Answer:
a).[tex]$3.99 \times 10^5 \ v/m$[/tex]
b). 2.9925 kV
Explanation:
Given :
For mass spectrometer
The magnetic field = B
B = 0.105 T
a). Given speed, v = [tex]$3.8 \times 10^6 \ m/s$[/tex]
We known
[tex]$\frac{E}{B}=v$[/tex]
∴ [tex]$E= 3.8 \times 10^6 \times 0.105$[/tex]
[tex]$=3.99 \times 10^5 \ v/m$[/tex]
b). Now spectrometer, d = 0.75 cm
[tex]$d=0.75 \times 10^{-2} \ m$[/tex]
We known
[tex]$E=\frac{V}{d}$[/tex]
[tex]$V = E\times d$[/tex]
[tex]$V = 3.99 \times 10^5 \times 0.75 \times 10^{-2}$[/tex]
[tex]$V = 2.9925 \times 10^3 \ V$[/tex]
= 2.9925 kV
A physics student sits in a chair. The chair pushes up on the student's body. Identify the other force of the interaction force pair.
Answer:
gravity pulls down on student the chair pushes up on the student's body with the same force gravity is pulling down on the student
how many electrons can occupy each sublevel?
Which of the following is NOT a reason
why gravity is important?
A It holds the planets in
orbit around the sun
B. It causes the ocean tides
C. It guides the growth of plants
D. None of the above
Answer:
I'm gonna say d
Explanation:
bc they all seem very important
hope this helped
A ball of mass m=10g, carrying a charge q =-20μe is suspended from a string of length L= 0.8m above a horizontal uniformly charged infinite plane sheet of charge density σ = 4μe/m^2. The ball is displaced from the vertical by an angle and allowed to swing from rest.
Required:
a. Obtain the equations of motion of the charged ball based on Newtonian laws of motion.
b. Assume the displaced angle θ is small and simplify the results obtained in part (a) to obtain the frequency of oscillations of the charged ball.
Answer:
a) [tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] = [tex]\frac{d^2 \theta}{d t^2}[/tex]
b) f = 2π [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]
Explanation:
a) To have the equations of motion, let's use Newton's second law.
Let's set a reference system where the x-axis is parallel to the path and the y-axis is in the direction of tension of the rope.
For this reference system the tension is in the direction of the y axis, we must decompose the weight and the electrical force.
Let's use trigonometry for the weight that is in the vertical direction down
sin θ = Wₓ / W
cos θ = W_y / w
Wₓ = W sin θ
W_y = W cos θ
we repeat for the electric force that is vertical upwards
F_{ex} = F_e sin θ
F_{ey} = F_e cos θ
the electric force is
F_e = q E
where the field created by an infinite plate is
E = [tex]\frac{ \sigma}{2 \epsilon_o}[/tex]
let's write Newton's second law
Y axis
T - W_y = 0
T = W cos θ
X axis
F_{ex} - Wₓ = m a (1)
we use that the acceleration is related to the position
a = dv / dt
v = dx / dt
where x is the displacement in the arc of the curve
substituting
a = d² x /dt²
we substitute in 1
q E sin θ - mg sin θ = m [tex]\frac{d^2 x}{dt^2}[/tex]
we have angular (tea) and linear (x) variables, if we remember that angles must be measured in radians
θ = x / R
x = R θ
we substitute
sin θ (q E - mg) = m \frac{d^2 R \ theta}{dt^2}
[tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] = [tex]\frac{d^2 \theta}{d t^2}[/tex]
this is the equation of motion of the system
b) for small oscillations
sin θ = θ
therefore the solution is simple harmonic
θ = θ₀ cos (wt + Ф)
if derived twice, we substitute
- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{\theta}{R } θ₀ cos (wt + Ф) = -w² θ₀ cos (wt + Ф)
w² = [tex]\frac{g}{R}[/tex] - [tex]\frac{q}{m} \frac{ \sigma }{2 \epsilon_o} \frac{1}{R}[/tex]
angular velocity is related to frequency
w = 2π f
f = 2π / w
f = 2π/w
f = 2π [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]
PLEASE CLICK ON THIS IMAGE I NEED HELP
Answer:
Second option
Explanation:
"Uniform" pretty much means the same thing happens.
Waves are ?
that can travel through matter.
Answer:
A wave can be thought of as a disturbance or oscillation that travels through space-time, accompanied by a transfer of energy. The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.
A physics student sits in a chair. The chair pushes up on the student's body. Identify the other force of the interaction force pair.
Answer:
The other force is the weight of the student.
Explanation:
With respect to Newton's third law of motion, for the student to sit and balance on the chair, there must be two equal and opposite forces involved. The student applies his/ her weight on the chair which acts downwards, while the chair applies an equal but opposite force to the weight of the student.
The force applied by the chair on the student's body is counter balanced by the student's weight. Note that, if the weight of the student is greater than the opposing force from the chair, the chair would collapse.
HELP me please cause I don't understand it.
Answer:
Force = 0.49 N (Approx)
Explanation:
Given:
Mass = 50 grams = 0.05 Kg
Acceleration = 9.81 m/s²
Find:
Force
Computation:
Force = Mass x Acceleration
Force = 0.05 x 9.81
Force = 0.4905
Force = 0.49 N (Approx)
4. When you are holding a book, energy is stored between the book and the Earth.
This type of energy is called
potential energy.
A. Elastic potential energy
B. Chemical potential energy
C. Gravitational potential energy
D. Kinetic energy
Answer:
gravitational potential energy
why type of volcano is built almost entirely from ejected lava fragments
Answer:
Shield volcanoes
Explanation:
What is the average kinetic energy of particles in a gas at a temperature of 245 Kelvins?
Answer:
2)
Explanation:
Which image best illustrates diffraction?
Answer: A
Explanation:
****PLEASE HELP**** WILL MARK BRAINLIEST
Assuming that voltage remains constant, what happens to the current in a
wire if the length of the wire increases?
O A. The current decreases.
OB. The current alternates between high and low values.
O C. The current increases.
O D. The current is not affected by a change in wire length.
Answer:
The Current decreases
Explanation:
HOPE THIS HELPS!
Which word best completes the sentence?
Select the word from the drop-down menu
He is quite
Choose...
despite never having left his smalL TOWEN
Answer:
it’s cosmopolitan
Explanation:
k12
what dose current equal?
An inclined plane consists of a 25 m length that raises an object 5 m above the ground. When pushing a 4500 N crate to the top of the ramp you exert 1000 N. What is the ideal Mechanical Advantage of this machine?
The IDEAL mechanical advantage of this ramp is (25m / 5m) = 5 .
But it's only giving you a real MA of (4500N/1000N) = 4.5 .
The friction between the crate and the surface of the ramp is robbing some of the work you do as you slide the crate up the ramp, which degrades the mechanical advantage.
A tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J
Answer:
3.9 × 10^7 J
Explanation:
Given that a tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J
Solution
Since the tank is half full, the height = 2.5m
Pressure = density × gravity × height
Pressure = 900 × 9.8 × 2.5
Pressure = 22050 Pascal
The cross sectional area of the pump will be area of a circle.
A = πr^2
A = π × 15^2
A = 706.858 m^2
Using the formula
Density = mass/volume
Mass = density × volume
Mass = 900 × 706.86 × 2.5
Mass = 1590.435
Energy = mgh
Energy = 1590.435 × 9.8 × 2.5
Energy = 38965657.8 J
Since the work done = energy
Therefore, the work done = 3.9 × 10^7 J
We intend to measure the open-loop gain (LaTeX: A_{open}A o p e n ) of an actual operational amplifier. The magnitude of LaTeX: A_{open}A o p e n is in the range of 106 V/V. However, the signal generator in measurement setup can supply minimal voltage of 1 mV, and the oscilloscope used at amplifier output can measure maximal voltage level of 10 V. Can you design a simple measurement setup using this signal generator and oscilloscope, and accurately measure the LaTeX: A_{open}A o p e n
Answer:
voltage divider, R₂ = 1000 R₁
measuring the output in the resistance R₁
Explanation:
Let's analyze the situation, in an op amp in open gain loop, the gain is maximum G = 10⁶ V / V
in this case the signal generator gives a minimum wave of 1 10⁻³ V, after passing through the amplified it becomes 10³ V which saturates the oscilloscope.
To solve this problem we must use a simple voltage divider, for this we use the fact that in a series circuit the voltage is the sum of the voltages of each element.
If we use two resistors whose relationship is
R₂ / R₁ = 10³
R₂ = 1000 R₁
When measuring the output in the resistance R₁ we have the desired divider, with a tolerance range, for the minimum output of the generator (1 10⁻³V) we have a reading of V = 1 V in the oscilloscope, for which we can use voltage up to 10V on the generator
Which statement best describes this situation
Answer:
what situation?
Explanation:
One of the smallest planes ever flown was the Bumble Bee II, which had a mass of 180 kg. If the pilot’s mass was 70 kg, what was the velocity of both plane and pilot if their momentum was 20,800 kg∙m/s to the west?
Answer:
83.2 m/s to the West
Explanation:
From the question given above, the following data were obtained:
Mass of plane = 180 Kg
Mass of pilot = 70 Kg
Momentum = 20800 Kg∙m/s West
Velocity =?
Next, we shall determine the total mass. This can be obtained as follow:
Mass of plane = 180 Kg
Mass of pilot = 70 Kg
Total mass =?
Total mass = Mass of plane + Mass of pilot
Total mass = 180 + 70
Total mass = 250 Kg
Finally, we shall determine the velocity. This can be obtained as follow:
Total mass = 250 Kg
Momentum = 20800 Kg∙m/s West
Velocity =?
Momentum = mass × Velocity
20800 = 250 × Velocity
Divide both side by 250
Velocity = 20800 / 250
Velocity = 83.2 m/s West
Thus, the velocity of both plane and pilot is 83.2 m/s to the West
a car moved 120km to the north. what is its displacement?
A 50 kg mass is sitting on a frictionless surface. An unknown constant force called force A pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s. If the 50 kg mass is now pushed by an unknown force B and reaches the velocity of 3 m/s in 4 seconds, compare the impulse delivered to the mass when acted upon by force A with the impulse delivered to the mass when acted on by force B? *
A) The impulse delivered to the mass when acted upon by force A is greater
B) The impulse delivered to the mass when acted upon by force B is greater
C) The impulse is the same in each case
D) We need to know the value of force A and force B in order to determine this
Answer:
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
3. If a spring extends by 3 cm when a 4 N weight is suspended from it, find the extension
when the weight is changed to
(a) 8 N
(b) 10 N
(c) 14 N
Each vertical line on the graph is 1 millisecond (0.001 s) of time. What is the period and
frequency of the sound waves?
Explanation:
Given that,
Each vertical line on the graph is 1 millisecond (0.001 s) of time.
We need to find the period and the frequency of the sound wave. The period of a wave is equal to the each vertical line on graph i.e. 0.001 s.
Let f be the frequency of the sound wave. So,
f = 1/T
i.e.
[tex]f=\dfrac{1}{0.001 }\\\\f=1000\ Hz[/tex]
So, the period and the frequency of the sound waves is 1 milliseond and 1000 Hz respectively.
a race car goes around a circular track of radius 150 m at speed of 10.0 m/s. How long does it take to complete one lap?
Answer:
94.25 seconds
Explanation:
Solve for period (T) using: v=(2*pi*r)/T
rearrange: vT=2*pi*r
rearrange: T=(2*pi*r)/v
Plug in values.
T=(2*pi*150)/10
T=94.25 seconds
If a race car goes around a circular track of a radius of 150 m at speed of 10.0 m/s ,then the time taken to complete the one lap would be 94.25 seconds.
What is speed?The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.
As given in the problem a race car goes around a circular track of radius 150 m at speed of 10.0 m/s.
vT = 2 × π × r
T = (2 × π × r)/ v
T = (2 × π× 150)/10
T = 94.25 seconds
Thus, the time taken to complete the one lap would be 94.25 seconds.
To learn more about speed here, refer to the link given below ;
https://brainly.com/question/7359669
#SPJ2
An 5kg object is released from rest near the surface of a planet. The vertical position of the object as a function of time is shown in the graph. All frictional forces are considered to be negligible. What is the closest approximation of the weight of the object.
a) 300N
b) 30N
c) 5N
d) 150N
Answer:
The correct option is b: 30 N.
Explanation:
First, we need to find the acceleration due to gravity (a):
[tex] y_{f} - y_{0} = v_{o}t - \frac{1}{2}a(\Delta t)^{2} [/tex] (1)
Where:
[tex]y_{f}[/tex]: is the final vertical position (obtained from the graph)
[tex]y_{0}[/tex]: is the initial vertical position (obtained from the graph)
v₀: is the initial speed = 0 (it is released from rest)
Δt: is the variation of time (from the graph)
From the graph, we can take the following values of height and time:
t₀ = 0 s → [tex]t_{f}[/tex] = 5 s
y₀ = 300 m → [tex]y_{f}[/tex] = 225 m
Now, by entering the above values into equation (1) and solving for "a" we have:
[tex] a = 2\frac{y_{0} - y_{f}}{(t_{f} - t_{0})^{2}} = 2\frac{300 m - 225 m}{(5 s - 0)^{2}} = 6 m/s^{2} [/tex]
Finally, the weight of the object is:
[tex] W = ma = 5 kg*6 m/s^{2} = 30 N [/tex]
Therefore, the correct option is b: 30 N.
I hope it helps you!
A wave with a frequency of 5Hz travels a distance of 40mm in 2 seconds.What is the speed of the wave
Answer:
20mm per second
Explanation:
If a 15 N box is lifted a distance of 3 m, how much work is done?
0 J
45 J
5 J
5 N
Answer:
W=45J
Explanation:
W=Fd
W=15(3)=45
W=45J
Who watching all star draft? Luka better get picked first ong
Planets don't collide into
the sun because they
A. Are moving
B. Have too much mass
C. Have their own gravity
D. Are more attracted to each other