Answer:
1.227 m
Explanation:
Given that
Minimum wavelength is 410 nm
Maximum wavelength is 750 nm
Grating is 7800 slits/cm
Distance is 3.2 m
To solve this question, we would use the formula
sin θ = λ/d
sin θ = (410*10^-9) / (0.01/7800)
Sin θ = 410*10^-9 / 1.282*10^-6
Sin θ = 0.32 and θ = 18.67 degrees
For the second wavelength = 750 nm
sin θ = [(0.32x750)/410]
sin θ = (240 / 410)
sin θ = 0.5853 or
θ = 35.8 degrees
And finally, the width of spectrum would be
3.2[tan 35.8 - tan 18.67]
3.2 * 0.3833
= 1.227 m
A student uses a spring scale attached to a textbook to compare the static and kinetic friction between the textbook and the top of a lab
table. If the scale measures 1,580 g while the student is pulling the sliding book along the table, which reading on the scale could have been
possible at the moment the student overcame the static friction? (1 point)
1,860 g
820 g
1,580 g
1,140 g
Answer:
1,860 g
Explanation:
In a system, the coefficient of static friction is usually higher than the coefficient of kinetic friction. This means that the kinetic friction is usually less than the static friction. From the question, since the book is already sliding, it means that kinetic friction is the friction in play. This means that before the reading on the scale that could have been possible at the moment the student overcame the static friction must be greater than the reading on the scale during sliding. The only option above 1580 g is 1860 g
As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins down with a constant angular acceleration. If the CD rotates clockwise (let's take clockwise rotation as positive) at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, the angular acceleration of the CD, as it spins to a stop at -20.1 rad/s 2. How many revolutions does the CD make as it spins to a stop?
Answer:
10.8rev
Explanation:
Using
Wf²-wf = 2 alpha x theta
0²- 56.36x56.36/ 2(-20.13) x theta
Theta = 68.09 rad
But 68.09/2π
>= 10.8 revolutions
Explanation:
Based on the passage, why is it important that different ethnic groups worked together on the strike? The groups needed to avoid speaking to one another because they wouldn’t understand. The different ethnic groups believed in being separate. The groups needed to trick the owners. They needed to be able to unite even though they spoke different languages.
Answer:D
Explanation:I got it right
Answer:
They needed to be able to unite even though they spoke different languages.
Explanation:
A ball is thrown from the ground so that it’s initial vertical and horizontal components of velocity are 40m/s and 20m/s respectively. Find the ball’s total time of flight and distance it traverses before hitting the ground.
Answer:
8 seconds
160 meters
Explanation:
Given in the y direction:
Δy = 0 m
v₀ = 40 m/s
a = -10 m/s²
Find: t
Δy = v₀ t + ½ at²
0 m = (40 m/s) t + ½ (-10 m/s²) t²
0 = 40t − 5t²
0 = 5t (8 − t)
t = 0 or 8
Given in the x direction:
v₀ = 20 m/s
a = 0 m/s²
t = 8 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (20 m/s) (8 s) + ½ (0 m/s²) (8 s)²
Δx = 160 m
A bus carrying 10 people has over turned on a remote hillside during an intense thunderstorm. What three factors could contribute to creating a delay in advanced care
Answer:
The three factors that can contribute to creating a delay in advanced care for the passengers in the overturned bus include:
1. Lack of communication: Since the accident happened on the remote hillside, there is a possibility that, there would be no communication network which could have afforded them the opportunity to call medical or technical team.
2. Steep Nature of the Hill: This is another factor which will affect the care which they could have received. Steeply area tends to be difficult for climbing in or out from.
3. Thunderstorm: This factor is another reason which could contribute to delay in receiving advance care. Thunderstorm create barriers for location f the area where the bus overturned or in other situation complicate the rescue efforts of the team sent out to rescue.
Explanation:
An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase the magnitude of the induced emf? A. Straighten the wire out to be flat B. Reduce the resistance of the wire of which the loop is made C. Turning the plane of the loop to be parallel to the magnetic field D. Reducing the diameter of the loop
Answer:
changing the magnetic field more rapidly
Explanation:
According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.
Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.
In a physics lab, Asha is given a 11.5 kg uniform rectangular plate with edge lengths 62.9 cm by 46.9 cm . Her lab instructor requires her to rotate the plate about an axis perpendicular to its plane and passing through one of its corners, and then prepare a report on the project. For her report, Asha needs the plate's moment of inertia ???? with respect to given rotation axis. Calculate ???? .
Answer:
6.9kgm²
Explanation:
For an axis through the center of the rectangle, I = m[(w²+L²)/12
Using the parallel axis theorem, the added value of I = mR² = m[(w²/4 + L²/4]
Adding the 2 expressions,
I = (m/3)*(w²+L²)
I =6.95 kg∙m²
The near point (the smallest distance at which an object can be seen clearly) and the far point (the largest distance at which an object can be seen clearly) are measured for six different people.
Near Point(cm) Far Point(cm)
Avishka 40 [infinity]
Berenice 30 300
Chadwick 25 500
Danya 25 [infinity]
Edouard 80 200
Francesca 50 [infinity]
Of the farsighted people, rank them by the power of the lens needed to correct their hyperopic vision. Rank these from largest to smallest power required.
1. Berenice
2. Avishka
3. Francesca
4. Edouard
Answer:
1. Berenice = 0.67 D
2. Avishka = 1.50 D
3. Francesca = 2.00 D
4. Edouard = 2.75 D
Explanation:
The farsighted people are those with near point greater than 25 cm.
They include Avishka, Berenice, Edouard and Francisca.
A converging lens is needed to correct farsightedness, or hyperopia, therefore, the focal length, f, is positive. The image formed is virtual and on the same side of the lens. Thus the image distance is negative
From the lens formula, 1/f = 1/v 1/u; but v is negative
Therefore, 1/f = 1/u - 1/v
But, power of a lens = 1/f in meters.
Therefore, P = 1/u - 1/v
For Avishka, u = 25 cm or 0.25 m, v = 0.4 m
P = 1/ 0.25 - 1/0.4 = 1.5 D
For Berenice, u = 0.25 m, v = 0.3 m
P = 1/0.25 - 1/0.30 =0.7 D
For Edouard, u = 0.25 m, v = 0.80 m
P = 1/0.25 - 1/0.80 =2.75 D
For Francesca, u = 0.25 m, v = 0.50 m
P = 1/0.25 - 1/0.5 = 2.0 D
When The farsighted people are those with a near point greater than 25cm.
Berenice is = 0.67 D
Avishka is = 1.50 D
Francesca is = 2.00 D
Edouard is = 2.75 D
What is Hyperopic Vision?
When The farsighted people are those with a near point greater than 25 cm. Then, They include Avishka, Berenice, Edouard, and also Francisca.
Also, A converging lens is needed to correct farsightedness, or hyperopia, thus, When the focal length, f, is positive. Also, The image formed is virtual and also on the same side of the lens. hence the image distance is negative.
Also, From the lens formula, 1/f = 1/v 1/u; but v is negative
Thus, 1/f = 1/u - 1/v
But, when the power of a lens = 1/f in meters.
Thus, P = 1/u - 1/v
For Avishka, u = 25 cm or 0.25 m, v = 0.4 m
After that, P = 1/ 0.25 - 1/0.4 = 1.5 D
Then, For Berenice, u = 0.25 m, v = 0.3 m
Now, P = 1/0.25 - 1/0.30 =0.7 D
For Edouard, u = 0.25 m, v is = 0.80 m
Then, P = 1/0.25 - 1/0.80 is =2.75 D
Now, For Francesca, u = 0.25 m, v is = 0.50 m
Therefore, P = 1/0.25 - 1/0.5 = 2.0 D
Find more information about Hyperopic Vision here:
https://brainly.com/question/25676535
If this is the only water being used in your house, how fast is the water moving through your house's water supply line, which has a diameter of 0.021 m (about 3/4 of an inch)?
Answer:
0.273m/s
Explanation:
first find out the meaning of 0.90×10−4m3/s
literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s
1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?
cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²
so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²
they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s
The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.
1. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe?
2. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now:______.
a. the same as before.
b. lower than before.
c. higher than before.
3. If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound that can be produced in the pipe?
4. What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?
4-1. Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present.
A. Only the odd multiples of the fundamental frequency.
B. Only the even multiples of the fundamental frequency.
C. All integer multiples of the fundamental frequency.
E. What length of open-closed pipe would you need to achieve the same fundamental frequency as the open pipe discussed in Part A?
A. Half the length of the open-open pipe.
B. Twice the length of the open-open pipe.
C. One-fourth the length of the open-open pipe.
D. Four times the length of the open-open pipe.
E. The same as the length of the open-open pipe.
F. What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?
F-1. Recall that possible frequencies of standing waves that can be generated in an open-closed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the third
harmonic.
Answer:
1) f = 214 Hz , 2) answer is c , 3) f = 428 Hz , 4) f₂ = 428 Hz , f₃ = 643Hz
Explanation:
1) A tube with both ends open, the standing wave has a maximum amplitude and a node in its center, therefore
L = λ / 2
λ = 2L
λ = 2 0.8
λ = 1.6 m
wavelength and frequency are related to the speed of sound (v = 343 m / s)
v =λ f
f = v / λ
f = 343 / 1.6
f = 214 Hz
2) In this case the air comes out through the open hole, so we can assume that the length of the tube is reduced
λ' = 2 L ’
as L ’<L₀
λ' <λ₀
f = v / λ'
f' > fo
the correct answer is c
3) in this case the length is L = 0.40 m
λ = 2 0.4 = 0.8 m
f = 343 / 0.8
f = 428 Hz
4) the different harmonics are described by the expression
λ = 2L / n n = 1, 2, 3
λ₂ = L
f₂ = 343 / 0.8
f₂ = 428 Hz
λ₃ = 2 0.8 / 3
λ₃ = 0.533 m
f₃ = 343 / 0.533
f₃ = 643 Hz
4,1) as we have two maximums at the ends, all integer multiples are present
the answer is C
E) the length of an open pipe created that has a wavelength of lam = 1.6 m is requested
in this pipe there is a maximum in the open part and a node in the closed part, so the expression
L = λ / 4
L = 1.6 / 4
L = 0.4 m
the answer is C
F) in this type of pipe the general expression is
λ = 4L / n n = 1, 3, 5 (2n + 1)
therefore only odd values can produce standing waves
λ₃ = 4L / 3
λ₃ = 4 0.4 / 3
λ₃ = 0.533
f₃ = 343 / 0.533
f₃ = 643 Hz
A fisherman in a stream 39 cm deep looks downward into the water and sees a rock on the stream bed. How deep does the stream appear to the fisherman
Answer:
30cm
Explanation:
assume that the eyes are substantially above the water so that sin(theta) is approximately theta.
( small angle approximation).
The point at which a ray leaving the fish hits the surface of the water is x to the side of the centreline and the depth of the water is d
x/d = sin( angle of incidence)
if the apparent depth of the water is h then
x/h = sin( angle of refraction)
and applying snells law
1 sin ( theta air) = 1.33 sin( theta water)
1 * x/h = 1.33 * x/d
d/h = 1.33
or h/d = 1/1.33
h/39 = 1.33
h = 39 /1.33 so that is the apparent depth of the stream assuming:-
1. Your eyes are almost directly overhead
and
2. your eyes are a significant distance above the surface of the water.
x/d = 1.33 x/h
h/d =39/1.3
= 30cm
Design a voltage divider to provide the following approximate voltages with respect to ground using a 30 V source: 8.18 V, 14.7 V, and 24.6 V. The current drain on the source must be limited to no more than 1 mA. The number of resistors, their values, and their wattage ratings must be specified. A schematic showing the circuit arrangement and resistor placement must be provided
Answer:
R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω , R₃ = 1.72 10³ Ω , R₄ = 5.4 10³ Ω 1/8 W resistor
Explanation:
For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)
Therefore we build a circuit with 4 resistors in series, in such a way that
V = i R
let the voltage
1st resistance
V = i R
R₁ = V / i
R₁ = 14.7 / 1 10⁻³
R₁ = 14.7 10³ Ω
power is
P = V i
P = 14.7 1 10⁻³
P = 14.7 10⁻³ W = 0.0147 W
a resistance of ⅛ W is indicated
2nd resistance
R₂ = 8.18 / 1 10⁻³
R₂ = 8.18 10³ Ω
Power
P = 8.18 1 10⁻³
P = 0.00818W
a 1/8 W resistor
3rd resistance
this resistance is calculated in such a way that
V₁ + V₂ + V₃ = 24.6
V₃ = 24.6 - V₁ -V₂
V₃ = 24.6 - 14.7 - 8.18
V₃ = 1.72 V
R₃ = 1.72 / 1 10⁻³
R₃ = 1.72 10³ Ω
power
P = Vi
P = 1.72 10⁻³
P = 0.00172 W
a resistance of ⅛ W
To obtain the voltage of 24.6 we use this three resistors together
4th resistance
The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage
30 = V₁ + V₂ + V₃ + V₄
V₄ = 30 - V₁ -V₂ -V₃
V₄ = 30 -14.7 - 8.18 - 1.72
V₄ = 5.4 V
R₄ = 5.4 / 1 10⁻³
R₄ = 5.4 10³ Ω
Power
P = V i
P = 5.4 10⁻³
P = 0.0054 W
⅛ W resistance
The values of these resistance are commercially
Let's check the consumption of the circuit
R_total = R₁ + R₂ + R₃ + R₄
R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³
R_total = 30 10³
the current circulating in the circuit is
i = V / R_total
i = 30/30 10³
i = 1 10⁻³ A
therefore it is within the order requirement.
for connections see attached diagram
Heat and thermodynamics Numerical
Answer:
K = 227.04 W/m.°C
Explanation:
First we need to find the heat required to melt the ice:
q = m H
where,
q = heat required = ?
m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg
H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg
Therefore,
q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)
q = 2839 J
Now, we find the heat transfer rate through rod:
Q = q/t
where,
t = time = (10 min)(60 s/1 min) = 600 s
Q = Heat Transfer Rate = ?
Therefore,
Q = 2839 J/600 s
Q = 4.73 W
From Fourier's Law of Heat Conduction:
Q = KA ΔT/L
where,
K = Thermal Conductivity = ?
A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²
L = Length of rod = 60 cm = 0.6 m
ΔT = Difference in temperature = 100°C - 0°C = 100°C
Therefore,
4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m
K = (4.73 W)/(0.0208 m.°C)
K = 227.04 W/m.°C
A force of pounds makes an angle of with a second force. The resultant of the two forces makes an angle of to the first force. Find the magnitudes of the second force and of the resultant.
This question is incomplete, the complete question is;
A force of 193 pounds makes an angle of 79°14' with a second force. The resultant of the two forces makes an angle of 27°0' to the first force. Find the magnitudes of the second force and of the resultant.
Answer:
magnitudes of the second force (vector) is 110.84 lb
the resultant force force has a magnitude of 239.85 lb
Explanation:
Given that;
Magnitude of resultant vector R = ?
Direction of resultant vector α = 27°0'
Magnitude of vector p = 193
Magnitude of vector Q = ?
Angle between two vectors ∅ = 79°14'
Using the formula
tan∝ = [ Qsin∅ / P + Qcos∅]
tan27°0' = [ Qsin79°14' / 193 + Qcos79°14' )
we cross multiply
(193tan27°0') + (Qcos79°14'tan27°0' ) = Qsin79°14'
Q = 193tan27°0' / (sin79°14' - cos79°14'tan27°0')
Q = 110.84 lb
Therefore magnitudes of the second force (vector) is 110.84 lb
Now
R = √( p² + Q² + 2PQcos∅ )
R = √( 193² + 110.84² + ( 2 × 193 × 110.84 × cos79°14'))
R = 239.85 lb
Therefore the resultant force force has a magnitude of 239.85 lb
Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).
Image is missing, so i have attached it
Answer:
19.04 × 10⁻⁴ T in the +x direction
Explanation:
We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.
Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)
We are given;
I = 2.4 A
R = 5 cm = 0.05 m
μ₀ is a constant = 4π × 10⁻⁷ H/m
B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)
B = 9.6 × 10⁻⁴ T
To get the direction of the field from each wire, we will use Flemings right hand rule.
From the diagram attached:
We can say the field at P from the top wire will point up/right
Also, the field at P from the bottom wire will point down/right
Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.
If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.
Let the angle be θ
Thus;
sin(θ) = (1.3/2)/5
θ = sin⁻¹(0.13) = 7.47⁰
The x component of each field would be:
9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T
Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction
The magnitude of the magnetic field at the point P will be "9.6 × 10⁻⁴ T".
Magnetic fieldThe region of the environment close to something like a magnetic entity or a current-carrying body wherein this same magnetic forces caused by the body as well as a current might well be sensed.
According to the question,
Current, I = 2.4 A
Radius, R = 5 cm or,
= 0.05 m
Constant, μ₀ = 4π × 10⁻⁷ H/m
We know the relation,
The magnetic field, B = [tex]\frac{\mu_0 I}{2 \pi R}[/tex]
By substituting the values in the above relation, we get
= [tex]\frac{4 \pi\times 10^{-7}\times 2.4}{2 \pi\times 0.05}[/tex]
= 9.6 × 10⁻⁴ T
Thus the above answer is appropriate.
Find out more information about magnetic field here:
https://brainly.com/question/14411049
At the pizza party you and two friends decide to go to Mexico City from El Paso, TX where y'all live. You volunteer your car if everyone chips in for gas. Someone asks how much the gas will cost per person on a round trip. Your first step is to call your smarter brother to see if he'll figure it out for you. Naturally he's too busy to bother, but he does tell you that it is 2015 km to Mexico City, there's 11 cents to the peso, and gas costs 5.8 pesos per liter in Mexico. You know your car gets 21 miles to the gallon, but we still don't have a clue as to how much the trip is going to cost (in dollars) each person in gas ($/person).
Answer:
cost_cost = $ 96
Explanation:
In this exercise we have units in the groin system and the SI system, to avoid problems let's reduce everything to the SI system
performance = 21 miles / gallon (1,609 km / 1 mile) (1 gallon / 3,785 l)
perfomance= 8,927 km / l
now let's use a direct rule of proportions (rule of three). If a liter travels 8,927 km, how many liters are needed to travel the 2015 km
#_gasoline = 2015 km (1l / 8.927 km) = 225.72 liters
Now let's find the total cost of fuel. Ns indicates that $ 0.11 = 1 peso and the liter of fuel costs 5.8 pesos
cost_litre = 5.8 peso ($ 0.11 / 1 peso) = $ 0.638
cost_gasoline = #_gasoline cost_litro
cost_gasoline = 225.72 0.638
cost_gasoline = $ 144
This cost is for the one way trip, the total round trip cost is
cost_total = 2 cost_gasoline
cost_total = $ 288
Now let's look for the cost in the vehicle, you and two people will go, for which a total of 3 people will go, so the cost per person is
cost_person = total_cost / #_people
cost_person = 288/3
cost_cost = $ 96
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
3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing Photo 2.
Answer:
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
Explanation:
I cannot find any attached photo, but we can proceed anyways theoretically.
The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point
i.e
[tex]E=\frac{F}{Q}[/tex]
But the force F
[tex]F= \frac{kQ1Q2}{r^2}[/tex]
But the electric field intensity due to a point charge Q at a distance r meters away is given by
[tex]E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }[/tex]
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
A 0.2-stone is attached to a string and swung in a circle of radius 0.6 m on a horizontal and frictionless surface. If the stone makes 150 revolutions per minute, the tension force of the string on the stone is:
Answer:
2960 N
Explanation:
Convert rev/min to rad/s:
150 rev/min × (2π rad/rev) × (1 min / 60 s) = 50π rad/s
Sum of forces in the centripetal direction:
∑F = ma
T = m v² / r
T = m ω² r
T = (0.2 kg) (50π rad/s)² (0.6 m)
T = 2960 N
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?
Using the sample in above question how many counts per second would be observed when the detector is 10 meters away from the sample?
Answer:
At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.
Explanation:
The number of particles (N) counts are inversely proportional to the distance between the source and the detector.
By using the below formula we can find the number of counts.
[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]
The number of count per second, when the distance is 3 meters.
[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]
Number of count per second when the distance is 10 meters.
[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]
A generator consists of 1000 turns of wire around a coil of area 10cm2 (0.0010m 2 ). Calculate the peak output voltage of this generator when operating at 100 cycles/second in a 0.50 tesla field.
Answer:
The peak output voltage of this generator is 314.2 V.
Explanation:
Given;
the number of turns of the coil, N = 1000 turns
area of the coil, A = 0.001 m²
angular frequency of the coil, f = 100 cycles/seconds
magnitude of the magnetic field, B = 0.5 T
The peak output voltage of this generator is given by;
E = NBAω
Where;
ω is the angular velocity = 2πf
E = NBA(2πf)
E = 1000 x 0.5 x 0.001(2 x π x 100)
E = 314.2 V
Therefore, the peak output voltage of this generator is 314.2 V.
A certain resistor dissipates 0.5 W when connected to a 3 V potential difference. When connected to a 1 V potential difference, this resistor will dissipate:
Answer:
0.056 WExplanation:
[tex]Power = IV[/tex]
From ohms law we know that
[tex]V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}[/tex]
Given data
P1 = 0.5 Watt
P2 = ?
V1= 3 Volts
V2= 1 Volt
Thus we can solve for the power dissipated as follows
[tex]P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}[/tex]
[tex]\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W[/tex]
The resistor will dissipate 0.056 Watt
A hammer is used to hit a nail into a board. Which statement is correct about the forces at play between the nail and the hammer?
O The nail exerts a much smaller force on the hammer in the opposite direction
O The nail exerts a much smaller force on the hammer in the same direction.
The nail exerts an equal force on the hammer in the same direction.
O The nail exerts an equal force on the hammer in the opposite direction.
Answer:
reviewing the final statements, the correct one is the quarter
The nail exerts an equal force on the hammer in the opposite direction.
Explanation:
This is an action-reaction problem or Newton's third law, which states that forces in naturals occur in pairs.
This is the foregoing, the hammer exerts a force on the nail of magnitude F and it will direct downwards, if we call this action and the nail exerts a force on the hammer of equal magnitude but opposite direction bone directed upwards, each force is applied in one of the bodies.
The difference in result that each force is that the force between the nail exerts a very high pressure (relation between the force between the nail area), instead the area of the hammer is much greater, therefore the pressure is small.
When reviewing the final statements, the correct one is the quarter
The nail exerts an equal force on the hammer in the opposite direction.
Air bags greatly reduces the chance og injury in a car accident.explain how they do si in terms of energy transfer
Answer:
in an accident, when the body collides with the air bags, the collision time of impact between the two bodies will increase due to the presence of air bags in the car. Larger is the impact time smaller is the transformation of energy between the body and air bag. That is why air bags greatly reduce the chance of injury in a car accident.
What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=2.93 eV.
Answer:
The maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.
Explanation:
Given;
work function of silver, Φ = 2.93 eV = 2.93 x 1.602 x 10⁻¹⁹ J = 4.6939 x 10⁻¹⁹ J
Apply Einstein Photo electric effect;
E = K.E + Ф
Where;
E is the energy of the incident light
K.E is the kinetic of electron
Ф is the work function of silver surface
For the incident light to have maximum wavelength, the kinetic energy of the electron will be zero.
E = Ф
hf = Ф
[tex]h\frac{c}{\lambda} = \phi[/tex]
where;
c is speed of light = 3 x 10⁸ m/s
h is Planck's constant, = 6.626 x 10⁻³⁴ J/s
λ is the wavelength of the incident light
[tex]\lambda = \frac{hc}{\phi}\\\\\lambda =\frac{6.626*10^{-34} *3*10^8}{4.6939*10^{-19}} \\\\\lambda = 4.235 *10^{-7} \ m\\\\\lambda = 423.5 *10^{-9} \ m\\\\\lambda = 423.5 \ nm[/tex]
Therefore, the maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.
Approximately how many calories are in one gram of carbohydrates?
Answer:
4 caloriesExplanation:
Carbohydrates are known as energy giving foods and they are contained in our diets no matter how small. Other components of foods are protein, fat and oil, vitamin, water etc. A calorie is a unit of energy. It tells us the amount of energy the weight of any food contains. Knowing the amount of calories of food we are taking in helps us to monitor our weight. Each component of food have their own calories depending on the weight of the food component.
For carbohydrate as a component of food, 1gram of carbohydrates contains 4 calories. With this conversion, we can therefore calculate the calories of carbohydrate we are taking in by taking the weight of the food we want to eat.
Hence, approximately 4 calories are in one gram of carbohydrates
One gram of carbohydrates contains approximately 4 calories.
how many calories are in one gram of carbohydrates?The amount of energy in macronutrients, like carbohydrates, is measured in calories per gram. In terms of nutrition, a "calorie" is the amount of energy needed to warm up one gram of water by one degree Celsius.
Carbohydrates are one of the three main types of nutrients that our body needs, along with proteins and fats. They give the body fuel. When you eat carbohydrates, your body breaks them down into a type of sugar called glucose. Cells in your body use this glucose for energy.
Read more about carbohydrates here:
https://brainly.com/question/336775
#SPJ6
A stonecutter's chisel has an edge area of 0.7 cm2. If the chisel is struck with a force of 42 N, what is the pressure exerted on the stone
Answer:
The pressure is [tex]P = 583333 \ N/m^2[/tex]
Explanation:
From the question we are told that
The area of the edge is [tex]A = 0.72 cm^2 = 0.72 *10^{-4}\ m[/tex]
The force is [tex]F = 42 \ N[/tex]
The pressure is mathematically represented as
[tex]P = \frac{F}{A}[/tex]
substituting values
[tex]P = \frac{42}{0.72*10^{-4}}[/tex]
[tex]P = 583333 \ N/m^2[/tex]
One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resultant magnetic field at the point y
Complete question is;
One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?
Answer:
B_net = 50 × 10^(-7) T
Explanation:
We are told that the 30 A wire lies on the x-plane while the 40 A wire is perpendicular to the xy plane and passes through the point (0,4,0).
This means that the second wire is 4 m in length on the positive y-axis.
Now, we are told to find the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis.
This means that the position we want to find is half the length of the second wire.
Thus, at this point the net magnetic field is given by;
B_net = √[(B1)² + (B2)²]
Where B1 is the magnetic field due to the first wire and B2 is the magnetic field due to the second wire.
Now, formula for magnetic field due to very long wire is;
B = (μ_o•I)/(2πR)
Thus;
B1 = (μ_o•I_1)/(2πR_1)
Also, B2 = (μ_o•I_2)/(2πR_2)
Now, putting the equation of B1 and B2 into the B_net equation, we have;
B_net = √[((μ_o•I_1)/(2πR_1))² + ((μ_o•I_2)/(2πR_2))²]
Now, factorizing out some common terms, we have;
B_net = (μ_o/2π)√[((I_1)/R_1))² + ((I_2)/R_2))²]
Now,
μ_o is a constant and has a value of 4π × 10^(−7) H/m
I_1 = 30 A
I_2 = 40 A
Now, as earlier stated, the point we are looking for is 2 metres each from wire 2 end and wire 1.
Thus;
R_1 = 2 m
R_2 = 2 m
So, let's calculate B_net.
B_net = ((4π × 10^(−7))/2π)√[(30/2)² + (40/2)²]
B_net = 50 × 10^(-7) T
A small wave pulse and a large wave pulse approach each other on a string; the large pulse is moving to the right.
Sometime after the pulses have met and passed each other, which of the following statements is correct? (More than one answer may be correct)
- the large pulse continues moving to the right
- the large pulse continues unchanged, moving to the right
- the small pulse is reflected and moves off to the right with a smaller amplitude
- the small pulse is reflected and moves off to the right with its original amplitude
- the two pulses combine into a single pulse moving to the right
Answer:
the large amplitude wave keeps moving to the right
the small amplitude wave continues to move to the left.
When checking the answers, the correct ones are 1, 2
Explanation:
The waves fulfill the principle of superposition, which states that the value of the function at a point is the algebraic sum of the waves at a given instant.
The two waves in this exercise travel in the opposite direction, so when they are close, the resulting wave is the sum of the two waves, having a complicated shape. But when the waves follow their movement, they give in the same way as the initial a,
the large amplitude wave keeps moving to the right
the small amplitude wave continues to move to the left.
When checking the answers, the correct ones are 1, 2
What is the frequency of the fundamental mode of vibration of a steel piano wire stretched to a tension of 440 N? The wire is 0.630 m long and has a mass of 5.69 g.
Answer:
220.698HzExplanation:
The fundamental frequency f₀ is expressed as f₀ =V/2L where;
V is the speed of the string = [tex]\sqrt{\frac{T}{M} }[/tex]
m is the mass of the string
L is the length of the string
T is the tension in the string
f₀ = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]
Given datas
m = 5.69g = 0.00569 kg
T = 440N
L = 0.630 m
Required
Fundamental frequency of the steel piano wire f₀
[tex]f_0 = \frac{1}{2(0.630)}\sqrt{\frac{440}{0.00569} } \\ \\f_0 = \frac{1}{1.26}\sqrt{77,328.65 } \\\\f_0 = \frac{1}{1.26} * 278.08\\\\f_0 = 220.698Hz[/tex]
Hence the frequency of the fundamental mode of vibration of the steel piano wire stretched to a tension of 440N is 220.698Hz
