Answer:
408 steps
Explanation:
Given :
[tex]Power =P= 746\ W\\mass=M\ =\ 70\ kg\\time\ =\ T\ =\ 1\ minute\\Height=H=16\ cm=0.16\ m[/tex]
Consider that n is the number of steps .
We know that
[tex]Power=P = \frac{work done}{time\ taken} \\[/tex]
that can be written as
P=[tex]\frac{n*MgH}{T}[/tex]
In terms of n ,
[tex]n=\frac{P*T}{MgH} \\Putting\ the\ value\ of\ P\ ,T\ M\ ,H\ and\ G=9.8\ ms^{2} \ in\ the\ previous \ equation\ we\ get[/tex][tex]n=\frac{ 746 W*60s}{ 70 kg *9.8 m/s2*0.16m} \\n=407.8\ steps[/tex]
~n=408 steps
a car slows down from - 27.7 m/s to -10.9 m/s while undergoing a displacement of -105 m .what is its acceleration?
Answer:
3.09 m/s²
Explanation:
Given:
Δx = -105 m
v₀ = -27.7 m/s
v = -10.9 m/s
Find: a
v² = v₀² + 2aΔx
(-10.9 m/s)² = (-27.7 m/s)² + 2a (-105 m)
a = 3.09 m/s²
A parallel-plate capacitor is formed from two 9.1 cm-diameter electrodes spaced 1.3 mm apart. The electric field strength inside the capacitor is 5.0×106N/C.
What is the charge (in nC) on each electrode?
Answer:
2.87nC
Explanation:
See attached file
A 975-kg pickup comes to rest from a speed of 87.5 km/h in a distance of 125 m. Suppose the pickup is initially traveling in the positive direction.
(a) If the brakes are the only thing making the car come to a stop, calculate the force (in newtons, in a component along the direction of motion of the car) that the brakes apply on the car .
(b) Suppose instead of braking that the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force, in newtons, exerted on the car in this case 33%
(c) What is the ratio of the force on the car from the concrete to the braking force?
Answer:
A) Force = 2303.925 N in the negative x-direction
B) F ≈ 143998.28 N
C) Ratio = 62.5
Explanation:
A) Since the brakes are the only thing making the van to come to a stop, then first of all, we will calculate the force (in a component along the direction of motion of the car) that the brakes will apply on the van.
Let's find the deceleration using Newton's law of motion formula;
v² = u² + 2as
where;
v = final velocity,
u = initial velocity,
s = displacement
a = acceleration
We are given;
u = 87.5 km/h = 24.3056 m/s
s = 125 m
v = 0 m/s
Thus;
0 = (24.3056)² + 2a(125)
- (24.3056)²= 250a
a = - 24.3056²/250
a = - 2.363 m/s²
Now, force = mass × acceleration
We are given mass = 975 kg
Thus;
Force = 975 x (-2.363)
Force = 2303.925 N in the negative x-direction
B) formula for kinetic energy is
KE = ½mv²
KE = ½(975)(24.3056)²
= 287996.568288 J
Work done on impact = F x 2
Thus;
2F = 287996.568288
F = 287996.568288/2
F ≈ 143998.28 N
C) Ratio = Force on car/braking force = 143998.284/2303.925 = 62.5
From their observations, Darcel and Chandra found that the speed of an object at the bottom of an incline depends on how the mass is distributed in the object. Now they consider the moments of inertia of the objects in the simulation, which are related to their mass distributions. Darcel asks Chandra what effect the moment of inertia has on the final speed. Which is the correct response?
Answer:
The larger the moment of inertia of the object, the slower it will be moving at the bottom of the incline
Explanation:
This is because Rotational kinetic energy varies proportionally to the moment oThe larger the moment of inertia of the object,therefore the slower it will be moving at the result is a slower speed
A 16.0-m uniform ladder weighing 520 N rests against a frictionless wall. The ladder makes a 65.0° angle with the horizontal.
(a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 850-N firefighter has climbed 4.20 m along the ladder from the bottom. Horizontal Force Magnitude = Direction = Vertical Force Magnitude = Direction =
(b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?
Answer:
we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.
Coefficient of static friction = 0.26
Explanation:
Given that:
length of the ladder = 16.0 m
weight of the ladder = 520 N
angle θ = 65.0°
(a) We are to find the horizontal and vertical forces the ground exerts on the base of the ladder when an :
Force = 850 N
distance of the climber from the base of the ladder = 4.20 m
The diagrammatic illustration representing what the given information entails can be seen from the attached file below.
Let consider the Ladder being at point A with the horizontal layer of the ground.
From the whole system; the condition for the equilibrium at the point A can be computed as :
[tex]N_2 (16 \ Sin\ 65) = 850(4.2 \ \times Cos \ 65 )+ 520 (\dfrac{16}{2}) Cos \ 65[/tex]
[tex]N_2 (14.50) = 850(1.7749 )+ 520 (8) \times 0.4226[/tex]
[tex]N_2 (14.50) = 1508.665+1758.016[/tex]
[tex]N_2 (14.50) = 3266.681[/tex]
[tex]N_2 =\dfrac{ 3266.681}{14.50}[/tex]
[tex]N_2 =225.28 \ N[/tex]
[tex]N_1 = mg+F\\[/tex]
where ;
w =mg
[tex]N_1 = 520+850[/tex]
[tex]N_1 = 1370 \ N[/tex]
Therefore; we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.
(b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?
the coefficient of static friction between ladder and ground when the firefighter is 9.40 m from the bottom can be calculated as:
[tex]N_2 (16 \ Sin\ 65) = 850(9.4 \ \times Cos \ 65 )+ 520 (\dfrac{16}{2}) Cos \ 65[/tex]
[tex]N_2 (14.50) = 850(3.9726 )+ 520 (8) \times 0.4226[/tex]
[tex]N_2 (14.50) =3376.71+1758.016[/tex]
[tex]N_2 (14.50) =5134.726[/tex]
[tex]N_2 =\dfrac{5134.726}{14.50}[/tex]
[tex]N_2 =354.12 \ N[/tex]
Therefore; the coefficient of the static friction is;
[tex]\mu = \dfrac{f_s}{N_1}[/tex]
[tex]\mu = \dfrac{354.12}{1370}[/tex]
[tex]\mu[/tex] = 0.26
Coefficient of static friction = 0.26
A 30 L electrical radiator containing heating oil is placed in a 50 m3room. Both the roomand the oil in the radiator are initially at 10◦C. The radiator with a rating of 1.8 kW is nowturned on. At the same time, heat is lost from the room at an average rate of 0.35 kJ/s.After some time, the average temperature is measured to be 20◦C for the air in the room,and 50◦C for the oil in the radiator. Taking the density and the specific heat of the oil to be950 kg/m3and 2.2 kJ/kg◦C, respectively, determine how long the heater is kept on. Assumethe room is well sealed so that there are no air leaks.
Answer:
Explanation:
Heat absorbed by oil
= mass x specific heat x rise in temperature
= 30 x 10⁻³ x 950 x 2.2 x 10³ x ( 50-10 )
= 25.08 x 10⁵ J
Heat absorbed by air
= 50 x 1.2 x 1.0054 x 10³ x ( 20-10 )
= 6.03 x 10⁵ J
Total heat absorbed = 31.11 x 10⁵ J
If time required = t
heat lost from room
= .35 x 10³ t
Total heat generated in time t
= 1.8 x 10³ t
Heat generated = heat used
1.8 x 10³ t = .35 x 10³ t + 31.11 x 10⁵
1.45 x 10³ t = 31.11 x 10⁵
t = 31.11 x 10⁵ / 1.45 x 10³
t = 2145.5 s
?
Which statement about energy transfer in a wave is true?
Particles of the medium absorb energy from the atmosphere.
Particles move from one location
Answer:
Check Explanation.
The two statements given aren't true.
Explanation:
Although the question seems incomplete, I will address the concept of energy transfer during a wave's propagation.
The particles involved in wave's propagation move back and forth perpendicularly to the way the wave is moving, but do not move (at least, no significant movement is noticeable) in the direction of the wave. The particles ‘participate’ in the wave propagation by bumping into one another and transferring energy. This is exactly why energy can be transferred, although the average position of the particles doesn’t change.
So, the particles of the medium do not absorb energy from the atmosphere and do not significantly move from one location to another.
Hope this Helps!!!
Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by
Answer:
the two beams must travel paths that differ by one-half of a wavelength.
A hard drive disk rotates at 7200 rpm. The disk has a diameter of 5.1 in (13 cm). What is the speed of a point 6.0 cm from the center axle
Answer:
The speed will be "3.4×10⁴ m/s²".
Explanation:
The given values are:
Angular speed,
w = 7200 rpm
i.e.,
= [tex]7200 \times \frac{2 \pi}{60}[/tex]
= [tex]753.6 \ rad/s[/tex]
Speed from the center,
r = 6.0 cm
As we know,
⇒ Linear speed, [tex]v=wr[/tex]
On putting the estimated values, we get
[tex]=753.6\times 0.06[/tex]
[tex]=45.216 \ m[/tex]
Now,
Acceleration on disk will be:
⇒ [tex]a=\frac{v^2}{r}[/tex]
[tex]=34074 \ m/s^2[/tex]
[tex]=3.4\times 10^4 \ m/s^2[/tex]
A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released if air resistance is negligible?
A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 35.0 minutes. How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.
A. 0.438c
B. 0.671c
C. 0.319c
D. 0.874c
E. 0.658c
The spaceship is traveling towards the Sun with velocity 0.319c.
What is special theory of relativity?
One of the most significant works in the history of physics is Albert Einstein's 1905 theory of special relativity. The theory of special relativity explains how speed affects space, time, and mass. Small amounts of mass (m) can be interchangeable with large amounts of energy (E), as defined by the classic equation E = mc², according to the theory, which offers a means for the speed of light to define the link between energy and matter.
Given parameters:
The radius of the orbit of Jupiter is 43.2 light-minutes.
that of the orbit of Mars is 12.6 light-minutes.
Hence, actual distance between them l₀= 43.2 - 12.6 light-minute = 30.6 light-minute.
Let the velocity of the spaceship is v.
Time taken to travel this actual distance = l₀ √(1+v²/c²)/v
According to the question
l₀ √(1+v²/c²)/v = 35.0
√(c²+v²) = 35.0 v
c² = 1224 v²
v = 0.319c
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How far apart (in mm) must two point charges of 90.0 nC (typical of static electricity) be to have a force of 3.80 N between them
Answer:
The distance between the two charges is =4.4mm
1. The uniform purely axial magnetic induction required by the experiment in a volume large enough to accommodate the Lorentz Tube is produced by the Helmholtz Coils. What is the magnetic induction due to a coil current 1.5 Ampere
Complete Question
The uniform purely axial magnetic induction required by the experiment in a volume large enough to accommodate the Lorentz Tube is produced by the Helmholtz Coils. What is the magnetic induction due to a coil current 1.5 Ampere? Convert the result in the still popular non-SI unit Gauss (1 Tesla = 10^4 Gauss).
B = N*mue*I/(2*r)
# of loops = 140
radius of the coil = 0.14m
Answer:
The magnetic induction is [tex]B = 2.639 \ Gauss[/tex]
Explanation:
From the question we are told that
The coil current is [tex]I = 1.5 \ A[/tex]
The number of loops is [tex]N = 140[/tex]
The magnetic field due to the current is mathematically represented as
[tex]B = \mu_o * N * I[/tex]
[tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting value
[tex]B = 4\pi * 10^{-7} * 140 * 1.5[/tex]
[tex]B = 2.639*19^{-4} \ T[/tex]
From question
(1 Tesla = [tex]10^4 \ Gauss[/tex]).
=> [tex]B = 2.693 *10^{-4} *10^4 = 2.63 \ Gauss[/tex]
=> [tex]B = 2.639 \ Gauss[/tex]
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of what, independent of the direction thrown.
Complete question is;
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of 31.1 m/s, independent of the direction thrown
Answer:
It is proved that the final speed is truly 31.1 m/s
Explanation:
From energy - conservation principle;
E_i = Initial potential energy + Initial Kinetic Energy
Or
E_i = U_i + K_i
Similarly, for final energy
E_f = U_f + K_f
So, expressing the formulas for potential and kinetic energies, we now have;
E_i = (m × g × y_i) + (½ × m × v_i²)
Similarly,
E_f = (m × g × y_f) + (½ × m × v_f²)
We are given;
y_i = 25 m
y_f = 0 m
v_i = 20 m/s
v_f = 31.1 m/s
So, plugging in relevant values;
E_i = m((9.8 × 25) + (½ × 20²))
E_i = 485m
Similarly;
E_f = m((9.8 × 0) + (½ × v_f²)
E_f ≈ ½m•v_f²
From energy conservation principle, E_i = E_f.
Thus;
485m = ½m•v_f²
m will cancel out to give;
½v_f² = 485
v_f² = 485 × 2
v_f² = 970
v_f = √970
v_f ≈ 31.1 m/s
If a motorcycle accelerates uniformly from rest at 5 m / s2, how long does it take for it to reach a speed of 20 m / s.
Answer:
4 seconds
Explanation:
Given:
v₀ = 0 m/s
v = 20 m/s
a = 5 m/s²
Find: t
v = at + v₀
20 m/s = (5 m/s²) t + 0 m/s
t = 4 s
The magnitude of the magnetic flux through the surface of a circular plate is 6.80 10-5 T · m2 when it is placed in a region of uniform magnetic field that is oriented at 43.0° to the vertical. The radius of the plate is 8.50 cm. Determine the strength of the magnetic field. mT A circular plate of radius r is lying flat. A field of arrows labeled vector B rising up and to the right pass through the plate.
Answer:
B = 4.1*10^-3 T = 4.1mT
Explanation:
In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:
[tex]\Phi_B=S\cdot B=SBcos\alpha[/tex] (1)
ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2
S: surface area of the circular plate = π.r^2
r: radius of the circular plate = 8.50cm = 0.085m
B: magnitude of the magnetic field = ?
α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°
You solve the equation (1) for B, and replace the values of the other parameters:
[tex]B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT[/tex]
The strength of the magntetic field is 4.1mT
A submarine heads toward a port, which broadcasts a signal at 175 MHz. If the submarine heads toward the port at 46 m/s, what change in frequency does the submarine observe in the port's signal
Answer:
Δf = 0.00003 MHz = 30 Hz
Explanation:
The apparent frequency due to the motion of receiver towards the source is given by Doppler's effect, while the source is stationary. Therefore, the formula is given as follows:
[tex]f = (\frac{c+v_{o} }{c})f_{0}[/tex]
where,
f = apparent frequency
c = speed of light = 3 x 10⁸ m/s
v₀ = velocity of observer = 46 m/s
f₀ = Original Frequency = 175 MHz = 175 x 10⁶ Hz
Therefore,
[tex]f = (\frac{300000000 m/s + 46 m/s}{300000000 m/s})(175 MHz)\\[/tex]
f = 175.00003 MHz
Therefore, the change in frequency is:
Δf = f - f₀
Δf = 175.00003 MHz - 175 MHz
Δf = 0.00003 MHz = 30 Hz
The resistance of a wire depends upon the material's resistivity and the length and cross‑sectional area of the wire. What will happen to the resistance of a wire if its length is multiplied by 4?
Answer:
R' = 4R
The resistance will become 4 times the initial value.
Explanation:
The resistance of a wire at room temperature, is given by the following formula:
R = ρL/A ----------- equation 1
where,
R = Resistance of wire
ρ = resistivity of the material
L = Length of wire
A = Cross-sectional area of wire
Now, if the length (L) is multiplied by 4, then resistance will become:
R' = ρ(4L)/A
R' = 4 (ρL/A)
using equation 1:
R' = 4R
The resistance will become 4 times the initial value.
Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 {\rm mi} away. He travels at a steady 47.0 mph. Beth leaves Los Angeles at 9:00 a.m. and drives a steady 55.0 mph. Who gets to San Francisco first? How long does the first to arrive have to wait for the second?
Complete question is;
Alan leaves Los Angeles at 8:AM to drive to San Francisco, 400 mi away. He travels at a steady 47 mph. Beth leaves Los Angeles at 9:00 AM and drives a steady 55.0 mph.
a. Who gets to San Francisco first?
b. How long does the first to arrive have to wait for the second?
Answer:
A) Beth will get there first
B) Beth will have to wait for 15 minutes for alan
Explanation:
We are given;
Distance between Los Angeles and San Francisco;d = 400 mi
Alan's velocity;v_a = 47 mph
Beth's velocity;v_b = 55 mph
we know that time is given by;
Time = distance/velocity
Time required for alan;
t_a = 400/47
t_a = 8.5106 hours = 8 hours 31 minutes
Time required for Beth;
t_b = 400/55
t_b = 7.2727 hours = 7 hours 16 minutes
So Alan will reach there at;
8:00 a.m + 8 hours 31 minutes = 4:31 p.m
Beth will reach there at;
9:00 a.m + 7 hours 16 minutes = 4:16 pm
Thus, beth will arrive 4:31 pm - 4:16pm = 15 minutes before alan
A) Beth will get there first
B) Beth will have to wait for 15 minutes for alan
A commuter train passes a passenger platform at a constant speed of 40.0 m/s. The train horn is sounded at its characteristic frequency of 320 Hz.
(a) What overall change in requency is detected by a person on the platform as the train moves from approaching to receding?
(b) What wavelength is detected by a person on the platform as the train approaches?
Answer:
a) -75 Hz
b) 0.11 [tex]m^{-1}[/tex]
Explanation:
a) Let us first find the frequency detected by the person on the platform.
We have to find the frequency observed by the person when the train was approaching and when the train was receding.
When the train was approaching:
[tex]f_o = \frac{v}{v - v_s} f_s[/tex]
where fo = frequency observed
fs = frequency from the source = 320 Hz
v = speed of sound = 343 m/s
vs = speed of the train = 40 m/s
Therefore:
[tex]f_o = \frac{343}{343 - 40} * 320\\\\f_o = \frac{343}{303} * 320\\\\f_o = 362 Hz[/tex]
The person on the platform heard the sound at a frequency of 362 Hz when the train was approaching.
When the train was receding:
[tex]f_o = \frac{v}{v + v_s} f_s[/tex]
[tex]f_o = \frac{343}{343 + 40} * 320\\\\f_o = \frac{343}{383} * 320\\\\f_o = 287 Hz[/tex]
The person on the platform heard the sound at a frequency of 287 Hz when the train was receding.
Therefore, the frequency change is given as:
Δf = 287 - 362 = -75 Hz
b) We can find the wavelength detected by the person on the platform as the train approaches by using the formula for speed:
[tex]v = \lambda f[/tex]
where λ = wavelength
f = frequency of the train as it approaches = 362 Hz
v = speed of train = 40 m/s
Therefore, the wavelength detected is:
40 = λ * 362
λ = 40 / 362 = 0.11 [tex]m^{-1}[/tex]
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. When each of the spheres has lost half its initial charge, the magnitude of the electrostatic force will be
Answer:
F' = F/4
Thus, the magnitude of electrostatic force will become one-fourth.
Explanation:
The magnitude of force applied by each charge on one another can be given by Coulomb's Law:
F = kq₁q₂/r² -------------- equation 1
where,
F = Force applied by charges
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between the charges
Now, in the final state the charges on both spheres are halved. Therefore,
q₁' = q₁/2
q₂' = q₂/2
Hence, the new force will be:
F' = kq₁'q₂'/r²
F' = k(q₁/2)(q₂/2)/r²
F' = (kq₁q₂/r²)(1/4)
using equation 1:
F' = F/4
Thus, the magnitude of electrostatic force will become one-fourth.
The magnitude of the electrostatic force will be F' = F/4
The magnitude of the electrostatic force:Here we used Coulomb's Law:
F = kq₁q₂/r² -------------- equation 1
Here
F = Force applied by charges
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between the charges
Now
q₁' = q₁/2
q₂' = q₂/2
So, the new force should be
F' = kq₁'q₂'/r²
F' = k(q₁/2)(q₂/2)/r²
F' = (kq₁q₂/r²)(1/4)
So,
F' = F/4
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1. In a Millikan type experiment, two horizontal plates are 2.5 cm apart. A latex sphere of
mass 1.5 x 10-15 kg remains stationary when the potential difference between the
plates is 460 V, with the upper plate positive. [2+2+2+2 = 8 marks]
a. Is the sphere charged negatively or positively?
b. What is the magnitude of the electric field intensity between the plates?
C. Calculate the magnitude of the charge on the latex sphere.
d. How many excess or deficit electrons does the sphere have?
Answer:
Explanation:
a. Is the sphere charged negatively or positively?
The sphere us negatively charged. In a Millikan type experiment, there will be two forces that will be acting on the sphere which are the electric force which acts upward and also the gravity which acts downward.
Because the upper plate is positively charged, there'll what an attractive curve with an upward direction which will be felt by the negatively charged sphere.
b. What is the magnitude of the electric field intensity between the plates?
The magnitude of the electric field intensity between the plates is 18400v/m.
C. Calculate the magnitude of the charge on the latex sphere.
The magnitude of the charge on the latex sphere hae been solved and attached
d. How many excess or deficit electrons does the sphere have?
There are 5 excess electrons that the sphere has.
Check the attachment for further explanation.
James drove his car down a mountain road at an average rate of 30 mph and retunred over the same road at an average rate of 20 mph. How far did he drive down the road before he turned around and drove back if his trip took 5 hours?
A. 1B. 2C. 3D. 4E. 5
Answer:
150 miles
Explanation:
5 times 30 is 150
Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!
Upon using Thomas Young’s double-slit experiment to obtain measurements, the following data were obtained. Use these data to determine the wavelength of light being used to create the interference pattern. Do this using three different methods.
The angle to the eighth maximum is 1.12°.
The distance from the slits to the screen is 302.0 cm.
The distance from the central maximum to the fifth minimum is 3.33 cm.
The distance between the slits is 0.000250 m.
The 3 equations I used were 1). d sin θ_m =(m)λ 2). delta x =λL/d and 3.) d(x_n)/L=(n-1/2)λ
but all my answers are different.
DID I DO SOMETHING WRONG!!!!!!!
Given info
d = 0.000250 meters = distance between slits
L = 302 cm = 0.302 meters = distance from slits to screen
[tex]\theta_8 = 1.12^{\circ}[/tex] = angle to 8th max (note how m = 8 since we're comparing this to the form [tex]\theta_m[/tex])
[tex]x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}[/tex] (n = 5 as we're dealing with the 5th minimum )
---------------
Method 1
[tex]d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}[/tex]
Make sure your calculator is in degree mode.
-----------------
Method 2
[tex]\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\[/tex]
-----------------
Method 3
[tex]\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\[/tex]
There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.
Inexperienced physics teachers often demonstrate the use of the electroscope by touching it with a charged glass rod at a single point. More experienced teachers typically drag the length of the rod across the top of the electroscope to increase the desired effect. Why does this help
Answer:
This is because The glass rod of the electroscope is an insulator therefore only charge transferred to the ball is at the point of contact on the rod. Thus, When the charge rod is dragged across the top of the electroscope, by the experienced teacher the more charge is transferred to electroscope thereby producing a greater effect
An X-Ray machine delivers a radiation dose of 5mRem/hr. at 3ft from the machine. How far will the X-Ray technician have to move to reduce his exposure to 2mRem/hr.? I1/I2 = (D2)2/(D1)2 -------> I1(D1)2 = I2(D2)2
Answer:
4.7ft
Explanation:
Pls see attached file
mention two similarities of citizen and aliens
Answer:
The main points of difference between a citizen and alien are: (a) A citizen is a permanent resident of a state, while an alien is a temporary resident, who comes for a specific duration of time as a tourist or on diplomatic assignment. ... Aliens do not possess such rights in the state where they reside temporarily
Explanation:
An 88.0 kg spacewalking astronaut pushes off a 645 kg satellite, exerting a 110 N force for the 0.450 s it takes him to straighten his arms. How far apart are the astronaut and the satellite after 1.40 min?
Answer:
The astronaut and the satellite are 53.718 m apart.
Explanation:
Given;
mass of spacewalking astronaut, = 88 kg
mass of satellite, = 645 kg
force exerts by the satellite, F = 110N
time for this action, t = 0.45 s
Determine the acceleration of the satellite after the push
F = ma
a = F / m
a = 110 / 645
a = 0.171 m/s²
Determine the final velocity of the satellite;
v = u + at
where;
u is the initial velocity of the satellite = 0
v = 0 + 0.171 x 0.45
v = 0.077 m/s
Determine the displacement of the satellite after 1.4 m
d₁ = vt
d₁ = 0.077 x (1.4 x 60)
d₁ = 6.468 m
According to Newton's third law of motion, action and reaction are equal and opposite;
Determine the backward acceleration of the astronaut after the push;
F = ma
a = F / m
a = 110 / 88
a = 1.25 m/s²
Determine the final velocity of the astronaut
v = u + at
The initial velocity of the astronaut = 0
v = 1.25 x 0.45
v = 0.5625 m/s
Determine the displacement of the astronaut after 1.4 min
d₂ = vt
d₂ = 0.5625 x (1.4 x 60)
d₂ = 47.25 m
Finally, determine the total separation between the astronaut and the satellite;
total separation = d₁ + d₂
total separation = 6.468 m + 47.25 m
total separation = 53.718 m
Therefore, the astronaut and the satellite are 53.718 m apart.
A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Neglect air resistance and use g=9.8m/sec2 as the acceleration of gravity.
Answer:
θ₀ = 84.78° (OR) 5.22°
Explanation:
This situation can be treated as projectile motion. The parameters of this projectile motion are:
R = Range of Projectile = 150 m
V₀ = Launch Speed of Projectile = 90 m/s
g = 9.8 m/s²
θ₀ = Launch angle (OR) Angle of Elevation = ?
The formula for range of a projectile is given as:
R = V₀² Sin 2θ₀/g
Sin 2θ₀ = Rg/V₀²
Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²
2θ₀ = Sin⁻¹ (0.18)
θ₀ = 10.45°/2
θ₀ = 5.22°
Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:
θ₀ = 90° - 5.22°
θ₀ = 84.78°
To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle with the same mass as an electron but the opposite charge ( e). Express your answer in vector form. Do not enter units in your expression.
Required:
a. If a positron is accelerated by a constant electric field, find the acceleration of the positron.
b. Assuming the positron started from rest, find the velocity of the positron
Answer:
a) a = + 1,758 10¹¹ m / s , b) = √ (2 1,758 10¹¹ E x)
Explanation:
a) A charged particle is an electric field undergoing force given by the expression
F = qE
where q is the charge of the paticle and E electric field.
In this case we are told that the particle is positron
q = + 1.6 10⁻¹⁹ C
let's calculate the force
F = + 1.6 10⁻¹⁹ E
we write the positive sign, to show that the particle accelerates in the same direction of the electric field
let's write Newton's second law to find the acceleration
F = ma
a = F / m
a = + 1.6 10-19 / 9.1 10-31 E
a = + 1,758 10¹¹ m / s
b) the velocity of the particle starting from rest
v² = v₀² + 2 a x
v = √ (2 1,758 10¹¹ E x)