the original [tex]f\; {\rm Hz}[/tex].
Part AFind the x- and y-components of the vector d⃗ = (4.0 km , 29 ∘ left of +y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.d⃗ = km Part BFind the x- and y-components of the vector v⃗ = (2.0 cm/s , −x-direction).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.v⃗ = cm/s Part CFind the x- and y-components of the vector a⃗ = (13 m/s2 , 36 ∘ left of −y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.a⃗ x = m/s2
Solution :
Part A .
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, d = [tex]\text{4 km 29}[/tex] degree left of [tex]y[/tex]-axis.
So the [tex]x[/tex] component is = -4 x sin (29°) = -1.939 km
[tex]y[/tex] component is = 4 x cos (29°) = 3.498 km
Part B
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{v = 2 cm/s}[/tex] , [tex]\text{-x direction}[/tex]
So the [tex]x[/tex] component is = -2 cm/s
[tex]y[/tex] component is = 0
Part C
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{a = 13 m/s, 36 degree}[/tex] left of [tex]y[/tex]-axis.
So the [tex]x[/tex] component is = -13 x sin (36°) = -7.6412 [tex]m/S^2[/tex]
[tex]y[/tex] component is = -13 x cos (36°) = -10.517 [tex]m/S^2[/tex]
The x- and y-components of the vectors is mathematically given as as follows for each Part respectively
x= -1.939 km, y= 3.498 km
x= -2 cm/s, 0
y=, x= -7.6412m/s^2, -10.517m/s^2
What are the x- and y-components of the vectors?
Question Parameters:
Generally, we follow a basic principle where
x component= Fsin\theta
y component= Fcos\theta
Therefore
For A
x component is
x= -4 x sin (29°)
x= -1.939 km
y component is
y= 4 x cos (29°)
y= 3.498 km
For B
x component is
x= -2 cm/s
y component is
y= 0
For C
x component is
x= -13 x sin (36°)
x= -7.6412m/s^2
y component is
y= -13 x cos (36°)
y= -10.517m/s^2
Read more about Cartession co ordinate
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A new car manufacturer advertises that their car can go from zero to sixty mph in 8 [s]. This is a description of
Answer:
Acceleration
Explanation:
The fact that new can go from zero to 60mph in 8 secs is a description of its pick-up or in physics, it's called acceleration.
Here initial velocity u= 0
final velocity v = 60 mph = 1m/minute.
or v =1609.344/60 = 26.82m/s
and time taken to do so is 8 sec
Acceleration a = (v-u)/t
a = (26.82-0)/8 = 3.35 m/s^2
Therefore, acceleration of the car a = 3.35 m/s^2.
A Catapult throws a payload in a circle with an arm that is 65.0 cm long. At a certain instant, the arm is rotating at 8.0 rad/s and the angular speed is increasing at 40.0 rad/s2. For this instant, find the magnitude of the acceleration of the payload.
Answer:
The acceleration of the payload is 26 m/s2.
Explanation:
length, L = 65 cm = 0.65 m
angular acceleration = 40 rad/s^2
The acceleration is given by
a = angular acceleration x length
a = 40 x 0.65
a = 26 m/s^2
A student claimed that thermometers are useless because a
thermometer always registers its own temperature. How would you
respond?
[
A tire is filled with air at 22oC to a gauge pressure of 240 kPa. After driving for some time, if the temperature of air inside the tire is 45oC, what fraction of the original volume of air must be removed to maintain the pressure at 240 kPa?
Answer:
7.8% of the original volume.
Explanation:
From the given information:
Temperature [tex]T_1[/tex] = 22° C = 273 + 22 = 295° C
Pressure [tex]P_1[/tex] = 240 kPa
Temperature [tex]T_2[/tex] = 45° C
At initial temperature and pressure:
Using the ideal gas equation:
[tex]P_1V_1 =nRT_1[/tex]
making V_1 (initial volume) the subject:
[tex]V_1 = \dfrac{nRT_1}{P_1}[/tex]
[tex]V_1 = \dfrac{nR*295}{240}[/tex]
Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:
the final volume [tex]V_2[/tex] can be computed as:
[tex]V_2 = \dfrac{nR*318}{240}[/tex]
Now, the change in the volume ΔV = V₂ - V₁
[tex]\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}[/tex]
[tex]\Delta V = \dfrac{23nR}{240}[/tex]
∴
The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:
[tex]= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}[/tex]
[tex]= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}[/tex]
[tex]= 0.078[/tex]
= 7.8% of the original volume.
if the tin is made of a metal which has a density of 7800 kg per metre cubic calculate the volume of the metal used to make tin and lead
Answer:
XL sleep usual Addison officer at home and ear is not a short time to be be free and ear is a short time to make a short time
Explanation:
so that I can take the class on Monday and ear is not a short time to be be free and ear is not a short time to be be free and ear is not a short time to be be free and ear is not a short time to time for a day or night and ear buds is Anshu and duster and duster fgor a day or night is Anshu and duster for a day or not a week of computer science from your computer and I am in the same as I am a short of ti and you can be the first time I will be be
suppose the tank is open to the atmosphere instead of being closed. how does the pressure vary along
Answer:
Pressure is more in the open container than the closed one.
Explanation:
The pressure due to the fluid at a depth is given by
Pressure = depth x density of fluid x gravity
So, when the container is open, the atmospheric pressure is also add up but when the container is closed only the pressure due to the fluid is there.
So, when the container is open, the pressure is atmospheric pressure + pressure due to the fluid.
hen the container is closed only the pressure due to the fluid is there.
An inductor of inductance 0.02H and capacitor of capatance 2uF are connected in series to an a.c. source of frequency 200 Hz- Calculate the Impedance in the circuit . TC
Explanation:
Given:
L = 0.02 H
C = [tex]2\:\mu \text{F}[/tex]
f = 200 Hz
The general form of the impedance Z is given by
[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]
Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to
[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]
[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]
Differences between angle of twist and angle of shear
Answer:
idek
Explanation:
What is the biggest planet in the solar system
Answer:
Jupiter
Explanation:
Answer:
The answer is Jupiter.
Explanation:
Jupiter is an orange/yellow colored planet.
A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m along the axis and above the center of the loop
Answer:
[tex]B=2.91\ \mu T[/tex]
Explanation:
Given that,
The current in the loop, I = 2 A
The radius of the loop, r = 0.4 m
We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :
[tex]B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}[/tex]
Put all the values,
[tex]B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T[/tex]
So, the required magnetic field is equal to [tex]2.91\ \mu T[/tex].
When a rigid body rotates about a fixed axis, all the points in the body have the same Group of answer choices linear displacement. angular acceleration. centripetal acceleration. tangential speed. tangential acceleration.
Answer:
angular acceleration.
Explanation:
Newton's law of universal gravitation states that the force of attraction (gravity) acting between the Earth and all physical objects is directly proportional to the Earth's mass, directly proportional to the physical object's mass and inversely proportional to the square of the distance separating the Earth's center and that physical object.
Generally, when a rigid body is made to rotate about a fixed axis, all the points in the body would typically have the same angular acceleration, angular displacement, and angular speed.
A 1.0 ball moving at 2.0 / perpendicular to a wall rebounds from the wall at 1.5 /. If the ball was in contact with the wall for 0.1 , what force did the wall impart onto the ball?
Answer:
5N
Explanation:
We have a simple problem of momentum here.
ΔMomentum= mΔv= FΔt
Solve for F
mΔv/Δt=F
Plug in givens
1*(2-1.5)/0.1=F
F=5N
The amount of force that the wall imparts on the ball is 5.0N
According to Newton's second law, the formula for calculating the force applied is expressed as:
[tex]F=ma[/tex]
m is the mass of the object
a is the acceleration of the object
Since acceleration is the change in velocity of an object, hence [tex]a=\frac{\triangle v}{t}[/tex]
The applied force formula becomes [tex]F=\frac{m\triangle v}{t}[/tex]
Given the following parameters
m = 1.0kg
[tex]\triangle v=2.0-1.5\\\triangle v=0.5m/s[/tex]
t = 0.1sec
Substitute the given parameter into the formula
[tex]F=\frac{1.0\times 0.5}{0.1}\\F=\frac{0.5}{0.1}\\F=5N[/tex]
Hence the amount of force that the wall imparts on the ball is 5.0N
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Container A and container B hold samples of the same ideal gas. The volume and the pressure of container A is equal to the volume and pressure of container B, respectively. If Container A has half as many molecules of the ideal gas in it as Container B does, then which of the following mathematical statements is correct regarding the absolute temperatures TA and TB in Container A and Container B. respectively?
A. TA = TB/2.
B. TA = 4TB.
C. TA = TB/4.
D. TA = 2TB.
E. TA = TB
Answer:
A. TA = TB/2.
Explanation:
Since container A has half as many molecules of the ideal gas in it as container B. Therefore, container A will have half the volume of gas as in container B:
[tex]V_A = \frac{1}{2}V_B[/tex]
Now, from Charle's Law:
[tex]\frac{V_A}{T_A}=\frac{V_B}{T_B}\\\\\frac{1}{2}\frac{V_B}{T_A}=\frac{V_B}{T_B}\\\\T_A = \frac{T_B}{2}[/tex]
Hence, the correct option is:
A. TA = TB/2.
If one lawn mower causes an 80-dB sound level at a point nearby, four lawnmowers together would cause a sound level of ____________ at that point. a.92 dB b.84 dB c.86 dB d.none of the above
Answer:
The intensity of 4 lawn movers is 86 dB.
Explanation:
Intensity of one lawnmower = 80 dB
Let the intensity is I.
Use the formula of intensity
[tex]dB = 10 log\left ( \frac{I}{Io} \right )\\\\80=10log\left ( \frac{I}{Io} \right )\\\\10^8 = \frac{I}{10^{-12}}\\\\I = 10^{-4} W/m^2[/tex]
Now the intensity of 4 lawn movers is
[tex]dB = 10 log\left ( \frac{4I}{Io} \right )\\\\dB=10log\left ( \frac{4\times10^{-4}}{10^{-12}} \right )\\\\dB = 86 dB\\[/tex]
A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now compressed so that its radius is R/2, and then it is entirely filled with the same fluid as before. As such, what is the density of the compressed sphere?
a. 8p
b. p/8
c. p/4
d. 4p
Answer:
a. 8p
Explanation:
We are given that
Radius of hollow sphere , R1=R
Density of hollow sphere=[tex]\rho[/tex]
After compress
Radius of hollow sphere, R2=R/2
We have to find density of the compressed sphere.
We know that
[tex]Density=\frac{mass}{volume}[/tex]
[tex]Mass=Density\times volume=Constant[/tex]
Therefore,[tex]\rho_1 V_1=\rho_2V_2[/tex]
Volume of sphere=[tex]\frac{4}{3}\pi r^3[/tex]
Using the formula
[tex]\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3[/tex]
[tex]\rho R^3=\rho_2\times \frac{R^3}{8}[/tex]
[tex]\rho_2=8\rho[/tex]
Hence, the density of the compressed sphere=[tex]8\rho[/tex]
Option a is correct.
Can you think of reasons why the charge on each ball decreases over time and where the charges might go
Answer:
By the principle of corona discharge.
Explanation:
The charge on each ball will decreases over time due to the electrical discharge in air.
According to the principle of corona discharge, when the curvature is small, the discharge of the charge takes placed form the pointed ends.
A 1.64 kg mass on a spring oscillates horizontal frictionless surface. The motion of the mass is described by the equation: X = 0.33cos(3.17t). In the equation, x is measured in meters and t in seconds. What is the maximum energy stored in the spring during an oscillation?
Answer:
[tex]K.E_{max}=0.8973J[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=1.64kg[/tex]
Equation of Mass
[tex]X=0.33cos(3.17t)[/tex]...1
Generally equation for distance X is
[tex]X=Acos(\omega t)[/tex]...2
Therefore comparing equation
Angular Velocity [tex]\omega=3.17rad/s[/tex]
Amplitude A=0.33
Generally the equation for Max speed is mathematically given by
[tex]V_{max}=A\omega[/tex]
[tex]V_{max}=0.33*3.17[/tex]
[tex]V_{max}=1.0461m/s[/tex]
Therefore
[tex]K.E_{max}=0.5mv^2[/tex]
[tex]K.E_{max}=0.5*1.64*(1.0461)^2[/tex]
[tex]K.E_{max}=0.8973J[/tex]
Topic: Chapter 10: Projectory or trajectile?
Projectile range analysis:
A projectile is launched from the ground at 10 m/s, at
an angle of 15° above the horizontal and lands 5.1 m away.
What other angle could the projectile be launched at, with the same velocity,
and land 5.1 m away?
90°
75°
45
50°
30°
Answer:
The other angle is 75⁰
Explanation:
Given;
velocity of the projectile, v = 10 m/s
range of the projectile, R = 5.1 m
angle of projection, 15⁰
The range of a projectile is given as;
[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]
To find another angle of projection to give the same range;
[tex]5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0[/tex]
Check:
sin(2θ) = sin(2 x 75) = sin(150) = 0.5
sin(2θ) = sin(2 x 15) = sin(30) = 0.5
As you move farther away from a source emitting a pure tone, the ___________ of the sound you hear decreases.
Answer:
frequency
Explanation:
The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.
So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is
[tex]f' = \frac{v + v_o}{v+ v_s} f[/tex]
where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.
Light with a wavelength of 5.0 · 10-7 m strikes a surface that requires 2.0 ev to eject an electron. Calculate the energy, in joules, of one incident photon at this frequency. _____ joules 4.0 x 10 -19 4.0 x 10 -49 9.9 x 10 -32 1.1 x 10 -48
Answer:
pretty sure its 6.2 x 10^-13
Explanation:
I looked it up I'm not a bigbrain but want to help
Part of your electrical load is a 60-W light that is on continuously. By what percentage can your energy consumption be reduced by turning this light off
Answer:
Following are the solution to the given question:
Explanation:
Please find the complete question in the attached file.
The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.
[tex]\to \frac{\$60}{\frac{30 \ days}{24\ hours}} = \$0.08 / kwh.[/tex]
Thus, [tex]\frac{\$0.08}{\$0.12} = 0.694 \ kW \times 0.694 \ kW \times 1000 = 694 \ W.[/tex]
The electricity used is continuously 694W over 30 days.
If just resistor loads (no reagents) were assumed,
[tex]\to I = \frac{P}{V}= \frac{694\ W}{120\ V} = 5.78\ A[/tex]
Energy usage reduction percentage = [tex](\frac{60\ W}{694\ W} \times 100\%)[/tex]
This bulb accounts for [tex]8.64\%[/tex] of the energy used, hence it saves when you switch it off.
