Answer:
a.432
B. 562
C.402
D. 316
A
Explanation:
Work is defined as the product of applied force and the distance through which the body is displaced on which the force is applied. Work was done by the workman will be 432 Nm.
What is work?
Work is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.
Work may be zero, positive and negative.it depend on the diection of body displaced . if the body is displaced in thw same direction of force it will be positive .
while if the displacement is in the opposite direction of force applied the work will be negative work . if their is no displacement of the body the work done will b e zero.
As given in question wall is stationary diplacement will be zero so in that work will also z
The given data in the problem is;
x is the displacement in the horizontal direection=22m
y is the displacement in the verical direection=9.6m
W is the weight of lumber = 45 N
[tex]\rm W=F\times y[/tex]
[tex]\rm W=F\times y \\\\\rm W=45\times 9.6\\\\ \rm W=432\;Nm[/tex]
Hence Work was done by the workman will be 432 Nm.
To learn more about the work refer to the link ;
https://brainly.com/question/3902440
A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 0.9 m from the axis of rotation, and he rotates with angular speed of 0.76 rad/sec. The moment of inertia of the student plus the stool is 5 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.33 m from the rotation axis.
Required:
a. Find the new angular speed of the student.
b. Find the kinetic energy of the student before and after the objects are pulled in.
Answer:
a) the new angular speed of the student is 0.9642 rad/s
b)
the kinetic energy of the student before the objects are pulled in is 1.9119 J
the kinetic energy of the student after the objects are pulled in is 2.4252 J
Explanation:
Given that;
mass of each object m = 1 kg
distance of objects from axis of rotation r = 0.9 m
Moment of inertia of each object initially [tex]I_{oi}[/tex]
[tex]I_{oi}[/tex] = mr² = 1kg ×(0.9m)² = 1 kg × 0.81 m² = 0.81 kg.m²
moment of inertia of each object finally [tex]I_{of}[/tex]
[tex]I_{of}[/tex] = mr² = 1kg × (0.33 m)² = 0.1089 kg.m²
Now
moment of inertia of student plus stool [tex]I_{}[/tex] = 5 kg.m²
initial angular speed ω₀ = 0.76 rad/sec
final angular speed ω = ?
Now using conservation of angular momentum;
([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀ = ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω
so we substitute
(5 + 2 (0.81) )0.76 = (5 + 2 (0.1089) )ω
5.0312 = 5.2178 ω
ω = 5.0312 / 5.2178
ω = 0.9642 rad/s
Therefore, the new angular speed of the student is 0.9642 rad/s
b)
K.E of student before = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀²
= (0.5) (5 + 2 (0.81) )(0.76)²
= 0.5 × 6.62 × 0.5776
= 1.9119 J
Therefore, the kinetic energy of the student before the objects are pulled in is 1.9119 J
KE of student finally = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω²
= (0.5) (5 + 2 (0.1089) ) (0.9642)²
= 0.5 × 5.2178 × 0.9296
= 2.4252 J
Therefore, the kinetic energy of the student after the objects are pulled in is 2.4252 J
Two students, each riding bicycles, start from the same apartment building and ride to the same building on campus, but each takes a different route. The first student rides 1300 m due east and then turns due north and travels another 1430 m before arriving at the destination. The second student heads due north for 1930 m and then turns and heads directly toward the destination.
(a) At the turning point, how far is the second student from the destination? ....m
(b) During the last leg of the trip, what direction (measured relative to due east) must the second student head? (Give your answer as a positive number from 0 to 180 degrees, either north or south of due east.) .... degrees south of east
Answer:
a) d= 1393 m
b) θ= 21º S of E.
Explanation:
a)
Since the second student goes due north, and the first student goes due east along 1300m till he turns directly northward, we conclude that when the second student turns, he is at 1300m west from the destination.Since he rode 1930 m due north, we can conclude also that the second student is 500 m past the destination in the north direction.So we can find the distance from the destination at the turning point, using the Pythagorean Theorem, taking the right triangle defined by the 1300 m segment due east, the 500 m segment due south, and which hypotenuse is the distance straight to the destination, as follows:[tex]d = \sqrt{(1300m)^{2} + (500m)^{2} } = 1393 m (1)[/tex]
b)
Taking the same right triangle than in (a), we can find the angle that makes the vector along the direction taken by the second student with the due east, applying the definition of tangent of an angle, as follows:[tex]tg \theta = \frac{500m}{1300m} = 0.385 (2)[/tex]
⇒ θ= tg⁻¹ (0.385) = 21º S of E.
Starting from the front door of your ranch house, you walk 55.0 m due east to your windmill, turn around, and then slowly walk 35.0 m west to a bench, where you sit and watch the sunrise. It takes you 30.0 s to walk from your house to the windmill and then 36.0 s to walk from the windmill to the bench.
Required:
a. For the entire trip from your front door to the bench, what is your average velocity?
b. For the entire trip from your front door to the bench, what is your average speed?
Answer:
Explanation:
Average velocity = Total displacement / total time
Average speed = total distance covered / total time
a )
For the entire trip from your front door to the bench
Total displacement = 55 - 35 = 20 m [ first displacement is positive and second displacement is negative , because second displacement is in opposite direction ]
Total displacement = 20 m
Total time = 30 + 36 = 66 s
Average velocity = 20 / 66
= .303 m / s
b )
For the entire trip from your front door to the bench
Total distance covered = 55 + 35 = 90 m
Total time = 30 + 36 = 66 s
Average speed = 90 / 66
= 1.36 m / s
PHYSICS QUESTION PLS HELP
The coaster starts at rest, so the kinetic energy (KE) at point A is 0. It is situated 33 m above ground, so its potential energy (PE) at A is
mgh = (3000 kg) (9.80 m/s²) (33 m) = 970,200 J
The total energy is the same, 970,200 J.
Assuming no energy is lost to friction or sound etc, energy is conserved throughout the coaster's motion, so the total energy should be the same at each point.
At point B, the coaster has dropped to a height of 10 m, so it has PE
mgh = (3000 kg) (9.80 m/s²) (10 m) = 294,000 J
which means it must have KE
970,200 J = KE + 294,000 J → KE = 676,200 J
which gives the coast a speed v at point B of
1/2 mv ² = 1/2 (3000 kg) v ² = 676,200 J → v ≈ 21.2 m/s
At point C, the coaster has a speed of 16.0 m/s, so it has KE
1/2 mv ² = 1/2 (3000 kg) (16.0 m/s)² = 384,000 J
and hence PE
970,200 J = 384,000 J + PE → PE = 586,200 J
This lets us determine the height h at C:
mgh = (3000 kg) (9.80 m/s²) h = 586,200 J → h ≈ 19.939 m
which means the loop has diameter h - 10 m ≈ 9.94 m.
At point D, the coaster is 15 m above the ground so its PE at D is
mgh = (3000 kg) (9.80 m/s²) (15 m) = 441,000 J
and so its KE is
970,200 J = KE + 441,000 J → KE = 529,200 J
and hence has speed v at D
1/2 mv ² = 1/2 (3000 kg) v ² = 529,200 J → v ≈ 18.9 m/s
An air-filled capacitor consists of two parallel plates, each with an area of A , separated by a distance d . A V potential difference is applied to these plates. What is the magnitude of the electric field between the plates
Answer:
E = V / d
Explanation:
In a charged capacitor an electric field is established that goes from the positive to the negative plate, this field is constant,
the potential difference is
D = E d
in this case they do not give the difference in potential V and the distance between the plates d
E = V / d
In a liquid with a density of 1500 kg/m3, longitudinal waves with a frequency of 410 Hz are found to have a wavelength of 7.80 m. Calculate the bulk modulus of the liquid.
Answer:
The bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²
Explanation:
Given;
density of the liquid, ρ = 1500 kg/m³
frequency of the wave, F = 410 Hz
wavelength of the sound, λ = 7.80 m
The speed of the wave is calculated as;
v = Fλ
v = 410 x 7.8
v = 3,198 m/s
The bulk modulus of the liquid is calculated as;
[tex]V = \sqrt{\frac{B}{\rho} } \\\\V^2 = \frac{B}{\rho}\\\\B = V^2 \rho\\\\B = (3,198 \ m/s)^2 \times 1500 \ kg/m^3\\\\B = 1.534 \ \times 10^{10} \ N/m^2[/tex]
Therefore, the bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²
a device that spreads light into different wavelengths is a what?
maybe a spectrograph ?
what is the difference between mass and weight
Answer:
The mass of an object is a measure of the object's inertial property, or the amount of matter it contains. The weight of an object is a measure of the force exerted on the object by gravity, or the force needed to support it. The pull of gravity on the earth gives an object a downward acceleration of about 9.8 m/s2.
Answer:
Explanation:
The mass is essentially "how much stuff" is in an object. ... Weight: There is a gravitational interaction between objects that have mass. If you consider an object interacting with the Earth, this force is called the weight. The unit for weight is the Newton (same as for any other force).
A car is traveling along a straight road at a velocity of +30.0 m/s when its engine cuts out. For the next 1.79 seconds, the car slows down, and its average acceleration is . For the next 4.03 seconds, the car slows down further, and its average acceleration is . The velocity of the car at the end of the 5.82-second period is +18.4 m/s. The ratio of the average acceleration values is = 1.53. Find the velocity of the car at the end of the initial 1.79-second interval.
Answer:
first value+2nd +3rd
Explanation:
thug life and there
how long it take a train to cover 630km having a speed of 30 km/hr
Answer:
21 hours
Explanation:
well 30 x 20 = 600 than 21 = 630
A 1 800-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.60 m before coming into contact with the top of the beam, and it drives the beam 13.6 cm farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.
Answer:
F = 614913.88 N
Explanation:
We are given;
Mass of pile driver; m = 1800 kg
Height of fall of pole driver; h = 4.6 m
Depth driven into beam; d = 13.6 cm = 0.136 m
Now, from energy equations and applying to this question, we can write that;
Workdone = Change in potential energy
Formula for workdone is; W = F × d
While the average potential energy here is; W = mg(h + d)
Thus;
Fd = mg(h + d)
Where F is the average force exerted by the beam on the pile driver while in bringing it to rest.
Making F the subject, we have;
F = mg(h + d)/d
F = 1800 × 9.81 × (4.6 + 0.136)/0.136
F = 614913.88 N
A Typical operating voltage of an electron microscope is 50 kV. A Typical experimental operating voltage range of a Scanning electron microscope is 1kV to 30kV. Higher voltages can penetrate and causes deformation on the sample. Lets assume it operates at 10kV. (i)What is the smallest distance that it could possibly resolve
Answer:
y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]
Explanation:
Let's solve this exercise in parts. Let's start by finding the wavelength of the electrons accelerated to v = 10 103 V, let's use the DeBroglie relation
λ= [tex]\frac{h}{p} = \frac{h}{mv}[/tex]
Let's use conservation of energy for speed
starting point
Em₀ = U = e V
final point
Em_f = K = ½ m v²
Em₀ = Em_f
eV = ½ m v²
v =[tex]\sqrt{\frac{2eV}{m} }[/tex]
we substitute
λ= [tex]\sqrt{ \frac{h^2 m}{2eV}}[/tex]
the diffraction phenomenon determines the minimum resolution, for this we find the first zero of the spectrum
a sin θ = m λ
first zero occurs at m = 1, also these experiments are performed at very small angles
sin θ = θ
θ = λ / a
This expression is valid for linear slits, in the microscope the slits are circular, when solving the polar coordinates we obtain
θ = 1.22 λ / D
where D is the diameter of the opening
we substitute
θ = [tex]\frac{1.22}{D}[/tex] \sqrt{ \frac{h^2 m}{2eV}}
this is the minimum angle that can be seen, if the distance is desired suppose that the distance of the microscope is L, as the angles are measured in radians
θ = y / L
when substituting
where y is the minimum distance that can be resolved for this acceleration voltage
y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]
The nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons).
(a) What is the force between the two alpha particles when they are 6.60 ✕ 10−15 m apart? N
(b) What is the initial magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u. m/s2
Answer:
a) F = 21.16 N, b) a = 3.17 10²⁸ m / s
Explanation:
a) The outside between the alpha particles is the electric force, given by Coulomb's law
F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]
in that case the two charges are of equal magnitude
q₁ = q₂ = 2q
let's calculate
F = [tex]9 \ 10^9 \ \frac{ (2 \ 1.6 \ 10^{-19} )^2 }{ (6.60 \ 10^{-15} )^2 }[/tex]
F = 21.16 N
this force is repulsive because the charges are of the same sign
b) what is the initial acceleration
F = ma
a = F / m
a = [tex]\frac{21.16}{4.0026 \ 1.67 \ 10^{-27} }[/tex]21.16 / 4.0025 1.67 10-27
a = 3.17 10²⁸ m / s
this acceleration is in the direction of moving away the alpha particles
Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 3 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Find the tension in each wire and the magnitude of each tension. (Use g = 9.8 m/s2 for the acceleration due to gravity. Round your answers to two decimal places.)
Answer:
Explanation:
Let the tension in ropes be T₁ and T₂ . Ropes are making angle of 52 and 40 degree with the horizontal . The vertical component of tension will add up together to balance the weight of the decoration
T₁ sin52 + T₂ sin40 = 3 x 9.8
.788 T₁ + .643 T₂ = 29.4
The horizontal component of tension will add up to zero because the decoration piece is at rest .
T₁ cos52 + T₂ cos40 = 0
.615 T₁ - .766 T₂ = 0
T₁ = 1.24 T₂
Substituting this value of T₁ in earlier equation , we have
.788 x 1.24 T₂ + .643 T₂ = 29.4
1.62 T₂ = 29.4
T₂ = 18.15 N
T₁ = 1.24 x 18.15 = 22.51 N .
An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta leading away from the heart. Since blood within the heart is essentially stationary, this pressure difference can be inferred from a measurement of the speed of blood flow in the aorta. Take the speed of sound in stationary blood to be c.
a. Sound sent by a transmitter placed directly inline with the aorta will be reflected back to a receiver and show a frequency shift with each heartbeat. If the maximum speed of blood in the aorta is v, what frequency will the receiver detect? Note that you cannot simply use the textbook Doppler Shift formula because the detector is the same device as the source, receiving sound after reflection.
b. Show that in the limit of low blood velocity (v <
f= 2fo v/c
Answer:
a) f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex] , b) Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo[tex]\frac{c+v}{c}[/tex]
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’ [tex]\frac{c}{ c-v}[/tex]
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀ [tex]\frac{c}{c-v}[/tex]
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex]
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
[tex]( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}[/tex]
f ’’ = fo [tex]( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )[/tex]
f ’’ = fo [tex]( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )[/tex]
leave the linear term
f ’’ = f₀ + f₀ 2[tex]\frac{v}{c}[/tex]
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
Explain what is happening in this picture
Answer:
in this video waves are coming up for the BOTTOM to the top of the sandbar
Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. (a) What acceleration would a proton experience if the gun's electric field were 2.95 × 104 N/C? (b) What speed would the proton attain if the field accelerated the proton through a distance of 1.26 cm?
Answer:
(A) the acceleration experienced by the proton 2.821 x 10¹² m/s²
(B) the speed of the proton is 2.67 x 10⁵ m/s
Explanation:
Given;
electric field experienced by the proton, E = 2.95 x 10⁴ N/C
charge of proton, Q = 1.6 x 10⁻¹⁹ C
mass of proton, m = 1.673 x 10⁻²⁷ kg
distance moved by the proton, d = 1.26 cm = 0.0126 m
(a)
The force experienced by the proton is calculated as;
F = ma = EQ
where;
a is the acceleration experienced by the proton
[tex]a = \frac{EQ}{m} \\\\a = \frac{2.95\times 10^4 \ \times \ 1.6\times 10^{-19}}{1.673 \times 10^{-27}} \\\\a = 2.821 \times 10^{12} \ m/s^2[/tex]
(b) the speed of the proton is calculated;
v² = u² + 2ad
v² = 0 + (2 x 2.821 x 10¹² x 0.0126)
v² = 7.109 x 10¹⁰
v = √7.109 x 10¹⁰
v = 2.67 x 10⁵ m/s
A girl jogs around a horizontal circle with a constant speed. She travels one fourth of a revolution, a distance of 25 m along the circumference of the circle, in 5.0 s. The magnitude of her acceleration is
Answer:
The centripetal acceleration of the girl is 2.468 m/s²
Explanation:
Given;
number of turns, = ¹/₄ Revolution
distance traveled by the girl, d = 25 m
time of motion, t = 5.0 s
The linear speed of the of the girl is calculated as;
[tex]v = \omega \ r\\\\v =(\frac{1}{4}rev \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1}{5 \ s} ) (25 \ m)\\\\v = (0.3142 \ \frac{rad}{s} )(25 \ m)\\\\v = 7.855 \ m/s[/tex]
The centripetal acceleration of the girl is calculated as;
[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{(7.855)^2}{25} \\\\a_c = 2.468 \ m/s^2[/tex]
Therefore, the centripetal acceleration of the girl is 2.468 m/s²
Temperature and salinity difference in ocean water cause...
Answer:
high and low tied maybe? not sure
At what height does a 3500-kg truck have a potential energy of 90,000 J gravitational potential energy relative to the ground?
Answer:
MGH=energy
3500*9.8*h=90000
h=90000/34300
h=2.62m
An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field towards the other end. If they move fast enough when they strike the positive electrode at the other end, they will give up their energy as X-Rays
(a) Through what potential difference should electrons be accelerated so that their speed is 1% of the speed of light?
(b) What potential difference would be needed to give the protons same kinetic energy as electrons?
(c) What speed would this potential difference give to the protons, both in m/s and as a % of the speed of light.
Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV = [tex]\frac{1}{2}[/tex] mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV = [tex]\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }[/tex]
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
[tex]K_e = K_p[/tex]
K_p = ½ m v_e²
K_p = [tex]\frac{1}{2}[/tex] 9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² = [tex]\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }[/tex]
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c = [tex]\frac{7 \ 10^4 }{ 3 \ 10^8}[/tex]
v/c= 2.33 10⁻⁴
What is the car's acceleration from 0 to 1 second?
A. 8 mph/s
B. 20 mph/s
C. 60 mph/s
D. 10 mph/s
Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 piston with a mass of 10 kg that can slide up and down vertically without friction. Both containers are placed on identical heaters and heated for equal amounts of time.
Required:
a. Will the final temperature of the gas in A be greater, less than, or equal to the temperature in B?
b. Show both processes on a single PV diagram.
c. What are the initial pressures in containers A and B?
d. Suppose the heaters have 25 W of power and are turned on for 15s. What is the final volume of container B?
Answer:
1) Final Temperature of the gas in A will be GREATER than the temperature in B
2) Diagram of both processes on a single PV has been uploaded below
3) The Initial pressures in containers A and B is 3039.87 J/liters
4) the final volume of container B is 923.36 cm³
Explanation:
Given that;
Temperature = 20°C = 293 K
mass of piston = 10 kg
Area = 100cm³
Volume V = 800 cm³ = 0.8 L
ideal gas constant R = 8.3 J/K·mol
1)
Final Temperature of the gas in A will b GREATER than the temperature in B
2)
Diagram of both processes on a single PV has been uploaded below,
3)
Initial pressures in containers A and B
PV = nRT
P = RT/V
we substitute
P = (8.3 × 293) / 0.8
P = 2431.9 / 0.8
P = 3039.87 J/liters
Therefore, The Initial pressures in containers A and B is 3039.87 J/liters
4)
Given that;
power = 25 W
time t = 15s
the final volume of container B = ?
we know that;
work done = power × time
work done = 25 × 15 = 375
Also work done = P( V₂ - V₁ )
so we substitute
375 = 3039.87 ( V₂ - 0.8 )
( V₂ - 0.8 ) = 375 / 3039.87
V₂ - 0.8 = 0.12336
V₂ = 0.12336 + 0.8
V₂ = 0.92336 Litres
V₂ = 923.36 cm³
Therefore, the final volume of container B is 923.36 cm³
Which would you choose to keep rods or cones? And why?
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 520 lines/mm , and the light is observed on a screen 1.4 m behind the grating.
Required:
What is the distance between the first-order red and blue fringes?
Answer:
0.143 m
Explanation:
Since
d = 1/N = 1/520 = 1.92 * 10^-3 mm
For red light;
θ = sin^-1 (1 * λred/d) = sin^-1 (1 * 656 * 10^-9/1.92 * 10^-6) = 19.98°
L = 1.4 * (tan 19.98) = 0.509 m
For blue light;
θ = sin^-1 (1 * λblue/d) = sin^-1 (1 * 486 * 10^-9/1.92 * 10^-6) = 14.66°
L = 1.4 * (tan 14.66°) = 0.366 m
Distance between the first-order red and blue fringes= 0.509 m - 0.366 m = 0.143 m
Acceleration figures for cars usually are given as the number of seconds needed to go from 0.0 to 97 km/h. Convert 97 km/h into m/s.
Answer:
26.9444m/s
pls brainliest
If the gravitational constant is extremely weak, how is the force of gravity on earth so strong?
A cylinder of gas is at room temperature (20°C). The air conditioner breaks down, and the temperature rises
to 46°C. What is the new pressure of the gas relative to its initial pressure (Pi)?
Answer: we divide this equation by pV and use {C}_{p}={C}_{V}+ . ... The temperature, pressure, and volume of the resulting gas-air mixture
Explanation:
The new pressure of the gas relative to its initial pressure P₁ is P₁ × 1.09
From the question given above, the following data were obtained:
Initial temperature (T₁) = 20 °C = 20 + 273 = 293 K
Initial pressure (P₁) = P₁
New temperature (T₂) = 46 °C = 46 + 273 = 319 K
New pressure (P₂) =?The new pressure of the gas can be obtained as follow:
P₁ / T₁ = P₂ / T₂
P₁ / 293 = P₂ / 319
Cross multiply
293 × P₂ = P₁ × 319
Divide both side by 293
P₂ = (P₁ × 319) / 293
P₂ = P₁ × 1.09Thus, the new pressure relative to it's initial pressure is P₁ × 1.09
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What is the value of the angle of inclination of the slide?
Answer:
63°
that's my answer
but then I am sorry if I'm wrong
Explanation:
90-27 = 63°
pls help everything is in the pic
Answer:
c
Explanation: