Answer:
The voltage is [tex]V_c = 9.92 \ V[/tex]
Explanation:
From the question we are told that
The voltage of the battery is [tex]V_b = 24 \ V[/tex]
The capacitance of the capacitor is [tex]C = 3.0 mF = 3.0 *10^{-3} \ F[/tex]
The resistance of the resistor is [tex]R = 100\ \Omega[/tex]
The time taken is [tex]t = 0.16 \ s[/tex]
Generally the voltage of a charging charging capacitor after time t is mathematically represented as
[tex]V_c = V_o (1 - e^{- \frac{t}{RC} })[/tex]
Here [tex]V_o[/tex] is the voltage of the capacitor when it is fully charged which in the case of this question is equivalent to the voltage of the battery so
[tex]V_c = 24 (1 - e^{- \frac{0.16}{100 * 3.0 *10^{-1}} })[/tex]
[tex]V_c = 9.92 \ V[/tex]
The Hermes spacecraft is traveling at 0.1c(1/10 the speed of light past Mars and shines a laser in front of the ship. You would see the light traveling at c (the speed of light )away from your ship. According to Einstein's special relativity how fast with a person on Mars observe the light to be traveling?
Answer:
So, according to Einstein's special relativity a person on Mars observe the light to be traveling at c = 3 x 10⁸ m/s.
Explanation:
The special theory of relativity has two main postulates:
1- VALIDITY OF PHYSICAL LAWS
The laws of physics such as Newton's Laws and Maxwell's Equations are valid in all inertial frame of references.
2- CONSTANCY OF SPEED OF LIGHT
The speed of light in vacuum is the same for all observers in uniform translational relative motion, and it is independent of the motion of the source or the observer. Thus, speed of light is a universal constant and its value is c = 3 x 10⁸ m/s.
So, according to Einstein's special relativity a person on Mars observe the light to be traveling at c = 3 x 10⁸ m/s.
A 20 g bullet strikes a 1.5 kg wood block and becomes embedded in it (all horizontal motion). The coefficient of friction between the block and the surface is 0.3 and the impact makes the block move a distance of 7 m before it comes to rest. What was the muzzle speed (in m/s) of the bullet? Report your answer with two significant figures.
Answer:
Explanation:
Kinetic frictional force acting on the block = μ mg
where μ is coefficient of friction , m is mass of block.
.3 x 1.5 x 9.8 = 4.41 N .
Let v be the velocity of bullet + block after collision
kinetic energy of composite mass after the strike
= 1 /2 x 1.52 x v²
this will be equal to work done by friction .
.76 v² = 4.41 x 7
v² = 40.62
v = 6.37 m /s
Now we can obtain muzzle speed of bullet by applying conservation of momentum .
Let this speed be u
initial momentum of bullet
= .02 x u
final momentum of composite mass
= 1.52 x 6.37
.02 u = 1.52 x 6.37
u = 484.12 m /s .
= 480 m /s ( in two significant figures )
A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg initially at rest. What is the initial momentum of the green train?
Answer:
I think answer is zero
bcz momentum=mass×velocity
body was initially at rest it means its velocity is zero
30×0=0
A generator rotates at 95 Hz in a magnetic field of 0.025 T. It has 550 turns and produces an rms voltage of 170 V and an rms current of 60.0 A.
Required:
a. What is the peak current produced?
b. What is the area of each turn of the coil?
Answer:
Peak current= 84.86 A
Area of each turn = 0.029 m^2
Explanation:
The peak value of current can be obtained from Irms= 0.707Io. Where Io is the peak current.
Hence;
Irms= 60.0A
Io= Irms/0.707
Io = 60.0/0.707
Io= 84.86 A
Vrms= 0.707Vo
Vo= Vrms/0.707= 170/0.707 = 240.45 V
From;
V0 = NABω
Where;
Vo= peak voltage
N= number of turns
B= magnetic field
A= area of each coil
ω= angular velocity
But ω= 2πf = 2×π×95= 596.9 rads-1
Substituting values;
A= Vo/NBω
A= 240.45/550×0.025×596.9
A= 0.029 m^2
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hits the center in 0.455 s. (Neglect any effects due to air resistance.)At what angle relative to the floor was the dart thrown?
Answer:
The angle is [tex]\theta = 15.48^o[/tex]
Explanation:
From the question we are told that
The distance of the dartboard from the dart is [tex]d = 3.66 \ m[/tex]
The time taken is [tex]t = 0.455 \ s[/tex]
The horizontal component of the speed of the dart is mathematically represented as
[tex]u_x = ucos \theta[/tex]
where u is the the velocity at dart is lunched
so
[tex]distance = velocity \ in \ the\ x-direction * time[/tex]
substituting values
[tex]3.66 = ucos \theta * (0.455)[/tex]
=> [tex]ucos \theta = 8.04 \ m/s[/tex]
From projectile kinematics the time taken by the dart can be mathematically represented as
[tex]t = \frac{2usin \theta }{g}[/tex]
=> [tex]usin \theta = \frac{g * t}{2 }[/tex]
[tex]usin \theta = \frac{9.8 * 0.455}{2 }[/tex]
[tex]usin \theta = 2.23[/tex]
=> [tex]tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}[/tex]
[tex]\theta = tan^{-1} [0.277][/tex]
[tex]\theta = 15.48^o[/tex]
At t=0 a 2150kg rocketship in outer space fires the engine which exerts a force=At2, and F(1.25s)=781.25N in the x direction. Find the impulse J during the interval t=2.00s and t=3.5s
Answer:
5.81 X 10^3 Ns
Explanation:
Given that
F = At² and F at t = 1.25 s is 781.25 N ?
A = F/t² at t = 1.25 s => F = 781.25/(1.25)² = 500 N/s²
d(Impulse) = Fdt
Impulse = ∫Fdt =∫At²dt evaluated in the interval 2.00 s ≤ t ≤ 3.50 s
Impulse = At³/3 = (500/3)(t³) = 166.7t³ between t = 2.00 s and t = 3.50 s
Impulse = 166.7[3.5³ - 2³] = 166.7[42.875 - 8] = 166.7[34.875] = 5813.7 Ns
5.81 X 10^3 N.s
a cylindrical jar is 10cm long and has a cross sectional area of 36cm. if it is completely filled with a fluid of relative density 0.2, calculate the mass of the fluid in the jar
Answer:
The mass of the fluid is 72 g.
Explanation:
The following data were obtained from the question:
Height (h) = 10 cm
Area of cross section (A) = 36cm²
Relative density = 0.2
Mass =..?
Next, we shall determine the volume of the cylinder. This can be achieved by doing the following:
Volume = Area x Height
Volume = 36 x 10
Volume = 360 cm³
Next, we shall determine the density of the liquid.
This can be obtained as follow:
Relative density = density of substance/density of water.
Relative density = 0.2
Density of water = 1 g/cm³
Density of fluid =...?
Relative density = density of substance/density of water.
0.2 = density of fluid / 1 g/cm³
Cross multiply
Density of fluid = 0.2 x 1 g/cm³
Density of fluid = 0.2 g/cm³
Finally, we shall determine the mass of fluid as follow:
Volume = 360 cm³
Density of fluid = 0.2 g/cm³
Mass of fluid =...?
Density = mass /volume.
0.2 g/cm³ = mass of fluid /360 cm³
Cross multiply
Mass of fluid = 0.2 g/cm³ x 360 cm³
Mass of fluid = 72 g
Therefore, the mass of the fluid in the jar is 72 g.
If theta is 30 degrees and there is no friction, what would be the block's acceleration down the incline, in meters per second squared?
Answer:
a= 4.9m/s²
Explanation:
Using Fnet= mgsintheta = ma
But a= gsintheta
a= 9.8xsin 30
= 4.9m/s²
An air bubble has a volume of 1.3 cm3 when it is released by a submarine 160 m below the surface of a freshwater lake. What is the volume of the bubble when it reaches the surface? Assume that the temperature and the number of air molecules in the bubble remain constant during the ascent.
Answer:
V2 = 21.44cm^3
Explanation:
Given that: the initial volume of the bubble = 1.3 cm^3
Depth = h = 160m
Where P2 is the atmospheric pressure = Patm
P1 is the pressure at depth 'h'
Density of water = ρ = 10^3kg/m^3
Patm = 1.013×10^5 Pa.
Patm = 101300Pa
g = 9.81m/s^2
P1 = P2+ρgh
P1 = Patm +ρgh
P1 = 1.013×10^5+10^3×9.81×160.
P1 = 101300+1569600
P1 = 1670900 Pa
For an ideal gas law
PV =nRT
P1V1/P2V2 = 1
V2 = ( P1/P2)V1
V2 = (P1/Patm)V1
V2 = ( 1670900 /101300 Pa) × 1.3
V2 = 1670900/101300
V2 = 16.494×1.3
V2 = 21.44cm^3
The volume of the bubble can be determined using ideal gas law. The volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].
The formula of the pressure of the static fluid
P1 = P2+ρgh
Where,
P1 - pressure at depth 'h'
P2 - atmospheric pressure = [tex]\bold {1.013x10^5 }[/tex] = 1670900 Pa
h - Depth = 160m
ρ - Density of water = [tex]\bold {10^3\ kg/m^3}[/tex]
g- gravitational acceleration = [tex]\bold {9.81\ m/s^2}[/tex]
The initial volume of the bubble = [tex]\bold {1.3\ cm^3}[/tex]
[tex]\bold {P1 = 1.013x10^5+10^3\times 9.81\times 160}\\\\\bold {P1 = 101300+1569600}\\\\\bold {P1 = 1670900\ Pa}[/tex]
For an ideal gas,
PV =nRT
[tex]\bold {\dfrac {P_1V_1}{P_2V_2 }= 1}[/tex]
[tex]\bold {V2 = \dfrac { P_1}{P_2V_1}}[/tex]
So,
[tex]\bold {V2 = \dfrac {1670900 }{101300 }\times 1.3}\\\\\bold {V2 =21.44\ cm^3}[/tex]
Therefore, the volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].
To know more air bubble volume,
https://brainly.com/question/10509397
what is transmission of heat?
Answer:
Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are
Conduction ConvectionRadiation.Example of Conduction:
Heating a metal
Example of Convection:
Sea Breeze
Example of Radiation:
Sun
Hope this helps ;) ❤❤❤
Answer:
Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.
There are three(3) different ways heat can transfer and they are:
a) Conduction (through direct contact).
b) Convection (through fluid movement).
c) Radiation (through electromagnetic waves).
Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).
2. Baking a pie in an oven(radiation).
Hope it helps!!Please mark me as the brainliest!!!Thanks!!!!❤❤❤
From mechanics, you may recall that when the acceleration of an object is proportional to its coordinate, d2xdt2=−kmx=−ω2x , such motion is called simple harmonic motion, and the coordinate depends on time as x(t)=Acos(ωt+ϕ), where ϕ, the argument of the harmonic function at t=0, is called the phase constant. Find a similar expression for the charge q(t) on the capacitor in this circuit. Do not forget to determine the correct value of ϕ based on the initial conditions described in the problem. Express your answer in terms of q0 , L, and C. Use the cosine function in your answer.
Answer:
q = q₀ sin (wt)
Explanation:
In your statement it is not clear the type of circuit you are referring to, there are two possibilities.
1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor
ΔV = Δ[tex]V_{C}[/tex]
we assume that the source has a voltage of the form
ΔV = ΔV₀o sin wt
The capacitance of a capacitor is
C = q / ΔV
q = C ΔV sin wt
the current in the circuit is
i = dq / dt
i = c ΔV₀ w cos wt
if we use
cos wt = sin (wt + π / 2)
we make this change by being a resonant oscillation
we substitute
i = w C ΔV₀ sin (wt + π/2)
With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current
2) Another possible circuit is an LC circuit.
In this case the voltage alternates between the inductor and the capacitor
V_{L} + V_{C} = 0
L di / dt + q / C = 0
the current is
i = dq / dt
they ask us for a solution so that
L d²q / dt² + 1 / C q = 0
d²q / dt² + 1 / LC q = 0
this is a quadratic differential equation with solution of the form
q = A sin (wt + Ф)
to find the constant we derive the proposed solution and enter it into the equation
di / dt = Aw cos (wt + Ф)
d²i / dt²= - A w² sin (wt + Ф)
- A w² + 1 /LC A = 0
w = √ (1 / LC)
To find the phase factor, for this we use the initial conditions for t = 0
in the case of condensate for t = or the charge is zero
0 = A sin Ф
Ф = 0
q = q₀ sin (wt)
Three identical capacitors are connected in series to a battery. If a total charge of Q flows from the battery, how much charge does each capacitor carry?
Answer:
Each of the capacitor carries the same charge, Q
Explanation:
When capacitors are connected in series, the battery voltage is divided equally across the capacitors. The total voltage across the three identical capacitors is calculated as follows;
[tex]V_T = V_1 + V_2 + V_3[/tex]
We can also calculate this voltage in terms of capacitance and charge;
[tex]V = \frac{Q}{C} \\\\V_T = V_1 + V_2 +V_3 \\\\(given \ total \ charge \ as \ Q, then \ the \ total \ voltage \ V_T \ can \ be \ written \ as)\\\\V_T = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} \\\\V_T = Q(\frac{1}{C_1 } +\frac{1}{C_2} + \frac{1}{C_3 })\\\\[/tex]
Therefore, each of the capacitor carries the same charge, Q
A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]f= -75 \ cm = - 0.75 \ m[/tex]
b
[tex]P = -1.33 \ diopters[/tex]
Explanation:
From the question we are told that
The image distance is [tex]d_i = -75 cm[/tex]
The value of the image is negative because it is on the same side with the corrective glasses
The object distance is [tex]d_o = \infty[/tex]
The reason object distance is because the object father than it being picture by the eye
General focal length is mathematically represented as
[tex]\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}[/tex]
substituting values
[tex]\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}[/tex]
=> [tex]f= -75 \ cm = - 0.75 \ m[/tex]
Generally the power of the corrective lens is mathematically represented as
[tex]P = \frac{1}{f}[/tex]
substituting values
[tex]P = \frac{1}{-0.75}[/tex]
[tex]P = -1.33 \ diopters[/tex]
Describe the orientation of magnetic field lines by drawing a bar magnet, labeling the poles, and drawing several lines indicating the direction of the forces.
Answer:
A field is a way of mapping forces surrounding any object that can act on another object at a distance without apparent physical connection. The field represents the object generating it. Gravitational fields map gravitational forces, electric fields map electrical forces, and magnetic fields map magnetic forces.
Explanation:
6. How would the measurements for potential difference and current change if a 200 Ω resistor was used in Circuit 1 instead of the 100 Ω resistor? Explain your answer.
Answer:
Explanation:
Resistance is defined as the opposition to the flow of an electric current in a circuit. This means that a higher amount of resistance tends to reduce the amount of current flowing through the resistance. The lower the current, the greater the possibility for the resistor to allow current to pass through it. if a 200 Ω resistor was used in Circuit 1 instead of the 100 Ω resistor, then the current in the circuit will tends to increase since we are replacing the load with a lesser resistor and a smaller resistance tends to allow more current to flow through it
For the potential difference, a decrease in the resistance value will onl decrease the potential difference flowing in the circuit according to ohm's law. According to the law the pd in a circuit is directly proportional to the current which means an increase in the resistance value will cause an increase in the corresponding pd and vice versa.
If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to:
5.19 x 10³Hz
Explanation:The capacitive reactance, [tex]X_{C}[/tex], which is the opposition given to the flow of current through the capacitor is given by;
[tex]X_C = \frac{1}{2\pi fC }[/tex]
Where;
f = frequency of the signal through the capacitor
C = capacitance of the capacitor.
Also, from Ohm's law, the voltage(V) across the capacitor is given by the product of current(I) and the capacitive reactance. i.e;
V = I x [tex]X_{C}[/tex] [Substitute the value of
=> V = I x [tex]\frac{1}{2\pi fC}[/tex] [Make f the subject of the formula]
=> f = [tex]\frac{I}{2\pi VC}[/tex] ---------------------(i)
From the question;
I = 3.33mA = 0.00333A
C = 8.50nF = 8.50 x 10⁻⁹F
V = 12.0V
Substitute these values into equation (i) as follows;
f = [tex]\frac{0.00333}{2 * 3.142 * 12.0 * 8.50 * 10^{-9}}[/tex] [Taking [tex]\pi[/tex] = 3.142]
f = 5.19 x 10³Hz
Therefore, the frequency is closest to f = 5.19 x 10³Hz
Tarik winds a small paper tube uniformly with 163 turns of thin wire to form a solenoid. The tube's diameter is 6.13 mm and its length is 2.49 cm . What is the inductance, in microhenrys, of Tarik's solenoid?
Answer:
The inductance is [tex]L = 40\mu H[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 163 \ turns[/tex]
The diameter is [tex]D = 6.13 \ mm = 6.13 *10^{-3} \ m[/tex]
The length is [tex]l = 2.49 \ cm = 0.0249 \ m[/tex]
The radius is evaluated as [tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{6.13 *10^{-3}}{2}[/tex]
[tex]r = 3.065 *10^{-3} \ m[/tex]
The inductance of the Tarik's solenoid is mathematically represented as
[tex]L = \frac{\mu_o * N^2 * A }{l }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]
A is the area which is mathematically evaluated as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * [ 3.065*10^{-3}]^2[/tex]
[tex]A = 2.952*10^{-5} \ m^2[/tex]
substituting values into formula for L
[tex]L = \frac{ 4\pi *10^{-7} * [163]^2 * 2.952*10^{-5} }{0.0249 }[/tex]
[tex]L = 40\mu H[/tex]
A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its location at time t = 3.01.
Answer:
New location at time 3.01 is given by: (7.49, 2.11)
Explanation:
Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:
[tex]V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11[/tex]
With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:
[tex]distance=v\,*\, t[/tex]
[tex]distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11[/tex]
Therefore, adding these displacements in component form to the original particle's position, we get:
New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)
Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart. The number of fringes per unit length on the screen can be doubled:______.
a. if the distance between the slits is doubled.
b. if the wavelength is changed to λ = λ/2.
c. if the distance between the slits is quadruple the original distance and the wavelength is changed to λ = 2λ.
d. if any of the above occurs.
e. only if the width of the slits is changed to w = w/2
Answer:
d. if any of the above occurs
Explanation:
That is The number of fringes per unit length on the screen can be doubled if
if the distance between the slits is doubled.
if the wavelength is changed to λ = λ/2.
And if the distance between the slits is quadruple the original distance and the wavelength is changed to λ = 2λ
10. A manufacturer knows from experience that the resistance of resistors she produces is normal with mean µ = 150Ω and the standard deviation σ = 5Ω. What percentage of the resistors will have resistance between 148 Ω and 152 Ω? Between 140 Ω and 160 Ω?
Answer:
Explanation:
Using the formula for finding the Z score
Z = x-µ/σ
x is the sample size
µ is the sample mean
σ is the standard deviation
For percentage of the resistors will have resistance between 148 Ω and 152 Ω, or is calculated as shown
P(148≤x152) = Z(152-150/5) - Z(148-150/5)
P(148≤x152) = Z(0.4)-Z( - 0.4)
P(148≤x152) = 0.6554-0.3446
The Z values are from the normal distribution table.
P(148≤x152) = 0.3108
The percentage of resistor that will have between 148 and 152 ohms is 0.3108×100% = 31.08%
Similarly for resistances between 140 Ω and 160 Ω
P(140≤x160) = Z(160-150/5) - Z(140-150/5)
P(140≤x160) = Z(2.0)-Z( - 2.0)
P(140≤x160) = 0.9775-0.02275
The Z values are from the normal distribution table.
P(140≤x160) = 0.9547
The percentage of resistor that will have between 140 and 160ohms is 0.9547×100% = 95.47%
The percentage of the resistors will have resistance between 148 Ω and 152 Ω is 31.08%
The percentage of the resistors will have resistance between 140 Ω and 160 Ω is 95.47%
Percentage of resistors:To solve for the probability we will use the standard score of the Z score, which is given by:
Z = (x - µ)/σ
where x is the sample size
µ is the sample mean = 150Ω
σ is the standard deviation = 5Ω
The probability of the resistors with resistance between 148 Ω and 152 Ω, will be:
P(148 ≤ 152) = Z((152-150)/5) - Z((148-150)/5)
P(148 ≤ 152) = Z(0.4)-Z( - 0.4)
P(148 ≤ 152) = 0.6554-0.3446
P(148 ≤ 152) = 0.3108
So, the percentage will be:
0.3108×100% = 31.08%
Similarly for resistances between 140 Ω and 160 Ω
P(140 ≤ 160) = Z((160-150)/5) - Z((140-150)/5)
P(140 ≤ 160) = Z(2.0)-Z( - 2.0)
P(140 ≤ 160) = 0.9775-0.02275
P(140 ≤ 160) = 0.9547
The percentage of resistor that will have between 140 and 160ohms is 0.9547×100% = 95.47%
Learn more about standard score:
https://brainly.com/question/25875323?referrer=searchResults
A person standing 180m from the foot of a high building claps hi
hand and hears the echo 0.03minutes later. What is the speed
sound in air at that temperature?
A) 331m/s
B) 240m/s C) 200m/s D) 300m/s
Answer:
C) 200 m/s
Explanation:
The sound travels a total distance of 360 m in 0.03 minutes.
v = (360 m) / (0.03 min × 60 s/min)
v = 200 m/s
Water flows through a cylindrical pipe of varying cross-section. The velocity is 3.0 m/s at a point where the pipe diameter is 1.0 cm. What is the flow rate R
Answer:
The flow rate is [tex]R =2.357 *10^{-4} \ m^3/s[/tex]
Explanation:
From the question we are told that
The velocity is [tex]v = 3.0 \ m/s[/tex]
The diameter of the pipe is [tex]d = 1.0 \ cm = 0.01 \ m[/tex]
The radius of the pipe is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.01}{2}[/tex]
[tex]r = 0.005 \ m[/tex]
The flow rate is mathematically represented as
[tex]R = v * A[/tex]
Where is the cross-sectional area of the pipe which is mathematically evaluated as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * (0.005)^2[/tex]
[tex]A = 7.855 * 10^{-5} \ m^2[/tex]
So
[tex]R = 3.0 * 7.855 *10^{-5}[/tex]
[tex]R = 2.357*10^{-4} \ m^3 /s[/tex]
what help in reversing direction of current of current
Answer:
To reverse the direction of an electric current, we simply reverse the voltage either automatically with the help of some switching circuitry or manually by changing the voltage source terminals connection.
Explanation:
For electric current to flow, there must be a potential difference, usually referred to as the voltage. The electric current flow is analogous to the flow of water under the action of a pump, through a series of pipe connections. The voltage is similar to the driving action of the pump, and current flows the same way water flows. The resistance due to drag on the pipe wall is equivalent to electric resistance. For current to flow in the reverse direction, the voltage or rather, the potential difference is changed, causing the current to flow in the opposite direction. This can be done by switching the terminals of the voltage source, or by automatic means. The automatic switching can be done with a transistor based circuitry.
A uniform electric field is created by two parallel plates separated by a distance of 0.04 m. What is the magnitude of the electric field established between the plates
Complete question:
A uniform electric field is created by two parallel plates separated by a
distance of 0.04 m. What is the magnitude of the electric field established
between the plates if the potential of the first plate is +40V and the second
one is -40V?
Answer:
The magnitude of the electric field established between the plates is 2,000 V/m
Explanation:
Given;
distance between two parallel plates, d = 0.04 m
potential between first and second plate, = +40V and -40V respectively
The magnitude of the electric field established between the plates is calculated as;
E = ΔV / d
where;
ΔV is change in potential between two parallel plates;
d is the distance between the plates
ΔV = V₁ -V₂
ΔV = 40 - (-40)
ΔV = 40 + 40
ΔV = 80 V
E = ΔV / d
E = 80 / 0.04
E = 2,000 V/m
Therefore, the magnitude of the electric field established between the plates is 2,000 V/m
The magnitude of the electric field developed that lies between the plates should be considered as the 2,000 V/m.
Calculation of the Electric Field:Since
The distance that lies between 2 parallel plates should be d = 0.04 m
The potential that lies between first and second plate should be like +40V and -40V
So, The magnitude of the electric field should be
E = ΔV / d
here,
ΔV represents the change in potential that lies between 2 parallel plates.
d represents the distance between the plates.
So,
ΔV = V₁ -V₂
ΔV = 40 - (-40)
ΔV = 40 + 40
ΔV = 80 V
And,
E = ΔV / d
E = 80 / 0.04
E = 2,000 V/m
Therefore, the magnitude of the electric field established between the plates is 2,000 V/m.
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7.00 kg piece of solid copper metal at an initial temperature T is placed with 2.00 kg of ice that is initially at -20.0°C. The ice is in an insulated container of negligible mass and no heat is exchanged with the surroundings. After thermal equilibrium is reached, there is 0.90 kg of ice and 1.10 kg of liquid water.
Required:
What was the initial temperature of the piece of copper?
Answer:
122°C
Explanation:
From the data Final temperature is 0 deg C since there is 0.9kg of ice and 1.10kg of liquid water.
That means that 1.10kg of the ice undergoes Heat of Fusion which is 3.34x10^5 J/kg...
Heat lost by copper = Heat gained by ice + Heat of fusion
-> (7.0kg)(390J/kg*C)(0-T) = (2.00kg)(2100J/kg*C)(0 - (-20) + (1.10kg)(3.34x10^5 J/kg)
-> T(2730) = 334001
-> T = 122°C
An insulated beaker with negligible mass contains liquid water with a mass of 0.285 kg and a temperature of 75.2 ∘C How much ice at a temperature of -22.8 ∘C must be dropped into the water so that the final temperature of the system will be 32.0 ∘C ? Take the specific heat of liquid water to be 4190 J/kg⋅K , the specific heat of ice to be 2100 J/kg⋅K , and the heat of fusion for water to be 3.34×105 J/kg
Answer:
Explanation:
We shall apply the theory of
heat lost = heat gained .
heat lost by water = mass x specific heat x temperature diff
= .285 x 4190 x ( 75.2 - 32 ) = 51587.28 J
heat gained by ice to attain temperature of zero
= m x 2100 x 22.8 = 47880 m
heat gained by ice in melting = latent heat x mass
= 334000m
heat gained by water at zero to become warm at 32 degree
= m x 4190 x 32 = 134080 m
Total heat gained = 515960 m
So
515960 m = 51587.28
m = .1 kg
= 100 gm
A particle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 6.00î m/s. (a) Find the vector position of the particle at any time t (where t is measured in seconds). ( t î + t2 ĵ) m (b) Find the velocity of the particle at any time t. ( î + t ĵ) m/s (c) Find the coordinates of the particle at t = 4.00 s. x = m y = m (d) Find the speed of the particle at t = 4.00 s. m/s
Answer:
a)
6t i + t² j m
b)
6 i + 2 t j m/s
c)
x = 24 m and y = 16 m
d)
10 m/s
Explanation:
Explanation is given in the attached document.
An inductor is connected to the terminals of a battery that has an emf of 12.0 V and negligible internal resistance. The current is 4.86 mA at 0.700 ms after the connection is completed. After a long time the current is 6.80 mA.
What are
(a) the resistance R of the inductor and
(b) the inductance L of the inductor?
Answer:
a) 1764.71 ohms
b) 1.73 H
Explanation:
From the question, we can identify the following parameters;
Vo =12 V , i = 4.86 mA, t =0.700 ms, io =6.80 mA
(a) Indcued emf V = L di/dt =0
From ohms law Vo = ioR
R = 12/6.80*0.001
R=1764.71 ohms
(b) For LR circuit
i =io (1-e^-t/T)
Time constant T = L/R
4.86 = 6.80 (1-e^-0.7*10^-3/T)
divide both side by 6.8
0.715 = 0.0007/T
L/R = 0.0007/0.715
L/R = 0.000979020979
Substitute R from above
L = 0.000979020979 * 1764.71
L =1.73 H
A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s at a distance of 120 m from its starting point. When the truck has travelled a distance of 60 m from its starting point, its speed is v1 m/s.
Which of the following statements concerning v1 is true?
a. v1< 12.5m/s
b. v1= 12.5m/s
c. v1 >12.5m/s
Answer:
the correct answer is c v₁> 12.5 m / s
Explanation:
This is a one-dimensional kinematics exercise, let's start by finding the link to get up to speed.
v² = v₀² + 2 a₁ x
as part of rest v₀ = 0
a₁ = v² / 2x
a₁ = 25² / (2 120)
a₁ = 2.6 m / s²
now we can find the velocity for the distance x₂ = 60 m
v₁² = 0 + 2 a1 x₂
v₁ = Ra (2 2,6 60)
v₁ = 17.7 m / s
these the speed at 60 m
we see that the correct answer is c v₁> 12.5 m / s
Which of the following is true about a planet orbiting a star in the uniform circular motion? B. The speed of the plant is always changing. C. The velocity of the plant is constant. D. The acceleration vector of the plant points towards the center of the circle.
Explanation:
If an object moves in a circular path, this type of motion is called uniform circular motion. The speed of the object is constant in this type of motion.As velocity is a vector quantity, it keeps on changing. It can be given by drawing tangent to the circle.Acceleration acts towards the center of circle. Hence, out of the given four options, the correct statement about a planet orbiting a star in uniform circular motion is (d) i.e. The acceleration vector of the plant points towards the center of the circle.Answer:
D. The acceleration vector of the plant points towards the center of the circle. (Works for Apex)
Explanation:
Velocity is always changing because to have velocity you require direction and the direction is always changing when orbiting therefore A and C are not true. And B is not true because the speed of the planet is not accelerating when orbiting around a star.