Answer:
In-car temperature is controlled in a variable displacement compressor system by varying the capacity of the refrigeration not by cycling the compressor on and off.
Explanation:
A Variable Displacement Compressor (VDC) works in a separate and rather more efficient manner than the fixed displacement compressor (FDC). The VDC automatically regulates the cooling effect of the refrigerator by changing its pumping capacity by means of a "wobbling piston" system.
During times of peak load, the refrigeration pumping capacity is increased, delivering a large flow rate of refrigerant, thus giving a high refrigerating effect. The reverse happens during times of load refrigeration load.
The fixed compressor tries to vary the temperature of the car by simply switching off the compressor, and turning it on during times of low and peak loads.
4. Chloe has a vertical velocity of 3 m/s when she leaves the 1 m diving board. At this instant, her center of gravity is 2.5 m above the water. How high above the water will Chloe go
Answer:
2.95m
Explanation:
Using h= 2.5+ v²/2g
Where v= 3m/s
g= 9.8m/s²
h= 2.95m
Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t = 0 and subsequently experiences a constant angular acceleration α = 1.3 rad/s2. Determine the velocity and acceleration of point A in terms of fixed i and j unit vectors at time t = 1.7 s.
Given that,
Angular velocity = 0.17 rad/s
Angular acceleration = 1.3 rad/s²
Time = 1.7 s
We need to calculate the angular velocity
Using angular equation of motion
[tex]\omega=\omega_{0}+\alpha t[/tex]
Put the value in the equation
[tex]\omega=0.17+1.3\times1.7[/tex]
[tex]\omega=2.38(k)\ m/s[/tex]
We need to calculate the angular displacement
Using angular equation of motion
[tex]\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}[/tex]
Put the value in the equation
[tex]\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}[/tex]
[tex]\theta=2.1675\times\dfrac{180}{\pi}[/tex]
[tex]\theta= 124.18^{\circ}[/tex]
We need to calculate the velocity at point A
Using equation of motion
[tex]v_{A}=v_{0}+\omega\times r[/tex]
Put the value into the formula
[tex]v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))[/tex]
[tex]v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i[/tex]
[tex]v_{A}=(-0.267j-0.393i)\ m/s[/tex]
We need to calculate the acceleration at point A
Using equation of motion
[tex]a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)[/tex]
Put the value in the equation
[tex]a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]
[tex]a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]
[tex]a_{A}=-0.146j-0.215i−0.636i+0.937j[/tex]
[tex]a_{A}=0.791j-0.851i[/tex]
[tex]a_{A}=-0.851i+0.791j\ m/s^2[/tex]
Hence, (a). The velocity at point A is [tex](-0.267j-0.393i)\ m/s[/tex]
(b). The acceleration at point A is [tex](-0.851i+0.791j)\ m/s^2[/tex]
Value of g in CGS system
Answer:
in CGS system G is denoted as gram
Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fish really is? Does the fish see the person's face closer or farther than it really is? Explain your answer.
Answer:
In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer
In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away
Explanation:
This exercise can be analyzed with the law of refraction that establishes that a ray of light when passing from one medium to another with a different index makes it deviate from its path,
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the refractive indices of the incident and refracted means and the angles are also for these two means.
In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer
1 sin θ₁ = 1.33 sin θ₂
θ₂ = sin⁻¹ ( 1/1.33 sin θ₁)
In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away
Answer:
The fish appears closer than it really is because light from the fish is refracted away from the normal as it enters the air pocket in the goggles. This is because air has a smaller index of refraction than water. The person will trace rays back to an image point in front of the actual fish. The fish will see the person's face exactly where it actually is because the light from the face is not refracted as it travels through water only, and does not change from one medium to another.
Explanation:
A proton moves at a speed 1.4 × 10^7 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.85 m. What is the field strength?
0.17T
Explanation:
When a charged particle moves into a magnetic field perpendicularly, it experiences a magnetic force [tex]F_{M}[/tex] which is perpendicular to the magnetic field and direction of the velocity. This motion is circular and hence there is a balance between the centripetal force [tex]F_{C}[/tex] and the magnetic force. i.e
[tex]F_{C}[/tex] = [tex]F_{M}[/tex] --------------(i)
But;
[tex]F_{C}[/tex] = [tex]\frac{mv^2}{r}[/tex] [m = mass of the particle, r = radius of the path, v = velocity of the charge]
[tex]F_{M}[/tex] = qvB [q = charge on the particle, B = magnetic field strength, v = velocity of the charge ]
Substitute these into equation (i) as follows;
[tex]\frac{mv^2}{r}[/tex] = qvB
Make B subject of the formula;
B = [tex]\frac{mV}{qr}[/tex] ---------------(ii)
Known constants
m = 1.67 x 10⁻²⁷kg
q = 1.6 x 10⁻¹⁹C
From the question;
v = 1.4 x 10⁷m/s
r = 0.85m
Substitute these values into equation(ii) as follows;
B = [tex]\frac{1.67 * 10 ^{-27} * 1.4 * 10^{7}}{1.6 * 10^{-19} * 0.85}[/tex]
B = 0.17T
Therefore, the magnetic field strength is 0.17T
a person lifts 4.5kg block up a vertical distance of 1.2meters and then carries it horizontally for 7.3meters. Calculate the Total Work done by the person and the block.
Answer:
The total work done by the person is given as = m g h
= 4.5kg x 9.8m/s²x1.2m
= 52.92J
This is the work done in moving the block in a vertical distance
However there is no work done when the block is moved in a horizontal direction since ko work is done against gravity.
Explanation:
A total electric charge of 2.00 nC is distributed uniformly over the surface of a metal sphere with a radius of 26.0 cm . The potential is zero at a point at infinity.
a) Find the value of the potential at 45.0 cm from the center of the sphere.
b) Find the value of the potential at 26.0 cm from the center of the sphere.
c) Find the value of the potential at 16.0 cm from the center of the sphere.
Answer:
a) 40 V
b) 69.23 V
c) 69.23 V
Explanation:
See attachment for solution
A small merry-go round is spinning about its center in a clockwise direction at 1.6 r a d s and with rotational kinetic energy of 580 J. Calculate the moment of inertia of the merry-go round.
Answer:
The moment of inertia of the merry-go round is 453.125 kg.m²
Explanation:
Given;'
angular velocity of the merry-go round, ω = 1.6 rad/s
rotational kinetic energy, K = 580 J
Rotational kinetic energy is given as;
K = ¹/₂Iω²
Where;
I is the moment of inertia of the merry-go round
[tex]I = \frac{2K}{\omega^2} \\\\I = \frac{2*580}{1.6^2} \\\\I = 453.125 \ kg.m^2[/tex]
Therefore, the moment of inertia of the merry-go round is 453.125 kg.m²
Since the small merry-go round is spinning about its center in a clockwise direction, its moment of inertia is equal to 453.13 [tex]Kgm^2[/tex]
Given the following data:
Angular velocity = 1.6 rad/sRotational kinetic energy = 580 Joules.To calculate the moment of inertia of the small merry-go round:
Mathematically, the rotational kinetic energy of an object is giving by the formula:
[tex]E_{rotational} = \frac{1}{2} Iw^2[/tex]
Where:
I is the moment of inertia.w is the angular speed.Making moment of inertia (I) the subject of formula, we have:
[tex]I = \frac{2E_{rotational}}{w^2}[/tex]
Substituting the given parameters into the formula, we have;
[tex]I = \frac{2(580)}{1.6^2}\\\\I = \frac{1160}{2.56}[/tex]
Moment of inertia (I) = 453.13 [tex]Kgm^2[/tex]
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A 2kg block is sitting on a hinged ramp such that you can increase the angle of the incline. The coefficient of static friction between the block and the ramp is 0.67 and the coefficient of kinetic friction is 0.25.
a. What angle do you have to tilt the ramp to get the block to slide?
b. What acceleration does the block experience at this angle when kinetic friction takes over?
Answer:
θ = 33.8
a = 3.42 m/s²
Explanation:
given data
mass m = 2 kg
coefficient of static friction μs = 0.67
coefficient of kinetic friction μk = 0.25
solution
when block start slide
N = mg cosθ .............1
fs = mg sinθ ...............2
now we divide equation 2 by equation 1 we get
[tex]\frsc{fs}{N} = \frac{sin \theta }{cos \theta }[/tex]
[tex]\frac{\mu s N }{N}[/tex] = tanθ
put here value we get
tan θ = 0.67
θ = 33.8
and
when block will slide then we apply newton 2nd law
mg sinθ - fk = ma ...............3
here fk = μk N = μk mg cosθ
so from equation 3 we get
mg sinθ - μk mg cosθ = ma
so a will be
a = (sinθ - μk cosθ)g
put here value and we get
a = (sin33.8 - 0.25 cos33.8) 9.8
a = 3.42 m/s²
How much electromagnetic energy is contained in each cubic meter near the Earth's surface if the intensity of sunlight under clear skies is 1000 W/m2
Answer:
344.8 x10^-8J/m³
Explanation:
Using=> energy intensity/ speed oflight
= 1000/2.9x10^8
= 344.8 x10^-8J/m³
The electromagnetic energy is 344.8 x10⁻⁸J/m³
We have to use the formula which says
Electromagnetic energy = energy intensity/ speed of light
We are given intensity as 1000 W/m²
Electromagnetic energy = 1000/2.9 x 10⁸
= 344.8 x 10⁻⁸J/m³
Therefore the electromagnetic energy is contained in each cubic meter will be 344.8 x 10⁻⁸J/m³
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An ideal gas in a cubical box having sides of length L exerts a pressure p on the walls of the box. If all of this gas is put into a box having sides of length 0.5L without changing its temperature, the pressure it exerts on the walls of the larger box will be...
p.
2p.
4p.
8p.
12p.
Answer:
2P
Explanation:
See attached file
key points that can be found in the realist philosophical position
Answer:
Key points that can be found in the realist philosophical position are as follows:
The view that we observe or identify is real, truly out there.The objects which are identified are independent of someone's perceptions, linguistic practices, conceptual scheme, and beliefs.Quantum mechanics is an example of philosophical realism that claims world is mind-independent.A dielectric material such as paper is placed between the plates of a capacitor holding a fixed charge. What happens to the electric field between the plates
Answer:
Majorly the electric field is reduced among other effect listed in the explanation
Explanation:
In capacitors the presence of di-electric materials
1. decreases the electric fields
2. increases the capacitance of the capacitors.
3. decreases the voltage hence limiting the flow of electric current.
The di-electric material serves as an insulator between the metal plates of the capacitors
Calculate the current through a 15.0-m long 20 gauge (having radius 0.405 mm) nichrome wire if it is connected to a 12.0-V battery. The resistivity of nichrome is 100 × 10-8 Ω ∙ m.
Given Information:
Radius of wire = r = 0.405 mm = 0.405×10⁻³ m
Length of wire = L = 15 m
Voltage = V = 12 V
Resistivity = ρ = 100×10⁻⁸ Ωm
Required Information:
Current = I = ?
Answer:
Current = I = 0.412 A
Explanation:
The current flowing through the wire can be found using Ohm's law that is
V = IR
I = V/R
Where V is the voltage across the wire and R is the resistance of the wire.
The resistance of the wire is given by
R = ρL/A
Where ρ is the resistivity of the wire, L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(0.405×10⁻³)²
A = 0.515×10⁻⁶ m²
So the resistivity of the wire is
R = ρL/A
R = (100×10⁻⁸×15)/0.515×10⁻⁶
R = 29.126 Ω
Finally, the current flowing through the wire is
I = V/R
I = 12/29.126
I = 0.412 A
Therefore, the current through a 15.0-m long 20 gauge nichrome wire is 0.412 A.
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a proton and an electron are situated 971 nm from each other and you study the forces that the particles exert on each other. As expected, the predictions of Coulomb's law are well confirmed.
You find that the forces are attractive and the magnitude of each force is:______
Answer:
The magnitude of each force is 2.45 x 10⁻¹⁶ N
Explanation:
The charge of proton, +q = 1.603 x 10⁻¹⁹ C
The charge of electron, -q = 1.603 x 10⁻¹⁹ C
Distance between the two charges, r = 971 nm = 971 x 10⁻⁹ m
Apply Coulomb's law;
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
where;
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²
q₁ and q₂ are the charges of proton and electron respectively
F is the magnitude of force between them
Substitute in the given values and solve for F
[tex]F = \frac{(8.99*10^9)(1.603*10^{-19})^2}{(971*10^{-9})^2} \\\\F = 2.45*10^{-16} \ N[/tex]
Therefore, the magnitude of each force is 2.45 x 10⁻¹⁶ N
In which direction does a bag at rest move when a force of 20 newtons is applied from the right?
ОА.
in the direction of the applied force
OB.
in the direction opposite of the direction of the applied force
OC. perpendicular to the direction of the applied force
OD
in a circular motion
Answer:
in the direction of the applied force
Explanation:
An ultrasound machine uses 1.64 × 105 watts of power. If it draws 12.0 amps of current, what is the resistance?
Answer:
R = 1138.9 Ω
Explanation:
Hello,
In this case, for the given power (P) and current (I), we can compute the resistance (R) via:
R = P / I²
Thus, we obtain:
R = 1.64x10⁵ W / (12.0 A)²
R = 1138.9 Ω
Best regards.
According to string theory, six space-time dimensions cannot be measured except as quantum numbers of internal particle properties because they are curled up in size of the order of
Answer:
Order of 10^(-35) m.
Explanation:
The string theory is a theoretical concept whereby the very small particles of particle physics are replaced by one dimensional objects which are called strings. This theory is also applicable to black hole physics, nuclear physics, cosmology, etc.
Now, according to string theory, six space-time dimensions cannot be measured except as quantum numbers of internal particle properties because they are curled up in size of the order of 10^(-35) m.
This is because the length of the scale is assumed to be on the order of the Planck length, or 10^(−35) meters which is the scale at which the effects of quantum gravity are usually believed to become very significant.
A uniform solid disk and a uniform ring are place side by side at the top of a rough incline of height h.
a) If they are released from rest and roll without slipping, determine the velocity vring of the ring when it reaches the bottom.
b) Verify your answer by calculating their speeds when they reach the bottom in terms of h.
Explanation:
velocity of disc [tex]=\sqrt((gh)/0.75)[/tex]
lets call (h) 1 m to make it simple.
= 3.614 m/s
[tex]\sqrt((4/3) x 1 x 9.8) = 3.614[/tex] m/s pointing towards this:
[tex]4×V_d=\sqrt(4/3hg)[/tex]
[tex]V_h=\sqrt(hg)[/tex]
velocity of hoop=[tex]\sqrt(gh)[/tex]
lets call (h) 1m to make it simple again.
[tex]\sqrt(9.8 x 1) = 3.13[/tex] m/s
[tex]\sqrt(gh) = sqrt(hg)
so [tex]4×V_d= \sqrt(4/3hg)V_h=\sqrt(hg)[/tex]
The disc is the fastest.
While i'm on this subject i'll show you this:
Solid ball [tex]=0.7v^2= gh[/tex]
solid disc [tex]= 0.75v^2 = gh[/tex]
hoop [tex]=v^2=gh[/tex]
The above is simplified from linear KE + rotational KE, the radius or mass makes no difference to the above formula.
The solid ball will be the faster of the 3, like above i'll show you.
solid ball: velocity [tex]=\sqrt((gh)/0.7)[/tex]
let (h) be 1m again to compare.
[tex]\sqrt((9.8 x 1)/0.7) = 3.741[/tex] m/s
solid disk speed [tex]=\sqrt((gh)/0.75)[/tex]
uniform hoop speed [tex]=\sqrt(gh)[/tex]
solid sphere speed [tex]=\sqrt((gh)/0.7)[/tex]
A slender rod of length L has a varying mass-per-unit-length from the left end (x=0) according to dm/dx=Cx where C has units kg/m2. Find the total mass in terms of C and L, and then calculate the moment of inertia of the rod for an axis at the left end note: you need the total mass in order to get the answer in terms of ML^2
Answer:
ML²/6
Explanation:
Pls see attached file
The total mass is M = CL²/2, and the moment of inertia is I = ML²/2,
Moment of inertia:The length of the rod is L. It has a non-uniform distribution of mass given by:
dm/dx = Cx
where C has units kg/m²
dm = Cxdx
the total mass M of the rod can be calculated by integrating the above relation over the length:
[tex]M =\int\limits^L_0 {} \, dm\\\\M=\int\limits^L_0 {Cx} \, dx\\\\M=C[x^2/2]^L_0\\\\M=C[L^2/2]\\\\[/tex]
Thus,
C = 2M/L²
Now, the moment of inertia of the small element dx of the rod is given by:
dI = dm.x²
dI = Cx.x²dx
[tex]dI = \frac{2M}{L^2}x^3dx\\\\I= \frac{2M}{L^2}\int\limits^L_0 {x^3} \, dx \\\\I= \frac{2M}{L^2}[\frac{L^4}{4}][/tex]
I = ML²/2
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⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a) If the coefficient of static friction is 0.5, what minimum force magnitude is required from the rope to start the crate moving? (b) If µk= 0.35, what is the magnitude of the initial acceleration of the crate?
Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
Which of the following technologies is based on the work of Ibn al-Haytham?
A. Telescopes to observe the visible light of distant stars
B. Radiation treatments for breast cancer
C. Radar to detect the movement of storms
O D. An orbiting observatory to detect X-rays from space objects
Answer:
The answer is A
Explanation:
Its A because he created a telescope to be able to observe stars.
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos 5t + π 8 where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos (5t + π / 8) where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm
Answer:(a) 7.392cm
(b) -15.32 cm/s
(c) -184cm/s²
(d) 0.4πs and 8.00cm
Explanation:The general equation of a simple harmonic motion (SHM) is given by;
x(t) = A cos (wt + Φ) --------------(i)
Where;
x(t) = position of the body at a given time t
A = amplitude or maximum displacement during oscillation
w = angular velocity
t = time
Φ = phase constant.
Given from question:
x(t) = 8.00 cos (5t + π / 8) ---------------(ii)
(a) At time t = 0;
The position, x(t), of the body (piston) is given by substituting the value of t = 0 into equation (ii) as follows;
x(0) = 8.00 cos (5(0) + π / 8)
x(0) = 8.00 cos (π /8)
x(0) = 8.00 x 0.924
x(0) = 7.392 cm
Therefore, the position of the piston at time t = 0 is 7.392cm
(b) To get the velocity, v(t), of the piston at t = 0, first differentiate equation (ii) with respect to t as follows;
v(t) = [tex]\frac{dx(t)}{dt}[/tex]
v(t) = [tex]\frac{d(8.00cos(5t + \pi / 8 ))}{dt}[/tex]
v(t) = 8 (-5 sin (5t + π / 8))
v(t) = -40sin(5t + π / 8) --------------------(iii)
Now, substitute t=0 into the equation as follows;
v(0) = -40 sin(5(0) + π / 8)
v(0) = -40 sin(π / 8)
v(0) = -40 x 0.383
v(0) = -15.32 cm/s
Therefore, the velocity of the piston at time t = 0 is -15.32 cm/s
(c) To find the acceleration a(t) of the piston at t = 0, first differentiate equation (iii), which is the velocity equation, with respect to t as follows;
a(t) = [tex]\frac{dv(t)}{dt}[/tex]
a(t) = [tex]\frac{d(-40sin (5t + \pi /8))}{dt}[/tex]
a(t) = -200 cos (5t + π / 8)
Now, substitute t = 0 into the equation as follows;
a(0) = -200 cos (5(0) + π / 8)
a(0) = -200 cos (π / 8)
a(0) = -200 x 0.924
a(0) = -184.8 cm/s²
Therefore, the acceleration of the piston at time t = 0 is -184cm/s²
(d) To find the period, T, first, let's compare equations (i) and (ii) as follows;
x(t) = A cos (wt + Φ) --------------(i)
x(t) = 8.00 cos (5t + π / 8) ---------------(ii)
From these equations it can be deduced that;
Amplitude, A = 8.00cm
Angular velocity, w = 5 rads/s
But;
w = [tex]\frac{2\pi }{T}[/tex] [Where T = period of oscillation]
=> T = [tex]\frac{2\pi }{w}[/tex]
=> T = [tex]\frac{2\pi }{5}[/tex]
=> T = 0.4π s
Therefore, the period and amplitude of the piston's motion are respectively 0.4πs and 8.00cm
Classify the bending of light as exhibited by the ray diagrams. According to your data, is light refracted away from or toward the normal as it passes at an angle into a medium with a higher index of refraction?
Answer:
the ray of light should approach normal
Explanation:
When light passes through two means of different refractive index, it fulfills the equation
n₁ sin θ₁ = n₂ sin θ₂
where index 1 and 2 refer to each medium
In this problem, they tell us that light passes to a medium with a higher index, which is why
n₁ <n₂
let's look for the angle in the second half
sinθ₂ = n₁ /n₂ sin θ₁
θ₂ = sin⁻¹ (n₁ /n₂ sin θ₁)
let's examine the angle argument the quantity n₁ /n₂ <1 therefore the argument decreases, therefore the sine and the angle decreases
Consequently the ray of light should approach normal
An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the speed remained the same, by what factor would its centripetal acceleration change?
Answer:
The centripetal acceleration changed by a factor of 0.5
Explanation:
Given;
first radius of the horizontal circle, r₁ = 500 m
speed of the airplane, v = 150 m/s
second radius of the airplane, r₂ = 1000 m
Centripetal acceleration is given as;
[tex]a = \frac{v^2}{r}[/tex]
At constant speed, we will have;
[tex]v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1[/tex]
a₂ = 0.5a₁
Therefore, the centripetal acceleration changed by a factor of 0.5
6a. A special lamp can produce UV radiation. Which two statements
describe the electromagnetic waves emitted by a UV lamp? *
They have a higher frequency than X-rays.
They have the same wave speed as visible light
They have a longer wavelength than microwaves.
They have a lower frequency than gamma rays.
They have a greater wave speed than radio waves.
Answer:
The correct options are:
B) They have the same wave speed as visible light
D) They have a lower frequency than gamma rays.
Explanation:
B) Ultraviolet rays, commonly known as UV rays, are a type of electromagnetic ways. As electromagnetic waves, in the layman's term, are all kinds of life that can be identified, all electromagnetic waves (UV rays, visible light, infrared, radio etc) all travel with the same velocity, that is the speed of light, given as v = 3 × 10⁸ m/s
D) The frequency of all electromagnetic rays can be found by electromagnetic spectrum (picture attached below).
We can clearly see in the picture that the frequencies of UV rays lie at about 10¹⁵ - 10¹⁶ Hz which is lower than the frequency of Gamma ray, which lie at about 10²⁰ Hz.
An electron starts from rest in a vacuum, in a region of strong electric field. The electron moves through a potential difference of 36 volts. What is the kinetic energy of the electron in electron volts (eV)
Answer:
The kinetic energy is [tex]KE = 5.67*10^{-18} \ J[/tex]
Explanation:
From the question we are told that
The potential difference is [tex]\Delta V = 36 \ volts[/tex]
The potential energy of the end is mathematically represented as
[tex]PEs = - q * \Delta V[/tex]
q is the charge on an electron with a constant value of [tex]q = 1.60 *10^{-19} \ C[/tex]
substituting values
[tex]PE = - 1.60*10^{-19} * 36[/tex]
[tex]PE = - 5.67*10^{- 18} \ J[/tex]
Now from the law of energy conservation
The [tex]PE_e = KEe[/tex]
Where [tex]KE _e[/tex] is the potential energy at the end
So
[tex]KE = 5.67*10^{-18} \ J[/tex]
The negative sign is not includes because kinetic energy can not be negative
A spherical shell rolls without sliding along the floor. The ratio of its rotational kinetic energy (about an axis through its center of mass) to its translational kinetic energy is:
Answer:
The ratio is [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]
Explanation:
Generally the Moment of inertia of a spherical object (shell) is mathematically represented as
[tex]I = \frac{2}{3} * m r^2[/tex]
Where m is the mass of the spherical object
and r is the radius
Now the the rotational kinetic energy can be mathematically represented as
[tex]RE = \frac{1}{2}* I * w^2[/tex]
Where [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{v}{r}[/tex]
=> [tex]w^2 = [\frac{v}{r}] ^2[/tex]
So
[tex]RE = \frac{1}{2}* [\frac{2}{3} *mr^2] * [\frac{v}{r} ]^2[/tex]
[tex]RE = \frac{1}{3} * mv^2[/tex]
Generally the transnational kinetic energy of this motion is mathematically represented as
[tex]TE = \frac{1}{2} mv^2[/tex]
So
[tex]\frac{RE}{TE} = \frac{\frac{1}{3} * mv^2}{\frac{1}{2} * m*v^2}[/tex]
[tex]\frac{RE}{TE} = \frac{2}{3}[/tex]
At what minimum speed must a roller coaster be traveling when upside down at the top of a 7.4 m radius loop-the-loop circle so the passengers will not fall out?
Answer:
v = 8.5 m/s
Explanation:
In order for the passengers not to fall out of the loop circle, the centripetal force must be equal to the weight of the passenger. Therefore,
Weight = Centripetal Force
but,
Weight = mg
Centripetal Force = mv²/r
Therefore,
mg = mv²/r
g = v²/r
v² = gr
v = √gr
where,
v = minimum speed required = ?
g = 9.8 m/s²
r = radius = 7.4 m
Therefore,
v = √(9.8 m/s²)(7.4 m)
v = 8.5 m/s
Minimum speed for a roller coaster while travelling upside down so that the person will not fall out = 8.5 m/s
For a roller coaster be traveling when upside down the Force balance equation can be written for a person of mass m.
In the given condition the weight of the person must be balanced by the centrifugal force.
and for the person not to fall out centrifugal force must be greater than or equal to the weight of the person
According to the Newton's Second Law of motion we can write force balance
[tex]\rm mv^2/r -mg =0 \\\\mg = mv^2 /r (Same\; mass) \\\\\\g = v^2/r\\\\v = \sqrt {gr}......(1)[/tex]
Given Radius of loop = r = 7.4 m
Putting the value of r = 7.4 m in equation (1) we get
[tex]\sqrt{9.8\times 7.4 } = \sqrt{72.594} = 8.5\; m/s[/tex]
For more information please refer to the link below
https://brainly.com/question/13259103
What happens when you increase the number of slits per millimeter (decrease the spacing between slits)?
Answer:
Increasing the number of slits not only makes the diffraction maximum sharper, but also much more intense. If a 1 mm diameter laser beam strikes a 600 line/mm grating, then it covers 600 slits and the resulting line intensity is 90,000 x that of a double slit. Such a multiple-slit is called a diffraction grating.