Answer:
λ = 52.5 nm
Explanation:
De Broglie's duality principle states that all matter has wave and particle characteristics, being related by the expression
p = h / λ
where the moment
p = mv
λ = h / mv
let's calculate
λ = 6.63 10⁻³⁴ / 1.675 10⁻²⁷ 7.54
λ = 5.25 10⁻⁸ m
Let's reduce anm
λ = 5.25 10⁻⁸ m (10⁹ nm / 1m)
λ = 52.5 nm
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]
P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism articulated to the ground at its vertex A, while vertex C is attached to the vertical cord fixed to the ground. If the coefficient of friction between the prism and the blocks is 0.4; determine the maximum angle that measures the inclination of the AC face with respect to the horizontal so that the system remains in equilibrium.
Answer:
21.8°
Explanation:
Let's call θ the angle between BC and the horizontal.
Draw a free body diagram for each block.
There are 4 forces acting on block D:
Weight force P pulling down,
Normal force N₁ pushing perpendicular to AB,
Friction force N₁μ pushing parallel up AB,
and tension force T pushing parallel up AB.
There are 4 forces acting on block E:
Weight force P pulling down,
Normal force N₂ pushing perpendicular to BC,
Friction force N₂μ pushing parallel to BC,
and tension force T pulling parallel to BC.
Sum of forces on D in the perpendicular direction:
∑F = ma
N₁ − P sin θ = 0
N₁ = P sin θ
Sum of forces on D in the parallel direction:
∑F = ma
T + N₁μ − P cos θ = 0
T = P cos θ − N₁μ
T = P cos θ − P sin θ μ
T = P (cos θ − sin θ μ)
Sum of forces on E in the perpendicular direction:
∑F = ma
N₂ − P cos θ = 0
N₂ = P cos θ
Sum of forces on E in the parallel direction:
∑F = ma
N₂μ + P sin θ − T = 0
T = N₂μ + P sin θ
T = P cos θ μ + P sin θ
T = P (cos θ μ + sin θ)
Set equal:
P (cos θ − sin θ μ) = P (cos θ μ + sin θ)
cos θ − sin θ μ = cos θ μ + sin θ
1 − tan θ μ = μ + tan θ
1 − μ = tan θ μ + tan θ
1 − μ = tan θ (μ + 1)
tan θ = (1 − μ) / (1 + μ)
Plug in values:
tan θ = (1 − 0.4) / (1 + 0.4)
θ = 23.2°
∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.
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.
Convert 76.2 kilometers to meters?
Answer
76200meters
Explanation:
we know that 1km=1000meters
to convert km into meters we we divide km by meters
=76.2/1000
=76200meters
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.
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
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].
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%.
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:
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 rectangular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing down. What is the direction of the induced current in the coil?
Answer:
There is no induced current on the coil.
Explanation:
Current is induced in a coil or a circuit, when there is a break of flux linkage. A break in flux linkage is caused by a changing magnetic field, and must be achieved by a relative motion between the coil and the magnet. Holding the magnet above the center of the coil will cause no changing magnetic filed since there is no relative motion between the coil and the magnet.
A small glass bead charged to 5.0 nC is in the plane that bisects a thin, uniformly charged, 10-cm-long glass rod and is 4.0 cm from the rod's center. The bead is repelled from the rod with a force of 910 N. What is the total charge on the rod?
Answer:
Explanation:
Let B= bead
Q = rod
the electric field at the glass bead pocation is
(Gauss theorem)
E = Q / (2 π d L εo)
the force is
F = q E = q Q / (2 π d L εo)
then
Q = 2 π d L εo F / q
Q = 2*3.14*4x10^-2*10^-1*8.85x10^-12*910x10^-4 / 5x10^-9 = 2.87x10^-8 C = 40.5 nC
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
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
What is the density of the unknown fluid in Figure below? ρwater = 1000 kgm−3
Answer:
2500 kg/m³
Explanation:
P = P
ρgh = ρgh
ρh = ρh
(1000 kg/m³) (8.9 cm) = ρ (3.5 cm)
ρ ≈ 2500 kg/m³
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]
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
A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle. What is the y-component of the velocity of the second ball?
Answer:
v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
p₀ = m v₁₀
final. After the crash
[tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}
Recall that velocities are a vector so it has x and y components
p₀ = p_{f}
we write this equation for each axis
X axis
m v₁₀ = m v_{1fx} + m v_{2fx}
Y Axis
0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
sin 23.3 = v_{2fy} / v_{2f}
cos 23.3 = v_{2fx} / v_{2f}
v_{2fy} = v_{2f} sin 23.3
v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
1.83 = v_{1f} cos θ + 1.15 cos 23.3
0 = - v_{1f} sin θ + 1.15 sin 23.3
1.83 = v_{1f} cos θ + 1.0562
0 = - v_{1f} sin θ + 0.4549
v_{1f} sin θ = 0.4549
v_{1f} cos θ = -0.7738
we divide these two equations
tan θ = - 0.5878
θ = tan-1 (-0.5878)
θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
v_{1f} cos (-30.45) = - 0.7738
v_{1f} = -0.7738 / cos 30.45
v_{1f} = -0.8976 m / s
the component and this speed is
v_{1fy} = v1f sin θ
v_{1fy} = 0.8976 sin (30.45)
v_{1fy} = - 0.4549 m / s
1. A 0.430kg baseball comes off a bar and goes straight up in the air. At a height of 10.0m, the baseball has a speed of 25.3m/s. Determine the mechanical energy at the height. Show all your work. 2. What is the baseball's mechanical energy when it is at a height of 8.0m? Explain?
Answer:
180 J
Explanation:
Mechanical energy = kinetic energy + potential energy
ME = KE + PE
ME = ½ mv² + mgh
ME = ½ (0.430 kg) (25.3 m/s)² + (0.430 kg) (9.8 m/s²) (10.0 m)
ME = 180 J
Mechanical energy is conserved, so it is 180 J at all points of the trajectory.
The baseball's mechanical energy when it is at a height of 8.0m is 180 J.
What is mechanical energy?The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time. Mechanical energy is always conserved.
Mechanical energy = kinetic energy + potential energy
Given is the mass of baseball m= 0.430 kg, height h =10m, speed v= 25.3m/s.
ME = KE + PE
ME = ½ mv² + mgh
Substitute the values, we get
ME = ½ (0.430 kg) (25.3 m/s)² + (0.430 kg) (9.8 m/s²) (10.0 m)
ME = 180 J
Thus, the baseball's mechanical energy when it is at a height of 8.0m is 180 J.
Learn more about mechanical energy.
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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
The hydrogen spectrum has a red line at 656 nm, and a blue line at 434 nm. What is the first order angular separation between the two spectral lines obtained with a diffraction grating with 5000 rulings/cm?
Answer:
Explanation:
grating element or slit width a = 1 x 10⁻² / 5000
= 2 x 10⁻⁶ m
angular width of first order spectral line of wavelength λ
= λ / a
for blue line angular width
= 434 x 10⁻⁹ / 2 x 10⁻⁶ radian
= 217 x 10⁻³ radian
for red line angular width
= 656 x 10⁻⁹ / 2 x 10⁻⁶ radian
= 328 x 10⁻³ radian
difference of their angular width
= 328 x 10⁻³ - 217 x 10⁻³
= 111 x 10⁻³ radian
Ans .
"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
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.
of
The radii a wheel are 25 cm
and 5cm respectively, it is found
that an effort of 40N is required
to raise slowly a load 16ON
160 N. Find the Mechanical
Adventage and Effeciency,
Answer:
Explanation:
Given that
Effort = 40N
Load = 16ON
M.A = load/effort
M.A= 160N/40N
M.A = 4
Velocity ratio = V.R =radius of the wheel/radius of the axel
= 25cm/5cm
= 5
Efficiency = mechanical advantage/velocity ratio × 100/1
= 4/5 × 100/1
= 0.8×100/1
= 80%
Hence, the mechanical advantage of the machine is 4 while the efficiency is 80%.
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.
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]
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
y=k/x, x is halved.
what happens to the value of y
Answer:
y is doubled
Explanation:
If x is halved, that means the value is doubled. Here is an exmaple:
y=1/2. If the denominater is doubled, y would equal y=1/1. So, the value of y has doubled from 0.5 to 1. Therefore, if the denominator is halved, the solution will be doubled.
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
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.
