Answer:
The width is [tex]w_c = 0.00422 \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 6.33*10^{-7} \ m[/tex]
The width of the slit is [tex]d = 0.3\ mm = 0.3 *10^{-3} \ m[/tex]
The distance of the screen is [tex]D = 1.0 \ m[/tex]
Generally the central maximum is mathematically represented as
[tex]w_c = 2 * y[/tex]
Here y is the width of the first order maxima which is mathematically represented as
[tex]y = \frac{\lambda * D}{d}[/tex]
substituting values
[tex]y = \frac{6.33*10^{-7} * 1.0}{ 0.30}[/tex]
[tex]y = 0.00211 \ m[/tex]
So
[tex]w_c = 2 *0.00211[/tex]
[tex]w_c = 0.00422 \ m[/tex]
Describe and name the different types of collision. In which are the linear momentum and kinetic energy conserved
Answer:
1. Elastic collision
2. Inelastic collision
Explanation:
Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision
the collision is described by the equation bellow
[tex]m1U1+ m2U2= m1V1+m2V2[/tex]
Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity
the collision is described by the equation bellow
[tex]m1U1+ m2U2= V(m1+m2)[/tex]
Astronomers discover an exoplanet (a planet of a star other than the Sun) that has an orbital period of 3.75
Earth years in its circular orbit around its sun, which is a star with a measured mass of 3.23×1030kg
. Find the radius of the exoplanet's orbit.
Answer:
[tex]r=4.24\times 10^{11}\ m[/tex]
Explanation:
Given that,
Orbital time period, T = 3.75 earth years
Mass of star, [tex]m=3.23\times 10^{30}\ kg[/tex]
We need to find the radius of the exoplanet's orbit. It is a concept of Kepler's third law of motion i.e.
[tex]T^2=\dfrac{4\pi^2}{GM}r^3[/tex]
r is the radius of the exoplanet's orbit.
Solving for r we get :
[tex]r=(\dfrac{T^2GM}{4\pi^2})^{1/3}[/tex]
We know that, [tex]1\ \text{earth year}=3.154\times 10^7\ \text{s}[/tex]
So,
[tex]r=(\dfrac{(3.75\times 3.154\times 10^7)^2\times 6.67\times 10^{-11}\times 3.23\times 10^{30}}{4\pi^2})^{1/3}\\\\r=4.24\times 10^{11}\ m[/tex]
So, the radius of the exoplanet's orbit is [tex]4.24\times 10^{11}\ m[/tex].
Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging, the blood flow and heartbeat of the child can be measured using an echolocation technique similar to that used by bats. For the purposes of these questions, please use 1500 m/s as the speed of sound in tissue. I need help with part B and C
To clearly see an image, the wavelength used must be at most 1/4 of the size of the object that is to be imaged. What frequency is needed to image a fetus at 8 weeks of gestation that is 1.6 cm long?
A. 380 kHz
B. 3.8 kHz
C. 85 kHz
D. 3.8 MHz
Answer:
380 kHz
Explanation:
The speed of sound is taken as 1500 m/s
The length of the fetus is 1.6 cm long
The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.
For this 1.6 cm baby, the wavelength must not exceed
λ = [tex]\frac{1}{4}[/tex] of 1.6 cm = [tex]\frac{1}{4}[/tex] x 1.6 cm = 0.4 cm =
0.4 cm = 0.004 m this is the wavelength of the required ultrasonic sound.
we know that
v = λf
where v is the speed of a wave
λ is the wavelength of the wave
f is the frequency of the wave
f = v/λ
substituting values, we have
f = 1500/0.004 = 375000 Hz
==> 375000/1000 = 375 kHz ≅ 380 kHz
Mention the importance of writing the physical quantities as vectors.
Answer:
Knowledge of vectors is important because many quantities used in physics are vectors. If you try to add together vector quantities without taking into account their direction you'll get results that are incorrect.
Explanation:
An example of the importance of vector addition could be the following:
Two cars are involved in a collision. At the time of the collision car A was travelling at 40 mph, car B was travelling at 60 mph. Until I tell you in which directions the cars were travelling you don't know how serious the collision was.
The cars could have been travelling in the same direction, in which case car B crashed into the back of car A, and the relative velocity between them was 20 mph. Or the cars could have been travelling in opposite directions, in which case it was a head on collision with a relative velocity between the cars of 100 mph!
Which of the following statements is true vibrations ?
Answer:
C. Neither ultrasonic nor infrasonic vibrations can be heard by humans.
Explanation:
The complete question is
Which of the following statements is true of vibrations? A. The frequency of infrasonic vibrations is much too high to be heard by humans. B. Ultrasonic vibrations have a frequency lower than the range for normal hearing. C. Neither ultrasonic nor infrasonic vibrations can be heard by humans. D. Infrasonic vibrations are used in sonar equipment and to detect flaws in steel castings.
Ultrasonic vibrations have frequencies higher than our range of hearing, while infrasonic vibrations have frequencies lower than our range of hearing. Ultrasonic vibrations or sound is used in sonar equipment, and is used for detecting hidden flaws in steel castings and structures. Both infrasonic and ultrasonic fall below and above our normal hearing range respectively, and are only audible to dogs, cats, and some other mammals.
Answer:
Answer is " Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate."
Explanation:
My options were:
A) Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body.
B) Resonance occurs as a result of sympathetic vibrations.
C) A non-vibrating object can begin to vibrate as a result of forced vibrations.
D) Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate.
A is correct
.
3. What color of laser light shines through a diffraction grating with a line density of 500 lines/mm if the third maxima from the central maxima (m=3) is at an angle of 45°?
Answer:
Wavelength is 471 nm
Explanation:
Given that,
Lines per unit length of diffraction grating is 500 lines/mm.
The third maxima from the central maxima (m=3) is at an angle of 45°
We need to find the color of laser light shines through a diffraction grating.
The condition for maxima is :
[tex]d\sin\theta=m\lambda[/tex]
d = 1/N, N = number of lines per mm
[tex]\lambda=\dfrac{1}{Nm}\sin\theta\\\\\lambda=\dfrac{10^{-3}}{500\times 3}\sin(45)\\\\\lambda=4.31\times 10^{-7}\\\\\text{or}\\\\\lambda=471\ nm[/tex]
PLEASE HELP WILL GIVE BRAINLIEST In an experiment, the hypothesis is that if leaf color is related to temperature, then exposing the plant to low temperatures will result in a leaf color change . This hypothesis is _____. 1. testable 2.falsifiable 3.a and b above 4.none of the above
Answer:
testable
Explanation:
high heat can cause browning and or welting
Answer:
A
Explanation:
If it is tested, it will really changed when the plant into low temperature
g A projectile is fired from the ground at an angle of θ = π 4 toward a tower located 600 m away. If the projectile has an initial speed of 120 m/s, find the height at which it strikes the tower
Answer:
The projectile strikes the tower at a height of 354.824 meters.
Explanation:
The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:
Horizontal motion
[tex]x = x_{o}+v_{o}\cdot t \cdot \cos \theta[/tex]
Vertical motion
[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2} \cdot g \cdot t^{2}[/tex]
Where:
[tex]x_{o}[/tex], [tex]x[/tex] - Initial and current horizontal position, measured in meters.
[tex]y_{o}[/tex], [tex]y[/tex] - Initial and current vertical position, measured in meters.
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
The time spent for the projectile to strike the tower is obtained from first equation:
[tex]t = \frac{x-x_{o}}{v_{o}\cdot \cos \theta}[/tex]
If [tex]x = 600\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]\theta = \frac{\pi}{4}[/tex], then:
[tex]t = \frac{600\,m-0\,m}{\left(120\,\frac{m}{s} \right)\cdot \cos \frac{\pi}{4} }[/tex]
[tex]t \approx 7.071\,s[/tex]
Now, the height at which the projectile strikes the tower is: ([tex]y_{o} = 0\,m[/tex], [tex]t \approx 7.071\,s[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])
[tex]y = 0\,m + \left(120\,\frac{m}{s} \right)\cdot (7.071\,s)\cdot \sin \frac{\pi}{4}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (7.071\,s)^{2}[/tex]
[tex]y \approx 354.824\,m[/tex]
The projectile strikes the tower at a height of 354.824 meters.
Monochromatic light falls on two very narrow slits 0.047 mm apart. Successive fringes on a screen 6.60 m away are 8.9 cm apart near the center of the pattern.
Determine the wavelength and frequency of the light.
Answer::
[tex]\lambda = 634 nm[/tex]
[tex]f = 4.73 *10^{14} \ Hz[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 0.047 \ mm = 0.047 *10^{-3} \ m[/tex]
The distance of the screen is [tex]D = 6.60 \ m[/tex]
The width of the fringe is [tex]y = 8.9 \ cm = 0.089 \ m[/tex]
Generally the width of the width of the fringes is mathematically represented as
[tex]y = \frac{\lambda * D }{d }[/tex]
=> [tex]\lambda = \frac{y * d }{D }[/tex]
=> [tex]\lambda = \frac{ 0.089 * (0.047 *10^{-3}) }{6.60 }[/tex]
=> [tex]\lambda = 634 *10^{-9}[/tex]
=> [tex]\lambda = 634 nm[/tex]
Generally the speed of light is mathematically represented as
[tex]c = f * \lambda[/tex]
=> [tex]f= \frac{ c}{\lambda }[/tex]
=> [tex]f= \frac{ 3.0 *10^{8}}{634 *10^{-9}}[/tex]
=> [tex]f = 4.73 *10^{14} \ Hz[/tex]
A steel ball attached to a spring moves in simple harmonic motion. The amplitude of the ball's motion is 11.0 cm, and the spring constant is 6.00 N/m. When the ball is halfway between its equilibrium position and its maximum displacement from equilibrium, its speed is 26.1 cm/s. (a) What is the mass of the ball (in kg)? kg (b) What is the period of oscillation (in s)? s (c) What is the maximum acceleration of the ball? (Enter the magnitude in m/s2.) m/s2
Answer:
a) m = 0.626 kg , b) T = 2.09 s , c) a = 1.0544 m / s²
Explanation:
In a spring mass system the equation of motion is
x = A cos (wt + Ф)
with w = √(k / m)
a) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф) (1)
give us that the speed is
v = 26.1 m / s
for the point
x = a / 2
the range of motion is a = 11.0 cm
x = 11.0 / 2
x = 5.5 cm
Let's find the time it takes to get to this distance
wt + Ф = cos⁻¹ (x / A)
wt + Ф = cos 0.5
wt + Ф = 0.877
In the exercise they do not indicate that the body started its movement with any speed, therefore we assume that for the maximum elongation the body was released, therefore the phase is zero f
Ф = 0
wt = 0.877
t = 0.877 / w
we substitute in equation 1
26.1 = -11.0 w sin (w 0.877 / w)
w = 26.1 / (11 sin 0.877))
w = 3.096 rad / s
from the angular velocity equation
w² = k / m
m = k / w²
m = 6 / 3,096²
m = 0.626 kg
b) angular velocity and frequency are related
w = 2π f
frequency and period are related
f = 1 / T
we substitute
w = 2π / T
T = 2π / w
T = 2π / 3,096
T = 2.09 s
c) maximum acceleration
the acceleration of defined by
a = dv / dt
a = - Aw² cos (wt)
the acceleration is maximum when the cosine is ±1
a = A w²
a = 11 3,096²
a = 105.44 cm / s²
we reduce to m / s
a = 1.0544 m / s²
Luz, who is skydiving, is traveling at terminal velocity with her body parallel to the ground. She then changes her body position to feet first toward the ground. What happens to her motion? She will continue to fall at the same terminal velocity because gravity has not changed. She will slow down because the air resistance will increase and be greater than gravity. She will speed up because air resistance will decrease and be less than gravity. She will begin to fall in free fall because she will have no air resistance acting on her.
Answer:
Option C - she will speed up because air resistance has reduced and be less than gravity
Explanation:
We are told that Luz is skydiving with terminal velocity and her body parallel to the ground. Now, at this point she will be experiencing a gravitational force acting downwards, and also air resistance as a result of the drag force on her body
Since the downward gravitational force on Luz is constant, she will fall with a net force of;
F_net = F_g - F_d
where;
F_net is the net force on Luz acting downwards
F_g is the gravitational force on Luz
F_d is the drag force on Luz
The drag force on her body is proportional to the surface area of attack.
We are now told that Luz changes her body position to feet first toward the ground. This means that the surface area of attack is reduced because the feet will consume less space than the frontal part of her body. Thus, the drag force will be lesser then before she changed her body position due to reduced air resistance on her body.
Now, from earlier, we saw that;
F_net = F_g - F_d
So, the lesser F_d is, the higher F_net becomes.
Thus, she will speed up because air resistance has reduced and be less than gravity.
Answer:
C
Explanation:
EDGE 2020
Did the kinetic frictional coefficient (for the wood/aluminum and felt/aluminum cases) vary with area of contact
Answer:
Explanation:
Friction is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose motion of one over the other.
While kinetic friction is the force that must be overcome so that a body can move with uniform speed over another.
Hence let consider one of the laws of friction which states that: '' Frictional force is independent of the area of the surfaces in contact.''
The value did not vary with area. This is because when calculating the kinetic fiction, the total contact area is not relevant and only the total weight of the system as well of as the block is put into consideration.
light bulb is connected to a 110-V source. What is the resistance of this bulb if it is a 100-W bulb
Answer:
121ohmsExplanation:
Formula used for calculating power P = current * voltage
P = IV
From ohms law, V = IR where R is the resistance. Substituting V = IR into the formula for calculating power, we will have;
P = IV
P =(V/R)V
P = V²/R
Given parameters
Power rating of the bulb P = 100 Watts
Source voltage V = 110V
Required
Resistance of the bulb R
Substituting the given parameters into the formula for calculating power to get Resistance R;
P = V²/R
100 = 110²/R
R = 110²/100
R = 110 * 110/100
R = 12100/100
R = 121 ohms
Hence, the resistance of this bulb is 121 ohms
What is the maximum wavelength of incident light for which photoelectrons will be released from gallium
Answer:
292 nm
Explanation:
The work function of gallium ∅ = 94.25 eV = 6.81 x 10^-19 J
at maximum wavelength, the energy of the photons is equal to its work function
Energy of the electron = hf
but hf = hc/λ
where h is the planck's constant = 6.63 × 10-34 m^2 kg/s
c is the speed of light = 3 x 10^8 m/s
λ is the wavelength that this occurs, which is the maximum wavelength
Equating, we have
hc/λ = ∅
substituting, we have
(6.63 × 10-34 x 3 x 10^8)/λ = 6.81 x 10^-19
(1.989 x 10^-25)/(6.81 x 10^-19) = λ
λ = 292.07 x 10^-9 = 292 nm
An engine causes a car to move 10 meters with a force of 100 N. The engine produces 10,000 J of energy. What is the efficiency of this engine?
Answer:
10%
Explanation:
Efficiency = work done / energy used
e = (10 m × 100 N) / (10,000 J)
e = 0.1
The efficiency is 0.1, or 10%.
A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the coefficient of contraction is 0.68 and the coefficient of velocity is 0.92, what is the discharge?
Answer:
The discharge rate is [tex]Q = 0.0192 \ m^3 /s[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 60 \ mm = 0.06 \ m[/tex]
The head is [tex]h = 6 \ m[/tex]
The coefficient of contraction is [tex]Cc = 0.68[/tex]
The coefficient of velocity is [tex]Cv = 0.92[/tex]
The radius is mathematically evaluated as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{ 0.06 }{2}[/tex]
[tex]r = 0.03 \ m[/tex]
The area is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * (0.03)^2[/tex]
[tex]A = 0.00283 \ m^2[/tex]
The discharge rate is mathematically represented as
[tex]Q = Cv *Cc * A * \sqrt{ 2 * g * h}[/tex]
substituting values
[tex]Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}[/tex]
[tex]Q = 0.0192 \ m^3 /s[/tex]
It is just as difficult to accelerate a car on a level horizontal surface on the Moon as it is here on Earth because
Answer:
Mass of the car is independent of gravity
Explanation:
Here, we want to state the reason why even though we have the acceleration due to gravity absent on the moon, it is still difficult to accelerate a car on a level horizontal level on the moon.
The answer to this is that the mass of the car that we want to accelerate is independent of gravity.
Had it been that gravity has an effect on the mass of the said car, then we might conclude that it will not be difficult to accelerate the car on a horizontal surface on the moon.
But due to the fact that gravity has no effect on the mass of the car to be accelerated, then the problem we have on earth with accelerating the car is the same problem we will have on the moon if we try to accelerate the car on a horizontal level surface.
1. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling without slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
2. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling with slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
Answer:
The hoop
Explanation:
Because it has a smaller calculated inertia of 2/3mr² compares to the disc
A mass M slides downward along a rough plane surface inclined at angle \Theta\:Θ= 32.51 in degrees relative to the horizontal. Initially the mass has a speed V_0\:V 0 = 6.03 m/s, before it slides a distance L = 1.0 m down the incline. During this sliding, the magnitude of the power associated with the work done by friction is equal to the magnitude of the power associated with the work done by the gravitational force. What is the coefficient of kinetic friction between the mass and the incline?
Answer: μ = 0.8885
Explanation: Force due to friction is calculated as: [tex]F_{f} = \mu.N[/tex]
At an inclined plane, normal force (N) is: N = mgcosθ, in which θ=32.51.
Power associated with work done by friction is [tex]P=F_{f}.x[/tex]. The variable x is displacement the object "spent its energy".
Power associated with work done by gravitational force is P = mghcosθ, where h is height.
The decline forms with horizontal plane a triangle as draw in the picture.
To determine force due to friction:
[tex]F_{f}.x=mghcos(\theta)[/tex]
[tex]F_{f}=\frac{mghcos(\theta)}{x}[/tex]
Replacing force:
[tex]\frac{m.g.h.cos(\theta)}{x} = \mu.m.g.cos(\theta)[/tex]
[tex]\mu=\frac{h}{x}[/tex]
Calculating h using trigonometric relations:
[tex]sin(32.51) = \frac{h}{1}[/tex]
h = sin(32.51)
Coefficient of Kinetic friction is
[tex]\mu=\frac{sin(32.51)}{1}[/tex]
μ = 0.8885
For these conditions, coefficient of kinetic friction is μ = 0.8885.
1. Rank the transformers on the basis of their rms secondary voltage. Rank from largest to smallest.
Vp = 240 V; Np = 1000 turns; Ns = 2000 turns
Vp = 480 V; Np = 4000 turns; Ns = 2000 turns
Vp = 480 V; Np = 2000 turns; Ns = 1000 turns
Vp = 120 V; Np = 500 turns; Ns = 2000 turns
Vp = 240 V; Np = 1000 turns; Ns = 500 turns
2. 100 A of rms current is incident on the primary side of each transformer. Rank the transformers on the basis of their rms secondary current. Rank from largest to smallest.
Vp = 240 V; Np = 1000 turns; Ns = 2000 turns
Vp = 480 V; Np = 2000 turns; Ns = 1000 turns
Vp = 240 V; Np = 1000 turns; Ns = 500 turns
Vp = 120 V; Np = 500 turns; Ns = 2000 turns
Vp = 480 V; Np = 4000 turns; Ns = 2000 turns
Answer:
1. Transformer 3> Transformer 1 and 2 > Transformer 4
2. Transformer 2,3,5 > Transformer 1 > Transformer 4
Explanation:
From;
Vs/Vp = Ns/Np
Where;
Vp= voltage in primary coil
Vs= voltage in secondary coil
Ns= number of turns in secondary coil
Np= number of turns in primary coil.
Vs= Ns×Vp/Np
Vs= 480 ×2000/4000
Vs= 240 V
Vs= 480 ×1000/2000
Vs=240 V
Vs= 120 × 2000/500
Vs= 480 V
Vs= 240 × 500/1000
Vs= 120 V
2. Ns/Np= Ip/Is
Is= Np×Ip/Ns
Is= 1000 × 100/2000
Is= 50 A
Is= 2000 × 100/1000
Is= 200 A
Is= 1000 × 100/500
Is= 200 A
Is= 500 × 100/2000
Is= 25 A
Is= 4000 × 100/2000
Is= 200 A
Which is one criterion that materials of a technological design should meet? They must be imported. They must be affordable. They must be naturally made. They must be locally produced.
Answer:
they must be affordable because they have to pay for it or they wont get the stuff they are bying.
Explanation:
need a brainliest please.
Answer: B, they must be affordable.
Explanation:
The electric field 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge?
Answer:
2.1×10¹⁸ C
Explanation:
Using,
E = kq/r²...................... Equation 1
Where E = Electric field, q = charge, r = distance, k = coulombs constant.
make q the subject of the equation
q = Er²/k.................. Equation 2
Given: E = 180000 N/C, r = 2.8 cm = 0.028 m
Constant: k = 9×10⁹ Nm²/C².
Substitute these values into equation 2
q = 180000(9×10⁹)/0.028²
q = 2.1×10¹⁸ C
Hence the object charge is 2.1×10¹⁸ C
Which scientist proposed a mathematical solution for the wave nature of light?
Answer:
Explanation:
Christian Huygens
Light Is a Wave!
Then, in 1678, Dutch physicist Christian Huygens (1629 to 1695) established the wave theory of light and announced the Huygens' principle.
A student wants to create a 6.0V DC battery from a 1.5V DC battery. Can this be done using a transformer alone
Answer:
Therefore, we need an invert, and a rectifier, along with the transformer to do the job.
Explanation:
A transformer, alone, can not be used to convert a DC voltage to another DC voltage. If we apply a DC voltage to the primary coil of the transformer, it will act as short circuit due to low resistance. It will cause overflow of current through winding, resulting in overheating pf the transformer.
Hence, the transformer only take AC voltage as an input, and converts it to another AC voltage. So, the output voltage of a transformer is also AC voltage.
So, in order to convert a 6 V DC to 1.5 V DC we need an inverter to convert 6 V DC to AC, then a step down transformer to convert it to 1.5 V AC, and finally a rectifier to convert 1.5 V AC to 1.5 V DC.
Therefore, we need an invert, and a rectifier, along with the transformer to do the job.
To get an idea of the order of magnitude of inductance, calculate the self-inductance in henries for a solenoid with 1500 loops of wire wound on a rod 13 cm long with radius 2 cm
Answer:
The self-inductance in henries for the solenoid is 0.0274 H.
Explanation:
Given;
number of turns, N = 1500 turns
length of the solenoid, L = 13 cm = 0.13 m
radius of the wire, r = 2 cm = 0.02 m
The self-inductance in henries for a solenoid is given by;
[tex]L = \frac{\mu_oN^2A}{l}[/tex]
where;
[tex]\mu_o[/tex] is permeability of free space = [tex]4\pi*10^{-7} \ H/m[/tex]
A is the area of the solenoid = πr² = π(0.02)² = 0.00126 m²
[tex]L = \frac{4\pi *10^{-7}(1500)^2*(0.00126)}{0.13} \\\\L = 0.0274 \ H[/tex]
Therefore, the self-inductance in henries for the solenoid is 0.0274 H.
The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called specular reflection). However, you've likely been told that when you look at something, you are seeing light reflected from the object that you are looking at. This is reflection of a different sort of diffuse reflection.
Suppose that the spotlight shines so that different parts of the beam reflect off of different two surfaces, one inclined at an angle alpha (from the horizontal) and one inclined at an angle beta. What would the angular separation between the rays reflected from the two surfaces?
Answer:
Explanation:
Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .
Now the reflecting surface is twisted so that is becomes inclined at angle alpha .
The reflected light will be deviated from its original direction by angle
2 x alpha .
Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle
2 x beta
Hence angle between these two reflected beam
= 2 beta - 2 alpha
= 2 ( β - α )
So, angular separation between the rays reflected from the two surfaces
= 2 ( β - α ) .
A sinusoidal electromagnetic wave is propagating in a vacuum in the +z-direction. If at a particular instant and at a certain point in space the electric field is in the +x-direction and has a magnitude of 4.00 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time?
Answer:
B = 1.33 10⁻⁸ T , the magnetic field must be in the y + direction
Explanation:
In an electromagnetic wave the electric and magnetic fields are in phase
c = E / B
B = E / c
let's calculate
B = 4.00 / 3 10⁸
B = 1.33 10⁻⁸ T
To determine the direction we use that the electric and magnetic fields and the speed of the wave are perpendicular.
If the wave advances in the + Z direction and the electric field is in the + x direction, the magnetic field must be in the y + direction
A resistance heater having 20.7 kW power is used to heat a room having 16 m X 16.5 m X 12.3 m size from 13.5 to 21 oC at sea level. The room is sealed once the heater is turned on. Calculate the amount of time needed for this heating to occur in min. (Write your answer in 3 significant digits. Assume constant specific heats at room temperature.)
Answer:
t = 23.6 min
Explanation:
First we need to find the mass of air in the room:
m = ρV
where,
m = mass of air in the room = ?
ρ = density of air at room temperature = 1.2041 kg/m³
V = Volume of room = 16 m x 16.5 m x 12.3 m = 3247.2 m³
Therefore,
m = (1.2041 kg/m³)(3247.2 m³)
m = 3909.95 kg
Now, we find the amount of energy consumed to heat the room:
E = m C ΔT
where,
E = Energy consumed = ?
C = Specific Heat of air at room temperature = 1 KJ/kg.⁰C
ΔT = Change in temperature = 21 °C - 13.5 °C = 7.5 °C
Therefore,
E = (3909.95 kg)(1 KJ/kg.°C)(7.5 °C)
E = 29324.62 KJ
Now, the time period can be calculated as:
P = E/t
t = E/P
where,
t = Time needed = ?
P = Power of heater = 20.7 KW
Therefore,
t = 29324.62 KJ/20.7 KW
t = (1416.65 s)(1 min/60 s)
t = 23.6 min
Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consider points along the line connecting the two sources.Required:a. At what distance from source A is there constructive interference between points A and B?b. At what distances from source A is there destructive interference between points A and B?
Answer:
a
[tex]z= 2.5 \ m[/tex]
b
[tex]z = (1 \ m , 4 \ m )[/tex]
Explanation:
From the question we are told that
Their distance apart is [tex]d = 5.00 \ m[/tex]
The wavelength of each source wave [tex]\lambda = 6.0 \ m[/tex]
Let the distance from source A where the construct interference occurred be z
Generally the path difference for constructive interference is
[tex]z - (d-z) = m \lambda[/tex]
Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0
so
[tex]z - (5-z) = 0[/tex]
=> [tex]2 z - 5 = 0[/tex]
=> [tex]z= 2.5 \ m[/tex]
Generally the path difference for destructive interference is
[tex]|z-(d-z)| = (2m + 1)\frac{\lambda}{2}[/tex]
=> [tex]|2z - d |= (0 + 1)\frac{\lambda}{2}[/tex]
=> [tex]|2z - d| =\frac{\lambda}{2}[/tex]
substituting values
[tex]|2z - 5| =\frac{6}{2}[/tex]
=> [tex]z = \frac{5 \pm 3}{2}[/tex]
So
[tex]z = \frac{5 + 3}{2}[/tex]
[tex]z = 4\ m[/tex]
and
[tex]z = \frac{ 5 -3 }{2}[/tex]
=> [tex]z = 1 \ m[/tex]
=> [tex]z = (1 \ m , 4 \ m )[/tex]
"In a Young's double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to"
Answer:
To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to a value 2D that is twotimes D
