Answer:
Explanation:
y_1 = (3 mm) sin(x - 3t)
comparing it with standard wave equation
y = A sin( ωt-kx )
we see
ω = -3 , k = -1
velocity = ω / k
= 3
y_2 = (6 mm) sin(2x - t)
we see
ω = -1 , k = -2
velocity = ω / k
= .5
y_3 = (1 mm) sin(4x - t)
we see
ω = -1 , k = -4
velocity = ω / k
= .25
y_4 = (2 mm) sin(x - 2t)
we see
ω = -2 , k = -1
velocity = ω / k
= 2
So greatest velocity to lowest velocity
y_1 = (3 mm) sin(x - 3t) , y_4 = (2 mm) sin(x - 2t) ,y_2 = (6 mm) sin(2x - t) , y_3 = (1 mm) sin(4x - t)
b )
Given the mass per unit length of wire the same , velocity is proportional to
√ T , where T is tension
so in respect of tension in the wire same order will exist for highest to lowest tension .
A 4.5 kg ball swings from a string in a vertical circle such that it has constant sum of kinetic and gravitational potential energy. Ignore any friction forces from the air or in the string. What is the difference in the tension between the lowest and highest points on the circle
Answer:
88.29 N
Explanation:
mass of the ball = 4.5 kg
weight of the ball will be = mass x acceleration due to gravity(9.81 m/s^2)
weight W = 4.5 x 9.81 = 44.145 N
centrifugal forces Tc act on the ball as it swings.
At the top point of the vertical swing,
Tension on the rope = Tc - W.
At the bottom point of the vertical swing,
Tension on the rope = Tc + W
therefore,
difference in tension between these two points will be;
Net tension = tension at bottom minus tension at the top
= Tc + W - (Tc - W) = Tc + W -Tc + W
= 2W
imputing the value of the weight W, we have
2W = 2 x 44.145 = 88.29 N
Cart A, with a mass of 0.20 kg, travels on a horizontal air trackat 3.0m/s and hits cart B, which has a mass of 0.40 kg and is initially traveling away from Aat 2.0m/s. After the collision the center of mass of the two cart system has a speed of:____________.A. zeroB. 0.33m/sC. 2.3m/sD. 2.5m/sE. 5.0m/s
Answer:
[tex]\large \boxed{\text{C. 2.3 m/s}}[/tex]
Explanation:
Data:
[tex]m_{\text{A} } = \text{0.20 kg};\,v_{\text{Ai}} = \text{3.0 m/s}\\m_{\text{B} } = \text{0.40 kg};\,v_{\text{Bi}} = \text{2.0 m/s}\\[/tex]
Calculation:
This is a perfectly inelastic collision. The two carts stick together after the collision and move with a common final velocity.
The conservation of momentum equation is
[tex]\begin{array}{rcl}m_{\text{A}}v_{\text{Ai}} +m_{\text{B}} v_{\text{Bi}}&=&(m_{\text{A}} + m_{\text{B}})v_{\text{f}}\\0.20\times 3.0 + 0.40\times 2.0 & = & (0.20 + 0.40)v_{\text{f}}\\0.60 + 0.80 & = & 0.60v_{\text{f}}\\1.40 & = & 0.60v_{\text{f}}\\v_{\text{f}}&=& \dfrac{1.40}{0.60}\\\\& = & \textbf{2.3 m/s}\\\end{array}\\\text{The centre of mass has a velocity of $\large \boxed{\textbf{2.3 m/s}}$}[/tex]
A 25 kg box is 220 N pulled at constant speed up a frictionless inclined plane by a force that is parallel to the incline. If the plane is inclined at an angle of 25o above the horizontal, the magnitude of the applied force is
Answer:
F = 103.54N
Explanation:
In order to calculate the magnitude of the applied force, you take into account that the forces on the box are the applied force F and the weight of the box W.
The box moves with a constant velocity. By the Newton second law you have that the sum of forces must be equal to zero.
Furthermore, you have that the sum of forces are given by:
[tex]F-Wsin\theta=0[/tex] (1)
F: applied force = ?
W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N
θ: degree of the incline = 25°
You solve the equation (1) for F:
[tex]F=Wsin\theta=(245N)sin(25\°)=103.54N[/tex] (2)
The applied force on the box is 103.54N
Potential difference of a battery is 2.2 V when it is connected
across a resistance of 5 ohm, if suddenly the potential difference
falls to 1.8V, its internal resistance will be
Answer:
1.1ohms
Explanation:
According to ohms law E = IR
If potential difference of a battery is 2.2 V when it is connected across a resistance of 5 ohm and if suddenly the voltage Falls to 1.8V then the current in the 5ohms resistor I = V/R = 1.8/5
I = 0.36A (This will be the load current).
Before we can calculate the value of the internal resistance, we need to know the voltage drop across the internal resistance.
Voltage drop = 2.2V - 1.8V = 0.4V
Then we calculate the internal resistance using ohms law.
According to the law, V = Ir
V= voltage drop
I is the load current
r = internal resistance
0.4 = 0.36r
r = 0.4/0.36
r = 1.1 ohms
A man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and he holds a brick in each hand.The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.0 k g times m squared. If by moving the bricks the man decreases the rotational inertia of the system to 2.0 k g times m squared, what is the resulting angular speed of the platform in rad/s
Answer:
resulting angular speed = 3.6 rev/s
Explanation:
We are given;
Initial angular speed; ω_i = 1.2 rev/s
Initial moment of inertia;I_i = 6 kg/m²
Final moment of inertia;I_f = 2 kg/m²
From conservation of angular momentum;
Initial angular momentum = Final angular momentum
Thus;
I_i × ω_i = I_f × ω_f
Making ω_f the subject, we have;
ω_f = (I_i × ω_i)/I_f
Plugging in the relevant values;
ω_f = (6 × 1.2)/2
ω_f = 3.6 rev/s
If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is the magnitude of the magnetic field at a distance of 19.8 cm from the wire
Answer:
The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Explanation:
Given;
first magnetic field at first distance, B₁ = 2.50 mT
first distance, r₁ = 12.6 cm = 0.126 m
Second magnetic field at Second distance, B₂ = ?
Second distance, r₂ = ?
Magnetic field for a straight wire is given as;
[tex]B = \frac{\mu I}{2 \pi r}[/tex]
Where:
μ is permeability
B is magnetic field
I is current flowing in the wire
r distance to the wire
[tex]Let \ \frac{\mu I}{2\pi} \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT[/tex]
Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
A charged particle q moves at constant velocity through a crossed electric and magnetic fields (E and B, which are both constant in magnitude and direction). Write the magnitude of the electric force on the particle in terms of the variables given. Do the same for the magnetic force
Answer:
The magnitude of the electric force on the particle in terms of the variables given is, F = qE
The magnitude of the magnetic force on the particle in terms of the variables given is, F = q (v x B)
Explanation:
Given;
a charged particle, q
magnitude of electric field, E
magnitude of magnetic field, B
The magnitude of the electric force on the particle in terms of the variables given;
F = qE
The magnitude of the magnetic force on the particle in terms of the variables given;
F = q (v x B)
where;
v is the constant velocity of the charged particle
Answer:
The magnitude of the electric force acting on a charged particle moving through an electric field = |qE|
The magnitude of the magnetic force of a charged particle moving at a particular velocity through a magnetic field = |qv × B|
Explanation:
The electric force acting on a charged particle, q, moving through an electric field, E, is given as a product of the charge on the particle (a scalar quantity) and the electric field (a vector quantity).
Electric force = qE
The magnitude of the electric force = |qE|
That is, magnitude of the product of the charge and the electric field vector.
The magnetic force acting on a charged particle, q, moving with a velocity, v, through a magnetic field, B is a vector product of qv [a product of the charge of the particle (a scalar quantity) and the velocity of the particle (a vector quantity)] and B (a vector quantity).
It is given mathematically as (qv × B)
The magnitude of the magnetic force is the magnitude of the vector product obtained.
Magnitude of the magnetic force = |qv × B|
Hope this Helps!!!
A 50 gram meterstick is placed on a fulcrum at its 50 cm mark. A 20 gram mass is attached at the 12 cm mark. Where should a 40 gram mass be attached so that the meterstick will be balanced in rotational equilibrium
Answer:
The 40g mass will be attached at 69 cm
Explanation:
First, make a sketch of the meterstick with the masses placed on it;
--------------------------------------------------------------------------
↓ Δ ↓
20 g.................50 cm.................40g
38 cm y cm
Apply principle of moment;
sum of clockwise moment = sum of anticlockwise moment
40y = 20 (38)
40y = 760
y = 760 / 40
y = 19 cm
Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm
12cm 50 cm 69cm
--------------------------------------------------------------------------
↓ Δ ↓
20 g.................50 cm.................40g
38 cm 19 cm
The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1 W. HooRU's surface area is 1.55 m2 and the emissivity of its surface is 0.287. Ignoring the radiation that HooRU absorbs from the cold universe, what is HooRU's temperature T?
Answer:
The temperature is [tex]T = 168.44 \ K[/tex]
Explanation:
From the question ewe are told that
The rate of heat transferred is [tex]P = 13.1 \ W[/tex]
The surface area is [tex]A = 1.55 \ m^2[/tex]
The emissivity of its surface is [tex]e = 0.287[/tex]
Generally, the rate of heat transfer is mathematically represented as
[tex]H = A e \sigma T^{4}[/tex]
=> [tex]T = \sqrt[4]{\frac{P}{e* \sigma } }[/tex]
where [tex]\sigma[/tex] is the Boltzmann constant with value [tex]\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.[/tex]
substituting value
[tex]T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }[/tex]
[tex]T = 168.44 \ K[/tex]
In 11.8 s, 151 bullets strike and embed themselves in a wall. The bullets strike the wall perpendicularly. Each bullet has a mass of 5 x 10^-3 kg and a speed of 1110 m/s.
Required:
a. What is the average change in momentum per second for the bullets?
b. Determine the average force exerted on the wall.
c. Assuming the bullets are spread out over an area of 3.0×10^−4m^2 obtain the average pressure they exert on this region of the wall.
Answer:
a. ΔP/Δt = 42.6 N
b. F = 42.6 N
c. P = 142042.4 Pa = 1.42 KPa
Explanation:
a.
First, we find the change in momentum of the bullets. For one bullet:
ΔP = m(Vf - Vi)
where,
ΔP = Change in Momentum = ?
m = mass of bullet = 5 x 10⁻³ kg
Vf = Final Speed = 1110 m/s
Vi = Initial Speed = 0 m/s (Since bullets are initially at rest)
Therefore,
ΔP = (3 x 10⁻³ kg)(1110 m/s - 0 m/s)
ΔP = 3.33 N.s
For 151 bullets:
ΔP = (151)(3.33 N.s)
ΔP = 502.83 N.s
Now, dividing this by time interval, Δt = 11.8 s
ΔP/Δt = 502.83 N.s/ 11.8 s
ΔP/Δt = 42.6 N
b.
According to Newton's Second Law, the force is equal to rate of change of linear momentum:
Average Force = F = ΔP/Δt
F = 42.6 N
c.
The pressure is given by:
Average Pressure = P = Average Force/Area
P = 42.6 N/ 3 x 10⁻⁴ m²
P = 142042.4 Pa = 1.42 KPa
Two narrow slits, illuminated by light consisting of two distinct wavelengths, produce two overlapping colored interference patterns on a distant screen. The center of the eighth bright fringe in one pattern coincides with the center of the third bright fringe in the other pattern. What is the ratio of the two wavelengths?
Answer:
The ration of the two wavelength is [tex]\frac{\lambda_1}{\lambda_2} = \frac{8}{3}[/tex]
Explanation:
Generally two slit constructive interference can be mathematically represented as
[tex]\frac{y}{L} = \frac{m * \lambda}{d}[/tex]
Where y is the distance between fringe
d is the distance between the two slit
L is the distance between the slit and the wall
m is the order of the fringe
given that y , L , d are constant we have that
[tex]\frac{m }{\lambda } = constant[/tex]
So
[tex]\frac{m_1 }{\lambda_1 } = \frac{m_2 }{\lambda_2 }[/tex]
So [tex]m_1 = 8[/tex]
and [tex]m_2 = 3[/tex]
=> [tex]\frac{m_2}{m_1} = \frac{\lambda_1}{\lambda_2}[/tex]
=> [tex]\frac{8}{3} = \frac{\lambda_1}{\lambda_2}[/tex]
So
[tex]\frac{\lambda_1}{\lambda_2} = \frac{8}{3}[/tex]
Light rays from stars bend toward smaller angles as they enter Earth's atmosphere. a. Explain why this happens using Snell's law and the speed of light. b. Where are the actual stars in relation to their apparent position as viewed from the Earth's surface?
Answer:
Following are the answer to this question:
Explanation:
In option (a):
The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle. Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.In option (b):
Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth. Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.A long cylindrical rod of diameter 200 mm with thermal conductivity of 0.5 W/m⋅K experiences uniform volumetric heat generation of 24,000 W/m3. The rod is encapsulated by a circular seeve having an outer diameter of 400 mm and a thermal conductivity of 4 W/m⋅K. The outer surface of the sleeve is exposed to cross flow air at 27∘C with a convection coefficient of 25 W/m2⋅K.
(a) Find the temperature at the interface between the rod and sleeve and on the outer surface.
(b) What is the temperature at the center of the rod?
Answer:
a, 71.8° C, 51° C
b, 191.8° C
Explanation:
Given that
D(i) = 200 mm
D(o) = 400 mm
q' = 24000 W/m³
k(r) = 0.5 W/m.K
k(s) = 4 W/m.K
k(h) = 25 W/m².K
The expression for heat generation is given by
q = πr²Lq'
q = π . 0.1² . L . 24000
q = 754L W/m
Thermal conduction resistance, R(cond) = 0.0276/L
Thermal conduction resistance, R(conv) = 0.0318/L
Using energy balance equation,
Energy going in = Energy coming out
Which is = q, which is 754L
From the attachment, we deduce that the temperature between the rod and the sleeve is 71.8° C
At the same time, we find out that the temperature on the outer surface is 51° C
Also, from the second attachment, the temperature at the center of the rod was calculated to be, 191.8° C
A heavy, 6 m long uniform plank has a mass of 30 kg. It is positioned so that 4 m is supported on the deck of a ship and 2 m sticks out over the water. It is held in place only by its own weight. You have a mass of 70 kg and walk the plank past the edge of the ship. How far past the edge do you get before the plank starts to tip, in m
Answer:
about 1 meter
Explanation:
The distance past the edge that the man will get before the plank starts to tip is; 0.4285 m
We are given;
Mass of plank; m = 30 kg
Length of plank; L = 6m
Mass of man; M = 70 kg
Since the plank has 2 supports which are the deck of the ship, then it means that, we can take moments about the right support before the 2m stick out of the plank.
Thus;
Moment of weight of plank about the right support;
τ_p = mg((L/2) - 2)
τ_p = 30 × 9.8((6/2) - 2)
τ_p = 294 N.m
Moment of weight of man about the right support;
τ_m = Mgx
where x is the distance past the edge the man will get before the plank starts to tip.
τ_m = 70 × 9.8x
τ_m = 686x
Now, moment of the board is counterclockwise while that of the man is clockwise. Thus;
τ_m = τ_p
686x = 294
x = 294/686
x = 0.4285 m
Read more at; https://brainly.com/question/22150651
If the number of loops in a coil around a moving magnet doubles, the emf created:_________
a. Doubles
b. Halves
c. Remains the same
Answer is a. Doubles
when the loops are increased in the coil then the magnetic field created doubles
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 4.50 m/s at point A and 5.00 m/s at point C. The cart takes 4.00 s to go from point A to point C, and the cart takes 1.60 s to go from point B to point C. What is the cart's speed at point B
Answer:
Vi = 4.8 m/s
Explanation:
First we need to find the magnitude of constant tangential acceleration. For that purpose we use the following formula between points A and C:
a = (Vf - Vi)/t
where,
a = constant tangential acceleration from A to C = ?
Vf = Final Velocity at C = 5 m/s
Vi = Initial Velocity at A = 4.5 m/s
t = time taken to move from A to C = 4 s
Therefore,
a = (5 m/s - 4.5 m/s)/4 s
a = 0.125 m/s²
Now, applying the same equation between points B and C:
a = (Vf - Vi)/t
where,
a = constant tangential acceleration from A to B = 0.125 m/s²
Vf = Final Velocity at C = 5 m/s
Vi = Initial Velocity at B = ?
t = time taken to move from B to C = 1.6 s
Therefore,
0.125 m/s² = (5 m/s - Vi)/1.6 s
Vi = 5 m/s - (0.125 m/s²)(1.6 s)
Vi = 4.8 m/s
A 0.260 m radius, 525 turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.) Find the magnetic field strength needed to induce an average emf of 10,000 V.
Answer:
B = 0.37T
Explanation:
In order to calculate the needed magnitude of the magnetic force you use the following formula, which calculate the induced emf of the solenoid when there is a change in the magnetic flux:
[tex]emf=-N\frac{\Delta \Phi_B}{\Delta t}=-N\frac{\Delta (BAcos\theta)}{\Delta t}[/tex] (1)
emf: induced voltage in the solenoid = 10,000V
N: turns of the solenoid = 525
ФB: magnetic flux
B: magnitude of the magnetic field = ?
A: cross-sectional area of the solenoid = π*r^2
r: radius of the cross-sectional area = 0.260m
Δt: interval time of the change of the magnetic flux = 4.17ms = 4.17*10^-3s
First, you have the magnetic field direction perpendicular to the plane of the solenoid, after, the angle between them is 90° (quarter of a revolution)
In the equation (1) the only parameter that changes on time is the angle, then, you can solve for B from the equation (1):
[tex]emf=-NBA\frac{cos(90\°)-cos(0\°)}{\Delta t}=\frac{NBA}{\Delta t}\\\\B=\frac{(\Delta t)(emf)}{NA}=\frac{(\Delta t)(emf)}{N(\pi r^2)}\\\\[/tex]
Finally, you replace the values of the parameters to calculate B:
[tex]B=\frac{(4.17*10^{-3}s)(10000V)}{(525)(\pi(0.260m)^2)}=0.37T[/tex]
The strength of the magnetic field is 0.37T
The temperature difference between the inside and the outside of an automobile engine is 434 C°. Express this temperature difference on:
a. The Fahrenheit scale.
b. The Kelvin scale.
Answer:
434°C=813.2 °F
434°C=707.15 K
Explanation:
Fahrenheit is a temperature scale that sets the freezing point of water to 32 degrees Fahrenheit (° F) and the boiling point to 212 ° F (at normal atmospheric pressure). The conversion from degrees celsius to gradis fahrenheit is done by: ℉ = ℃ * 1.80+ 32.00
So, being 434°C: ℉ = 434℃ * 1.80+ 32.00= 813.2
Then: 434°C=813.2 °F
Kelvin is the temperature unit of the International System. It is one of the seven basic temperature units. Its symbol in the international system of units is K. Zero on the Celsius scale or degrees Celsius (0 ° C) is defined as the equivalent of 273.15 K. Then the conversion of Census degrees to Kelvin degrees is done by: K = ℃ + 273.15
Being 434 °C: K=434 °C+273.15=707.15
Then: 434°C=707.15 K
You are working as a summer intern for the Illinois Department of Natural Resources (DNR) and are assigned to some initial work on water resources project. The department will be overseeing the construction of a dam to create a large fresh water lake that will be approximately 14 meters deep. A horizontal pipe 1.2 meters long and 3 cm in diameter will pass through the dam at a depth of 7 meters to allow for release of the water in emergencies and for sampling. In normal situations, a plug will secure the pipe opening.
1)What will be the total force on the left side of the plug?FLEFT =___________ N2)What is the total force on the right side of the plug?FRIGHT =__________ N3)What is the gauge pressure on the plug?Pgauge =___________ Pa4)If the plug were to be removed, how much water would flow out in 3 hours??VOL =_____________ m3
Answer:
1) F = 71.6 N , 2) F = 120.2 N , 3) P_m= 68.600 Pa, 4) V = 2.4210-5 m³
Explanation:
This is a problem of fluid mechanics, to find the force we must use its definition
P = F / A
F = P A
The area of the circular pipe is
A = π r² = π d²/4
The pressure is given by the expression
P = P_atm + ρ g h
1) the force on the outer side is
P = P_atm
we substitute in the expression of force
F = P_atm π d² / 4
let's calculate
F = 1,013 10⁵ π 0.03²/4
F = 7.16 10¹ N
F = 71.6 N
2) the force on the inner side
the pressure
P = P_atm + ρ g h
P = 1,015 10⁵ + 1,000 9.8 7
P = 1,701 10⁵ Pa
F = 1,701 10⁵ π 0.03² / 4
F = 1,202 10²
F = 120.2 N
3) manometric pressure
Pm = ρ g h
P m = 1000 9.8 7
P_m= 68.600 Pa
4) In this part they ask for the volume that comes out in time t= 3 h
to calculate this volume we can use the flow ratio
Q = A v
V t = A v
V = A v / t
sent is the velocity of the water that comes out, to calculate it we use the Bernoulli equation
we will use index 1 for the lake surface and ionice 2 apa the position of the plug
P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂
As the lake has much more capacity than the pipeline, the velocity of the surface of the lake is peeling, in this case we approximate it steel
(P₁-P₂) + ρ G (y₁ -y₂) = ρ g v₂²
1000 9.8 v₂² = ρ g h + 1000 9.8 (7-0)
9800 v₂² = 1000 9.8 7 + 68600
v₂ = √ (137200)
v₂ = 370.4 m / s
t = 3 h (3600s / h) = 10800 s
we substitute in the volume equation
V = π d²/4 370.4 / 10800
V = 2.4210-5 m³
What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?
(Given, the universal gas constant = 8.315 J/(mol.k))
Answer:
U = 12,205.5 J
Explanation:
In order to calculate the internal energy of an ideal gas, you take into account the following formula:
[tex]U=\frac{3}{2}nRT[/tex] (1)
U: internal energy
R: ideal gas constant = 8.135 J(mol.K)
n: number of moles = 10 mol
T: temperature of the gas = 100K
You replace the values of the parameters in the equation (1):
[tex]U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J[/tex]
The total internal energy of 10 mol of Oxygen at 100K is 12,205.5 J
Set the battery to a value between 0.0 V and 1.5 V. Now drag the voltage meter toward the capacitor and move the red and black leads to measure the voltage. Determine the potential difference between the two plates and whether the top plate is at higher or lower voltage than the bottom plate
Answer:
the positive terminal has higher potential(voltage) than the negative. Any terminal at the positive terminal has higher potential
Explanation:
ΔV =Vtop - Vbottom
A swimmer of mass 64.38 kg is initially standing still at one end of a log of mass 237 kg which is floating at rest in water. He runs toward the other end of the log and dives off with a horizontal speed of 3.472 m/s relative to the water. What is the speed of the log relative to water after the swimmer jumps off
Answer:
0.9432 m/s
Explanation:
We are given;
Mass of swimmer;m_s = 64.38 kg
Mass of log; m_l = 237 kg
Velocity of swimmer; v_s = 3.472 m/s
Now, if we consider the first log and the swimmer as our system, then the force between the swimmer and the log and the log and the swimmer are internal forces. Thus, there are no external forces and therefore momentum must be conserved.
So;
Initial momentum = final momentum
m_l × v_l = m_s × v_s
Where v_l is speed of the log relative to water
Making v_l the subject, we have;
v_l = (m_s × v_s)/m_l
Plugging in the relevant values, we have;
v_l = (64.38 × 3.472)/237
v_l = 0.9432 m/s
One kind of baseball pitching machine works by rotating a light and stiff rigid rod about a horizontal axis until the ball is moving toward the target. Suppose a 144 g baseball is held 82 cm from the axis of rotation and released at the major league pitching speed of 87 mph.
Required:
a. What is the ball's centripetal acceleration just before it is released?
b. What is the magnitude of the net force that is acting on the ball just before it is released?
Answer:
a. ac = 1844.66 m/s²
b. Fc = 265.63 N
Explanation:
a.
The centripetal acceleration of the ball is given as follows:
ac = v²/r
where,
ac = centripetal acceleration = ?
v = speed of ball = (87 mph)(1 h/ 3600 s)(1609.34 m / 1 mile) = 38.9 m/s
r = radius of path = 82 cm = 0.82 m
Therefore,
ac = (38.9 m/s)²/0.82 m
ac = 1844.66 m/s²
b.
The centripetal force is given as:
Fc = (m)(ac)
Fc = (0.144 kg)(1844.66 m/s²)
Fc = 265.63 N
A solenoidal coil with 23 turns of wire is wound tightly around another coil with 310 turns. The inner solenoid is 20.0 cm long and has a diameter of 2.20 cm. At a certain time, the current in the inner solenoid is 0.130 A and is increasing at a rate of 1800 A/s. For this time, calculate:
(a) the average magnetic flux through each turn of the inner solenoid;
(b) the mutual inductance of the two solenoids;
(c) the emf induced in the outer solenoid by the changing current in the inner solenoid.
Answer:
Explanation:
From the given information:
(a)
the average magnetic flux through each turn of the inner solenoid can be calculated by the formula:
[tex]\phi _ 1 = B_1 A[/tex]
[tex]\phi _ 1 = ( \mu_o \dfrac{N_i}{l} i_1)(\pi ( \dfrac{d}{2})^2)[/tex]
[tex]\phi _ 1 = ( 4 \pi *10^{-7} \ T. m/A ) ( \dfrac{310}{20*10^{-2} \ m }) (0.130 \ A) ( \pi ( \dfrac{2.20*10^{-2} \ m }{2})^ 2[/tex]
[tex]\phi_1 = 9.625 * 10^{-8} \ Wb[/tex]
(b)
The mutual inductance of the two solenoids is calculated by the formula:
[tex]M = 23 *\dfrac{9.625*10^{-8} \ Wb}{0.130 \ A}[/tex]
M = [tex]1.703 *10^{-5}[/tex] H
(c)
the emf induced in the outer solenoid by the changing current in the inner solenoid can be calculate by using the formula:
[tex]\varepsilon = -N_o \dfrac{d \phi_1}{dt}[/tex]
[tex]\varepsilon = -M \dfrac{d i_1}{dt}[/tex]
[tex]\varepsilon = -(1.703*10^{-5} \ H) * (1800 \ A/s)[/tex]
[tex]\varepsilon = -0.030654 \ V[/tex]
[tex]\varepsilon = -30.65 \ V[/tex]
A block slides down a ramp with friction. The friction experienced by the block is 21 N. The mass of the block is 8 kg. The ramp is 6 meters long (meaning, the block slides across 6 meters of ramp with friction). The block is originally 2 meters vertically above the ground (the bottom of the ramp). What is the change in energy of the block due to friction, expressed in Joules
Complete Question
The complete question is shown on the first uploaded image
Answer:
the change in energy of the block due to friction is [tex]E = -126 \ J[/tex]
Explanation:
From the question we are told that
The frictional force is [tex]F_f = 21 \ N[/tex]
The mass of the block is [tex]m_b = 8 \ kg[/tex]
The length of the ramp is [tex]l = 6 \ m[/tex]
The height of the block is [tex]h = 2 \ m[/tex]
The change in energy of the block due to friction is mathematically represented as
[tex]\Delta E = - F_s * l[/tex]
The negative sign is to show that the frictional force is acting against the direction of the block movement
Now substituting values
[tex]\Delta E = -(21)* 6[/tex]
[tex]\Delta E = -126 \ J[/tex]
While sitting in your car by the side of a country road, you see your friend, who happens to have an identical car with an identical horn, approaching you. You blow your horn, which has a frequency of 260 Hz; your friend begins to blow his horn as well, and you hear a beat frequency of 5.0 Hz. Part A How fast is your friend approaching you
Answer:
-6.49 m/s
Explanation:
This is doppler effect.
The equation is;
F_l = [(v + v_l)/(v + v_s)]F_s
Where;
F_l is frequency observed by the listener
v is speed of sound
v_l is speed of listener
v_s is speed of source of the sound
F_s is frequency of the source of the sound
In this question, the source of the sound is the moving vehicle.
Thus;
F_l = F_beat + F_s
We are given beat frequency (f_beat) as 5 Hz while source frequency (F_s) as 260 Hz.
So,
F_l = 5 + 260
F_l = 265 Hz
Since listener is sitting by car, thus; v_l = 0 m/s
Thus,from our doppler effect equation, let's make v_s the subject;
v_s = F_s[(v + v_l)/F_l] - v
Speed of sound has a value of v = 344 m/s
Thus;
v_s = 260[(344 + 0)/265] - 344
v_s = -6.49 m/s
This value is negative because the source is moving towards the listener
A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mass. It has a rotation rate of 8.0 rev/s. Two antennas deploy in the plane of rotation extending from the center of mass of the satellite. Each antenna can be approximated as a rod of mass 150.0 kg and length 6.6 m. What is the new rotation rate of the satellite (in rev/s)
Answer:
6.3 rev/s
Explanation:
The new rotation rate of the satellite can be found by conservation of the angular momentum (L):
[tex] L_{i} = L_{f} [/tex]
[tex] I_{i}*\omega_{i} = I_{f}*\omega_{f} [/tex]
The initial moment of inertia of the satellite (a solid sphere) is given by:
[tex] I_{i} = \frac{2}{5}m_{s}r^{2} [/tex]
Where [tex]m_{s}[/tex]: is the satellite mass and r: is the satellite's radium
[tex] I_{i} = \frac{2}{5}m_{s}r^{2} = \frac{2}{5}1900 kg*(4.6 m)^{2} = 1.61 \cdot 10^{4} kg*m^{2} [/tex]
Now, the final moment of inertia is given by the satellite and the antennas (rod):
[tex] I_{f} = I_{i} + 2*I_{a} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}m_{a}l^{2} [/tex]
Where [tex]m_{a}[/tex]: is the antenna's mass and l: is the lenght of the antenna
[tex] I_{f} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}150.0 kg*(6.6 m)^{2} = 2.05 \cdot 10^{4} kg*m^{2} [/tex]
So, the new rotation rate of the satellite is:
[tex] I_{i}*\omega_{i} = I_{f}*\omega_{f} [/tex]
[tex]\omega_{f} = \frac{I_{i}*\omega_{i}}{I_{f}} = \frac{1.61 \cdot 10^{4} kg*m^{2}*8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kg*m^{2}} = 6.3 rev/s[/tex]
Therefore, the new rotation rate of the satellite is 6.3 rev/s.
I hope it helps you!
Two metal spheres are hanging from nylon threads. When you bring the spheres close to each other, they tend to attract. Based on this information alone, discuss all the possible ways that the spheres could be charged. Is it possible that after the spheres touch, they will cling together? Explain.
Explanation:
In the given question, the two metal spheres were hanged with the nylon thread.
When these two spheres were brought close together, they attracted each other. The attraction between these spheres is the result of the opposite charges between them.
The possible ways by which these two metal spheres can be charged are by induction that is touching the metal or by rubbing them.
During induction, the same charges are transferred to each sphere. In this case, either both the spheres will be negatively charged or positively charged.
It is not possible that after the sphere touch each other they will cling together because the same charge repels each other and during touching, if one sphere is neutral, then the charged one will transfer the same charge. And as we know that same charge repel each other therefore they will repel each other.
An empty bottle has a mass of 35.00 grams. When filled with water, it has a mass of 98.44 grams. Of the same bottle is filled with a different fluid, the mass is 89.22 grams. What is the specific gravity of this other fluid
Answer:
Specific gravity of other fluid = .854 (Approx)
Explanation:
Given:
Mass of water = 35 g
Mass of filled bottle with water = 98.44 g
Mass of filled bottle with fluid = 89.22 g
Computation:
Mass of water = 98.44g - 35g = 63.44g
Density of water = 1000 g/L
Volume of bottle = 63.44/1000 = 0.06344L
Mass of other liquid = 89.22g - 35g = 54.22g
Density of other liquid = 54.22g/0.06344L = 854.665826 g/L
Water has a specific gravity = 1
So , specific gravity of other fluid
1000 / 854.665826 = 1 / specific gravity of other fluid
Specific gravity of other fluid = .854 (Approx)
A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point circled A, at a potential of 1.50 103 V, to point circled B, at 4.00 103 V. What is its speed at point circled B?
Answer:
[tex]v_B=3.78\times 10^5\ m/s[/tex]
Explanation:
It is given that,
Charge on helium nucleus is 2e and its mass is [tex]6.63\times 10^{-27}\ kg[/tex]
Speed of nucleus at A is [tex]v_A=6.2\times 10^5\ m/s[/tex]
Potential at point A, [tex]V_A=1.5\times 10^3\ V[/tex]
Potential at point B, [tex]V_B=4\times 10^3\ V[/tex]
We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :
increase in kinetic energy = increase in potential×charge
[tex]\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s[/tex]
So, the speed at point B is [tex]3.78\times 10^5\ m/s[/tex].