Answer:
The change in entropy is 1.6 W/K.
Explanation:
Thickness, d = 0.5 cm
Area, A = 1 m^2
T = 25°C
T' = - 10°C
Coefficient of thermal conductivity of glass, K = 0.8 W/mK
The change in entropy is given by
S = Q/T
Here,
[tex]S =\frac{Q}{T}\\\\S = \frac{K A (T - T')}{d(T - T')}\\\\S = \frac{0.8\times 1}{0.5} = 1.6 W/K[/tex]
Click Stop Using the slider set the following: coeff of restitution to 1.00 A velocity (m/s) to 6.0 A mass (kg) to 6.0 B velocity (m/s) to 0.0 Calculate what range can the mass of B be to cause mass A to bounce off after the collision. Calculate what range can the mass of B be to cause mass A to continue forward after the collision. Check your calculations with the simulation. What are the ranges of B mass (kg)
Answer:
[tex]M_b=6kg[/tex]
Explanation:
From the question we are told that:
Coefficient of restitution [tex]\mu=1.00[/tex]
Mass A [tex]M_a=6kg[/tex]
Initial Velocity of A [tex]U_a=6m/s[/tex]
Initial Velocity of B [tex]U_b=0m/s[/tex]
Generally the equation for Coefficient of restitution is mathematically given by
[tex]\mu=\frac{V_b-V_a}{U_a-U_b}[/tex]
[tex]1=\frac{v_B}{6}[/tex]
[tex]V_b=6*1[/tex]
[tex]V_b=6m/s[/tex]
Generally the equation for conservation of linear momentum is mathematically given by
[tex]M_aU_a+M_bU_b=M_aV_a+M_bV_b[/tex]
[tex]6*6+=M_b*6[/tex]
[tex]M_b=6kg[/tex]
1.a machine gun fires a ball with an initial velocity of 600m/s with an elevation of 30° with respect to the ground neglecting air resistance calculate:
a.the maximum height that can be reached?
b.the time of flight of the bullet?
c.the maximum horizontal displacement of the ired bullet?
Answer:
See explanation
Explanation:
a) maximum height of a projectile = u sin^2θ/2g
H= 600 × (sin 30)^2/2 × 10
H= 7.5 m
b) Time of flight
t= 2u sinθ/g
t= 2 × 600 sin 30/10
t= 60 seconds
Range
R= u^2sin2θ/g
R= (600)^2 × sin2(30)/10
R= 31.2 m
What happens to the acceleration if you triple the force that you apply to the painting with your hand? (Use the values from the example given in the previous part of the lecture.) Submit All Answers Answer: Not yet correct, tries 1/5 3. A driver slams on the car brakes, and the car skids to a halt. Which of the free body diagrams below best matches the braking force on the car. (Note: The car is moving in the forward direction to the right.] (A) (B) (C) (D) No more tries. Hint: (Explanation) The answer is A. The car is moving to the right and slowing down, so the acceleration points to the left. The only significant force acting on the car is the braking force, so this must be pointing left because the net force always shares the same direction as the object's acceleration. 4. Suppose that the car comes to a stop from a speed of 40 mi/hr in 24 seconds. What was the car's acceleration rate (assuming it is constant). Answer: Submit Al Answers Last Answer: 55 N Only a number required, Computer reads units of N, tries 0/5. 5. What is the magnitude (or strength) of the braking force acting on the car? [The car's mass is 1200 kg.) Answer: Submit Al Answers Last Answer: 55N Not yet correct, tries 0/5
Answer:
2) when acceleration triples force triples, 3) a diagram with dynamic friction force in the opposite direction of movement of the car
4) a = 2.44 ft / s², 5) fr = 894.3 N
Explanation:
In this exercise you are asked to answer some short questions
2) Newton's second law is
F = m a
when acceleration triples force triples
3) Unfortunately, the diagrams are not shown, but the correct one is one where the axis of movement has a friction force in the opposite direction of movement, as well as indicating that the car slips, the friction coefficient of dynamic.
The correct answer is: a diagram with dynamic friction force in the opposite direction of movement of the car
4) let's use the scientific expressions
v = v₀ - a t
as the car stops v = 0
a = v₀ / t
let's reduce the magnitudes
v₀ = 40 mile / h ([tex]\frac{5280 ft}{1 mile}[/tex]) ([tex]\frac{1 h}{3600 s}[/tex]) = 58.667 ft / s
a = 58.667 / 24
a = 2.44 ft / s²
5) let's use Newton's second law
fr = m a
We must be careful not to mix the units, we will reduce the acceleration to the system Yes
a = 2.44 ft / s² (1 m / 3.28 ft) = 0.745 m / s²
fr = 1200 0.745
fr = 894.3 N
A body starts from rest and accelerates uniformly at 5m/s. Calculate the time taken by the body to cover a distance of 1km
Answer:
20 seconds
Explanation:
We are given 2 givens in the first statement
v0=0 and a=5
And we are trying to find time needed to cover 1km or 1000m.
So we use
x-x0=v0t+1/2at²
Plug in givens
1000=0+2.5t²
solve for t
t²=400
t=20s
Two guitar strings, of equal length and linear density, are tuned such that the second harmonic of the first string has the same frequency as the third harmonic of the second string. The tension of the first string is 510 N. Calculate the tension of the second string.
Answer:
The tension in the second string is 226.7 N.
Explanation:
Length is L, mass per unit length = m
T = 510 N
Let the tension in the second string is T'.
second harmonic of the first string = third harmonic of the second string
[tex]2 f = 3 f'\\\\2\sqrt{\frac{T}{m}} = 3 \sqrt {\frac{T'}{m}}\\\\4 T = 9 T'\\\\4\times 510 = 9 T'\\\\T' = 226.7 N[/tex]
state the story of archimedes
Answer:
Archimedes was born about 287 BCE in Syracuse on the island of Sicily. He died in that same city when the Romans captured it following a siege that ended in either 212 or 211 BCE. One story told about Archimedes' death is that he was killed by a Roman soldier after he refused to leave his mathematical work.
Choose the force diagram that best represents a ball thrown upward by Peter, at the
top of its path.
Diagram A
Diagram B
Diagram C
Diagram D
Answer:Diagram A
Explanation:
Since the air resistance is to be neglected, only the gravitational force acts on the ball ( and has acted all the way from the throw upward). Diagram A reflects this fact correctly indicating the gravity acting on the ball downward.
A 165-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,015 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable?
Answer:
22.1 years
Explanation:
Since the current in the wire is I = nevA where n = electron density = 8.50 × 10²⁸ electrons/cm³ × 10⁶ cm³/m³= 8.50 × 10³⁴ electrons/m³, e = electron charge = 1.602 × 10⁻¹⁹ C, v = drift velocity of electrons and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.00 cm = 2 × 10⁻² m
Making v subject of the formula, we have
v = I/neA
So, v = I/neπd²/4
v = 4I/neπd²
Since I = 1,015 A, substituting the values of the other variables into the equation, we have
v = 4I/neπd²
v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π ×(2 × 10⁻² m)²]
v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π × 4 × 10⁻⁴ m²]
v = (1,015 A)/[42.779 × 10¹¹ electronsC/m]
v = 23.73 × 10⁻¹¹ m/s
v = 2.373 × 10⁻¹⁰ m/s
Since distance d = speed, v × time, t
d = vt
So, the time it takes one electron to travel the full length of the cable is t = d/v
Since d = distance moved by free charge = length of transmission line = 165 km = 165 × 10³ m and v = drift velocity of charge = 2.373 × 10⁻¹⁰ m/s
t = 165 × 10³ m/2.373 × 10⁻¹⁰ m/s
t = 69.54 × 10⁷ s
t = 6.954 × 10⁸ s
Since we have 365 × 24 hr/day × 60 min/hr × 60 s/min = 31536000 s in a year = 3.1536 × 10⁷ s
So, 6.954 × 10⁸ s = 6.954 × 10⁸ s × 1yr/3.1536 × 10⁷ s = 2.21 × 10 yrs = 22.1 years
It will take one electron 22.1 years to travel the full length of the cable
A point charge of -3.0 x 10-C is placed at the origin of coordinates. Find the clectric field at the point 13. X= 5.0 m on the x-axis.
Answer:
-1.0778×10⁻¹⁰ N/C
Explanation:
Applying,
E = kq/r²................ equation 1
Where E = elctric field, q = charge, r = distance, k = coulomb's law
From the question,
Given: q = -3.0×10 C, r = 5.0 m
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values in equation 1
E = (-3.0×10)(8.98×10⁹)/5²
E = -1.0778×10⁻¹⁰ N/C
Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C
It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor.
1. What is the passenger's apparent weight before the elevator starts moving?
2. What is the passenger's apparent weight whilethe elevator is speeding up?
3. What is the passenger's apparent weight afterthe elevator reaches its cruising speed?
Answer:
1. 588 N
2. 738 N
3. 588 N
Explanation:
time, t = 4 s
initial velocity, u = 0
final velocity, v = 10 m/s
mass, m= 60 kg
1.
Weight of passenger before starts
W =m g = 60 x 9.8 = 588 N
2.
When the elevator is speeding up
v = u + a t
10 = 0 + a x 4
a = 2.5 m/s2
Now the weight is
W' = m (a + g) = 60 (9.8 + 2.5) = 738 N
3.
When he reaches the cruising speed, the weight is
W = 588 N
Một mặt phẳng vô hạn tích điện đều, mật độ σ = 4.10-9 C/cm2 , đặt thẳng đứng trong không khí. Một quả cầu nhỏ có khối lượng 8 g, mang điện tích q = 10-8 C treo gần vào mặt phẳng, sao cho dây treo lúc đầu song song với mặt phẳng. Lấy g = 9,8m/s2 . Khi cân bằng, dây treo quả cầu hợp với mặt phẳng 1 góc bằng bao nhiêu?
Answer:
The angle is 16 degree.
Explanation:
A uniformly charged infinite plane, density σ = 4.10-9 C/cm2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2 . In equilibrium, by what angle does the string hanging from the ball make an angle with the plane?
Surface charge density, σ = 4 x 10^-5 C/m^2
charge, q = 10^-8 C
mass, m = 0.008 kg
The electric field due to the plate is
[tex]E= \frac{\sigma }{2\varepsilon 0}[/tex]
Let the angle make with the vertical is A and T is the tension in the string.
[tex]T sin A = q E....(1)\\\\T cos A = m g .... (2)\\\\Divie (1) by (2)\\\\tan A =\frac{q E}{m g}\\\\tan A = \frac{10^{-8}\times 4\times 10^{-5}}{2\times 8.85\times 10^{-12}\times 0.008\times9.8}\\\\tan A = 0.288\\\\A = 16 degree\\[/tex]
vector A has a magnitude of 8 unit make an angle of 45° with posetive x axis vector B also has the same magnitude of 8 unit along negative x axis find the magnitude of A+B?
Answer:
45 × 8 units = A + B as formular
* A ball is projected horizontally from the top of
a building 19.6m high.
a, How long when the ball take to hit the ground?
b, If the line joining the point of projection to
the point where it hits the ground is 45
with the horizontal. What must be the
initial velocity of the ball?
c,with what vertical verocity does the ball strike
the grounds? (9= 9.8 M152)
Explanation:
Given
Ball is projected horizontally from a building of height [tex]h=19.6\ m[/tex]
time taken to reach ground is given by
[tex]\text{Cosidering vertical motion}\\\Rightarrow h=ut+0.5at^2\\\Rightarrow 19.6=0+0.5\times 9.8t^2\\\Rightarrow t^2=4\\\Rightarrow t=2\ s[/tex]
(b) Line joining the point of projection and the point where it hits the ground makes an angle of [tex]45^{\circ}[/tex]
From the figure, it can be written
[tex]\Rightarrow \tan 45^{\circ}=\dfrac{h}{x}\\\\\Rightarrow x=h\cdot 1\\\Rightarrow x=19.6[/tex]
Considering horizontal motion
[tex]\Rightarrow x=u_xt\\\Rightarrow 19.6=u_x\times 4\\\Rightarrow u_x=4.9\ m/s[/tex]
(c) The vertical velocity with which it strikes the ground is given by
[tex]\Rightarrow v^2-u_y^2=2as\\\Rightarrow v^2-0=2\times 9.8\times 19.6\\\Rightarrow v=\sqrt{384.16}\\\Rightarrow v=19.6\ m/s[/tex]
Thus, the ball strikes with a vertical velocity of [tex]19.6\ m/s[/tex]
Explanation:
Given
Ball is projected horizontally from a building of height
time taken to reach ground is given by
(b) Line joining the point of projection and the point where it hits the ground makes an angle of
From the figure, it can be written
Considering horizontal motion
(c) The vertical velocity with which it strikes the ground is given by
Thus, the ball strikes with a vertical velocity of
Astronauts in space move a toolbox from its initial position ????????→=<15,14,−8>m to its final position ????????→=<17,14,−1>m. The two astronauts each push on the box with a constant force. Astronaut 1 exerts a force ????1→=<18,7,−12>???? and astronaut 2 exerts a force ????2→=<16,−10,16>????.
Required:
What is the total work performed on the toolbox?
If both forces are measured in Newtons, then the net force is
F = (18, 7, -12) N + (16, -10, 16) N = (34, -3, 4) N
The toolbox undergoes a displacement (i.e. change in position) in the direction of the vector
d = (17, 14, -1) m - (15, 14, -8) m = (2, 0, -9) m
The total work done by the astronauts on the toolbox is then
F • d = (34, -3, 4) N • (2, 0, -9) m = (68 + 0 - 36) N•m = 32 J
The work done by the two astronauts is equal to 96 J.
What is work done?work done?Work done is defined as the product of force applied and the distance moved by the force.
Work done = Force × DistanceThe forces applied = 18+16 N, 7+ -10 N, and -12 + 16N
Forces = 34 N, -3 N, and 4N
Distances = (17 - 15, 14 - 14, -1 - - 8) m
Distances = 2, 0, 7
Work done = 34 × 2 + -3 × 0 + 4 × 7
Work done = 96 J
Therefore, the work done by the two astronauts is equal to 96 J.
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How do you find the product of gamma decay?
Answer:
The mass and atomic numbers don't change
Explanation:
An excited atom relaxes to the ground state emitting a photon...called a gamma ray.
The answer is that the mass and atomic numbers don't change.
In gamma decay, the product refers to the nucleus resulting from the emission of a gamma ray. Gamma decay occurs when an excited atomic nucleus releases excess energy in the form of a high-energy photon called a gamma ray.
To find the product of gamma decay, you need to identify the nucleus before and after the decay process. The product nucleus is determined by the parent nucleus that undergoes gamma decay.
During gamma decay, the number of protons and neutrons in the nucleus remains unchanged. Therefore, the identity of the element remains the same, but the energy state of the nucleus is altered.
The product nucleus is typically represented by the same chemical symbol as the parent nucleus, followed by a superscript indicating the mass number (total number of protons and neutrons) and a subscript indicating the atomic number (number of protons).
For example, if a parent nucleus with an atomic number of Z and a mass number of A undergoes gamma decay, the product nucleus will have the same atomic number Z and mass number A.
It's important to note that gamma decay does not involve the emission or absorption of any particles, only the release of electromagnetic radiation (gamma ray).
Thus, the product nucleus remains unchanged in terms of atomic number and mass number.
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what is conservation energy?
Explanation:
Conservation of energy, principle of physics according to which the energy of interacting bodies or particles in a closed system remains constant
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Three 15-Ω and two 25-Ω light bulbs and a 24 V battery are connected in a series circuit. What is the current that passes through each bulb?
1) 0.18 A
2) 0.25 A
3) 0.51 A
4) 0.74 A
5) The current will be 1.6 A in the 15-Ω bulbs and 0.96 A in the 25-Ω bulbs.
Answer:
I = 0.25 A
Explanation:
Given that,
Three 15 ohms and two 25 ohms light bulbs and a 24 V battery are connected in a series circuit.
In series combination, the equivalent resistance is given by :
[tex]R=R_1+R_2+R_3+....[/tex]
So,
[tex]R=15+15+15+25+25\\\\=95\ \Omega[/tex]
The current each resistor remains the same in series combination. It can be calculated using Ohm's law i.e.
V = IR
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{24}{95}\\\\I=0.25\ A[/tex]
So, the current of 0.25 A passes through each bulb.
190 students sit in an auditorium listening to a physics lecture. Because they are thinking hard, each is using 125 W of metabolic power, slightly more than they would use at rest. An air conditioner with a COP of 5.0 is being used to keep the room at a constant temperature. What minimum electric power must be used to operate the air conditioner?
Answer:
W = 4.75 KW
Explanation:
First, we will calculate the heat to be removed:
Q = (No. of students)(Metabolic Power of Each Student)
Q = (190)(125 W)
Q = 23750 W = 23.75 KW
Now the formula of COP is:
[tex]COP = \frac{Q}{W}\\\\W = \frac{Q}{COP}\\\\W = \frac{23.75\ KW}{5}\\\\[/tex]
W = 4.75 KW
A stopped organ pipe of length L has a fundamental frequency of 220 Hz. For which of the following organ pipes will there be a resonance if a tuning fork of frequency 660 Hz is sounded next to the pipe?
a. a stopped organ pipe of length L
b. a stopped organ pipe of length 2L
c. an open organ pipe of length L;
d. an open organ pipe of length 2L.
Answer:
Option (a), (d) are correct.
Explanation:
Frequency, f = 220 Hz
Resonant frequency = 660 Hz
The next frequency of stopped organ pipe is
2f, 3 f, 4 f ....
= 2 x 220 , 3 x 220 , 4 x 220 ....
= 440 Hz, 660 Hz, 880 Hz
So, the option (a) is correct.
The next resonant frequency of open organ pipe is
3 f, 5 f,...
= 3 x 220, 5 x 220 , ..
= 660 Hz, 1100 Hz,...
So, option (d) is correct.
The following two waves are sent in opposite directions on a horizontal string so as to create a standing wave in a vertical plane: y1(x, t) = (8.20 mm) sin(4.00πx - 430πt) y2(x, t) = (8.20 mm) sin(4.00πx + 430πt), with x in meters and t in seconds. An antinode is located at point A. In the time interval that point takes to move from maximum upward displacement to maximum downward displacement, how far does each wave move along the string?
Answer:
Explanation:
From the information given:
The angular frequency ω = 430 π rad/s
The wavenumber k = 4.00π which can be expressed by the equation:
k = ω/v
∴
4.00 = 430 /v
v = 430/4.00
v = 107.5 m/s
Similarly: k = ω/v = 2πf/fλ
We can say that:
k = 2π/λ
4.00 π = 2π/λ
wavelength λ = 2π/4.00 π
wavelength λ = 0.5 m
frequency of the wave can now be calculated by using the formula:
f = v/λ
f = 107.5/0.5
f = 215 Hz
Also, the Period(T) = 1/215 secs
The time at which particle proceeds from point A to its maximum upward displacement and to its maximum downward displacement can be computed as t = T/2;
Thus, the distance(x) covered by each wave during this time interval(T/2) will be:
x = v * t
x = v * T/2
x = λ/2
x = 0.5/2
x = 0.25 m
Based on the information in the table, what
is the acceleration of this object?
t(s) v(m/s)
0.0
9.0
1.0
4.0
2.0
-1.0
3.0
-6.0
A. -5.0 m/s2
B. -2.0 m/s2
C. 4.0 m/s2
D. 0.0 m/s2
Answer:
Option A. –5 m/s²
Explanation:
From the question given above, the following data were obtained:
Initial velocity (v₁) = 9 m/s
Initial time (t₁) = 0 s
Final velocity (v₂) = –6 m/s
Final time (t₂) = 3 s
Acceleration (a) =?
Next, we shall determine the change in the velocity and time. This can be obtained as follow:
For velocity:
Initial velocity (v₁) = 9 m/s
Final velocity (v₂) = –6 m/s
Change in velocity (Δv) =?
ΔV = v₂ – v₁
ΔV = –6 – 9
ΔV = –15 m/s
For time:
Initial time (t₁) = 0 s
Final time (t₂) = 3 s
Change in time (Δt) =?
Δt = t₂ – t₁
Δt = 3 – 0
Δt = 3 s
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Change in velocity (Δv) = –15 m/s
Change in time (Δt) = 3 s
Acceleration (a) =?
a = Δv / Δt
a = –15 / 3
a = –5 m/s²
Thus, the acceleration of the object is
–5 m/s².
what is time taken by radio wave to go and return back from communication satellite to earth??
Answer:
Radio waves are used to carry satellite signals. These waves travel at 300,000 km/s (the speed of light). This means that a signal sent to a satellite 38,000 km away takes 0.13 s to reach the satellite and another 0.13 s for the return signal to be received back on Earth.
Explanation:
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The thrust F of a screw propeller is known to depend upon the diameter d, Speed of advance v, fluid density e, revolution per second N, and the coefficient of viscosity M, of the fluid. Find the expression for F, in terms of the quantities
Answer:
[tex]{ \bf{F = { \tt{ \frac{4}{3} \pi {r}^{3}v gM}}}}[/tex]
Two projectiles A and B are fired simultaneously from a level, horizontal surface. The projectiles are initially 62.2 m apart. Projectile A is
fired with a speed of 19.5 m/s at a launch angle 30° of while projectile B is fired with a speed of 19.5 m/s at a launch angle of 60°. How long
it takes one projectile to be directly above the other?
Let the point where A is launched act as the origin, so that the horizontal positions at time t of the respective projectiles are
• A : x = (19.5 m/s) cos(30°) t
• B : x = 62.2 m + (19.5 m/s) cos(60°) t
These positions are the same at the moment one projectile is directly above the other, which happens for time t such that
(19.5 m/s) cos(30°) t = 62.2 m + (19.5 m/s) cos(60°) t
Solve for t :
(19.5 m/s) (cos(30°) - cos(60°)) t = 62.2 m
t = (62.2 m) / ((19.5 m/s) (cos(30°) - cos(60°))
t ≈ 8.71 s
When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What maximum wavelength of light could be used to ejectelectrons from aluminum
Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
Ф is the minimum energy needed to eject electron the aluminum metal
E is the energy of the incident light
The energy of the incident light is calculated as follows;
[tex]E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J[/tex]
The work function of the aluminum metal is calculated as;
Ф = E - K.E
Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)
Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J
Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;
[tex]\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm[/tex]
In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-called cathode rays. One remarkable series of experiments with cathode rays, conducted by J. J. Thomson around 1897, led to the discovery of the electron.
With the idea that cathode rays were charged particles, Thomson used a cathode-ray tube to measure the ratio of charge to mass, q/m, of these particles, repeating the measurements with different cathode materials and different residual gases in the tube.
Part A
What is the most significant conclusion that Thomson was able to draw from his measurements?
He found a different value of q/m for different cathode materials.
He found the same value of q/m for different cathode materials.
From measurements of q/m he was able to calculate the charge of an electron.
From measurements of q/m he was able to calculate the mass of an electron.
Part B
What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.
Express your answer in terms of e, m, d, v0, L, and E0.
Part C
Now imagine that you place your entire apparatus inside a region of magnetic field of magnitude B0 (Figure 2) . The magnetic field is perpendicular to E⃗ 0 and directed straight into the plane of the figure. You adjust the value of B0 so that no deflection is observed on the screen.
What is the speed v0 of the electrons in this case?
Express your answer in terms of E0 and B0.
Answer:
a) He found the same value of q/m for different cathode materials.
b) y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex] , c) v = [tex]\frac{E_o}{B_o}[/tex]
Explanation:
In Thomson's experiments he was able to measure the deflection of the light beam under the effect of the magnetic field and with these results find the e / m relationship, which in all cases is the same, therefore the most important conclusion is that the value e E / m is constant for all materials.
b) In the part of the plates the electrons are accelerated by the electric field,
F = ma
- e E = m a
a = - (e/m) E₀
the distance traveled is
X axis
x = v₀ t
the separation of the plates is x = d
t = vo / d
Y axis
y = v_{oy} t + ½ to t²
y = ½ a t²
y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex]
c) In this case there is a magnetic field B₀ and the electrons have no deflection
F = - e E + e v x B
if there is no deviation F = 0
e E = e v B
v = [tex]\frac{E_o}{B_o}[/tex]
A body of mass 2kg is released from from a point 100m above the ground level. calculate kinetic energy 80m from the point of released.
Answer:
1568J
Explanation:
Since the problem states 80 m from the point of drop, the height relative to the ground will be 100-80=20m.
Use conservation of Energy
ΔUg+ΔKE=0
ΔUg= mgΔh=2*9.8*(20-100)=-1568J
ΔKE-1568J=0
ΔKE=1568J
since KEi= 0 since the object is at rest 100m up, the kinetic energy 20meters above the ground is 1568J
Why don’t you see tides ( like those of the ocean ) in your swimming pool ?
Which indicates the first law of thermodynamics
Answer:
(d)
Explanation:
because dU = Q -W so ,that the option d(D) is correct
g A spherical container of inner diameter 0.9 meters contains nuclear waste that generates heat at the rate of 872 W/m3. Estimate the total rate of heat transfer from the container to its surroudings ignoring radiation.
Answer: The total rate of heat transfer from the container to its surroundings ignoring radiation is 332.67 W.
Explanation:
Given: Inner diameter = 0.9 m
q = 872 [tex]W/m^{3}[/tex]
Now, radii is calculated as follows.
[tex]r = \frac{diameter}{2}\\= \frac{0.9}{2}\\= 0.45 m[/tex]
Hence, the rate of heat transfer is as follows.
[tex]Q = q \times V[/tex]
where,
V = volume of sphere = [tex]\frac{4}{3} \pi r^{3}[/tex]
Substitute the values into above formula as follows.
[tex]Q = q \times \frac{4}{3} \pi r^{3}\\= 872 W/m^{3} \times \frac{4}{3} \times 3.14 \times (0.45 m)^{3}\\= 332.67 W[/tex]
Thus, we can conclude that the total rate of heat transfer from the container to its surroundings ignoring radiation is 332.67 W.
