Answer:
λ₂ = 357.3 nm
Explanation:
The expression for double-slit interference is
d sin θ = m λ constructive interference
d sin θ = (m + ½) λ destructive interference.
The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits
d sin θ = m λ₁
now ask for destructive interference for m = 4
d sin θ = (m + ½) λ₂
we match these two expressions
m λ₁ = (m + ½) λ₂
λ₂ = ( m / m + ½) λλ₁
let's calculate
λ₂ =[tex]\frac{4}{(4.000 +0.5) \ 401}[/tex]
λ₂ = 357.3 nm
A diffraction grating has 6000 lines per centimeter ruled on it. What is the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe when the grating is illuminated with a beam of light of wavelength 500 nm
Explanation:
Hope it Will help he hsuejwoamxgehanwpalasmbwfwfqoqlmdbehendalmZbgevzuxwllw. yeh we pabdvddxhspapalw. X
The angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
What is diffraction?Waves spreading outward around obstructions are known as diffraction. Sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as very small moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities all exhibit diffraction.
Given:
The number of lines = 6000 per cm,
The Wavelength, λ = 500 nm = 500 × 10 ⁻⁹ m
Calculate the diffraction grating,
[tex]d = 1 / no\ of\ lines[/tex]
d = 10⁻² / 6000 m,
Calculate the second-order maxima angle and third-order maxima angle by the formula given below,
[tex]dsin\theta_1 = n_1 \lambda[/tex]
[tex]sin\theta_1 = n_1\lambda / d[/tex]
[tex]\theta _1 = sin^{-1}[2\times 500\times 10 ^{-9}/10^{-2}\times 6000][/tex]
θ₁ = sin⁻¹(0.6)
θ₁ = 36.87°
Similarly, for θ₂,
θ₂ = sin⁻¹(3 × 500 × 10 ⁻⁹ / 10⁻² × 6000)
θ₂ = sin⁻¹(0.9)
θ₂ = 64.16°
Calculate the separation as follows,
θ₂ - θ₁ = 64.16° - 36.87°
θ₂ - θ₁ = 27.29°
Therefore, the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
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Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k = 8.99 × 1012N/C.
Answer:
Electric field at A = 9.28 x 10¹² N/C
Explanation:
Given:
K = 8.99 x 10¹² N/C
Missing information:
Length = 11 cm = 11 x 10⁻² m
q = 12.5 C
Find:
Electric field at A
Computation:
Electric field = Kq / r²
Electric field at A = [(8.99 x 10¹²)(12.5)] / [11 x 10⁻²]²
Electric field at A = 9.28 x 10¹² N/C
A refrigerator has a coefficient of performance equal to 4.00. The refrigerator takes in 110 J of energy from a cold reservoir in each cycle. (a) Find the work required in each cycle. J (b) Find the energy expelled to the hot reservoir. J
Answer:
The correct answer is:
(a) 27.5 Joules
(b) 141.5 Joules
Explanation:
Given:
Energy,
[tex]Q_c = 110 \ J[/tex]
Coefficient of performance refrigerator,
[tex]Cop(refrig)=4[/tex]
(a)
As we know,
⇒ [tex]Cop(refrig) = \frac{Q_c}{Work}[/tex]
or,
⇒ [tex]Work=\frac{Q_c}{Cop(refrig)}[/tex]
[tex]=\frac{110}{4}[/tex]
[tex]=27.5 \ Joules[/tex]
(b)
⇒ [tex]Heat \ expelled = Heat \ removed +Work \ done[/tex]
or,
⇒ [tex]Q_h = Q_c+Work[/tex]
[tex]=114+27.5[/tex]
[tex]=141.5 \ Joules[/tex]
a bullet is dropped from the same height when another bullet is fired horizontally. they will hit the ground
Answer:
it will drop simultaneously
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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A mass is tired to spring and begins vibration periodically the distance between it's lowest position is 48cm what is the Amplitude of the vibration
Answer:
The amplitude of vibration of the spring is "24 cm"
The periodic vibrating body's motion follows a sinusoidal path. This sinusoidal path is illustrated in the attached picture.
From the picture, it can be clearly seen that the amplitude of the periodic vibration motion is the distance from its mean position to the highest point.
Since the distance of both the highest and the lowest points from the mean position is the same. Therefore, the distance between the lowest and the highest point must be equal to two times the amplitude of the wave.
Amplitude = 24 cm
The voltage in an EBW operation is 45 kV. The beam current is 50 milliamp. The electron beam is focused on a circular area that is 0.50 mm in diameter. The heat transfer factor is 0.87. Calculate the average power density in the area in watt/mm2.
Answer:
[tex]P_d=6203.223062W/mm^2[/tex]
Explanation:
From the question we are told that:
Voltage [tex]V=45kV[/tex]
Current [tex]I=50mAmp[/tex]
Diameter [tex]d=0.50mm[/tex]
Heat transfer factor [tex]\mu= 0.87.[/tex]
Generally the equation for Power developed is mathematically given by
[tex]P=VI\\\\P=45*10^3*50*10^{-3}[/tex]
[tex]P=2.250[/tex]
Therefore
Power in area
[tex]P_a=1400*0.87[/tex]
[tex]P_a=1218watt[/tex]
Power Density
[tex]P_d=\frac{P_a}{Area}[/tex]
[tex]P_d=\frac{1218}{\pi(0.5^2/4)}[/tex]
[tex]P_d=6203.223062W/mm^2[/tex]
A resistor is submerged in an insulated container of water. A voltage of 12 V is applied to the resistor resulting in a current of 1.2 A. If this voltage and current are maintained for 5 minutes, how much electrical energy is dissipated by the resistor
Explanation:
Given:
[tex]\Delta t = 5\:\text{min} = 300\:\text{s}[/tex]
[tex]V = 12 V[/tex]
[tex]I = 1.2 A[/tex]
Recall that power P is given by
[tex]P = VI[/tex]
so the amount of energy dissipated [tex]\Delta E[/tex] is given by
[tex]\Delta E = VI\Delta t = (12\:\text{V})(1.2\:\text{A})(300\:\text{s})[/tex]
[tex]\:\:\:\:\:\:\:= 4320\:\text{W} = 4.32\:\text{kW}[/tex]
Your little sister (mass 25 kg) is sitting in her little red wagon (mass
8.5 kg) at rest. You begin pulling her forward, accelerating her with a
constant force for 2.35 s to a speed of 1.8 m/s. Calculate the impulse
you imparted to the wagon and its passenger.
Answer:
p = 60.6N*s
Explanation:
v_f = v_0+a*t
a = (v_f-v_0)/t
a = (1.8m/s)/2.35s
a = 0.77m/s²
F = m*a
F = (25kg+8.5kg)*0.77m/s²
F = 25.8N
^p = F*t
p = 25.8N*2.35s
p = 60.6N*s
what effect does the force of gravity have on a stone thrown vertically upwards
Answer:
rock go down
Explanation:
what comes up must come down.
If a conducting loop of radius 10 cm is onboard an instrument on Jupiter at 45 degree latitude, and is rotating with a frequency 2 rev/s; What is the maximum emf induced in this loop? If its resistance is 0.00336 ohms, how much current is induced in this loop? And what is the maximum power dissipated in the loop due to its rotation in Jupiter's magnetic field?
Answer:
a) fem = - 2.1514 10⁻⁴ V, b) I = - 64.0 10⁻³ A, c) P = 1.38 10⁻⁶ W
Explanation:
This exercise is about Faraday's law
fem = [tex]- \frac{ d \Phi_B}{dt}[/tex]
where the magnetic flux is
Ф = B x A
the bold are vectors
A = π r²
we assume that the angle between the magnetic field and the normal to the area is zero
fem = - B π 2r dr/dt = - 2π B r v
linear and angular velocity are related
v = w r
w = 2π f
v = 2π f r
we substitute
fem = - 2π B r (2π f r)
fem = -4π² B f r²
For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T
we reduce the magnitudes to the SI system
f = 2 rev / s (2π rad / 1 rev) = 4π Hz
we calculate
fem = - 4π² 428 10⁻⁶ 4π 0.10²
fem = - 16π³ 428 10⁻⁶ 0.010
fem = - 2.1514 10⁻⁴ V
for the current let's use Ohm's law
V = I R
I = V / R
I = -2.1514 10⁻⁴ / 0.00336
I = - 64.0 10⁻³ A
Electric power is
P = V I
P = 2.1514 10⁻⁴ 64.0 10⁻³
P = 1.38 10⁻⁶ W
Vector a has a magnitude of 8 and makes an angle of 45 with positive x axis vector B has also the same magnitude of 8 units and direction along the
Answer:
prove that Sin^6 ϴ-cos^6ϴ=(2Sin^2ϴ-1)(cos^2ϴ+sin^4ϴ)
please sove step by step with language it is opt maths question
The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.
Answer:
15.88
is the correct answer
A car is traveling at 118 km/h when the driver sees an accident 85 m ahead and slams on the brakes. What minimum constant deceleration is required to stop the car in time to avoid a pileup
Answer:
The constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
Explanation:
From the question, the car is traveling at 118 km/h, that is the initial velocity, u = 118km/h
The distance between the car and the accident at the moment when the driver sees the accident is 85 m, that is s = 85 ,
Since the driver slams on the brakes and the car will come to a stop, then the final velocity, v = 0 km/h = 0 m/s
First, convert 118 km/h to m/s
118 km/h = (118 × 1000) /3600 = 32.7778 m/s
∴ u = 32.7778 m/s
Now, to determine the deceleration, a, required to stop,
From one of the equations of motion for linear motion,
v² = u² + 2as
Then
0² = (32.7778)² + 2×a×85
0 = 1074.3841 + 170a
∴ 170a = - 1074.3841
a = - 1074.3841 / 170
a = - 6.3199
a ≅ - 6.32 m/s²
Hence, the constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
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There is more than 1 answer,
The picture is down
Answer:
test her prototype and collect data about its flight
Define hydropower or hydroelectric power ?
No Spam..
[tex]\:[/tex]
Hydroelectric power, also called hydropower is the electricity produced from generators driven by turbines that convert the potential energy of falling or fast-flowing water into mechanical energy.
Answer:
Hydroelectric power/hydropower - electricity produced by a hydraulic source, specifically energy generated falling or flowing water
Describe sound and record
Answer:
record is information created, received and maintained as evidence and information by an organization or person.in simpler terms it's a collection of of fields probably of different data types.
sound is however something loud or soft.which can be defined as vibrations that travel through the air or another medium.
I hope this helps
A painter sets up a uniform plank so that he can paint a high wall. The plank is 2 m long and weighs 400 N. The two supports holding up the plank are placed 0.2 m from either end. Show that the upwards force on each of the planks is 200 N. Draw a sketch.
The upward force on each supporting plank is 200 N
The given parameters include;
weight of the plank, W₁ = 400 Nlength of the plank, l = 2 mupward force of each supporting plank, = W₂ and W₃To show that the upward force of each supporting plank is 200 N, make the following sketch.
W₂ W₃
↑ ↑
-----------------------------------------------------------------------
0.2m ↓ 0.2m
400 N
The two supporting planks keeps the 2m plank in equilibrium position. If the plank is in equilibrium position the sum of the upward forces equals sum of the downward force.W₂ + W₃ = 400 N
But the distance of each supporting plank from the end is equal, (0.2m).
Then, W₂ = W₃
2W₂ = 400 N
W₂ = 400N/2
W₂ = 200 N
W₃ = 200 N
Therefore, the upward force on each supporting plank that keeps the plank in equilibrium position is 200 N.
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A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted. What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.30 cm?
A.
52.3 kN
B.
62.3 kN
C.
72.3 kN
D.
42.3 N
Answer:
cobina
Explanation:
me 2
Light takes 1.2 sec to get from the moon to the Earth. Assume you are looking at the moon with noticeable earth shine. If the Sun burned out, you would eventually see the crescent of the moon disappear. The earth shine part of the moon would disappear Answer 2.4 s after the crescent disappeared.
Answer:
1.2 seconds
Explanation:
Answer to the following question is 1.2 seconds
Because light from the moon takes 1.2 seconds to reach Earth, the light released from the crescent immediately before it vanishes will also take 1.2 seconds to reach Earth. As a result, the earth-shine portion of the moon will vanish 1.2 seconds after the crescent has vanished.
herical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.010-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass tran
Answer: Below is the complete question
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)
answer:
mass transfer coefficient = 9.56 * 10^-5 m/s
Explanation:
Candy density = 1950 kg/m^3
Candy diameter = 1 cm
Velocity of water = 1 m/s
water density = 1000 kg/m^3
Viscosity of water = 1 * 10^-3 kg/m/s
diffusion coefficient of candy in water = 2 * 10^-9 m^2/s
solubility of candy = 2 kg/m^3
Determine the mass transfer coefficient ( m/s )
( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3
where : Re = vdp / μ , Sh = KLd / Deff
attached below is the remaining solution .
mass transfer coefficient = 9.56 * 10^-5 m/s
A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand.
(a) At what angle was the ball thrown if its initial speed was 12.0 m/ s, assuming that the smaller of the two possible angles was used?
(b) What other angle gives the same range, and why would it not be used?
(c) How long did this pass take?
Answer:
a) θ = 14.23º, b) θ₂ = 75.77, c) t = 0.6019 s
Explanation:
This is a missile throwing exercise.
a) the reach of the ball is the distance traveled for the same departure height
R = [tex]\frac{v_o^2 \ sin 2 \theta }{g}[/tex]
sin 2θ = [tex]\frac{Rg}{v_o^2}[/tex]
sin 2θ = 7.00 9.8 / 12.0²
2θ = sin⁻¹ (0.476389) = 28.45º
θ = 14.23º
the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º
θ ’= 90 -14.23
θ’= 75.77º
b) the two angles that give the same range are
θ₁ = 14.23
θ₂ = 75.77
the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.
C) the time of the pass can be calculated with the expression
x = v₀ₓ t
t = x / v₀ₓ
t = 7 / 11.63
t = 0.6019 s
A circular parallel-plate capacitor whose plates have a radius of 25 cm is being charged with a current of 1.3 A. What is the magnetic field 11 cm from the center of the plates
The magnetic field at 11 cm from the center of the plates is 2.364 x 10⁻⁷ T.
Given;
radius of the circular plate, d = 25 cm = 0.25 m
current in the plate, I = 1.3 A
distance from the center of the circular plate, r = 11 cm = 0.11 m
To find:
magnetic field (B)The magnetic field from the given distance is calculated as from Biot Savart equation:
[tex]B = \frac{\mu_o I}{2\pi r} \\\\where;\\\\\mu_o \ is \ permeability \ of \ free \ space \ 4\pi \times 10^{-7} \ T.m/A\\\\B = \frac{(4\pi \times 10^{-7} ) \times (1.3)}{2\pi \times 0.11} \\\\B = 2.364 \ \times 10^{-6} \ T[/tex]
Therefore, the magnetic field 11 cm from the center of the plates is 2.364 x 10⁻⁷ T.
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1 A thing ring has a mass of 6kg and a radius of 20cm. calculate the rotational inertia.
Answer:
2400kgm²
Explanation:
Rotational inertia=mass x radius²
Which technological device makes an energy conversion in the same way that a human ear makes an energy conversion?
a.) a loudspeaker
b.) a headphone
c.) a light bulb
d.) a microphone
I think it's c because of the concept of mechanical energy to electrical energy but I'm not sure
Answer:
I THINK C
Explanation:
BECAUSE A Light Emitting Diode (LED) glows even when a weak electric current passes through it.
You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction between the box and the inclined plane is 0.680. a) Determine the static frictional force which holds the box in place. b) You slowly raise one end of the track, slowly increasing the incline of the angle. Determine the maximum angle that the incline can make with the horizontal so that the box just remains at rest. Ms 680 u Fgsin 281 Ffg Mgm r 680 55 4 8
Answer:
[tex]\theta=34 \textdegree[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=55kg[/tex]
Angle [tex]\theta =28.0[/tex]
Coefficient of static friction [tex]\alpha =0.680[/tex]
Generally, the equation for Newtons second Law is mathematically given by
For
[tex]\sum_y=0[/tex]
[tex]N=mgcos \theta[/tex]
for
[tex]\sum_x=0[/tex]
[tex]F_{s}=mgsin\theta[/tex]
Where
[tex]F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta[/tex]
[tex]F_{s}=0.68*55*9.8*cos 28[/tex]
[tex]F_{s}=323.62N[/tex]
Therefore
[tex]\alpha mgcos \theta=mg sin \theta[/tex]
[tex]\theta=tan^{-1}(0.68)[/tex]
[tex]\theta=34 \textdegree[/tex]
(a) The static frictional force which holds the box in place is 323.62 N.
(b) The maximum angle that the incline can make with the horizontal is 34.2⁰.
Net forceThe net force applied to keep the box at rest must be zero in order for the box to remain in equilibrium position. Apply Newton's second law of motion to determine the net force.
∑F = 0
Static frictional forceThe static frictional force is calculated as follows;
Fs = μFncosθ
Fs = 0.68 x (55 x 9.8) x cos28
Fs = 323.62 N
Maximum angle the incline can makeFn(sinθ) - μFn(cosθ) = 0
mg(sinθ) - μmg(cosθ) = 0
μmg(cosθ) = mg(sinθ)
μ(cosθ) = (sinθ)
μ = sinθ/cosθ
μ = tanθ
θ = tan⁻¹(μ)
θ = tan⁻¹(0.68)
θ = 34.2⁰
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A river is 87. meters wide and its current flows northward at 6 meters per second. A boat is launched with a velocity of 1.0 meters per second eastward from the west bank of the river. Determine the magnitude and direction of the boat’s resultant velocity as it crosses the river.
Answer:
explained
Explanation:
If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 1. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 2. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways.
A boat is trying to cross a river. Due to the velocity of river the path traveled by boat is diagonal. The velocity of boat v boat is in positive y direction. The velocity of river v river is in positive x direction. The resultant diagonal velocity v total which makes an angle of theta with the horizontal x axis is towards north east direction.
Figure 1. A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.
An airplane is trying to fly straight north with velocity v sub p. Due to wind velocity v sub w in south west direction making an angle theta with the horizontal axis, the plane’s total velocity is thirty eight point 0 meters per seconds oriented twenty degrees west of north.
Figure 2. An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the direction of its total velocity (solid arrow).
In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 1 and Figure 2. These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means.
How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Chapter 3.2 Vector Addition and Subtraction: Graphical Methods and Chapter 3.3 Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple—they add like ordinary numbers. For example, if a field hockey player is moving at 5 m/s
straight toward the goal and drives the ball in the same direction with a velocity of 30 m/s
relative to her body, then the velocity of the ball is 35 m/s
relative to the stationary, profusely sweating goalkeeper standing in front of the goal.
In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity (
The figure shows components of velocity v in horizontal vx and in vertical y axis v y. The angle between the velocity vector v and the horizontal axis is theta.
Figure 3. The velocity, v, of an object traveling at an angle θ to the horizontal axis is the sum of component vectors and
These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known.
Which of the following is a form of mechanical energy?
A. Chemical energy
B. Gravitational potential energy
C. Thermal energy
D. Nuclear energy
Answer:
B
Explanation:
no reason for this answer
need help with question after number 40
Answer:
4 device 3 e e and 4 divide 3 + 5 barabar 11 33 size to 46 size 35 and size 49 browser
