Answer:
The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction
Explanation:
In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:
[tex]\vec{F_B}=q\vec{v}\ X\ \vec{B}[/tex] (1)
v: speed of the proton = 9.9*10^5 m/s
q: charge of the proton = 1.6*10^-19C
B: magnetic field = ?
FB: magnetic force on the proton = 1.6*10^-13N
When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:
^j X (-^i) = -(-^k)=^k
To obtain the magnitude of the magnetic field you use:
[tex]F_B=qvBsin90\°=qvB\\\\B=\frac{F_B}{qv}=\frac{1.6*10^{-13}N}{(1.6*10^{-19}C)(9.9*10^5m/s)}\\\\B=1.01T[/tex]
The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction
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
Which best describes the relationship between heat, internal energy, and thermal energy?
Internal energy is heat that flows, and heat is the part of thermal energy that can be transferred.
Internal energy is thermal energy that flows, and thermal energy is the part of heat that can be transferred.
Thermal energy is heat that flows, and heat is the part of internal energy that can be transferred.
Heat is thermal energy that flows, and thermal energy is the part of internal energy that can be transferred.
Mark this and return
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Answer:
I think it is the 4th answer choice
Explanation:
Heat is thermal energy that flows in the direction of high temp to low temp, and internal energy is the "energy contained in a system", and thermal energy is a part of that.
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.
How did the magnet’s density measurement using the Archimedes’ Principle compare to the density measurement using the calculated volume? Which method might be more accurate? Why?
Answer:
The two methods will yield different results as one is subject to experimental errors that us the Archimedes method of measurement, the the density measurement method will be more accurate
Explanation:
This is because the density method using the calculated volume will huve room for less errors that's occur in practical method i.e Archimedes method due to human error
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]
A very large sheet of a conductor carries a uniform charge density of on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Complete Question
A very large sheet of a conductor carries a uniform charge density of [tex]4.00\ pC/mm^2[/tex] on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Answer:
The electric field is [tex]E = 4.5198 *10^{5} \ N/C[/tex]
Explanation:
From the question we are told that
The charge density is [tex]\sigma = 4.00pC /mm^2 = 4.00 * 10^{-12 } * 10^{6} = 4.00 *10^{-6}C/m[/tex]
The position outside the surface is [tex]a = 3.00 \ mm = 0.003 \ m[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{\sigma}{\epsilon _o }[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with values [tex]\epsilon _o = 8.85 *10^{-12} F/m[/tex]
substituting values
[tex]E = \frac{4.0*10^{-6}}{8.85 *10^{-12} }[/tex]
[tex]E = 4.5198 *10^{5} \ N/C[/tex]
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.
In the child's game of tetherball, a rope attached to the top of a tall pole is tied to a ball. Players hit the ball in opposite directions in an attempt to wrap the ball and rope around the pole. Assume the rope has negligible mass and that resistive forces, such as air resistance and friction, can be neglected. As the ball wraps around the pole between hits, how does the angular speed of the ball change
Answer:
The angular speed of the ball will increase
Explanation:
the angular speed of the ball will increase because the force of hit by the players will sum up in opposite direction to increase the angular speed
A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 1 kilograms is tied to the middle of the clothesline, it sags a distance of 4 meters. What is the magnitude of the tension on the ends of the clothesline
Answer:
The tension on the clotheslines is [tex]T = 8.83 \ N[/tex]
Explanation:
The diagram illustrating this question is shown on the first uploaded image
From the question we are told that
The distance between the two poles is [tex]d = 12 \ m[/tex]
The mass tie to the middle of the clotheslines [tex]m = 1 \ kg[/tex]
The length at which the clotheslines sags is [tex]l = 4 \ m[/tex]
Generally the weight due to gravity at the middle of the clotheslines is mathematically represented as
[tex]W = mg[/tex]
let the angle which the tension on the clotheslines makes with the horizontal be [tex]\theta[/tex] which mathematically evaluated using the SOHCAHTOA as follows
[tex]Tan \theta = \frac{ 4}{6}[/tex]
=> [tex]\theta = tan^{-1}[\frac{4}{6} ][/tex]
=> [tex]\theta = 33.70^o[/tex]
So the vertical component of this tension is mathematically represented a
[tex]T_y = 2* Tsin \theta[/tex]
Now at equilibrium the net horizontal force is zero which implies that
[tex]T_y - mg = 0[/tex]
=> [tex]T sin \theta - mg = 0[/tex]
substituting values
[tex]T = \frac{m*g}{sin (\theta )}[/tex]
substituting values
[tex]T = \frac{1 *9.8}{2 * sin (33.70 )}[/tex]
[tex]T = 8.83 \ N[/tex]
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
What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3
Answer:
3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²
- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²
Explanation:
Complete Question
3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?
- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.
The image of this setup attached to this question as obtained from online is attached to this solution.
Solution
3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.
Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²
The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²
- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.
I₂ = I₁ cos² θ
where
I₂ = intensity of light that passes through the second polarizer = ?
I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²
θ = angle between the major axis of the first and second polarizer = 30°
I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²
In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through
I₃ = I₂ cos² θ
I₃ = intensity of light that passes through the third/additional polarizer = ?
I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²
θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)
I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²
Hope this Helps!!!
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 5.0-Ω resistor and a 9.0-Ω resistor are connected in parallel. A 4.0-Ω resistor is then connected in series with this parallel combination. An ideal 6.0-V battery is then connected across the series-parallel combination of the three resistors. What is the current through (a) the 4.0-Ω resistor? (b) the 5.0-Ω resistor? (c) the 9.0-Ω resistor?
Answer:
Explanation:
The current through the resistor is 0.83 A
.
Part b
The current through resistor is 0.53 A
.
Part c
The current through resistor is 0.30 A
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
Learn more about force here: https://brainly.com/question/14282312
A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length r. If I1 is the moment of inertia of this object with respect to an axis passing through the center of the rod and perpendicular to it and I2 is the moment of inertia with respect to an axis passing through one of the masses, it follows that:
a. I1 > I2
b. I2 > I1.
c. I1 = I2.
Answer:
B: I2>I1
Explanation:
See attached file
A simple pendulum of length 1.62 m has a mass of 117 g attached. It is drawn back 38.0 degrees and then released. What is the maximum speed of the mass
Answer:
The maximum speed of the mass is 4.437 m/s.
Explanation:
Given;
length of pendulum, L = 1.62 m
attached mass, m = 117 g
angle of inclination, θ = 38°
This mass was raised to a height of
h = 1.62 - cos38° = 1.0043 m
Apply the principle of conservation of mechanical energy
PE = KE
mgh = ¹/₂mv²
v = √(2gh)
v = √(2 * 9.8 * 1.0043)
v = 4.437 m/s.
Therefore, the maximum speed of the mass is 4.437 m/s.
A proton of mass and a charge of is moving through vacuum at a constant velocity of 10000 directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of E =3.62e+3 N/C . The region of uniform electric field is 5 mm wide in the east-west direction.
Required:
How far (in meters) will the proton have been deflected towards the south by the time it exits the region of uniform electric field.
Complete Question
A proton of mass mp= 1.67×10−27 kg and a charge of qp= 1.60×10−19 C is moving through vacuum at a constant velocity of 10,000 m/s directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of E = 3.62e+3 N/C . The region of uniform electric field is 5 mm wide in the east-west direction. How far (in meters) will the proton have been deflected towards the south by the time it exits the region of uniform electric field. You may neglect the effects of friction and gravity, and assume that the electric field is zero outside the specified region. Answer is to be in units of meters
Answer:
[tex]s = 0.039 \ m[/tex]
Explanation:
From the question we are told that
The mass of the proton is [tex]m = 1.67 *10^{-27} \ g[/tex]
The charge of on the proton is [tex]q = 1.60 *10^{-19} \ C[/tex]
The speed of the proton is [tex]v = 10000 \ m/s[/tex]
The magnitude of the electric field is [tex]E = 3.62*10^{3 } \ N/C[/tex]
The width covered by the electric field [tex]d = 5mm = 5 *10^{-3} \ m[/tex]
Generally the acceleration of the proton due to the electric toward the south (at the point where the force on the proton is equal to the electric force due to the electric field) is mathematically represented as
[tex]a = \frac{q* E}{m}[/tex]
Substituting values
[tex]a = \frac{1.60*10^{-19 } * 3.26 *10^{3}}{ 1.67*10^{-27}}[/tex]
[tex]a = 3.12*10^{11} \ m/s^2[/tex]
Generally the time it will take the proton to cross the electric field is mathematically represented as
[tex]t = \frac{d}{v}[/tex]
Substituting values
[tex]t = \frac{5 *10^{-3}}{10000}[/tex]
[tex]t = 5 *10^{-7} \ s[/tex]
Generally the the distance covered by the proton toward the south is
[tex]s = ut + \frac{1}{2} * a*t^2[/tex]
Here u = 0 m/s this because before the proton entered the electric field region the it velocity towards the south is zero
So
[tex]s = \frac{1}{2} * a*t^2[/tex]
Substituting values
[tex]s = \frac{1}{2} * 3.12 *10^{11}*(5 *10^{-7})^2[/tex]
[tex]s = 0.039 \ m[/tex]
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:
How does an atom of rubidium-85 become a rubidium ion with a +1 charge?
Answer:
C. The atom loses 1 electron to have a total of 36.
Explanation:
Cations have a positive charge. Cations lose electrons.
The number of electrons in a Rubidium atom is 37. If the atom loses 1 electron, then it has 36 left.
Monochromatic coherent light shines through a pair of slits. If the wavelength of the light is decreased, which of the following statements are true of the resulting interference pattern? (There could be more than one correct choice.)
a. The distance between the maxima decreases.
b. The distance between the minima decreases.
c. The distance between the maxima stays the same.
d. The distance between the minima increases.
e. The distance between the minima stays the same.
Answer:
he correct answers are a, b
Explanation:
In the two-slit interference phenomenon, the expression for interference is
d sin θ= m λ constructive interference
d sin θ = (m + ½) λ destructive interference
in general this phenomenon occurs for small angles, for which we can write
tanθ = y / L
tan te = sin tea / cos tea = sin tea
sin θ = y / La
un
derestimate the first two equations.
Let's do the calculation for constructive interference
d y / L = m λ
the distance between maximum clos is and
y = (me / d) λ
this is the position of each maximum, the distance between two consecutive maximums
y₂-y₁ = (L 2/d) λ - (L 1 / d) λ₁ y₂ -y₁ = L / d λ
examining this equation if the wavelength decreases the value of y also decreases
the same calculation for destructive interference
d y / L = (m + ½) κ
y = [(m + ½) L / d] λ
again when it decreases the decrease the distance
the correct answers are a, b
Five identical cylinders are each acted on by forces of equal magnitude. Which force exerts the biggest torque about the central axes of the cylinders
Answer:
From the image, the force as shown in option A will exert the biggest torque on the cylinder about its central axes.
Explanation:
The image is shown below.
Torque is the product of a force about the center of rotation of a body, and the radius through which the force acts. For a given case such as this, in which the cylinders are identical, and the forces are of equal magnitude, the torque at the maximum radius away from the center will exert the maximum torque. Also, the direction of the force also matters. To generate the maximum torque, the force must be directed tangentially away from the circle formed by the radius through which the force acts away from the center. Option A satisfies both condition and hence will exert the most torque on the cylinder.
A commercial diffraction grating has 500 lines per mm. Part A When a student shines a 480 nm laser through this grating, how many bright spots could be seen on a screen behind the grating
Answer:
The number of bright spot is m =4
Explanation:
From the question we are told that
The number of lines is [tex]s = 500 \ lines / mm = 500 \ lines / 10^{-3} m[/tex]
The wavelength of the laser is [tex]\lambda = 480 nm = 480 *10^{-9} \ m[/tex]
Now the the slit is mathematically evaluated as
[tex]d = \frac{1}{s} = \frac{1}{500} * 10^{-3} \ m[/tex]
Generally the diffraction grating is mathematically represented as
[tex]dsin\theta = m \lambda[/tex]
Here m is the order of fringes (bright fringes) and at maximum m [tex]\theta = 90^o[/tex]
So
[tex]\frac{1}{500} * sin (90) = m * (480 *10^{-3})[/tex]
=> [tex]m = 4[/tex]
This implies that the number of bright spot is m =4
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°
A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What will be the wavelength if the tension is decreased to 600 N and the frequency is kept constant
Answer:
The wavelength will be 33.9 cm
Explanation:
Given;
frequency of the wave, F = 1200 Hz
Tension on the wire, T = 800 N
wavelength, λ = 39.1 cm
[tex]F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}[/tex]
Where;
F is the frequency of the wave
T is tension on the string
μ is mass per unit length of the string
λ is wavelength
[tex]\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\[/tex]
when the tension is decreased to 600 N, that is T₂ = 600 N
[tex]T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm[/tex]
Therefore, the wavelength will be 33.9 cm
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.
An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about _____ years.
Answer:
An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about 2 years.
Explanation:
Given;
orbital period of 3 years, P = 3 years
To calculate the years of an orbital with a semi-major axis, we apply Kepler's third law.
Kepler's third law;
P² = a³
where;
P is the orbital period
a is the orbital semi-major axis
(3)² = a³
9 = a³
a = [tex]a = \sqrt[3]{9} \\\\a = 2.08 \ years[/tex]
Therefore, An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about 2 years.
You’re driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. Your reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10 m/s2.
a. How much distance is between you and the deer when you come to a stop?
b. What is the maximum speed you could have and still not hit the deer?
Answer:
(a) Distance between deer and car = 5 m
(b) Vmax = 21.92 m/s
Explanation:
a.
First we calculate distance covered during response time:
s₁ = vt --------- equation 1
where,
s₁ = distance covered during response time = ?
v = speed of car = 20 m/s
t = response time = 0.5 s
Therefore,
s₁ = (20 m/s)(0.5 s)
s₁ = 10 m
Now, we calculate the distance covered by the car during deceleration. Using 3rd equation of motion:
2as₂ = Vf² - Vi²
s₂ = (Vf² - Vi²)/2a ------ eqation 2
where,
a = deceleration = - 10 m/s²
s₂ = Distance covered during deceleration = ?
Vf = Final Velocity = 0 m/s (since car finally stops)
Vi = Initial Velocity = 20 m/s
Therefore,
s₂ = [(0 m/s)² - (20 m/s)²]/2(-10 m/s²)
s₂ = (400 m²/s²)/(20 m/s²)
s₂ = 20 m
thus, the total distance covered by the car before coming to rest is given as:
s = s₁ + s₂
s = 10 m + 20 m
s = 30 m
Now, the distance between deer and car, when it comes to rest, can be calculated as:
Distance between deer and car = 35 m - s = 35 m - 30 m
Distance between deer and car = 5 m
b.
Since, the distance covered by the car in total must be equal to 35 m at maximum. Therefore,
s₁ + s₂ = 35 m
using equation 1 and equation 2 from previous part:
Vi t + (Vf² - Vi²)/2a = 35 m
Vi(0.5 s) + [(0 m/s)² - Vi²]/2(-10 m/s²) = 35 m
0.5 Vi + 0.05 Vi² = 35
0.05 Vi² + 0.5 Vi - 35 = 0
solving this quadratic equation, we get:
Vi = - 31.92 m/s (OR) Vi = 21.92 m/s
For maximum velocity:
Vmax = 21.92 m/s
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
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 boat that has a speed of 6km / h must cross a 200m wide river perpendicular to the current that carries a speed of 1m / s. Calculate a) the final speed of the boat b) displacement experienced by the boat in the direction of the current when making the journey
Answer:
a) 1.94 m/s
b) 120 m
Explanation:
Convert km/h to m/s:
6 km/h = 1.67 m/s
a) The final speed is found with Pythagorean theorem:
v = √((1.67 m/s)² + (1 m/s)²)
v = 1.94 m/s
b) The time it takes the boat to cross the river is:
t = (200 m) / (1.67 m/s)
t = 120 s
The displacement in the direction of the current is:
x = (1 m/s) (120 s)
x = 120 m