Answer:
I can help
Explanation:
If 2 balls had the same volume but ball a has twice as much mass as babil which one will have the greater density
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
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
Four point charges have the same magnitude of 2.4×10^−12C and are fixed to the corners of a square that is 4.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.
Answer:
7.2N/C
Explanation:
Pls see attached file
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
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:
The air is partially removed at the upper part of the tube and the valve is closed. The height of the water above its pool surface is 0.7 m.The height of the liquid above its pool surface is 0.33 m.The difference in the heights of the pool surfaces is 0.22 m.
Required:
Find the density of the unknown liquid.
Answer:
2121.2kg/m^3 is the density of the test liquid on the left
Explanation:
See attached file
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
A ranger needs to capture a monkey hanging on a tree branch. The ranger aims his dart gun directly at the monkey and fires the tranquilizer dart. However, the monkey lets go of the branch at exactly the same time as the ranger fires the dart. Will the monkey get hit or will it avoid the dart?
Answer:
Yes the monkey will get hit and it will not avoid the dart.
Explanation:
Yes, the monkey will be hit anyway because the dart will follow a hyperbolic path and and will thus fall below the branches, so if the monkey jumps it will be hit.
No, the monkey will not avoid the dart because dart velocity doesn't matter. The speed of the bullet doesn’t even matter in this case because a faster bullet will hit the monkey at a higher height and while a slower bullet will simply hit the monkey closer to the ground.
A woman living in a third-story apartment is moving out. Rather than carrying everything down the stairs, she decides to pack her belongings into crates, attach a frictionless pulley to her balcony railing, and lower the crates by rope.
Required:
How hard must she pull on the horizontal end of the rope to lower a 49 kg crate at steady speed?
Answer:
T = 480.2N
Explanation:
In order to find the required force, you take into account that the sum of forces must be equal to zero if the object has a constant speed.
The forces on the boxes are:
[tex]T-Mg=0[/tex] (1)
T: tension of the rope
M: mass of the boxes 0= 49kg
g: gravitational acceleration = 9.8m/s^2
The pulley is frictionless, then, you can assume that the tension of the rope T, is equal to the force that the woman makes.
By using the equation (1) you obtain:
[tex]T=Mg=(49kg)(9.8m/s^2)=480.2N[/tex]
The woman needs to pull the rope at 480.2N
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.
Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. He tells you that the force is always pointed away a definite point in space, which we can call the force center. The magnitude of the force turns out to be proportional to B/r3, where r is the distance from the force center to any other point. Your friend says that it has been determined that the constant of proportionality has been determined to be B= 2 (in units to be determined later), so that the magnitude of the force on a particle (in newtons) can be written as 2r^3, when the particle is at a distance r from the force center.
Required:
Write an expression of potential energy.
Answer:
U = 1 / r²
Explanation:
In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related
F = - dU / dr
this derivative is a gradient, that is, a directional derivative, so we must have
dU = - F. dr
the esxresion for strength is
F = B / r³
let's replace
∫ dU = - ∫ B / r³ dr
in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product
let's evaluate the integrals
U - Uo = -B (- / 2r² + 1 / 2r₀²)
To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)
U = B / 2r²
we substitute the value of B = 2
U = 1 / r²
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.
An electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen and directed toward the right. What is the direction of the magnetic force on the electron?
Answer: the direction of the magnetic force on the electron will be moving out of the screen, perpendicular to the magnetic field.
Explanation:
The magnetic force F on a moving electron at right angle to a magnetic field is given by the formula:
F = BqVSinØ
If an electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen and directed toward the right. Then, the direction of the magnetic force on the electron will be perpendicular to the magnetic field
According to the Fleming's left - hand rule, the direction of the magnetic force on the electron will be moving out of the plane of the screen.
Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of 38.0° with the normal to the surface, while in the glass it makes an angle of 26.0° with the normal. What is the index of refraction of the glass?
Answer:
n_glass = 1.404
Explanation:
In order to calculate the index of refraction of the light you use the Snell's law, which is given by the following formula:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex] (1)
n1: index of refraction of vacuum = 1.00
θ1: angle of the incident light respect to normal of the surface = 38.0°
n2: index of refraction of glass = ?
θ2: angle of the refracted light in the glass respect to normal = 26.0°
You solve the equation (1) for n2 and replace the values of all parameters:
[tex]n_2=n_1\frac{sin\theta_1}{sin\theta_2}=(1.00)\frac{sin(38.0\°)}{sin(26.0\°)}\\\\n_2=1.404[/tex]
The index of refraction of the glass is 1.404
What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 18.0 m/s in the same direction
What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 18.0 m/s in the same direction? Assume the deer remains on the car.
Answer:28.29m/s
Explanation:
In this situation, linear momentum is conserved. And since the deer remains on the car after collision, the linear momentum is given as;
([tex]m_{C}[/tex] x [tex]u_{C}[/tex]) + ([tex]m_{D}[/tex] x [tex]u_{D}[/tex]) = ([tex]m_{C}[/tex] + [tex]m_{D}[/tex]) v -----------------(i)
Where;
[tex]m_{C}[/tex] = mass of car
[tex]u_{C}[/tex] = initial velocity of car before collision
[tex]m_{D}[/tex] = mass of deer
[tex]u_{D}[/tex] = initial velocity of the deer before collision
v = common velocity with which the car and the deer move after collision
From the question;
[tex]m_{C}[/tex] = 900kg
[tex]u_{C}[/tex] = +30.0m/s (direction of the motion of the car taken positive)
[tex]m_{D}[/tex] = 150kg
[tex]u_{D}[/tex] = +18.0m/s (relative to the direction of the car, the velocity of the deer is also positive )
Substitute these values into equation (i) as follows;
(900 x 30.0) + (150 x 18.0) = (900 + 150)v
27000 + 2700 = 1050v
29700 = 1050v
v = [tex]\frac{29700}{1050}[/tex]
v = 28.29m/s
Therefore, the velocity of the car after hitting the deer is 28.29m/s. This is also the velocity of the deer after being hit by the car.
a wave with a high amplitude______?
. . . is carrying more energy than a wave in the same medium with a lower amplitude.
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.
Find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level. Take the zero of potential energy at sea level.
Answer:
P = 1470980 J
Explanation:
We have,
Mass of the hiker is 79 kg
It is required to find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level.
It is given by :
[tex]P=mgh\\\\P=79\times 9.8\times 1900\\\\P=1470980\ J[/tex]
So, the potential energy of 1470980 J is associated with a hiker.
A current carrying wire is oriented along the y axis It passes through a region 0.45 m long in which there is a magnetic field of 6.1 T in the z direction The wire experiences a force of 15.1 N in the x direction.1. What is the magnitude of the conventional current inthe wire?I = A2. What is the direction of the conventional current in thewire?-y+y
Answer:
The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.
Explanation:
- To find the direction of the conventional current in the wire you use the following formula:
[tex]\vec{F}=i\vec{l}\ X\ \vec{B}[/tex] (1)
i: current in the wire = ?
F: magnitude of the magnetic force on the wire = 15.1N
B: magnitude of the magnetic field = 6.1T
l: length of the wire that is affected by the magnetic field = 0.45m
The direction of the magnetic force is in the x direction (+^i) and the direction of the magnetic field is in the +z direction (+^k).
The direction of the current must be in the +y direction (+^j). In fact, you have:
^j X ^k = ^i
The current and the magnetic field are perpendicular between them, then, you solve for i in the equation (1):
[tex]F=ilBsin90\°\\\\i=\frac{F}{lB}=\frac{15.1N}{(0.45m)(6.1T)}=5.5A[/tex]
The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.
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.
Find out more information about momentum here:
https://brainly.com/question/25121535
Find the pressure difference (in kPa) on an airplane wing if air flows over the upper surface with a speed of 125 m/s, and along the bottom surface with a speed of 109 m/s. [Express answer in TWO decimal places]
Answer:
P= 2414.9 Pa
Explanation:
given
density of air , p = 1.29 kg/m³
speed of air over the upper surface , v₁ = 125 m/s
speed of air over the lower surface , v₂ = 109 m/s
the pressure difference on an airplane wing , P = 0.5 × p × ( v₁² - v₂²)
P = 0.5 × 1.29 × ( 125² - 109²)
P= 0.645(3744)
P = 2414.9 Pa
the pressure difference on an airplane wing is 2414.9 Pa
According to the model in which active galactic nuclei are powered by supermassive black holes, the high luminosity of an active galactic nucleus primarily consists of
Answer:
the high luminosity of an active galactic nucleus primarily consists of light emitted by hot gas in an accretion disk that swirls around the black hole
Wanda exerts a constant tension force of 12 N on an essentially massless string to keep a tennis ball (m = 60 g) attached to the end of the string traveling in uniform circular motion above her head at a constant speed of 9.0 m/s. What is the length of the string between her hand and the tennis ball? You may ignore gravity in this problem (assume the motion of the tennis ball and string happen in a purely horizontal plane). A. 41 m B. 0.24 m C. 3.2 cm D. 0.41 m
Answer:
r = 0.405m = 40.5cm
Explanation:
In order to calculate the length of the string between Wanda and the ball, you take into account that the tension force is equal to the centripetal force over the ball. So, you can use the following formula:
[tex]F_c=ma_c=m\frac{v^2}{r}[/tex] (1)
Fc: centripetal acceleration (tension force on the string) = 12N
m: mass of the ball = 60g = 0.06kg
r: length of the string = ?
v: linear speed of the ball = 9.0m/s
You solve for r in the equation (1) and replace the values of the other parameters:
[tex]r=\frac{mv^2}{F_c}=\frac{(0.06kg)(9.0m/s)^2}{12N}=0.405m[/tex]
The length of the string between Wanda and the ball is 0.405m = 40.5cm
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 position of a particle is r(t)= (4.0t'i+ 2.4j- 5.6tk) m. (Express your answers in vector form.) (a) Determine its velocity (in m/s) and acceleration (in m/s2) as functions of time. (Use the following as necessary: t. Assume t is seconds, r is in meters, and v is in m/s, Do not include units in your answer.) v(t)= ________m/s a(t)= ________m/s2 (b) What are its velocity (in m/s) and acceleration (in m/s2) at time t 0? v(0) =_______ m/s a(0)=_______ m/s2
The position of a particle is r(t)= (4.0t²i+ 2.4j- 5.6tk) m. (Express your answers in vector form.)
(a) Determine its velocity (in m/s) and acceleration (in m/s²) as functions of time. (Use the following as necessary: t. Assume t is seconds, r is in meters, and v is in m/s, Do not include units in your answer.)
v(t)= ________m/s
a(t)= ________m/s²
(b) What are its velocity (in m/s) and acceleration (in m/s²) at time t 0?
v(0) =_______ m/s
a(0)=_______ m/s²
Answer:
(a)
v(t)= [tex]8ti - 5.6k[/tex] m/s
a(t)= 8i m/s²
(b)
v(0) = -5.6k m/s
a(0)= 8i m/s²
Explanation:From the question, the position of the particle is given by;
r(t)= (4.0t²i+ 2.4j- 5.6tk) -----------------(i)
(a)
(i)To get the velocity, v(t), of the particle, we'll take the first derivative of the position of the particle (given by equation (i)) with respect to time, t, as follows;
v(t) = [tex]\frac{dr(t)}{dt}[/tex] = [tex]\frac{d(4.0t^2i + 2.4j - 5.6tk)}{dt}[/tex]
v(t) = [tex]\frac{dr(t)}{dt}[/tex] = [tex]8ti +0j - 5.6k[/tex]
v(t) = [tex]8ti - 5.6k[/tex] --------------------(ii)
(ii) To get the acceleration, a(t), of the particle, we'll take the first derivative of the velocity of the particle (given by equation (ii)) with respect to time, t, as follows;
a(t) = [tex]\frac{dv(t)}{dt}[/tex] = [tex]\frac{d(8ti - 5.6k)}{dt}[/tex]
a(t) = 8i --------------------(iii)
(b)
(i) To get the velocity of the particle at time t = 0, substitute the value of t = 0 into equation (ii) as follows;
v(t) = [tex]8ti - 5.6k[/tex]
v(0) = 8(0)i - 5.6k
v(0) = 0 - 5.6k
v(0) = -5.6k
(ii) To get the acceleration of the particle at time t = 0, substitute the value of t = 0 into equation (iii) as follows;
a(t) = 8i
a(0) = 8i
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
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
If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will Group of answer choices
Answer:
P' = 4 P
Therefore, the power dissipated by the circuit will becomes four times of its initial value.
Explanation:
The power dissipation by an electrical circuit is given by the following formula:
Power Dissipation = (Voltage)(Current)
P = VI
but, from Ohm's Law, we know that:
Voltage = (Current)(Resistance)
V = IR
Substituting this in formula of power:
P = (IR)(I)
P = I²R ---------------- equation 1
Now, if we double the current , then the power dissipated by that circuit will be:
P' = I'²R
where,
I' = 2 I
Therefore,
P' = (2 I)²R
P' = 4 I²R
using equation 1
P' = 4 P
Therefore, the power dissipated by the circuit will becomes four times of its initial value.