Answer:
A simple arrangement by means of which e.m.f,s. are compared is known as?
(a)Voltmeter
(b)Potentiometer
(c)Ammeter
(d)None of the above
Explanation:
A locomotive is pulling three train cars along a level track with a force of 100,000N. The car next to the locomotive has a mass of 80,000kg, next one, 50,000kg, and the last one, 70,000 kg. you can neglect the friction on the cars being pulled.
A) what if the magnitude of the force between that the 80,000-kg car exerts on the 50,000-kg car?
B) what is the magnitude of the force that the 50,000-kg car exerts on the 70,000-kg car?
Answer:
a) 60000 N
b) 35000 N
Explanation:
Force from locomotive = 100000 N
mass of first car = 80000 kg
mass of second car = 50000 kg
mass of third car = 70000 kg
friction is neglected in this system
Total mass of the cars = 80000 + 50000 + 70000 = 200000 kg
All the car in the system will accelerate at the same rate since they are pulled by the same force
We know that force F = ma
where
a is the acceleration of the cars
m is the total mass in the system
from this we can say that
a = F/m
a = 100000/200000 = 0.5 m/s^2
a) The total mass involved in this case = mass of the last two cars after the 80000 kg car = 50000 + 70000 = 120000 kg
therefore force exerted F = ma
F = 0.5 x 120000 = 60000 N
b) The total mass in this case = mass of the third car only = 70000 kg
F = ma
F = 70000 x 0.5 = 35000 N
An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and inductor, or a capacitor. At time t = 0 the voltage is zero and increasing toward a maximum. At time t = T/4 the current in the unknown element is equal to zero, and at time t = T/2 the current is I = -I max, where Imax is the current amplitude. What is the unknown element?
a. a resistor
b. an inductor or a capacitor
c. an inductor
d. a capacitor
A 6.7 cm diameter circular loop of wire is in a 1.27 T magnetic field. The loop is removed from the field in 0.16 ss . Assume that the loop is perpendicular to the magnetic field.
Required:
What is the average induced emf?
Answer:
The induced emf is [tex]\epsilon = 0.0280 \ V[/tex]
Explanation:
From the question we are told
The diameter of the loop is [tex]d = 6.7 cm = 0.067 \ m[/tex]
The magnetic field is [tex]B = 1.27 \ T[/tex]
The time taken is [tex]dt = 0.16 \ s[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = - N * \frac{\Delta \phi}{dt}[/tex]
Where N = 1 given that it is only a circular loop
[tex]\Delta \phi = \Delta B * A[/tex]
Where [tex]\Delta B = B_f - B_i[/tex]
where [tex]B_i[/tex] is 1.27 T given that the loop of wire was initially in the magnetic field
and [tex]B_f[/tex] is 0 T given that the loop was removed from the magnetic field
Now the area of the of the loop is evaluated as
[tex]A = \pi r^2[/tex]
Where r is the radius which is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.067}{2}[/tex]
[tex]r = 0.0335 \ m[/tex]
So
[tex]A = 3.142 * (0.0335)^2[/tex]
[tex]A = 0.00353 \ m^2[/tex]
So
[tex]\Delta \phi = (0- 127)* (0.00353)[/tex]
[tex]\Delta \phi = -0.00448 Weber[/tex]
=> [tex]\epsilon = - 1 * \frac{-0.00448}{0.16}[/tex]
=> [tex]\epsilon = 0.0280 \ V[/tex]
12. A concave lens has a focal length of 10 cm. An object 2.5 cm high is placed 30 cm from the lens. Determine the position and size of the image. (3)
Answer:
I think 9.5
Explanation:
............
which one is more powerful hydrogen bomb or atom bomb and why?
Hydrogen bomb is more powerful than atom bomb
Hydrogen has a calorie value of 150000KJ .It is very much than nuclear bomb or atom bombScientists also told that Hydrogen bomb is more powerful.But both bombs are destructive.A liquid is poured into a vessel to a depth of 16cm when viewed from the top, the bottom appears to be raised 4cm. What is the refractive index of the liquid?
Answer:
Solution
Verified by Toppr
Correct option is
C
3 cm
RI=apparent depthreal depth
Substituting, 34=apparentdepth12
Therefore, apparent depth=412×3=9
The height by which it appears to be raised is 12−9=3cm
Was this answer helpful?
71
0
SIMILAR QUESTIONS
A coin is placed at the bottom of a glass tumbler and then water is added. It appeared that the depth of the coin has been reduced because
Medium
View solution
>
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
A heat engine operates between 200 K and 100 K. In each cycle it takes 100 J from the hot reservoir, loses 25 J to the cold reservoir, and does 75 J of work. This heat engine violates the second law but not the first law of thermodynamics. Why is this true?
Answer:
It does not violate the first law because the total energy taken is what is used 100J = 25J + 75J
But violates 2nd lawbecause the engine has a higher energy after doing work than the initial for e.g A cold object in contact with a hot one never gets colder, transferring heat to the hot object and making it hotter confirming the second law
(4) Use the preliminary observations to answer these questions; Compared to no polarizer or analyzer in the optical path, by what percent does the light intensity decrease when (a) The polarizer is introduced into the optical path? (b) The both polarizer and analyzer are introduced into the optical path?
Answer:
a) I = I₀/2, b) I = I₀/2 cos² θ
Explanation:
To answer these questions, let's analyze a little the way of working of a polarized
* When a non-polarized light hits a polarizer, the electric field that is not in the direction of the polarizer is absorbed, so the transmitted light is
i = I₀ / 2
and is polarized in the direction of the polarizer
* when a polarized light reaches the analyzer it must comply with Malus's law
I = I₁ cos² θ
where the angle is between the polarized light and the analyzer.
With this, let's answer the questions
a) When a polarizer is placed in the non-polarized light path, half of it is absorbed and only the light that has polarization in the direction of the polarizer is transmitted with an intensity of
I = I₀/2
b) when a polarizer and an analyzer are fitted, the intensity of the light transmitted by the analyzer is
I = I₀/2 cos² θ
where the final value depends on the angle between the polarizer and the analyzer.
Let's look at two extreme cases
θ = 0 I = Io / 2
θ = 90º I = 0
What is the displacement of the object after 3 seconds?
Answer:
3 meters
Explanation:
An object on a level surface experiences a horizontal force of 12.7 N due to kinetic friction. The coefficient of kinetic friction is 0.42.
What is the mass of the object? (Express your answer to two significant figures)kg
Answer:
The mass of the object is 3.08 kg.
Explanation:
The horizontal force is12.7 N and the coefficient of the kinetic fraction are 0.42. Now we have to compute the mass of the object. Thus, use the below formula to find the mass of the object.
Let the mass of the object = m.
The coefficient of kinetic friction, n = 0.42
Therefore,
Force, F = n × mg
12.7 = 0.42 × 9.8 × m
m = 3.08 kg
The mass of the object is 3.08 kg.
A horizontal circular platform rotates counterclockwise about its axis at the rate of 0.945 rad/s. You, with a mass of 69.7 kg, walk clockwise around the platform along its edge at the speed of 1.01 m/s with respect to the platform. Your 20.7 kg poodle also walks clockwise around the platform, but along a circle at half the platform's radius and at half your linear speed with respect to the platform. Your 17.7 kg mutt, on the other hand, sits still on the platform at a position that is 3/4 of the platform's radius from the center. Model the platform as a uniform disk with mass 93.1 kg and radius 1.93 m. Calculate the total angular momentum of the system.
Answer:
317.22
Explanation:
Given
Circular platform rotates ccw 93.1kg, radius 1.93 m, 0.945 rad/s
You 69.7kg, cw 1.01m/s, at r
Poodle 20.2 kg, cw 1.01/2 m/s, at r/2
Mutt 17.7 kg, 3r/4
You
Relative
ω = v/r
= 1.01/1.93
= 0.522
Actual
ω = 0.945 - 0.522
= 0.42
I = mr^2
= 69.7*1.93^2
= 259.6
L = Iω
= 259.6*0.42
= 109.4
Poodle
Relative
ω = (1.01/2)/(1.93/2)
= 0.5233
Actual
ω = 0.945- 0.5233
= 0.4217
I = m(r/2)^2
= 20.2*(1.93/2)^2
= 18.81
L = Iω
= 18.81*0.4217
= 7.93
Mutt
Actual
ω = 0.945
I = m(3r/4)^2
= 17.7(3*1.93/4)^2
= 37.08
L = Iω
= 37.08*0.945
= 35.04
Disk
I = mr^2/2
= 93.1(1.93)^2/2
= 173.39
L = Iω
= 173.39*0.945
= 163.85
Total
L = 109.4+ 7.93+ 36.04+ 163.85
= 317.22 kg m^2/s
A chemist must dilute 55.6 ml of 1.48 M aqueous silver nitrate (AgNO3)solution until the concentration falls to 1.00 M. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in milliliters. Round your answer to 3 significant digits.
Answer:
82.2 mL
Explanation:
The process of adding water to a solution to make it more dilute is known as dilution. The formula for dilution is;
C1V1=C2V2
Where;
C1= concentration of stock solution
V1= volume of stock solution
C2= concentration of dilute solution
V2= volume of dilute solution
V2= C1V1/C2
V2= 1.48 × 55.6/ 1.0
V2= 82.2 mL
A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object and the screen. In terms of f, what is the smallest value d can have for an image to be in focus on the screen?
Answer:
2f
Explanation:
The formula for the object - image relationship of thin lens is given as;
1/s + 1/s' = 1/f
Where;
s is object distance from lens
s' is the image distance from the lens
f is the focal length of the lens
Total distance of the object and image from the lens is given as;
d = s + s'
We earlier said that; 1/s + 1/s' = 1/f
Making s' the subject, we have;
s' = sf/(s - f)
Since d = s + s'
Thus;
d = s + (sf/(s - f))
Expanding this, we have;
d = s²/(s - f)
The derivative of this with respect to d gives;
d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²
Equating to zero, we have;
(2s/(s - f)) - s²/(s - f)² = 0
(2s/(s - f)) = s²/(s - f)²
Thus;
2s = s²/(s - f)
s² = 2s(s - f)
s² = 2s² - 2sf
2s² - s² = 2sf
s² = 2sf
s = 2f
A uniform bar has two small balls glued to its ends. The bar is 2.10 m long and with mass 3.70 kg , while the balls each have mass 0.700 kg and can be treated as point masses.
Required:
Find the moment of inertia of this combination about an axis
a. perpendicular to the bar through its center.
b. perpendicular to the bar through one of the balls.
c. parallel to the bar through both balls.
d. parallel to the bar and 0.500 m from it.
Answer:
Explanation:
a )
moment of inertia in the first case will be sum of moment of inertia of two balls + moment of inertia of bar
= 2 x .700 x (2.1 / 2 )² + 3.7 x 2.1² / 12
= 1.5435 + 1.35975
= 2.90325 kg m²
b )
moment of inertia required
= moment of inertia of bar + moment of inertia of the other ball
= 3.70 x (2.1² / 3 ) + .7 x 2.1²
= 5.439 + 3.087
= 8.526 kg m²
c )
In this case moment of inertia of the combination = 0 as distance of masses from given axis is zero .
d )
masses = 3.7 + .7 = 4.4 kg
distance from axis = .5 m
moment of inertia about given axis
= 4.4 x .5²
= 1.1 kg m².
Question 2
A) A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released? (1 point)
The spring exerts a restoring force to the right and compresses even further
The spring exerts a restoring force to the left and returns to its equilibrium position
The spring exerts a restoring force to the right and returns to its equilibrium position
The spring exerts a restoring force to the left and stretches beyond its equilibrium position
1. Which example best describes a restoring force?
B) the force applied to restore a spring to its original length
2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?
C) The spring exerts a restoring force to the left and returns to its equilibrium position.
3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?
D) 1 m
4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?
D)It is a vector quantity.
5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?
A) It decreases in magnatude.
At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to:_____
(a) 3 times that of the Celsius and
(b) 1/5 times that of the Celsius
Answer:
C = 26.67° and F = 80°C = -20° and F = -4°Explanation:
Find:
3 times that of the Celsius and 1/5 times that of the CelsiusComputation:
F = (9/5)C + 32
3 times that of the Celsius
If C = x
So F = 3x
So,
3x = (9/5)x + 32
15x = 9x +160
6x = 160
x = 26.67
So, C = 26.67° and F = 80°
1/5 times that of the Celsius
If C = x
So F = x/5
So,
x/5 = (9/5)x + 32
x = 9x + 160
x = -20
So, C = -20° and F = -4°
Your favorite radio station broadcasts at a frequency of 91.5 MHz with a power of 11.5 kW. How many photons does the antenna of the station emit in each second?
Answer:
Number of photons emit per second = 1.9 × 10²⁹ (Approx)
Explanation:
Given:
Frequency = 91.5 MHz
Power = 11.5 Kw = 11,500 J/s
Find:
Number of photons emit per second
Computation:
Total energy with frequency (E) = hf
Total energy with frequency (E) = 6.626×10⁻³⁴ × 91.5×10⁶
Total energy with frequency (E) = 6.06×10⁻²⁶ J
Number of photons emit per second = 11,500 / 6.06×10⁻²⁶
Number of photons emit per second = 1897.689 × 10²⁶
Number of photons emit per second = 1.9 × 10²⁹ (Approx)
Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (Assume the small-angle approximation is valid here.)
(a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.97 cm, find the wavelength of the incident light.
(b) At what angle does the next set of bright fringes appear?
Answer:
a
[tex]\lambda = 1.667 nm[/tex]
b
[tex]\theta = 0.8681^o[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 0.220 \ mm = 0.00022 \ m[/tex]
The is distance of the screen from the slit is [tex]D = 2.60 \ m[/tex]
The distance between the central bright fringe and either of the adjacent bright [tex]y = 1.97 cm = 1.97 *10^{-2}\ m[/tex]
Generally the condition for constructive interference is
[tex]d sin \tha(\theta ) = n \lambda[/tex]
From the question we are told that small-angle approximation is valid here.
So [tex]sin (\theta ) = \theta[/tex]
=> [tex]d \theta = n \lambda[/tex]
=> [tex]\theta = \frac{n * \lambda }{d }[/tex]
Here n is the order of maxima and the value is n = 1 because we are considering the central bright fringe and either of the adjacent bright fringes
Generally the distance between the central bright fringe and either of the adjacent bright is mathematically represented as
[tex]y = D * sin (\theta )[/tex]
From the question we are told that small-angle approximation is valid here.
So
[tex]y = D * \theta[/tex]
=> [tex]\theta = \frac{ y}{D}[/tex]
So
[tex]\frac{n * \lambda }{d } = \frac{y}{D}[/tex]
[tex]\lambda =\frac{d * y }{n * D}[/tex]
substituting values
[tex]\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }[/tex]
[tex]\lambda = 1.667 *10^{-6}[/tex]
[tex]\lambda = 1.667 nm[/tex]
In the b part of the question we are considering the next set of bright fringe so n= 2
Hence
[tex]dsin (\theta ) = n \lambda[/tex]
[tex]\theta = sin^{-1}[\frac{ n * \lambda }{d} ][/tex]
[tex]\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ][/tex]
[tex]\theta = 0.8681^o[/tex]
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.
https://brainly.com/question/13552918
#SPJ2
An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and inductor, or a capacitor. At time t = 0 the voltage is zero. At time t = T/4 the current in the unknown element is equal to zero, and at time t = T/2 the current is I = -Imax, where Imax is the current amplitude. What is the unknown element?
Answer:
Capacitor, is the right answer.
Explanation:
The unknown element is a Capacitor.
Below is the calculation that proves that it is a capacitor.
We know that for the Capacitor
i = Imax × sin(wt+(pi/2)).
i = Imax × sin ((2 × pi/T) × (T/4) + (pi/2))
i = Imax × sin(3.142) = 0 A
at, t = T/2
wt = (2 × pi/T) × (T/2) = pi
wt + (pi/2) = pi + (pi/2) = ( 3 × pi/2) =
i = Imax × sin(3 × pi/2) = -Imax
Which is in a correct agreement with capacitor therefore, the answer is a Capacitor.
A stone is dropped from the upper observation deck of a tower, 50 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the distance (in meters) of the stone above ground level at time t. h(t) = (b) How long does it take the stone to reach the ground? (Round your answer to two decimal places.) s (c) With what velocity does it strike the ground? (Round your answer to one decimal place.) m/s (d) If the stone is thrown downward with a speed of 9 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)
Answer:
A. Using displacement =Ut + 1/2gt²
=> 0 + 1/2 (-9.8)t²
= -4.9t²
So
h(t) = 50+ displacement
= 50 - 4.9t²
B. To reach the ground
h(t) = 0
So
50-4.9t²= 0
t = √ (50/4.9)
= 3.2s
C. Using
V = u+ gt
U= 0
V= - 9.8(3.2)
= 31.4m/s
D. If u = -9m/s
Then s = ut + 1/2gt²
5t- 1/2gt²
But distance from the ground is
=.> 50-5t- 4.8t²= 0
So t solving the quadratic equation
t= 3.58s
(a) The distance of the stone above the ground level at time t is [tex]h(t) = 50 - 4.9t^2[/tex]
(b) The time taken for the stone to strike the ground is 3.19 s.
(c) The velocity of the stone when it strikes the ground is 31.4 m/s.
(d) The time taken for the stone to reach the ground when thrown at the given speed is 2.41 s.
The given parameters;
height above the ground, h₀ = 50 mThe distance of the stone above the ground level at time t is calculated as;
[tex]h(t) = h_0 - ut - \frac{1}{2} gt^2\\\\h(t) = 50 - 0 -0.5\times 9.8t^2\\\\h(t) = 50 - 4.9t^2[/tex]
The time taken for the stone to strike the ground is calculated as;
[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 50}{9.8} } \\\\t = 3.19 \ s[/tex]
The velocity of the stone when it strikes the ground is calculated as;
[tex]v =u + gt\\\\v = 0 + 3.2 \times 9.8\\\\v = 31.4 \ m/s[/tex]
The time taken for the stone to reach the ground when thrown at speed of 9 m/s is calculated as;
[tex]50 = 9t + \frac{1}{2} (9.8)t^2\\\\50 = 9t + 4.9t^2\\\\4.9t^2 + 9t - 50 = 0\\\\a = 4.9 \, \ b = 9, \ \ c = -50\\\\solve \ the \ quadratic \ equation\ using \ formula \ method\\\\t = \frac{-b \ \ + /- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-9 \ \ + /- \ \ \sqrt{(9)^2 - 4(4.9 \times -50)} }{2(4.9)} \\\\t = 2.41 \ s \ \ or \ \ - 4.24 \ s[/tex]
Thus, the time taken for the stone to reach the ground when thrown at the given speed is 2.41 s.
Learn more here:https://brainly.com/question/9527588
A sample of gas is enclosed in a container of fixed volume. Identify which of the following statements are true. Check all that apply.If the container is heated, the gas particles will lose kinetic energy and temperature will increase.
Answer:
B. If the container is cooled, the gas particles will lose kinetic energy and temperature will decrease.
C. If the gas particles move more quickly, they will collide more frequently with the walls of the container and pressure will increase.
E. If the gas particles move more quickly, they will collide with the walls of the container more often and with more force, and pressure will increase.
#FreeMelvin
2. The nuclear model of the atom held that
a. electrons were randomly spread through "a sphere of uniform positive
electrification."
b. matter was made of tiny electrically charged particles that were smaller than the
atom
C. matter was made of tiny, indivisible particles.
d. the atom had a dense, positively charged nucleus.
Answer:
the atom had a dense, positively charged nucleus.
Explanation:
Ernest Rutherford, based on the experiment carried out by two of his graduate students, established the authenticity of the nuclear model of the atom.
According to the nuclear model, an atom is made up of a dense positive core called the nucleus. Electrons are found to move round this nucleus in orbits. This is akin to the movement of the planets round the sun in the solar system.
Figure (3) shows a car travelling along the route PQRST in 30 minutes. What is the average speed of the car in km/hour?
Answer:
60 km/hour.
Explanation:
We'll begin by calculating the total distance traveled by the car. This is illustrated below:
Total distance traveled = sum of distance between PQRST
Total distance = 10 + 5 + 10 + 5
Total distance = 30 km
Next, we shall convert 30 mins to hour. This can obtained as follow:
Recall:
60 mins = 1 hour
Therefore,
30 mins = 30/60 = 0.5 hour.
Finally, we shall determine the average speed of the car as follow:
Distance = 30 km
Time = 0.5 hour
Speed =?
Speed = distance /time
Speed = 30/0.5
Speed = 60 km/hour
Therefore, the speed of the car is 60 km/hour.
What is the force that attracts objects with mass toward each other?
Explanation:
gravitional force attracts objects with mass toward each other.
You measure the power delivered by a battery to be 4.26 W when it is connected in series with two equal resistors. How much power will the same battery deliver if the resistors are now connected in parallel across it
Answer:
The power delivered by the battery is 17.04 W
Explanation:
Power through a circuit is given as
P = IV ....1
where P is the power
I is the current through the circuit
V is the voltage through the circuit
The voltage in a circuit is given as
V = IR ....2
Let us take the value of each resistor as equal to R
when connected in series, the total resistance will be
[tex]R_{t}[/tex] = R + R = 2R
If we assume constant voltage through the circuit, then from equation 2, the current in this case is
I = V/2R
If the resistors are connected in parallel, then the total resistance will be
[tex]\frac{1}{R_{t} }[/tex] = [tex]\frac{1}{R}[/tex] +
[tex]R_{t}[/tex] = R/2
The current in this case will be increased since the resistance is reduced
I = 2V/R
comparing the two situations, we can see that the current increased when connected in parallel to a ratio of
[tex]\frac{2V}{R}[/tex] ÷ [tex]\frac{V}{2R}[/tex] =
This means that the current increased 4 times
From equation 1, we can see that electrical power is proportional to the current at a constant voltage, therefore, the power will also increase by four times to
P = 4 x 4.26 = 17.04 W
The magnetic flux that passes through one turn of a 8-turn coil of wire changes to 5.0 Wb from 8.0 Wb in a time of 0.098 s. The average induced current in the coil is 140 A. What is the resistance of the wire
Answer:
Resistance is 1.75 ohms
Explanation:
Magnetic flux:
[tex]{ \phi{ = NBA}}[/tex]
N is number of turns, N = 8
B is magnetic flux
A is area of projection.
From faradays law:
[tex]E = - \frac{ \triangle \phi}{t} [/tex]
where E is the Electro motive force.
But E = IR
where I is current and R is resistance:
[tex]IR = \frac{( \phi_{1} - \phi _{2}) }{t} \\ \\ 140 \times R = \frac{8 \times (8 - 5)}{0.098} \\ \\ R = \frac{24}{0.098 \times 140} \\ \\ resistance = 1.75 \: ohms[/tex]
In a photoelectric-effect experiment it is observed that no current flows unless the wavelength is less than 540 nmnm . Part A What is the work function of this material
Answer:
Φ = 36.84 × 10^(-20) J
Explanation:
In the photoelectric effect, the energy of the incoming photon is usually used in part to extract the photoelectron from the material (work function) and then the rest is converted into kinetic energy of the photoelectron which is given by the formula;
K_max = hf - Φ
where;
hf represents the energy of the incoming photon
h is the Planck's constant
f is the light frequency
Φ is the work function of the material
K_max is the maximum kinetic energy of the photoelectrons.
From the question, we are told that no current flows unless the wavelength is less than 540 nm. This means that when the wavelength has this value, the maximum kinetic energy of the photoelectrons is zero i.e K_max = 0. Thus the energy of the incoming photons is just enough to extract the photoelectrons from the material.
Thus,
hf - Φ = 0
hf = Φ - - - (1)
We are given the wavelength as;
λ = 540 nm = 540 × 10^(-9) m
Now, let's find the frequency of the light by using the relationship between frequency and wavelength. The equation is;
f = c/λ
Where c is speed of light = 3 × 10^(8) m/s
f = (3 × 10^(8))/(540 × 10^(-9))
f = 5.56 × 10^(14) Hz
Thus, from equation 1,we can now find the work function;
Φ = hf
h is Planck's constant and has a value of 6.626 × 10^(-34) J.s
Thus;
Φ = 6.626 × 10^(-34) × 5.56 × 10^(14)
Φ = 36.84 × 10^(-20) J
A radar pulse returns 3.0 x 10-4 seconds after it is sent out, having been reflected by an object. What is the distance between the radar antenna and the object
Answer:
The distance is [tex]D = 45000 \ m[/tex]
Explanation:
From the question we are told that
The time taken is [tex]t = 3.0 *10^{-4 } \ s[/tex]
Generally the speed of the radar is equal to the speed of light and this has a value
[tex]c = 3.0*10^{8} \ m /s[/tex]
Now the distance covered by the to and fro movement of the radar is mathematically evaluated as
[tex]d = c * t[/tex]
=> [tex]d = 3.0*10^{8} * 3.0*10^{-4}[/tex]
=> [tex]d = 90000 \ m[/tex]
Therefore the distance between the radar antenna and the object is
[tex]D = \frac{d}{2}[/tex]
[tex]D = \frac{ 90000}{2}[/tex]
[tex]D = 45000 \ m[/tex]
The distance between the radar antenna and the object will be 45000 m.
What is a radar antenna?A radar antenna is a device that sends out radio waves and listens for their reflections. The ability of an antenna to identify the exact direction in which an item is placed determines its performance.
The given data in the problem is;
t is the time= 3.0 x 10⁻⁴
d is the distance between the radar antenna and the object=?
c is the peed of light=3×10⁸ m/sec
The radar's speed is usually equal to the speed of light, and this has a value. The distance covered by the radars to and fro movement is now calculated mathematically as
[tex]\rm d= c \times t \\\\ \rm d= 3.0 \times 10^8 \times 3.0 \times 10^{-4} \\\\ d=90000 \ m[/tex]
As a result, the radar antenna's distance from the target is
[tex]\rm D=\frac{d}{2} \\\\ \rm D=\frac{90000}{2} \\\\ \rm D=\ 45000 \ m[/tex]
Hence the distance between the radar antenna and the object will be 45000 m.
To learn more about the radar antena refer to the link;
https://brainly.com/question/24067190
"Two waves of the same frequency have amplitudes 1.00 and 2.00. They interfere at a point where their phase difference is 60.0°. What is the resultant amplitude?"
Answer:
The resultant amplitude of the two waves is 2.65.
Explanation:
Given;
amplitude of the first wave, A₁ = 1
amplitude of the second wave, A₂ = 2
phase difference of the two amplitudes, θ = 60.0°.
The resultant amplitude of two waves after interference is given by;
[tex]A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2Cos \theta} \\\\A = \sqrt{1^2 + 2^2 + 2(1)(2)Cos 60} \\\\A= 2.65[/tex]
Therefore, the resultant amplitude of the two waves is 2.65.