La longitud del faldón de la rampa es de 5.4 m.
La pendiente expresada en porcentaje sigue la siguiente ecuación:
[tex]m=\frac{y}{x}*100[/tex] (1)
Donde:
y es la elevacion de la rampa (faldón)x es la longitud de la ramapa (20 m)Sabemos que la pendiente es de 27%. Por lo tanto, usando la ecuación 1, despejamos y.
[tex]27=\frac{y}{20}*100[/tex]
[tex]y=\frac{27*20}{100}[/tex]
[tex]y=5.4\: m[/tex]
La longitud del faldón es 5.4 m
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A certain car traveling 33.0mph skids to a stop in 39m from the point where the brakes were applied. In approximately what distance would the car stop had it been going 66.0mph
Answer: 156.02 metre.
Explanation:
Give that a certain car traveling 33.0mph skids to a stop in 39m from the point where the brakes were applied.
Let us use third equation of motion,
V^2 = U^2 + 2as
Since the car is decelerating, V = 0
And acceleration a will be negative.
U = 33 mph
S = 39 m
Substitute both into the formula
0 = 33^2 - 2 × a × 39
0 = 1089 - 78a
78a = 1089
a = 1089 / 78
a = 13.96 m/h^2
If we assume that the car decelerate at the same rate.
the distance the car will stop had it been going 66.0mph will be achieved by using the same formula
V^2 = U^2 + 2as
0 = 66^2 - 2 × 13.96 × S
4356 = 27.92S
S = 4356 / 27.92
S = 156.02 m
Therefore, the car would stop at
156.02 m
what is used to measure the amount of sunshine
Answer:
Campbell-Stokes sunshine recorders or modern sunshine sensors.
Explanation:
Answer:
modern sunshine sensorssunshine recordersCampbell-Stokes
Một thang máy chuyển động thẳng đứng hướng xuống dưới chậm dần đều với gia tốc a= -4m/s2. Trần thang máy treo một vật nhỏ bằng một sợi dây mảnh, khoảng cách từ vật tới sàn thang máy h= 2m. Thang máy đang chuyển động thì dây đứt. Tính thời gian từ lúc dây đứt đến khi vật chạm sàn thang máy ? Lấy g= 10m/s2
Answer:
ejkshdkashalsflasfaksg
Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen if the spacing between the first and second dark fringes is to be 4.2 mm?
Answer:
The distance is [tex]D = 2.6 \ m[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 476.1 \ nm = 476.1 *10^{-9} \ m[/tex]
The distance between the slit is [tex]d = 0.29 \ mm = 0.29 *10^{-3} \ m[/tex]
The between the first and second dark fringes is [tex]y = 4.2 \ mm = 4.2 *10^{-3} \ m[/tex]
Generally fringe width is mathematically represented as
[tex]y = \frac{\lambda * D }{d}[/tex]
Where D is the distance of the slit to the screen
Hence
[tex]D = \frac{y * d}{\lambda }[/tex]
substituting values
[tex]D = \frac{ 4.2 *10^{-3} * 0.29 *10^{-3}}{ 476.1 *10^{-9} }[/tex]
[tex]D = 2.6 \ m[/tex]
In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:_____.
Answer:
In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal: speed of light(c)
Explanation:
Generally the ratio of the E(electric field ) and the B(magnetic field ) is equal to the speed of the electromagnetic wave i.e the speed of light (c) the value is
[tex]c = 3.0 *10^{8} \ m/s[/tex]
UVC light used in sterilizers, has wavelengths between 100 to 280 nm. If a certain UVC wave has a wavelength of 142.9 nm, what is the energy of one of its photons in J
Answer:
The energy of one of its photons is 1.391 x 10⁻¹⁸ J
Explanation:
Given;
wavelength of the UVC light, λ = 142.9 nm = 142.9 x 10⁻⁹ m
The energy of one photon of the UVC light is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f is frequency of the light
f = c / λ
where;
c is speed of light = 3 x 10⁸ m/s
λ is wavelength
substitute in the value of f into the main equation;
E = hf
[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{142.9*10^{-9}} \\\\E = 1.391*10^{-18} \ J[/tex]
Therefore, the energy of one of its photons is 1.391 x 10⁻¹⁸ J
One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious space camper and you launch a serious rocket: it reaches an altitude of 211 km . What gain Δ???? in gravitational potential energy does the launch accomplish? The mass and radius of the Moon are 7.36×1022 kg and 1740 km, respectively.
Answer:
ΔP.E = 6.48 x 10⁸ J
Explanation:
First we need to calculate the acceleration due to gravity on the surface of moon:
g = GM/R²
where,
g = acceleration due to gravity on the surface of moon = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of moon = 7.36 x 10²² kg
R = Radius of Moon = 1740 km = 1.74 x 10⁶ m
Therefore,
g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²
g = 2.82 m/s²
now the change in gravitational potential energy of rocket is calculated by:
ΔP.E = mgΔh
where,
ΔP.E = Change in Gravitational Potential Energy = ?
m = mass of rocket = 1090 kg
Δh = altitude = 211 km = 2.11 x 10⁵ m
Therefore,
ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)
ΔP.E = 6.48 x 10⁸ J
A metal sphere A of radius a is charged to potential V. What will be its potential if it is enclosed by a spherical conducting shell B of radius b and the two are connected by a wire?
Answer:
The potential will be Va/b
Explanation:
So Let sphere A charged Q to potential V.
so, V= KQ/a. ....(1
Thus, spherical shell B is connected to the sphere A by a wire, so all charge always reside on the outer surface.
therefore, potential will be ,
V ′ = KQ/b = Va/b... That is from .....(1), KQ=Va]
1. Why do you see colors when you look at reflected light from a CD or DVD disk, or when you look at a soap bubble or oil film on water?
2. What do you think causes the colors on the artwork panels on the side of HLS2 (Health Sciences building) which change with time of day and the angle from which you view them?
Explanation:
1.The light reflected from the CD/DVD or soap bubble or oil film forms an interference with the surrounding light. The inference both constructive and destructive making some color appear and some disappear.
2.As light behaves as wave it will interfere differently at different angles. At certain angle it will interfere constructively and at certain angle it will interfere destructively making some color brighter and some disappear. So, at different angles the color are different.
Interference pattern is responsible for the formation of different colour when a light reflected from CD or soap bubble.
We can see colors when we look at reflected light from a CD or DVD disk, or a soap bubble or oil film on water because of the interference pattern. The colors that we see on the CD are created due to the reflection of white light from ridges in the metal. When light passes through something with many small ridges or scratches, we often see rainbow colors and interesting patterns.
These patterns are called interference patterns. White light is made up of 7 colors i.e. red, orange, yellow, green, blue, indigo, violet. The CD converts or separates the white light into 7 colors so we can conclude that interference pattern is responsible for the formation of different colour when a light reflected from CD OR soap bubble.
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A long bar slides on two contact points and is in motion with velocity ν. A steady, uniform, magnetic field B is present. The induced current through resistor R is:
Answer:
The induced current in the resistor is I = BLv/R
Explanation:
The induced emf ε in the long bar of length, L in a magnetic field of strength, B moving with a velocity, v is given by
ε = BLv.
Now, the current I in the resistor is given by
I = ε/R where ε = induced emf in circuit and R = resistance of resistor.
So, the current I = ε/R.
substituting the value of ε the induced emf, we have
I = ε/R
I = BLv/R
So, the induced current through the resistor is given by I = BLv/R
Describe how you expect the waveform and the sound you hear changes when you hit the tuning fork harder.
Answer:
In a tuning fork, two basic qualities of sound are considered, they are
1) The pitch of the waveform: This pitch depends on the frequency of the wave generated by hitting the tuning fork.
2) The loudness of the waveform: This loudness depends on the intensity of the wave generated by hitting the tuning fork.
Hitting the tuning fork harder will make it vibrate faster, increasing the number of vibrations per second. The number of vibration per second is proportional to the frequency, so hitting the tuning fork harder increase the frequency. From the explanation on the frequency above, we can say that by increasing the frequency the pitch of the tuning fork also increases.
Also, hitting the tuning fork harder also increases the intensity of the wave generated, since the fork now vibrates faster. This increases the loudness of the tuning fork.
how does a system naturally change over time
Answer:
The movement of energy and matter in a system differs from one system to another. On the other hand, in open system both the matter and energy move into and out of the system. Therefore, matter and energy in a system naturally change over time will decrease in entropy.
Explanation:
Answer:
Decrease in entropy
Explanation:
Various systems which exist in nature possess energy and matter that move through these system continuously. The movement of energy and matter in a system differs from one system to another.
In a closed system for example, only energy flows in and out of the system while matter does not enter or leave the system.
On the other hand, in open system both the matter and energy move into and out of the system.
You have three resistors: R1 = 1.00 Ω, R2 = 2.00 Ω, and R3 = 4.00 Ω in parallel. Find the equivalent resistance for the combination
Answer:
4 / 7
Explanation:
1/total resistance = 1/1 + 1/2 + 1/4
= 1¾
total resistance = 1 ÷ 1¾
= 4/7
A 10kg block with an initial velocity of 10 m/s slides 1o m across a horizontal surface and comes to rest. it takes the block 2 seconds to stop. The stopping force acting on the block is about
Answer:
-50N
Explanation:
F=ma=m(Vf-Vi)/t
m=10kgVf=0m/sVi=10m/st=2sF=(10)(-10)/(2)=-50N
So the force acting on the block is -50N, where the negative sign simply tells us that the force is opposite to the direction of movement.
Discuss the phase change condition due to reflection of light from a surface. Summarize equations of interference for thin film.
Answer:
if this surface has a higher index than in the medium where the light travels, the reflected wave has a phase change of 180º
Explanation:
When a ray of light falls on a surface if this surface has a higher index than in the medium where the light travels, the reflected wave has a phase change of 180º this can be explained by Newton's third law, the light when arriving pushes the atoms of the medium that is more dense, and these atoms respond with a force of equal magnitude, but in the opposite direction.
When the fractional index is lower than that of the medium where the reflacted beam travels, notice a change in phase.
Also, when light penetrates the medium, it modifies its wavelength
λ = λ₀ / n
We take these two aspects into account, the condition for contributory interference is
d sin θ = (m + 1/2) λ
for destructive interference we have
d sin θ = m λ
in general this phenomenon is observed at 90º
2 d = (m +1/2) λ° / n
2nd = (m + ½) λ₀
Can a car moving with a negative velocity moves faster than a car moving with a positive velocity? explain.
Answer:
Yes.
Explanation:
This is because "negative velocity" just means it is in the negative in relation to the point of 0. Negative velocity doesn't equal a decrease in velocity. For example lets say you were parked next to a cone (this cone represents zero) if you accelerate forwards then that would be positive acceleration. If you were to accelerate backwards, this would be in the negative direction, aka negative velocity.
SUMMARY:
A negative velocity means that the object which has the negative velocity is moving in the opposite direction of an object moving at a positive velocity. This is a question of frame of reference. The possibility for the velocity is what makes it different to the speed. Speed is only positive.
of
The radii a wheel are 25 cm
and 5cm respectively, it is found
that an effort of 40N is required
to raise slowly a load 16ON
160 N. Find the Mechanical
Adventage and Effeciency,
Answer:
Explanation:
Given that
Effort = 40N
Load = 16ON
M.A = load/effort
M.A= 160N/40N
M.A = 4
Velocity ratio = V.R =radius of the wheel/radius of the axel
= 25cm/5cm
= 5
Efficiency = mechanical advantage/velocity ratio × 100/1
= 4/5 × 100/1
= 0.8×100/1
= 80%
Hence, the mechanical advantage of the machine is 4 while the efficiency is 80%.
A planar electromagnetic wave is propagating in the x direction. At a certain point P and at a given instant, the magnitude of the electric field of the wave is 0.082 V/m . What is the magnetic vector of the wave at point P at that instant?
A) (0.27 nT)k
B) (-0.27 nT)k
C) (0.27 nTİ
D) (6.8 nT)k
E) (-6.8 nT))
Answer:
b
Explanation:
A mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 6 feet below the equilibrium position, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 1 2 the instantaneous velocity. Find the equation of motion x(t) if the mass is driven by an external force equal to f(t)
Answer:I don’t know
Explanation:
Please help!
Much appreciated!
Answer:
your question answer is 22°
Give examples of motion in which the directions of the velocity and acceleration vectors are (a) opposite, (b) the same, and (c) mutually perpendicular
Answer:
a) When moving body applies brake then velocity and acceleration would be in opposite direction
b) When body starts to increase velocity then velocity and acceleration would be in same direction
c) When body is circulating then velocity and acceleration would be perpendicular to each other
Explanation:
a) When body applies brake then its velocity starts decreasing, in this case its acceleration would try to stop the moving body. So direction of velocity would be same as direction of motion of body but direction of acceleration would be in opposite direction
b) When body starts to increase velocity, its acceleration would make the body to move faster. So direction of velocity would be the direction of motion of body and acceleration would also be in same direction
c) When body moves in circular path then its acceleration would be towards centre of circle and velocity would try to snap the body out of circle to straight line which in tangent to circle.
On the way to school, the bus speeds up from 20 m/s to 36 m/s in 4 seconds. What distance does the bus cover in this time frame
Answer:
Explanation:
initial velocity u = 20 m /s
final velocity v = 36 m /s
time taken t = 4 s .
acceleration = (v - u) / t
= (36 - 20) / 4
a = 4 m / s ²
from the formula
v² - u² = 2 a s , s is distance covered .
putting the values
36² - 20² = 2 x 4 x s
1296 - 400 = 8 x s
s = 112 m .
Answer:112
Explanation:
Two hoops, staring from rest, roll down identical incline planes. The work done by nonconservative forces is zero. The hoops have the same mass, but the larger hoop has twice the radius. Which hoop will have the greater total kinetic energy at the bottom
Answer:
They both have the same total K.E at the bottom
Explanation:
This Is because If assuming no work is done by non conservative forces, total mechanical energy must be conserved
So
K1 + U1 = K2 + U2
But If both hoops start from rest, and and at the bottom of the incline the level for gravitational potential energy is zero for reference
thus
K1 = 0 , U2 = 0
ΔK = ΔU = m g. h
But if the two inclines have the same height, and both hoops have the same mass m,
So difference in kinetic energy, must be the same for both hoops.
A rope, under a tension of 153 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by . where at one end of the rope, is in meters, and is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation?
Complete question is;
A rope, under a tension of 153 N and fixed at both ends, oscillates in a second harmonic standing wave pattern. The displacement of the rope is given by
y = (0.15 m) sin[πx/3] sin[12π t].
where x = 0 at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c)the mass of the rope? (d) If the rope oscillates in a third - harmonic standing wave pattern, what will be the period of oscillation?
Answer:
A) Length of rope = 4 m
B) v = 24 m/s
C) m = 1.0625 kg
D) T = 0.11 s
Explanation:
We are given;
T = 153 N
y = (0.15 m) sin[πx/3] sin[12πt]
Comparing this displacement equation with general waveform equation, we have;
k = 2π/λ = π/2 rad/m
ω = 2πf = 12π rad/s
Since, 2π/λ = π/2
Thus,wavelength; λ = 4 m
Since, 2πf = 12π
Frequency;f = 6 Hz
A) We are told the rope oscillates in a second-harmonic standing wave pattern. So, we will use the equation;
λ = 2L/n
Since second harmonic, n = 2 and λ = L = 4 m
Length of rope = 4 m
B) speed is given by the equation;
v = fλ = 6 × 4
v = 24 m/s
C) To calculate the mass, we will use;
v = √T/μ
Where μ = mass(m)/4
Thus;
v = √(T/(m/4))
Making m the subject;
m = 4T/v²
m = (4 × 153)/24²
m = 1.0625 kg
D) Now, the rope oscillates in a third harmonic.
So n = 3.
Using the formula f = 1/T = nv/2L
T = 2L/nv
T = (2 × 4)/(3 × 24)
T = 0.11 s
A flat loop of wire consisting of a single turn of cross-sectional area 7.30 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.50 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 2.60
Answer:
-0.73mA
Explanation:
Using amphere's Law
ε =−dΦB/ dt
=−(2.6T)·(7.30·10−4 m2)/ 1.00 s
=−1.9 mV
Using ohms law
ε=V =IR
I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA
Rank these electromagnetic waves on the basisof their speed (in vacuum).
Rank from fastest to slowest. To rankitems as equivalent, overlap them.
yellow light
FM radio wave
green light
X-ray
AM radio wave
infrared wave
Answer:
All electromagnetic waves travel at the same speed in a vacuum
Explanation:
All the wave listed in the question are electromagnetic waves. The speed of electromagnetic waves (collectively called light) in a vacuum is fixed. Its value is 3×10^8 ms^-1. This is a constant for all electromagnetic waves irrespective of their frequency.
Hence for any electromagnetic wave, its speed is 3×10^8 ms^-1, this will be the common velocity of all the electromagnetic waves listed in the question in a vacuum thus we can not rank them according to speed.
A diffraction grating with 161 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles in the first-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm
Answer:
[tex]\theta_1 = 0.400^o[/tex]
[tex]\theta_2 =0.378^o[/tex]
Explanation:
From the question we are told that
The number of slits per cm is k = [tex]161\ slits\ per\ cm = 161 \ slits\ per\ 0.01 m[/tex]
The order of the maxima is n = 1
The wavelength are [tex]\lambda_1 = 434 nm = 434 *10^{-9} \ m \ \ \ , \lambda_2 = 410nm = 410 *10^{-9} \ m[/tex]
The spacing between the slit is mathematically represented as
[tex]d = \frac{ 0.01}{k}[/tex]
=> [tex]d = \frac{ 0.01}{161}[/tex]
=> [tex]d = 6.211 *10^{-5} \ m[/tex]
Generally the condition for constructive interference is
[tex]n\lambda = d \ sin \theta[/tex]
At [tex]\lambda_1[/tex]
[tex]\theta _1 = sin^{-1} [ \frac{1 * 434 *10^{-9}}{6.211 *10^{-5}} ][/tex]
[tex]\theta_1 = 0.400^o[/tex]
At [tex]\lambda_2[/tex]
[tex]\theta _2 = sin^{-1} [ \frac{1 * 410 *10^{-9}}{6.211 *10^{-5}} ][/tex]
[tex]\theta_2 =0.378^o[/tex]
Astronomers discover an exoplanet (a planet of a star other than the Sun) that has an orbital period of 3.75
Earth years in its circular orbit around its sun, which is a star with a measured mass of 3.23×1030kg
. Find the radius of the exoplanet's orbit.
Answer:
[tex]r=4.24\times 10^{11}\ m[/tex]
Explanation:
Given that,
Orbital time period, T = 3.75 earth years
Mass of star, [tex]m=3.23\times 10^{30}\ kg[/tex]
We need to find the radius of the exoplanet's orbit. It is a concept of Kepler's third law of motion i.e.
[tex]T^2=\dfrac{4\pi^2}{GM}r^3[/tex]
r is the radius of the exoplanet's orbit.
Solving for r we get :
[tex]r=(\dfrac{T^2GM}{4\pi^2})^{1/3}[/tex]
We know that, [tex]1\ \text{earth year}=3.154\times 10^7\ \text{s}[/tex]
So,
[tex]r=(\dfrac{(3.75\times 3.154\times 10^7)^2\times 6.67\times 10^{-11}\times 3.23\times 10^{30}}{4\pi^2})^{1/3}\\\\r=4.24\times 10^{11}\ m[/tex]
So, the radius of the exoplanet's orbit is [tex]4.24\times 10^{11}\ m[/tex].
The maximum gauge pressure in a hydraulic system is 15 atm. What is the largest mass that could be lifted by this system if the diameter of the piston is 65 cm
Answer:
The maximum force that can be lifted by this system is 51,478.4 kg
Explanation:
Given;
maximum gauge pressure of the hydraulic system, Hp = 15 atm = 1.52 x 10⁶ N/m²
diameter of the piston, d = 65 cm = 0.65 m
The maximum gauge pressure of the piston is given as;
[tex]Hp = \frac{F}{A}[/tex]
Where;
F is the maximum force of the piston
A is the area of the piston
[tex]A = \pi (\frac{0.65}{2} )^2\\\\A = 0.3319 \ m^2[/tex]
F = Hp x A
F = 1.52 x 10⁶N/m² x 0.3319m²
F = 504488 N
Force is given as;
F = mg
m = F/g
m = 504488/9.8
m = 51,478.4 kg
Therefore, the maximum force that can be lifted by this system is 51,478.4 kg
an alternating voltage of 100V, 50HZ Is Applied across an impedance of (20-j30) calculate the resistance, the capacitance, current, the phase angle between current and voltage
The resistance R = 20 Ω
The capacitance C = 106.1 μF
The current, I is 2.773 A at 56.31°.
The phase angle of the between the current and the voltage is 56.31° leading.
Since the impedance Z = 20 - j30 Ω, the resistance, R is the real part of the impedance. So R = ReZ = 20 Ω
So, the resistance R = 20 Ω
To find the capacitance, we need first to find the reactance of the capacitor X. Since the impedance Z = 20 - j30, the reactance of the capacitor X. is the imaginary part of the impedance. So X = ImZ = 30 Ω.
Now the reactance of the capacitor X = 1/ωC where ω = angular frequency of the circuit = 2πf where f = frequency of the circuit = 50 Hz and C = capacitance
So, C = 1/ωX = 1/2πfX
Substituting the values of the variables into the equation, we have
C = 1/2πfX
C = 1/(2π × 50 Hz × 30 Ω)
C = 1/3000π
C = 1/9424.778
C = 1.061 × 10⁻⁴ F
C = 106.1 × 10⁻⁶ F
C = 106.1 μF
So, the capacitance is 106.1 μF
The current I = V/Z where V = voltage = 100 V at 0° and Z = impedance.
The magnitude of Z = √(20² + (-30)²)
= √(400 + 900)
= √1300
= 36.06 Ω
and its angle Φ = tan⁻¹(ImZ/ReZ)
= tan⁻¹(-30/20)
= tan⁻¹(-1.5) = -56.31°
So, V = 100 ∠ 0° and Z = 36.06 ∠ -56.31°
So, the current, I = V/Z = (100 ∠ 0°)/36.06 ∠ -56.31°
= 100/36.06 ∠(0° - (-56.31° ))
= 2.773 ∠ 56.31° A
So, the current is 2.773 A at 56.31°.
Since the current is 2.773 A at 56.31°, the phase angle of the between the current and the voltage is 56.31° leading.
So, the phase angle of the between the current and the voltage is 56.31° leading.
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