Answer:
A. New space missions show that Pluto is much larger than originally thought.
Explanation:
The new definition of a planet that was adopted in 2006, defined planet as an object that orbits the sun, with sufficient mass to be round, not a satellite of another object, and has removed debris and small objects from the area around its orbit.
This new definition of a planet that was adopted in 2006, classified Pluto as "dwarf planet", because Pluto meets planetary criteria except that it has not cleared debris from its orbital neighborhood.
However, new Horizons spacecraft flew by Pluto in 2015, revealed that Pluto is much larger than originally thought
Therefore, the correct option is "A"
A. New space missions show that Pluto is much larger than originally thought.
Answer: it means that we now technically have over 100 planets
Explanation:
it’s not New space missions show that Pluto is much larger than originally thought!!!
A brick is dropped with zero initial speed from the roof of a building and strikes the ground in 1.90 s. How tall is the building?
Answer:
17.69 m
Explanation:
The time it takes the brick to strike the ground is 1.90 seconds.
We can apply one of Newton's equation of linear motion to find the height of the building:
[tex]s = ut + 0.5gt^2[/tex]
where s = distance (in this case height)
u = initial velocity = 0 m/s
t = time = 1.90 s
g = acceleration due to gravity = 9.8 m/s^2
Therefore:
s = (0 * 1.9) + (0.5 * 9.8 * 1.9 * 1.9)
s = 0 + 17.68
s = 17.69 m
The height of the building is 17.69 m.
An electric motor drawing 10 amps at 110 V in steady state produces shaft power at 9.7 Nm and 1000 RPM. For a first Law analysis and considering the motor as the control volume How much heat will be produced from the motor (in Watts)
Answer:
The heat rate produced from the motor is 84.216 watts.
Explanation:
The electric motor receives power from electric current and releases power in the form of mechanical energy (torque) and waste heat and can be considered an stable-state system. The model based on the First Law of Thermodynamics for the electric motor is:
[tex]\dot W_{e} - \dot W_{T} -\dot Q = 0[/tex]
Where:
[tex]\dot Q[/tex] - Heat transfer from the electric motor, measured in watts.
[tex]\dot W_{e}[/tex] - Electric power, measured in watts.
[tex]\dot W_{T}[/tex] - Mechanical power, measured in watts.
The heat transfer rate can be calculated in terms of electric and mechanic powers, that is:
[tex]\dot Q = \dot W_{e} - \dot W_{T}[/tex]
The electric and mechanic powers are represented by the following expressions:
[tex]\dot W_{e} = i \cdot V[/tex]
[tex]\dot W_{T} = T \cdot \omega[/tex]
Where:
[tex]i[/tex] - Current, measured in amperes.
[tex]V[/tex] - Steady-state voltage, measured in volts.
[tex]T[/tex] - Torque, measured in newton-meters.
[tex]\omega[/tex] - Angular speed, measured in radians per second.
Now, the previous expression for heat transfer rate is expanded:
[tex]\dot Q = i \cdot V - T \cdot \omega[/tex]
The angular speed, measured in radians per second, can be obtained by using the following expression:
[tex]\omega = \frac{\pi}{30}\cdot \dot n[/tex]
Where:
[tex]\dot n[/tex] - Rotational rate of change, measured in revolutions per minute.
If [tex]\dot n = 1000\,rpm[/tex], then:
[tex]\omega = \left(\frac{\pi}{30} \right)\cdot (1000\,rpm)[/tex]
[tex]\omega \approx 104.720\,\frac{rad}{s}[/tex]
Given that [tex]i = 10\,A[/tex], [tex]V = 110\,V[/tex], [tex]T = 9.7\,N\cdot m[/tex] and [tex]\omega \approx 104.720\,\frac{rad}{s}[/tex], the heat transfer rate from the electric motor is:
[tex]\dot Q = (10\,A)\cdot (110\,V) -(9.7\,N\cdot m)\cdot \left(104.720\,\frac{rad}{s} \right)[/tex]
[tex]\dot Q = 84.216\,W[/tex]
The heat rate produced from the motor is 84.216 watts.
The only factor connecting horizontal and vertical components of projectile motion is _____.
Answer:
VelocityExplanation:
When a body is launched in air and allowed to fall freely under the influence of gravity, the motion experienced by the body is known as a projectile motion. The body is launched at a particular velocity and at an angle theta to the horizontal. The velocity of the body ca be resolved towards the horizontal component and the vertical component.
Along the horizontal Ux = Ucos(theta)
Along the vertical Uy = Ucos(theta)
Ux and Uy are the velocities of the body along the horizontal and vertical components respectively.
This means that the only factor connecting horizontal and vertical components of projectile motion is its velocity since we are able to calculate the velocity of the body along both components irrespective of its initial velocity.
Cass is walking her dog (Oreo) around the neighborhood. Upon arriving at Calina's house (a friend of Oreo's), Oreo turns part mule and refuses to continue on the walk. Cass yanks on the chain with a 67 N force at an angle of 30° above the horizontal. Determine the horizontal and vertical components of the tension force.
Answer:
Horizontal component: [tex]F_x = 58\ N[/tex]
Vertical component: [tex]F_y = 33.5\ N[/tex]
Explanation:
To find the horizontal and vertical components of the force, we just need to multiply the magnitude of the force by the cosine and sine of the angle with the horizontal, respectively.
Therefore, for the horizontal component, we have:
[tex]F_x = F * cos(angle)[/tex]
[tex]F_x = 67 * cos(30)[/tex]
[tex]F_x = 58\ N[/tex]
For the vertical component, we have:
[tex]F_y = F * sin(angle)[/tex]
[tex]F_y = 67 * sin(30)[/tex]
[tex]F_y = 33.5\ N[/tex]
So the horizontal component of the tension force is 58 N and the vertical component is 33.5 N.
A noisy channel needs to transfer 87 kbps, but has a SNR of 11 dB (decibels). Calculate the minimum Bandwidth required , in kHz, according to Shannon.
Answer:
24KHz
Explanation:
See attached file
A mass on a spring vibrates in simple harmonic motion at a frequency of 3.26 Hz and an amplitude of 5.76 cm. If the mass of the object of 0.218 kg, what is the spring constant
Answer:
91.48N/m
Explanation:
In a spring-mass system undergoing a simple harmonic motion, the inverse of the frequency f, of oscillation is proportional to the square root of the mass m, and inversely proportional to the square root of the spring constant, k. This can be expressed mathematically as follows;
[tex]\frac{1}{f}[/tex] = [tex]2\pi\sqrt{\frac{m}{k} }[/tex] -----------(i)
From the question;
f = 3.26 Hz
m = 0.218kg
Substitute these values into equation (i) as follows;
[tex]\frac{1}{3.26}[/tex] = [tex]2\pi\sqrt{\frac{0.218}{k} }[/tex] [Square both sides]
([tex]\frac{1}{3.26}[/tex])² = ([tex]2\pi[/tex])²([tex]\frac{0.218}{k}[/tex])
([tex]\frac{1}{10.6276}[/tex]) = [tex]4\pi[/tex]²([tex]\frac{0.218}{k}[/tex]) [Take [tex]\pi[/tex] to be 3.142]
([tex]\frac{1}{10.6276}[/tex]) = [tex]4(3.142)[/tex]²([tex]\frac{0.218}{k}[/tex])
([tex]\frac{1}{10.6276}[/tex]) = [tex]39.488[/tex]([tex]\frac{0.218}{k}[/tex])
([tex]\frac{1}{10.6276}[/tex]) = ([tex]\frac{8.608}{k}[/tex]) [Switch sides]
([tex]\frac{8.608}{k}[/tex]) = ([tex]\frac{1}{10.6276}[/tex]) [Re-arrange]
([tex]\frac{k}{8.608}[/tex]) = ([tex]\frac{10.6276}{1}[/tex]) [Cross-multiply]
k = 8.608 x 10.6276
k = 91.48N/m
Therefore, the spring constant of the spring is 91.48N/m
If two radio telescope dishes are wired together in the right way, the "D" used in determining the angular resolution is determined by
Answer:
D is determined by distance between the telescopes.
Explanation:
The bulldog and skateboard have a combined mass of 20 kg. In case B (the middle of the three pictures of the bulldog and the well), the bulldog and skateboard have a KE of 380 J at the bottom of the well. How deep is the well in meters?
Answer:
h = 1.94 m
Explanation:
When the bull dog and skate board reach the bottom of the well, all of its potential energy is converted to the kinetic energy:
Kinetic Energy Gained by Bull Dog and Skate Board = Potential Energy Lost by Bull Dog and Skate Board
K.E = P.E
K.E = mgh
h = K.E/mg
where,
h = depth of well = ?
K.E = Kinetic Energy at bottom = 380 J
m = mass of bull dog and skate board = 20 kg
g = 9.8 m/s²
Therefore,
h = 380 J/(20 kg)(9.8 m/s²)
h = 1.94 m
To understand the behavior of the electric field at the surface of a conductor, and its relationship to surface charge on the conductor. A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.
1) Which of the following describes the electricfield inside this conductor?
A) It is in the same direction as the original external field.
B) It is in the opposite direction from that of the original external field.
C) It has a direction determined entirely by the charge on its surface.
D) It is always zero.
2) The charge density inside the conductor is:______.
a) 0.
b) non-zero;but uniform.
c) non-zero;non-uniform.
d) infinite.
3) Assume that at some point just outside the surface of the conductor, the electric field has magnitudeE and is directed toward the surface of the conductor. What is the charge density eta on the surface of the conductor at that point?
Answer:
1) Option D is correct.
The electric field inside a conductor is always zero.
2) Option A is correct.
The charge density inside the conductor is 0.
3) Charge density on the surface of the conductor at that point = η = -E ε₀
Explanation:
1) The electric field is zero inside a conductor. Any excess charge resides entirely on the surface or surfaces of a conductor.
Assuming the net electric field wasn't zero, current would flow inside the conductor and this would build up charges on the exterior of the conductor. These charges would oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero.
2) Since there are no charges inside a conductor (they all reside on the surface), it is logical that the charge density inside the conductor is also 0.
3) Surface Charge density = η = (q/A)
But electric field is given as
E = (-q/2πε₀r²)
q = -E (2πε₀r²)
η = (q/A) = -E (2πε₀r²)/A
For an elemental point on the surface,
A = 2πrl = 2πr²
So,
η = -E ε₀
Hope this Helps!!!
A 75kg passenger at the bottom of a roller coaster loop that has a radius of 20m. If the roller coaster car is moving 10m/s, what is the apparent weight of the passenger? g
Answer:
The apparent weight of the passenger is 360 N
Explanation:
Given;
The mass of the passenger, m = 75 kg
radius of the loop, r = 20 m
velocity of the roller coaster, v = 10 m/s
Centripetal force acting on this passenger is given as;
[tex]F = \frac{mv^2}{r}[/tex]
where;
F is the centripetal force acting on the passenger
m is the mass of the passenger
v is the velocity of the passenger
r is the radius of the track
[tex]F = \frac{mv^2}{r} \\\\F = \frac{75*10^2}{20} \\\\F = 375 \ N[/tex]
Real weight of the passenger,
W = mg
where;
g is acceleration due to gravity
W = 75 x 9.8
W = 735 N
Apparent weight of the passenger = Real weight - Centripetal force
Apparent weight of the passenger = 735 N - 375 N
Apparent weight of the passenger = 360 N
Therefore, the apparent weight of the passenger is 360 N
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12 + 35 t + 1
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1
Answer:
Velocity = 131 m/s
Speed = 131 m/s
Explanation:
Equation of motion, s = f(t) = 12t² + 35 t + 1
To get velocity of the particle, let us find the first derivative of s
v (t) = ds/dt = 24t + 35
At t = 4
v(4) = 24(4) + 35
v(4) = 131 m/s
Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s
PLEASE! PLEASE! PLEASE! HELP! I need this by Monday! It's a project DUE on Monday! I'm giving 75 points for the Brainiest :-)
Answer:
Please see below for all the numbers to be entered in the table:
Explanation:
Coaster World: F = 160 N; D = 40 m; T = 10 s; W = 160 * 40 = 6400 J; V = 40/10 = 4 m/s
Wally: F = 800 N; D = 10 m; T = 3.5 s W = 800*10 = 8000 J; V = 10/3.5 = 2.86 m/s
Elijah: F = 1400 N; D = 800 m; T = 40 m = 2400 s; W = 1400*800 = 112000 J; V = 800/2400 = 0.33 m/s
George: F = 600 N; D = 80 m; T = 50 m = 3000 s; W = 600 * 80 = 48000 j so he should get paid: 48,000/1000= $48; V = 80/3000 = 0.027 m/s
Answer:
Down below
Explanation:
Coaster World: F = 160 N; D = 40 m; T = 10 s; W = 160 * 40 = 6400 J; V = 40/10 = 4 m/s
Wally: F = 800 N; D = 10 m; T = 3.5 s W = 800*10 = 8000 J; V = 10/3.5 = 2.86 m/s
Elijah: F = 1400 N; D = 800 m; T = 40 m = 2400 s; W = 1400*800 = 112000 J; V = 800/2400 = 0.33 m/s
George: F = 600 N; D = 80 m; T = 50 m = 3000 s; W = 600 * 80 = 48000 j so he should get paid: 48,000/1000= $48; V = 80/3000 = 0.027 m/s
1. A current of 0.001 A can be felt by the human body. 0.005 A can produce a pain response. 0.015 A can cause a loss of muscle control. In the procedures of this lesson, over 0.030 A of current traveled in the three-battery circuit. Why was this circuit safe to handle with dry hands?
Answer:
It was safe to handle the circuit with dry hands because dry skin body resistance is very high, measuring up to 500,000 ohms.
Explanation:
Given;
Current of 0.001 A to be felt
Current of 0.005 A can produce a pain response
Current of 0.015 A can cause a loss of muscle control
Total current that traveled in the three-battery circuit = 0.03 A
Thus, we can conclude that, it was safe to handle the above mentioned circuit with dry hands because dry skin body resistance is very high, measuring up to 500,000 ohms.
A grating having 5000 lines/cm is used with light of wavelength 633 nm. How many total maxima (count central maxima plus all those on either side of the central maxima) are produced
Answer:
The total number of maxima produced is [tex]m_T = 7[/tex] maxima
Explanation:
From the question we are told that
The number of lines per cm is [tex]n = 5000 \ lines/cm[/tex]
The wavelength of the light is [tex]\lambda = 633 nm = 633 *10^{-9} \ m[/tex]
Now the distance between the lines is mathematically evaluated as
[tex]d = \frac{1}{n}[/tex]
substituting values
[tex]d = \frac{1}{5000}[/tex]
[tex]d = \frac{1 *10^{-2}}{5000}[/tex] N/B - this statement convert it from cm to m
[tex]d = 2 *10^{ -6} \ m[/tex]
Generally the condition for diffraction i mathematically represented as
[tex]dsin(\theta ) = m \lambda[/tex]
at maximum [tex]\theta = 90 ^o[/tex]
[tex]d sin (90) = \lambda m[/tex]
here m is the number of maxima
Thus making m the subject we have
[tex]m = \frac{d sin (90)}{ \lambda }[/tex]
So [tex]m = \frac{2*10^{-6} sin (90)}{ 633 *10^{-9}}[/tex]
[tex]m = 3.2[/tex]
=> m =3
Now the total number of maxima would include the bright fringe(3) and dark fringe (3) plus the central maxima (1)
Thus
[tex]m_T = 3 + 3 +1[/tex]
[tex]m_T = 7[/tex] maxima
Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time moving with a total distance of 0.12 m. Samantha pushes the sandpaper against the surface with a normal force of 2.6 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the normal force during the sanding process
Answer:
W = 12.96 J
Explanation:
The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:
F = μR
where,
F = Friction Force = ?
μ = 0.92
R = Normal Force = 2.6 N
Therefore,
F = (0.92)(2.6 N)
F = 2.4 N
Now, the displacement is given as:
d = (0.12 m)(45)
d = 5.4 m
So, the work done will be:
W = F d
W = (2.4 N)(5.4 m)
W = 12.96 J
1. A base-ball of mass 0.3kg approaches the bat at a speed of 30 miles/hour and when the ball hits the bat for 0.5 s, it started to move away from the bat at a speed of 60 miles/hour. Find the impulse
Answer:
I = 27kg.mi/h
Explanation:
In order to calculate the impulse of the ball, you use the following formula:
[tex]I=m\Delta v[/tex] [tex]=m(v-v_o)[/tex] (1)
m: mass of the ball = 0.3kg
v: speed of the ball after the bat hit it = 60mi/h
vo: speed of the ball before the bat hit it = 30mi/h
You replace the values of all parameters in the equation (1):
[tex]I=(0.3kg)(60mi/h-(-30mi/h))=27kg\frac{mi}{h}[/tex]
where the minus sign of the initial velocity means that the motion of the ball is opposite to the final direction of such a motion.
The imulpse of the ball is 27 kg.miles/hour
what is meaning of convection
Answer:
Convection is heat transfer through the movement of liquids and gases.
What is the role of the part in the diagram labeled Y?
modulate, amplify, and send out waves
O capture, amplify, and demodulate waves
change the amplitude and frequency of waves
O change the pulse and phase of waves
Question is incomplete and image is not attached ti the question. The required image is attached below, so the complete question is:
The diagram shows a device that uses radio waves.
What is the role of the part in the diagram labeled Y?
modulate, amplify, and send out waves capture, amplify, and demodulate waves change the amplitude and frequency of waves change the pulse and phase of wavesAnswer:
2. capture, amplify, and demodulate waves
Explanation:
The part Y labeled in the diagram refers to radio receiver which capture, amplify and demodulate the radio waves.
The radio receiver seperates required radio frequency signals through antenna and consist of an amplifier that amplify or increase the power of receiving signal. At the end, demodulators present in receivers recover the information from the modulated wave.
Hence, the correct option is 2.
Answer:
B
Explanation:
edge 2020
A 25-kilogram object is placed on a compression spring, and it creates a displacement of 0.15 meters. What is the weight of an object that creates a displacement of 0.23 m on the same spring? Enter your answer as a number rounded to the nearest tenth, such as: 42.5
Answer:
I hope it is correct ✌️
A woman who weighs 500 N stands on an 8 m long board that weighs 100 N. The board is supported at each end. The support force at the right end is 3 times bigger than the support force at the left end. How far from the right end is the woman standing
Answer:
[tex]1.6\; \rm m[/tex].
Explanation:
Let [tex]x[/tex] denote the distance (in meters) between the person and the right end of the board.
To keep the calculations simple, consider another unknown: let [tex]y[/tex] denote the support force (in Newtons) on the left end. The support force on the right end of this board would be [tex]3 \, y[/tex] (also in Newtons.)
Now there are two unknowns. At least two equations will be required for finding the exact solutions. For that, consider this board as a lever, but with two possible fulcrums. Refer to the two diagrams attached. (Not to scale.)
In the first diagram, the support at the left end of the board is considered as the fulcrum. In the second diagram, the support at the right end of the board is considered as the fulcrum.Calculate the torque in each situation. Note that are four external forces acting on this board at the same time. (Two support forces and two weights.) Why does each of the two diagrams show only three? In particular, why is the support force at each "fulcrum" missing? The reason is that any force acting on the lever at the fulcrum will have no direct impact on the balance between torques elsewhere on the lever. Keep in mind that the torque of each force on a lever is proportional to [tex]r[/tex], the distance between the starting point and the fulcrum. Since that missing support force starts right at the fulcrum, its [tex]r[/tex] will be zero, and it will have no torque in this context.
Hence, there are three (non-zero) torques acting on the "lever" in each diagram. For example, in the first diagram:
The weight of the board acts at the center of the board, [tex](1/2) \times 8\; \rm m = 4\; \rm m[/tex] from the fulcrum. This force will exert a torque of [tex]\tau(\text{weight of board}) = 4\; \rm m \times (-100\; \rm N) = (-400\; \rm N \cdot m)[/tex] on this "lever". The negative sign indicates that this torque points downwards.The weight of the person acts at [tex]x\;\rm m[/tex] from the right end of the board, which is [tex](8 - x)\; \rm m[/tex] from the fulcrum at the other end of this board. This force will exert a torque of [tex]\tau(\text{weight of person}) = (8 - x)\; {\rm m \times (-500\; \rm N)} = (-500\, \mathnormal{(8 - x)})\; \rm N \cdot m[/tex] on this "lever". This torque also points downwards.The support on the right end of the board acts at [tex]8\; \rm m[/tex] from the fulcrum (i.e., the left end of this board.) This force will exert a torque of [tex]\tau(\text{support, right}) = 8\; {\rm m} \times (3\, \mathnormal{y})\; {\rm N} = (24\, y)\; \rm N \cdot m[/tex] on the "lever". This torque points upwards.If the value of [tex]x[/tex] and [tex]y[/tex] are correct, these three torques should add up to zero. That is:
[tex]\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\, (8 - x))}_{\text{person}} + \underbrace{24\, y}_{\text{support}} = 0[/tex].
That gives the first equation of this system. Similarly, a different equation can be obtained using the second diagram:
[tex]\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\,x)}_{\text{person}} + \underbrace{8\, y}_{\text{support}} = 0[/tex].
Combine these two equations into a two-by-two system. Solve the system for [tex]x[/tex] and [tex]y[/tex]:
[tex]\left\lbrace\begin{aligned}&x = 1.6\\ &y = 150\end{aligned}\right.[/tex].
In other words, the person is standing at about [tex]1.6\; \rm m[/tex] from the right end of the board. The support force at the left end of the board is [tex]150\; \rm N[/tex].
4. How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a space in between such that the cut edges are closest to each other
Answer:
Explanation:
Pls see diagram in attached file
A diver wants to jump from a board, the initial height is 10 meters and he wants to reach a horizontal distance of 2 meters. What minimum speed must he have when jumping from the board to achieve his goal?
Answer:
1.4 m/s
Explanation:
The minimum speed will be when the diver's initial velocity is horizontal.
First, find the time it takes for the diver to fall 10 meters.
Given:
Δy = 10 m
v₀ᵧ = 0 m/s
aᵧ = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
10 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 1.43 s
Now find the initial horizontal velocity.
v = (2 m) / (1.43 s)
v = 1.4 m/s
What is the electric potential (relative to infinity) due to these charges at the center of this square
Answer:
Zero
Explanation:
See attached file pls
A motorcyclist changes his speed from 20 km / h to 100 km / h in 3 seconds, maintaining a constant acceleration in that time interval. If the mass of the motorcycle is 200 kg and that of its rider is 80 kg, what is the value of the net force to accelerate the motorcycle? Help!
Answer:
2000 N
Explanation:
20 km/h = 5.56 m/s
100 km/h = 27.78 m/s
F = ma
F = m Δv/Δt
F = (200 kg + 80 kg) (27.78 m/s − 5.56 m/s) / (3 s)
F = 2074 N
Rounded to one significant figure, the force is 2000 N.
A 60-watt light bulb carries a current of 0.5 ampere. The total charge passing through it in one hour is:
Answer:
Total charge = 1800C
Explanation:
Q= IT
I = currentt = timeQ = chargeParallel rays of monochromatic light with wavelength 583 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm. If the intensity at the center of the central maximum is 5.00×10^−4W/m^2. What is the intensity at a point on the screen that is 0.900 mm from the center of the central maximum?
Answer:
I = 2.18 10⁻⁴ W / m²
Explanation:
The two-slit interference pattern is described by the expression for constructive interference.
d sin θ = m λ
If we also want to know the distribution of intensities we must perform the su of the electric field of the two waves, and find the intensity as the square of the velvet field, obtaining the expression
I = I_max cos² ((π d /λ L) y)
where d is the separation of the slits, λ the wavelength, L the distance to the screen e and the separation of the interference line with respect to the central maximum
let's reduce the magnitudes to the SI system
λ = 583 nm = 583 10⁻⁹ m
L = 75.0 cm = 75.0 10⁻² m
d = 0.640 mm = 0.640 10⁻³ m
y = 0.900 mm = 0.900 10⁻³ m
let's calculate the intensity of this line
I = 5 10⁻⁴ cos² ((π 0.640 10⁻³ /583 10⁻⁹ 0.75 10⁻²) 0.900 10⁻³)
I = 5 10⁻⁴ cos2 (413.84)
I = 5 10⁻⁴ 0.435
I = 2.18 10⁻⁴ W / m²
A spherical balloon contains a charge +Q uniformly distributed over its surface. When it has a diameter D, the electric field at its surface has magnitude E. If the balloon is not blown up to thrice this diameter without changing the charge, the electric field at its surface is?
Answer:
E = 1/9 E₀
Explanation:
In this exercise we are told that the electric field is Eo when the diameter of the balloon is D, the expression
we are asked to shorten the electric field when the diameter is 3D with the same eclectic charge
For this we can use the gauss law to find the field in the new diameter, for this we create a Gaussian surface in the form of a sphere
Ф = ∫ E. dA = [tex]q_{int}[/tex] /ε₀
In this case the lines of the electric field and the radii of the sphere are parallel, therefore the scalar product is reduced to the algebraic product and the charge inside the sphere is the initial charge Q
A = 4π r²
E 4π r² = Q /ε₀
E = 1 /4πε₀ Q / r²
the value of the indicated distance is 3 times the initial diamete
r = 3 D / 2
we substitute
E = 1/4 πε₀ Q (2/ 3D)²
for the initial conditions
E₀ = 1 / 4πε₀ Q (2/D)²
subtitled in the equation above
E = 1/9 E₀
A student in her physics lab measures the standing-wave modes of a tube. The lowest frequency that makes a resonance is 30 Hz. As the frequency is increased, the next resonance is at 90 Hz.
What will be the next resonance after this?
Answer:
The next resonance will be 150 Hz.
Explanation:
The frequency of the sound produced by a tube, both open and closed, is directly proportional to the speed of propagation. Hence, to produce the different harmonics of a tube, the wave propagation speed must be increased.
The frequency of the sound produced by a tube, both open and closed, is inversely proportional to the length of the tube. The greater the length of the tube, the frequency is lower.
Frecuency of the standing sound wave modes in a open-closed tube is:
fₙ=n*f₁ where m is an integer and f₁ is the first frecuency (30 Hz)
The next resonance is at 90 Hz. This means that it occurs when n = 3:
f₃=3*30 Hz= 90 Hz
This means that the next resonance occurs when n = 5:
f₅=5*30 Hz= 150 Hz
The next resonance will be 150 Hz.
A car has a mass of 1200 kg and an acceleration of 4 m/s^2. If the friction on the car is 200 N, how much force is the thrust providing?
Answer:
5000N
Explanation:
According to Newton's second law of motion, the net force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) of the body caused by the force. i.e
∑F = m x a -------------(i)
From the question, the net force is the combined effect of the thrust (F) and the friction force (Fₓ). i.e
∑F = F + Fₓ -------------(ii)
Where;
Fₓ = -200N [negative sign because the friction force opposes motion]
Combine equations(i) and (ii) together to get;
F + Fₓ = m x a
F = ma - Fₓ -------------(iii)
Where;
m = mass of car = 1200kg
a = acceleration of the car = 4m/s²
Now substitute the values of m, a and Fₓ into equation (iii) as follows;
F = (1200 x 4) - (-200)
F = 4800 + 200
F = 5000N
Therefore, the force the thrust is providing is 5000N
The speed of light in a material is 1.74 x 108 m's. What is the index of refraction of this material?
Answer:
The index of refraction of this material is 1.7241
Explanation:
Recall that the index of refraction (n) of a medium is defined as the quotient between the speed of light in vacuum divided the speed of light in the medium whose index of refraction is being calculated. In mathematical terms:
[tex]n=\frac{c}{v}[/tex]
Therefore, in our case, since we know the speed of light in the medium ([tex]v=1.74\,\,10^8\,\,m/s[/tex]) and the speed of light in vacuum ([tex]c=3\,\,10^8\,\,m/s[/tex]), we can estimate the index of refraction of the medium:
[tex]n=\frac{c}{v} \\n=\frac{3\,\,10^8}{1.74\,\,10^8} \\n=1.7241[/tex]