Answer:
With the equation:
sin T = l / a (*)
note that the width of the central diffraction maximum is inversely proportional to the width of the slit. If we increase the width size, a, the angle T at which the intensity first becomes zero decreases, resulting bin a narrower central band.
Explanation:
When you place an egg in water, it sinks. If you add salt to the water, after some time the egg floats. Choose the correct explanation
A. Adding salt to the water increases its density When the density of the water matches that of the egg, the egg becomes neutrally buoyant and floats
B. Adding salt to the water decreases Its density When the densily of the water matches that of the egg, the egg becomes neutrally buoyant and floats
C. Adding salt to the water decreases its volume When the volume of the water matches that of the egg, the egg becomes neutrally buoyant and floats
D. Adding salt to the water increases ts volume When the volume of the water matches that of the egg, the egg becomes neutraly buoyant and floats
Answer:
B
Explanation:
This happens because adding salt to the water decreases Its density. When the density of the water matches that of the egg, the egg becomes neutrally buoyant and floats.
The weight of the egg becomes equal to the upward buoyant force by the water on to the egg and hence, the egg floats.
Answer: A) Adding salt the to water increases its density. When the density of the water matches that of the egg, the egg becomes neutrally buoyant and floats.
Explanation:
The efficiency of a carnot cycle is 1/6. If on reducing the temperature of the sink 75 degree Celsius, the efficiency becomes 1/3, determine the initial and final temperature between which the cycle is working.
Answer:
375 and 450
Explanation:
The computation of the initial and the final temperature is shown below:
In condition 1:
The efficiency of a Carnot cycle is [tex]\frac{1}{6}[/tex]
So, the equation is
[tex]\frac{1}{6} = 1 - \frac{T_2}{T_1}[/tex]
For condition 2:
Now if the temperature is reduced by 75 degrees So, the efficiency is [tex]\frac{1}{3}[/tex]
Therefore the next equation is
[tex]\frac{1}{3} = 1 - \frac{T_2 - 75}{T_1}[/tex]
Now solve both the equations
solve equations (1) and (2)
[tex]2(1 - T_2/T_1) = 1 - (T_2 - 75)/T_1\\\\2 - 1 = 2T_2/T_1 - (T_2 - 75)/T_1\\\\ = (T_2 + 75)/T_1T_1 = T_2 + 75\\\Now\ we\ will\ Put\ the\ values\ into\ equation (1)\\\\1/6 = 1 - T_2/(T_2 + 75)\\\\1/6 = (75)/(T_2 + 75)[/tex]
T_2 + 450 = 75
T_2 = 375
Now put the T_2 value in any of the above equation
i.e
T_1 = T_2 + 75
T_1 = 375 + 75
= 450
The normal force equals the magnitude of the gravitational force as a roller coaster car crosses the top of a 58-m-diameter loop-the-loop. What is the car's speed at the top?
Answer:
16.87 m/s
Explanation:
To find the speed of the car at the top, when the normal force is equal the gravitational force, we just need to equate both forces:
[tex]N = P[/tex]
[tex]m*a_c = mg[/tex]
[tex]a_c[/tex] is the centripetal acceleration in the loop:
[tex]a_c = v^2/r[/tex]
So we have that:
[tex]mv^2/r = mg[/tex]
[tex]v^2/r = g[/tex]
[tex]v^2 = gr[/tex]
[tex]v = \sqrt{gr}[/tex]
So, using the gravity = 9.81 m/s^2 and the radius = 29 meters, we have:
[tex]v = \sqrt{9.81 * 29}[/tex]
[tex]v = \sqrt{284.49} = 16.87\ m/s[/tex]
The speed of the car is 16.87 m/s at the top.
A solid spherical conductor has a radius of 12 cm. The electric field at 24 from the center of the sphere has a magnitude of 640 N/C. What is the charge density (in C/m2) on the sphere
Answer:
Charge density on the sphere = 2.2 × 10⁻⁸ C/m²
Explanation:
Given:
Radius of sphere (r) = 12 cm = 0.12 m
Distance from the electric field R = 24 cm = 0.24 m
Magnitude (E) = 640 N/C
Find:
Charge density on the sphere
Computation:
Charge on the sphere (q) = (1/K)ER² (K = 9 × 10⁹)
Charge on the sphere (q) = [1/(9 × 10⁹)](640)(0.24)²
Charge on the sphere (q) = 4 × 10⁻⁹ C
Charge density on the sphere = q / [4πr²]
Charge density on the sphere = [4 × 10⁻⁹] / [4(3.14)(0.12)²]
Charge density on the sphere = [4 × 10⁻⁹] / [0.18]
Charge density on the sphere = 2.2 × 10⁻⁸ C/m²
Between which of the following surfaces with the force of friction be smallest
Answer:
Smother places........
A 300-foot cable weighing 5 pounds per foot is hanging from a winch 300 feet above ground level. Find the work (in ft-lb) done in winding up the cable when there is a 300-pound load attached to the end of the cable.
Answer:
315,000 ft·lb
Explanation:
At 300 ft and 5 lb/ft, the weight of the cable is (300 f)(5 lb/ft) = 1500 lb. The work done to raise it is equivalent to the work done to raise the cable's center of mass. Since the cable is of uniform density, its center of mass is half the cable length below the winch.
total work done = work to raise cable + work to raise load
= (1500 lb)(150 ft) +(300 lb)(300 ft) = 315,000 ft·lb
When one person was talking in a small room, the sound intensity level was 60 dB everywhere within the room. Then, there were 14 people talking in similar manner simultaneously in the room, what was the resulting sound intensity level?
A. 64 dB
B. 60 dB
C. 69 dB
D. 79 dB
E. 71 dB
Answer:
E= 71dB
Explanation:
See attached file for step by step calculation
A lunar module weighs 12 metric tons on the surface of the Earth. How much work is done in propelling the module from the surface of the moon to a height of 80 miles
Answer:
149.05J
Explanation:
Hello,
Data;
Weight = 12N
x = 80miles
Radius of the moon = 1079.4mi
W = mg
But gravity at moon = ⅙ the gravity on earth
12 = ⅙ × m
M = 2kg
Mass of the module = 2kg
K = F.x
F(x) = k / x²
2 = k / (1079.4)²
k = 2.33×10⁶
Work = ∫f.dx
work = ∫₁₀₇₉.₄¹¹⁵⁹ . (2.33×10⁶/x²).dx
Work = (-2.33×10⁶) ×(¹/₁₁₅₉ - ¹/₁₀₇₉)
work = 149.05J
An elevator filled with passengers has a mass of 1,700 kilograms and accelerates upward from rest at a rate of 1 meters/seconds 2 for 1.8 seconds. Calculate the tension in the cable (in Newtons) supporting the elevator during this time.
Answer:
The tension in the cable is 18371.9 newtons.
Explanation:
Physically speaking, the tension can be calculated with the help of the Second Newton's Law. The upward acceleration means that magnitude of tension must be greater than weight of elevator, whose equation of equilibrium is described below:
[tex]\Sigma F = T - m\cdot g = m \cdot a[/tex]
Where:
[tex]T[/tex] - Tension in the cable, measured in newtons.
[tex]m[/tex] - Mass of the elevator, measured in kilograms.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
[tex]a[/tex] - Net acceleration of the elevator, measured in meters in square second.
Now, tension is cleared and resultant expression is also simplified:
[tex]T = m \cdot (a + g)[/tex]
If [tex]m = 1700\,kg[/tex], [tex]a = 1\,\frac{m}{s^{2}}[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the tension in the cable is:
[tex]T = (1700\,kg)\cdot \left(1\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]T = 18371.9\,N[/tex]
The tension in the cable is 18371.9 newtons.
Two cars start moving from the same point. One travels south at 28 mi/h and the other travels west at 21 mi/h. At what rate is the distance between the cars increasin
Complete question:
Two cars start moving from the same point. One travels south at 28 mi/h and the other travels west at 21 mi/h. At what rate is the distance between the cars increasing four hours later.
Answer:
The rate at which the distance between the cars is increasing four hours later is 35 mi/h.
Explanation:
Given;
speed of one car, dx/dt = 28 mi/h South
speed of the second car, dy/dt = 21 mi/h West
The distance between the cars is the line joining west to south, which forms a right angled triangle with the two positions.
Apply Pythagoras theorem to evaluate this distance;
let the distance between the cars = z
x² + y² = z² -------- equation (1)
Differentiate with respect to time (t)
[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}[/tex] ----- equation (2)
Since the speed of the cars is constant, after 4 hours their different distance will be;
x: 28(4) = 112 mi
y: 21(4) = 84 mi
[tex]z = \sqrt{x^2 + y^2} \\\\z = \sqrt{112^2 + 84^2} \\\\z = 140 \ mi[/tex]
Substitute in the value of x, y, z, dx/dt, dy /dt into equation (2) and solve for dz/dt
[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \\\\2(112)(28) + 2(84)(21) = 2(140)\frac{dz}{dt} \\\\9800 = 280\frac{dz}{dt} \\\\\frac{dz}{dt} = \frac{9800}{280} \\\\\frac{dz}{dt} = 35 \ mi/h[/tex]
Therefore, the rate at which the distance between the cars is increasing four hours later is 35 mi/h
1. How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. the line density of the grating were doubled? Refer to the diffraction grating equations in your answer.
Answer:
a) The distance between the ineas doubles, intensity decreases with distance
b) The distance between the ineas doubles
Explanation:
The diffraction pattern of a grid is given as a percentage
d sin θ = m λ
where d give the distance between two consecutive lines, θ it is at an angle, λ the wavelength and m is an integer that determines the order of diffraction, let's not forget that the entire spectrum is at a value of m and then it is repeated.
Let's apply this to our case
a) distance from grid to observation screen doubles
Here we have two effect:
* the energy of the source is constant, it must be distributed over a surface, therefore the intensity decreases with distance
* The other factor can be found using trigonometry
tan θ= y / L
where y is the distance from the central maximum to the line under study and L is the distance to the screen
In general, diffraction experiments cover very small angles
tan θ = sin θ/ cos θ = sin θ
we substitute
sinθ = y / L
we subtitle into the diffraction equation
d y / L = m λ
y = L / d m λ
L = 2 L₀
y = 2 L₀ m λ / d
we see that by doubling the distance to the screen the lines we are seeing are separated by double
b) When the density of lines doubles, it means that in the same distance I have twice as many lines, therefore the distance between two consecutive lines is reduced by half
d = d₀o / 2
y = (L m λ) / d
y = (L m λ/ d₀) 2
we see that The distance between the ineas doubles
If you place a 50 Volt power source across a 10 Ohm resistor, what will the current
be?
Answer:
use voltage=current*resistance
v=i*r
50=i*10
50/10=i
i=5
hope it's clear
Given the following data:
Voltage = 50 Volt
Resistance = 10 Ohm
To determine the current flowing through the circuit, we would apply Ohm's law:
Mathematically, Ohm's law is given by the formula;
[tex]V = IR[/tex]
Where;
V is voltage measured in voltage.
I is current measured in amperes.
R is resistance measured in ohms.
Making I the subject of formula, we have:
[tex]I = \frac{V}{R}[/tex]
Substituting the given parameters into the formula, we have;
[tex]I = \frac{50}{10}[/tex]
Current, I = 5 Ampere.
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how does current change under different polarity?
Answer:
Due to flipping of polarity
Explanation:
During the changing of polarity, the current on the one side is maximum as the polarity change then the current is gradually reducing toget from another end.
You push a hockey puck that is initially at rest on slick ice by applying a constant force until the puck reaches a final velocity of 1 m/s. On the second attempt, you want the hockey puck to reach the same final velocity by applying a force that is twice as large.
1. Therefore, you must exert the force for a time interval that is
A. shorter than the time interval of your first attempt.
B. longer than the time interval of your first attempt.
C. the same as the time interval of your first attempt.
2. After the hockey puck has reached the final velocity, you suddenly stop pushing it. The hockey puck:
A. stops abruptly
B. reduces speed gradually
C. continues at constant velocity
D. increases speed gradually
E. reduces speed abruptly
Answer:
1. A
2. B or C
Explanation:
1.
F=ma, meaning that if you use two times more force on a constant mass, the acceleration must double. Acceleration is change in velocity, which means that if you are aiming for the same final velocity the change must happen in half of the time. Therefore, the correct answer is choice A.
2.
By Newton's first law, an object in motion will stay in motion unless an external force acts on it. Since there is nothing pushing the puck in the other direction, the puck will either keep on going for at a constant velocity or will reduce its speed gradually, depending on whether or not this ice is considered to be frictionless. Hope this helps!
(1) You must exert the force for a time interval that is shorter than the time interval of your first attempt.
(2) The hockey puck reduces speed abruptly.
According to Newton's second law of motion; the force applied to an object is directly proportional to the mass and acceleration of the object.
F = ma
[tex]F = \frac{mv}{t}[/tex]
The force applied to an object is directly proportional to the velocity of the object and inversely proportional to the time of motion of the object.To double the force, you must halve the time interval.Thus, to apply a force that is twice as large as the first while maintaining the same velocity, you must exert the force for a time interval that is shorter than the time interval of your first attempt.
(2) The force applied to an object is direct directly proportional to the velocity of the object. Once you stop applying force to the hockey puck, it moves for a short with initial momentum gained before it will stop.
Thus, the magnitude of the velocity (speed) will drop sharply once the force on the object is removed.
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a proton moves in a circle of radius 0.4 when it enters a region with a magnetic field of 1.0t which points into the plane the speed of the proton is
Answer:
4 x 10⁷m/s
Explanation:
When a charged particle moves in a curved path in a magnetic field, it experiences some magnetic force, and in the absence of any other force, which supplies the centripetal force needed to keep the particle in balance.
Let the magnetic force be [tex]F_{M}[/tex]
Let the centripetal force be [tex]F_{C}[/tex]
=> [tex]F_{M}[/tex] = [tex]F_{C}[/tex] --------------(i)
We know that;
[tex]F_{M}[/tex] = qvBsinθ
Where;
q = charge on the particle
v = speed of the particle
B = magnetic field intensity
θ = angle between the speed and magnetic field vectors
Also;
[tex]F_{C}[/tex] = [tex]\frac{mv^2}{r}[/tex]
Where;
m = mass of the particle
v = velocity/speed of the particle
r = radius of the circular path of motion.
From equation (i)
qvBsinθ = [tex]\frac{mv^2}{r}[/tex] [divide both sides by v]
qBsinθ = [tex]\frac{mv}{r}[/tex] [make v subject of the formula]
v = qrBsinθ / m --------------------(ii)
From the question;
B = 1.0T
r = 0.4m
θ = 90° [since magnetic field is always perpendicular to velocity]
q = 1.6 x 10⁻¹⁹C [charge of a proton]
m = 1.6 x 10⁻²⁷kg [mass of a proton]
Substitute these values into equation(ii) as follows;
v = (1.6 x 10⁻¹⁹ x 0.4 x 1.0 x sin90°) / (1.6 x 10⁻²⁷)
v = 4 x 10⁷ m/s
Therefore the speed of the proton is 4 x 10⁷m/s
What is Ohm's Law, and how does it work in real life.
Explanation:
Ohms law states that the electrical current present in a metallic conductor is directly proportional to the potential difference between the metallic conductor and inversely proportional to the resistance therefore if the voltage is increased resistance also increases provided that temperature and other physical properties remains constant V=IR
Three solid, uniform, cylindrical flywheels, each of mass 65.0 kg and radius 1.47 m, rotate independently around a common axis through their centers. Two of the flywheels rotate in one direction at 8.94 rad/s, but the other one rotates in the opposite direction at 3.42 rad/s.
Required:
Calculate the magnitude of the net angular momentum of the system.
Answer:
the angular momentum is 1015.52 kg m²/s
Explanation:
given data
mass of each flywheel, m = 65 kg
radius of flywheel, r = 1.47 m
ω1 = 8.94 rad/s
ω2 = - 3.42 rad/s
to find out
magnitude of the net angular momentum
solution
we get here Moment of inertia that is express as
I = 0.5 m r² .................1
put here value and we get
I = 0.5 × 65 × 1.47 × 1.47
I = 70.23 kg m²
and
now we get here Angular momentum that is express as
L = I × ω ...........................2
and Net angular momentum will be
L = 2 × I x ω1 - I × ω2
put here value and we get
L = 2 × 70.23 × 8.94 - 70.23 × 3.42
L = 1015.52 kg m²/s
so
the angular momentum is 1015.52 kg m²/s
The magnitude of the net angular momentum of the system will be "1015.52 kg.m²/s".
MomentumAccording to the question,
Flywheel's mass, m = 65 kg
Flywheel's radius, r = 1.47 m
ω₁ = 8.94 rad/s
ω₂ = 3.42 rad/s
We know,
The moment of inertia (I),
= 0.5 m r²
By substituting the values,
= 0.5 × 65 × 1.47 × 1.47
= 70.23 kg.m²
hence, The angular momentum be:
→ L = I × ω or,
= 2 × I × ω₁ - l × ω₂
= 2 × 70.23 × 8.94 - 70.23 × 3.42
= 1015.52 kg.m²/s
Thus the above answer is correct.
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a certain volume of dry air at NTP is allowed to expand five times of it original volume under adiabatic condition.calculate the final pressure.(air=1.4)
Answer:
Final pressure 0.105atm
Explanation:
Let V1 represent the initial volume of dry air at NTP.
under adiabatic condition: no heat is lost or gained by the system. This does not implies that the constant temperature throughout the system , but rather that no heat gained or loss by the system.
Adiabatic expansion:
[tex]\frac{T_1}{T_2} =(\frac{V_1}{V_2} )^{\gamma -1}[/tex]
273/T2=(5V1/V1)^(1.4−1)
273/T2=5^0.4
Final temperature T2=143.41 K
Also
P1/P2=(V2/V1)^γ
1/P2=(5V1/V1)^1.4
Final pressure P2=0.105atm
If there is a potential difference V between the metal and the detector, what is the minimum energy Emin that an electron must have so that it will reach the detector
Answer:
Emin= eV
Explanation:
This minimum amount of energy is the work function and so for an electron to reach the detector the energy must b more than work function and the stopping potential so
Emin=eV
What is surface tension??
Answer:
Surface tension is the tendency of liquid surfaces to shrink into the minimum surface area possible. Surface tension allows insects (e.g. water striders), usually denser than water, to float and slide on a water surface.
Explanation:
Answer:
It is the tension of the surface film of a liquid caused by the attraction of the particles in the surface layer by the bulk of the liquid, which tends to minimize surface area.
A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation between the plates is increased, what will happen to the charge on the capacitor and the electric potential across it
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. [tex]Q[/tex], the amount of charge stored in this capacitor, will stay the same.
The formula [tex]\displaystyle Q = C\, V[/tex] relates the electric potential across a capacitor to:
[tex]Q[/tex], the charge stored in the capacitor, and[tex]C[/tex], the capacitance of this capacitor.While [tex]Q[/tex] stays the same, moving the two plates apart could affect the potential [tex]V[/tex] by changing the capacitance [tex]C[/tex] of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
[tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex],
where
[tex]\epsilon[/tex] is the permittivity of the material between the two plates.[tex]A[/tex] is the area of each of the two plates.[tex]d[/tex] is the distance between the two plates.Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of [tex]\epsilon[/tex]. Neither will that change the area of the two plates.
However, as [tex]d[/tex] (the distance between the two plates) increases, the value of [tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex] will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula [tex]\displaystyle Q = C\, V[/tex] can be rewritten as:
[tex]V = \displaystyle \frac{Q}{C}[/tex].
The value of [tex]Q[/tex] (charge stored in this capacitor) stays the same. As the value of [tex]C[/tex] becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
Consider a wave exhibiting simple harmonic oscillation with a period of 6 s. If 4 s after the wave begins it exhibits a displacement of 0.5 m, what is the amplitude of the wave
Answer:
A = 1.0m
Explanation:
In order to calculate the amplitude of the wave you take into account the general formula for a simple harmonic motion, which is given by:
[tex]x=Acos(\omega t)[/tex] (1)
A: amplitude
x: position = 0.5m
t: time = 4s
w: angular frequency of the wave
You calculate the angular frequency of the wave by using the information about the period of the wave:
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{6s}=1.047\fra{rad}{s}[/tex]
Next, you solve the equation (1) for A, and replace the values of the other parameters:
[tex]A=\frac{x}{cos(\omega t)}=\frac{0.5m}{cos(1.047rad/s(4s))}=-1[/tex]
You take the absolute value of the amplitude.
The amplitude of the wave is 1.0 m
Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of the blocks, with a mass of 1.0 kg accelerates downward at 34g. What is the mass of the other box?
Answer:
1/7 kgExplanation:
Find the diagram attached for better understanding of the question.
Given the mass of one of the blocks to be 1.0kg and accelerates downward at 3/4g.
g = acceleration due to gravity.
Let the block accelerating downward be M, mass of the other body be 'm' and the acceleration of the body M be 'a'.
M = 1.0 kg and a = 3.4g
According to newton's second law; [tex]\sum fy = ma_y[/tex]
For body of mass m;
T - mg = ma ... (1)
For body of mass M;
Mg - T = Ma ... (2)
Adding equation 1 ad 2;
+Mg -mg = ma + Ma
Ma-Mg = -mg-ma
M(a-g) = -m(a+g)
Substituting M = 1.0 kg and a = 3/4g into the resulting equation;
3/4 g-g = -m(3/4 g+g)
3/4 g-g = -m(7/4 g)
-g/4 = -m(7/4 g)
1/4 = 7m/4
28m = 4
m = 1/7 kg
Therefore the mass of the other box is 1/7 kg
An experiment is set up to test the angular resolution of an optical device when red light (wavelength ????r ) shines on an aperture of diameter D . Which aperture diameter gives the best resolution? D=(1/2)????r D=????r D=2????r
Explanation:
As per Rayleigh criterion, the angular resolution is given as follows:
[tex]\theta=\frac{1.22 \lambda}{D}[/tex]
From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.
Fermat's principle is of least time rather than of least distance. Would least distance apply for reflection? For refraction? Why are your answers different?
Answer:
Fermat's principle states that the path taken by a ray between two given points is the path that can be traversed in the least time.
Thus this least distance being the same as least time will only apply to reflection alone because in reflection light travels In the same direction so least distance will also mean least time. But for refraction light travels in different mediums at different speeds so least distance and least time paths will definately not be the same
Rope BCA passes through a pulley at point C and supports a crate at point A. Rope segment CD supports the pulley and is attached to an eye anchor embedded in a wall. Rope segment BC creates an angle of ϕ = 51.0 ∘ with the floor and rope segment CD creates an angle θ with the horizontal. If both ropes BCA and CD can support a maximum tensile force Tmax = 120 lb , what is the maximum weight Wmax of the crate that the system can support? What is the
Answer:
Wmax = 63.65 ≈ 64 lb
Explanation:
A solid wooden door, 90 cm wide by 2.0 m tall, has a mass of 35 kg. It is open and at rest. A small 500-g ball is thrown perpendicular to the door with a speed of 20 m/s and hits the door 60 cm from the hinged side, causing it to begin turning. The ball rebounds along the same line with a speed of 16.0 m/s relative to the ground.
Required:
How much energy is lost during this collision?
a. 15J
b. 16J
c. 13J
d. 4.8J
e. 30J
Answer:
the kinetic energy lost in the collison is a) 30 J
Explanation:
given data
mass of door m1 = 35 kg
width a = 90 cm = 0.9 m
the mass of ball m2 = 500 g = 0.5 kg
initial speed of ball u = 20 m/s
final speed of ball v = 16 m/s
r = 60 cm = 0.6 m
soluion
we will consider here final angular speed of the door = w
so now we use conservation of angular momentum that is
Li = Lf ........................1
that is express as
m2 × u × r = I × w + m2 × v × r
put here value and we get
0.5 × 20 × 0.6 = [tex](m1 \times \frac{a^2}{12})[/tex] × w + 0.5 × 16 × 0.6
solve it we get
w = 0.508 rad/s
so that here
the kinetic energy lost in the collison,
KE = KE initial - KE final ..................2
put here value
KE = 0.5 × m2 × u² - (0.5 × I × w² + 0.5 × m2 × v²)
KE = 0.5 × (0.5 × 20² - (35 × 0.9² ÷ 12) × 0.508² - 0.5 × 16²) J
KE = 30 J
the kinetic energy lost in the collison is a) 30 J
The sound level of one person talking at a certain distance from you is 61 dB. If she is joined by 5 more friends, and all of them are talking at the same time as loudly as she is, what sound level are you being exposed to?
Answer:
Explanation:
For sound level in decibel scale the relation is
dB = 10 log I / I₀ where I₀ = 10⁻¹² and I is intensity of sound whose decibel scale is to be calculated .
Putting the given values
61 = 10 log I / 10⁻¹²
log I / 10⁻¹² = 6.1
I = 10⁻¹² x 10⁶°¹
[tex]=10^{-5.9}[/tex]
intensity of sound of 5 persons
[tex]I=5\times 10^{-5.9}[/tex]
[tex]dB=10log\frac{5 X 10^{-5.9}}{10^{-12}}[/tex]
= 10log 5 x 10⁶°¹
= 10( 6.1 + log 5 )
= 67.98
sound level will be 67.98 dB .
The first Leyden jar was probably discovered by a German clerk named E. Georg von Kleist. Because von Kleist was not a scientist and did not keep good records, the credit for the discovery of the Leyden jar usually goes to physicist Pieter Musschenbroek from Leyden, Holland. Musschenbroek accidentally discovered the Leyden jar when he tried to charge a jar of water and shocked himself by touching the wire on the inside of the jar while holding the jar on the outside. He said that the shock was no ordinary shock and his body shook violently as though he had been hit by lightning. The energy from the jar that passed through his body was probably around 1 J, and his jar probably had a capacitance of about 1 nF.A) Estimate the charge that passed through Musschenbroek's body.
B) What was the potential difference between the inside and outside of the Leyden jar before Musschenbroek discharged it?
Answer:
a) q = 4.47 10⁻⁵ C
b) ΔV = 4.47 10⁴ V
Explanation:
A Leyden bottle works as a condenser that accumulates electrical charge, so we can use the formula of the energy stored in a capacitor
U = Q² / 2C
Q = √ (2UC)
let's reduce the magnitudes to the SI system
c = 1 nF = 1 10⁻⁹ F
let's calculate
q = √ (2 1 10⁻⁹-9)
q = 0.447 10⁻⁴ C
q = 4.47 10⁻⁵ C
b) for the potential difference we use
C = Q / ΔV
ΔV = Q / C
ΔV = 4.47 10⁻⁵ / 1 10⁻⁹
ΔV = 4.47 10⁴ V
a wave with a high amplitude______?
. . . is carrying more energy than a wave in the same medium with a lower amplitude.