Answer:
Car 3
Explanation:
A rectangular block of metal of resistivity rho has dimensions d x 2d x 3d . A potential difference V is to be applied between two opposite faces of the block.
(A) To which two faces of the block should the potential difference be applied to give the maximum current density?
(B) What is the maximum current density j(subscript max) from part A?
(C) To which two faces of the block should the potential difference be applied to give the maximum current?
1. to the faces that area a distance d apart
2. to the faces that are a distance 2d apart
3. to the faces that are a distance 3d apart
Answer:
a) The potential difference should be applied to the d dimension face.
b) The maximum current density j = V/3ρd
c) 3. to the faces that are a distance 3d apart
Explanation:
a) Current density is the ratio of current flowing through a conductor, and cross-sectional area of the conductor. mathematically, it is written as
j = I/A
where I is the electric current, and
A is the area of the conductor.
From the equation, we can see that reducing the area of the conductor will increase the current density for a given amount of current passing through the conductor. The face d wide will give the least cross-sectional area of current flow.
b) current density can be gotten from
j = σE ....equ 1
where σ is the conductivity of the conductor which is the inverse of resistivity ρ. this means that
σ = 1/ρ ....equ 2
where ρ is the resistivity of the conductor
E is the electric field and is the volt through the conductor per unit length of the conductor
in this case, the maximum current density will be when the length is length 3d, and the volt is the potential difference V
therefore,
E = V/3d ....equ 3
substituting equ 2 and equ 3 in equ1, we'll have
the maximum current density j = V/3ρd
c) To get the maximum current, the potential difference should be applied to the faces that are 3d wide apart because the resistance of a conductor varies inversely as the cross-sectional area. The maximum current will be gotten when the resistance is at its minimum, and the minimum resistance will be gotten with the most cross-sectional area. The 3d wide face will give the maximum cross-sectional area.
Determine the smallest distance x to a position where 450-nm light reflected from the top surface of the glass interferes constructively with light reflected from the silver coating on the bottom. The light changes phase when reflected at the silver coating.
A wedge of glass of refractive index 1.64 has a silver coating on the bottom, as shown in the image attached below.
Determine the smallest distance x to a position where 450-nm light reflected from the top surface of the glass interferes constructively with light reflected from the silver coating on the bottom. The light changes phase when reflected at the silver coating.
Answer:
the smallest distance x = 2.74 × 10⁻³ m or 2.74 mm
Explanation:
From the given information:
The net phase change is zero because both the light ray reflecting from the air-glass surface and silver plate undergo a phase change of [tex]\dfrac{\lambda}{2}[/tex] , as such the condition for the constructive interference is:
nΔy = mλ
where;
n = refractive index
Δy = path length (inside the glass)
So, from the diagram;
[tex]\dfrac{y}{x}=\dfrac{10^{-5} \ m}{0.2 \ m}[/tex]
[tex]\dfrac{y}{x} = 5 \times 10^{-5}[/tex]
[tex]y = 5 \times 10^{-5} x[/tex]
Now;
Δy can now be = 2 ( 5 × 10⁻⁵ [tex]x[/tex])
Δy =1 × 10⁻⁴[tex]x[/tex]
From nΔy = mλ
n( 1 × 10⁻⁴[tex]x[/tex] ) = mλ
[tex]x = \dfrac{m \lambda}{n \times 1 \times 10^{-4} }[/tex]
when the thickness is minimum then m = 1
Thus;
[tex]x = \dfrac{1 \times 450 \times 10^{-9} \ m}{1.64 \times 1 \times 10^{-4} }[/tex]
x = 0.00274 m
x = 2.74 × 10⁻³ m or 2.74 mm
Answer: B. The surface of the coating is rough, so light that shines on it gets scattered in many directions.
Explanation: On Edge!!!!!!!!!!!!!!!!!!!!
Two charges, +9 µC and +16 µC, are fixed 1 m apart, with the second one to the right. Find the magnitude and direction of the net force (in N) on a −7 nC charge when placed at the following locations.
(a) halfway between the two
(b) half a meter to the left of the +9 µC charge
(c) half a meter above the +16 µC charge in a direction perpendicular to the line joining the two fixed charges (Assume this line is the x-axis with the +x-direction toward the right. Indicate the direction of the force in degrees counterclockwise from the +x-axis.)
Answer:
A) 1.76U×10⁻³N
B) 2.716×10⁻³N
C) 264.5⁰
Explanation:
See detailed workings for (a), (b), (c) attached.
A car is moving along a road at 28.0 m/s with an engine that exerts a force of
2,300.0 N on the car to balance the drag and friction so that the car maintains a
constant speed. What is the power output of the engine?
Answer:
Power = Force × Distance/time
Power = Force × Velocity
Power = 2,300.0 N × 28.0 m/s²
Power = 64400 Nm/s
Explanation:
First show the formula of Power
Re-arrange formula and used to work out Power
Pretty simple stuff!
Hope this Helps!!
A helicopter blade starts to rotate from rest with a constant angular acceleration of \alpha=\:α = 0.98 radians/second2. How many revolutions will the blade make before it is rotating at 500 revolutions per minute (RPM)?
Answer:
n = 223 revolutions
Explanation:
It is given that,
The angular acceleration of a helicopter blade, [tex]\alpha =0.98\ rad/s^2[/tex]
Initial speed of the helicopter blade, [tex]\omega_i=0[/tex]
The final speed of the blade, [tex]\omega_f=500\ rpm=500\times \dfrac{2\pi}{60}\ rad/s=52.35\ rad/s[/tex]
We need to find the number of revolutions. Firstly we will find the angle turned by the blade. Let the angle is [tex]\theta[/tex]. So,
[tex]\omega_f^2-\omega_i^2=2\alpha \theta[/tex]
[tex]\theta=\dfrac{\omega_f^2}{2\alpha}[/tex]
[tex]\theta=1398.22\ rad [/tex]
Let there are n number of revolutions made by the blade. So,
[tex]n=\dfrac{\theta}{2\pi}\\\\n=\dfrac{1398.22}{2\pi}\\\\n=222.53\ rev[/tex]
or
n = 223 rev
So, there are 223 revolutions.
why was the observation of the double-slit interference pattern more convincing evidence for the wave theory of light than the observation of diffraction
Answer:
The double slit experiment showed for the first time that light can be interfered, producing bands of light and dark fringes on a screen. This phenomenon was a unique and typical characteristic of waves.
Explanation:
Th double slit experiment by Thomas Young proved, and sealed for the first time the wave nature of light; showing that light just as any other wave can produce interference which was a unique, typical phenomenon of waves. The Interference of light was shown by allowing light to pass through narrow slits and superimpose on a wall or screen, at a distance away from the slit, producing a clear pattern of light and dark fringes. This was the first experiment to proof that darkness can be produced by the addition of light on light. Interference is accompanied by a spatial redistribution of the optical intensity without violation of power conservation. The phenomenon of interference proved the intuitive ideas of Huygens regarding the wave nature of light against Newton's particle nature of light theory.
An inventor claims to have developed a food freezer that, in steady-state conditions, requires a power input of 0.25 kW to extract energy by heat transfer at a rate of 3050 J/s from the freezer contents, which are at a temperature of 270 K. Determine if this claim is real considering an ambient temperature of 293 K. (a) Can the freezer operation in such conditions
Answer:
The inventors claim is not real
a) No the the freezer cannot operate in such conditions
Explanation:
From the question we are told that
The power input is [tex]P_i = 0.25 kW = 0.25 *10^{3} \ W[/tex]
The rate of heat transfer [tex]J = 3050 J/s[/tex]
The temperature of the freezer content is [tex]T = 270 \ K[/tex]
The ambient temperature is [tex]T_a = 293 \ K[/tex]
Generally the coefficient of performance of a refrigerator at idea conditions is mathematically represented as
[tex]COP = \frac{T }{Ta - T}[/tex]
substituting values
[tex]COP = \frac{270 }{293 - 270}[/tex]
[tex]COP =11.7[/tex]
Generally the coefficient of performance of a refrigerator at real conditions is mathematically represented as
[tex]COP = \frac{J}{P_i}[/tex]
substituting values
[tex]COP = \frac{3050}{0.25 *10^{3}}[/tex]
[tex]COP = 12.2[/tex]
Now given that the COP of an ideal refrigerator is less that that of a real refrigerator then the claims of the inventor is rejected
This is because the there are loss in the real refrigerator cycle that are suppose to reduce the COP compared to an ideal refrigerator cycle where there no loss that will reduce the COP
A skater spins at 3rev/s when she stretches her arms outward. If she keeps her fists on her chest she can spin at 4.5rev/s and her body inertia is 3kg.m2. What is her body inertia when she stretches her arms outward?
Answer:
Body inertia I = 4.5 kg/m^2
Explanation:
Here, we want to calculate the body inertia when the arms are stretched outwards.
We know from the question that angular momentum is conserved
Thus;
I * 3 = 4.5 * 3
I = 4.5 kg/m^2
Red light is bent the least of all colors as it passes through a prism. What does this tell you about red light? It has a short wavelength. It has a long wavelength. It has a high intensity. It has a low intensity.
Answer:
Longest wavelength, lowest intensity
Explanation:
Answer:
It has a long wavelength
Explanation:
GRADPOINT
A cylindrical capacitor is made of two thin-walled concentric cylinders. The inner cylinder has radius 5 mm , and the outer one a radius 11 mm . The common length of the cylinders is 160 m . What is the potential energy stored in this capacitor when a potential difference 6 V is
Answer:
The potential energy is [tex]PE = 2.031 *10^{-7} \ J[/tex]
Explanation:
From the question we are told that
The inner radius is [tex]r_i = 5 \ mm = 0.005 \ m[/tex]
The outer radius is [tex]r_o = 11 \ mm = 0.011 \ m[/tex]
The common length is [tex]l = 160 \ m[/tex]
The potential difference is [tex]V = 6 \ V[/tex]
Generally the capacitance of the cylindrical capacitor is mathematically represented as
[tex]C = \frac{2 \pi * k * \epsilon_o }{ ln \frac{ r_o }{r_i} } * l[/tex]
Where [tex]\epsilon _o[/tex] is the permitivity of free space with the values [tex]\epsilon _o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
and k is the dielectric constant of the dielectric material here the dielectric material is free space so k = 1
Substituting values
[tex]C = \frac{2* 3.142 * 1 * 8.85*10^{-12} }{ ln \frac{ 0.011}{0.005} } * 160[/tex]
[tex]C = 1.129 *10^{-8} \ F[/tex]
The potential energy stored is mathematically represented as
[tex]PE = \frac{1}{2} * C * V ^2[/tex]
substituting values
[tex]PE = 0.5 * 1.129 *10^{-8} * (6)^2[/tex]
[tex]PE = 2.031 *10^{-7} \ J[/tex]
A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm light from a He-Ne laser. Because the lecture hall is very large, the interference pattern will be projected on a wall that is 4.0 m from the slits. For easy viewing by all students in the class, the professor wants the distance between the m
What slit separation is required in order to produce the desired interference pattern?
d=________m
Note: if the professor wants the distance between the m = 0 and m = 1 maxima to be 25 cm
Answer:
d = 1.0128×10⁻⁵m
Explanation:
given:
length L = 4.0m
maximum distance between m = 0 and m = 1 , y = 25cm = 0.25m
wavelength λ = 633nm = 633×10⁻⁹m
note:
dsinθ = mλ (constructive interference)
where d is slit seperation, θ is angle of seperation , m is order of interference , and λ is wavelength
for small angle
sinθ ≈ tanθ
[tex]d (\frac{y}{L}) =[/tex] mλ
[tex]d (\frac{y}{L}) = (1)(633nm)[/tex]
[tex]d(\frac{0.25}{4} ) = (1)(633nm)[/tex]
d = 1.0128×10⁻⁵m
The voltage and power ratings of a particular light bulb, which are its normal operating values, are 110 V and 60 W. Assume the resistance of the filament of the bulb is constant and is independent of operating conditions. If the light bulb is operated at a reduced voltage and the power drawn by the bulb is 36 W. What is the operating voltage of the bulb?
a. 78 V
b. 72 V
c. 66 V
d. 90 V
e. 85 V
Answer:
c. 66 V
Explanation:
p =IV
I =P/V
P1/V1=P2/V2
60/110=36/V2
0.55 = 36/V2
V2 =36/0.55 = 65.5V
V2 = 66V
A scientist is testing the seismometer in his lab and has created an apparatus that mimics the motion of the earthquake felt in part (a) by attaching the test mass to a spring. If the test mass weighs 13 N, what should be the spring constant of the spring the scientist use to simulate the relative motion of the test mass and the ground from part (a)?
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]a_{max} = 0.00246 \ m/s^2[/tex]
b
[tex]k =722.2 \ N/m[/tex]
Explanation:
From the question we are told that
The amplitude is [tex]A = 1.8 \ cm = 0.018 \ m[/tex]
The period is [tex]T = 17 \ s[/tex]
The test weight is [tex]W = 13 \ N[/tex]
Generally the radial acceleration is mathematically represented as
[tex]a = w^2 r[/tex]
at maximum angular acceleration
[tex]r = A[/tex]
So
[tex]a_{max} = w^2 A[/tex]
Now [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{2 * \pi }{T}[/tex]
Therefore
[tex]a_{max} = [\frac{2 * \pi}{T} ]^2 * A[/tex]
substituting values
[tex]a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018[/tex]
[tex]a_{max} = 0.00246 \ m/s^2[/tex]
Generally this test weight is mathematically represented as
[tex]W = k * A[/tex]
Where k is the spring constant
Therefore
[tex]k = \frac{W}{A}[/tex]
substituting values
[tex]k = \frac{13}{0.018}[/tex]
[tex]k =722.2 \ N/m[/tex]
An electron initially at rest is accelerated over a distance of 0.210 m in 33.3 ns. Assuming its acceleration is constant, what voltage was used to accelerate it
Answer:
V = 451.47 volts
Explanation:
Given that,
Distance, d = 0.21 m
Initial speed, u = 0
Time, t = 33.3 ns
Let v is the final velocity. Using second equation of motion as :
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
a is acceleration, [tex]a=\dfrac{v-u}{t}[/tex] and u = 0
So,
[tex]d=\dfrac{1}{2}(v-u)t[/tex]
[tex]v=\dfrac{2d}{t}\\\\v=\dfrac{2\times 0.21}{33.3\times 10^{-9}}\\\\v=1.26\times 10^7\ m/s[/tex]
Now applying the conservation of energy i.e.
[tex]\dfrac{1}{2}mv^2=qV[/tex]
V is voltage
[tex]V=\dfrac{mv^2}{2q}\\\\V=\dfrac{9.1\times 10^{-31}\times (1.26\times 10^7)^2}{2\times 1.6\times 10^{-19}}\\\\V=451.47\ V[/tex]
So, the voltage is 451.47 V.
A person can survive a feet-first impact at a speed of about 12 m/s (27 mi/h) on concrete, 15 m/s (34 mi/h) on soil, and 34 m/s (76 mi/h) on water. What is the reason for the different values for different surfaces
Answer:
Different surfaces have different impact force during collision which depends on the time it takes the person to come to rest after collision.
Explanation:
Given;
speed on concrete = 12 m/s (27 mi/h)
speed on soil = 15 m/s (34 mi/h)
speed on water = 34 m/s (76 mi/h)
The impact force on this person during collision is rate of change of momentum;
[tex]F = \frac{\delta P}{\delta t}[/tex]
During collision, the force exerted on this person depends on how long the collision lasts; that is, how long it takes for this person to come to rest after collision with each of the surfaces.
The longer the time of collision, the smaller the force exerted by each.
It takes shorter time for the person to come to rest on concrete surface than on soil surface, also it takes shorter time for the person to come to rest on soil surface than on water surface.
As a result of the reason above, the force exerted on the person during collision by the concrete surface is greater than that of soil surface which is greater than that of water surface.
Sally who weighs 450 N, stands on a skate board while roger pushes it forward 13.0 m at constant velocity on a level straight street. He applies a constant 100 N force.
Work done on the skateboard
a. Rodger Work= 0J
b. Rodger work= 1300J
c. sally work= 1300J
d. sally work= 5850J
e. rodger work= 5850J
Answer:
b. Rodger work = 1300 J
Explanation:
Work done: This can be defined as the product of force and distance along the direction of the force.
From the question,
Work is done by Rodger using a force of 100 N in pushing the skateboard through a distance of 13.0 m.
W = F×d............. Equation 1
Where W = work done, F = force, d = distance.
Given: F = 100 N, d = 13 m
Substitute these values into equation 1
W = 100(13)
W = 1300 J.
Hence the right option is b. Rodger work = 1300 J
The first step to merging is entering the ramp and _____.
A. honking to indicate your location
B. matching your speed
C. signaling your intent
D. telling your passengers where you're going
Answer:
B. matching your speed
Explanation:
To merge safely, you must identify a gap in traffic and match your speed to the speed of the gap. Before you make your move to fill the gap, you should signal your intent.*
_____
* At least one resource says "The first step ... is to make sure you're traveling at the same speed ..." Then it goes on to say "Use your indicator. Do it early ...." The accompanying animation shows blinkers being activated on the ramp before the merge lane is entered. Apparently, "the first step" is not necessarily the first thing you do.
Answer:
It's C "signaling your intent"
Explanation:
The key thing to look at is they are asking the rest of the first step and that;s C
A block of mass 27.00 kg sits on a horizontal surface with, coefficient of kinetic
friction 0.50 and a coefficient of static friction 0.65. How much force is required to
get the block moving?
Answer:
The force is [tex]F = 172 \ N[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m_b = 27.0 \ kg[/tex]
The coefficient of static friction is [tex]\mu_s = 0.65[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.50[/tex]
The normal force acting on the block is
[tex]N = m * g[/tex]
substituting values
[tex]N = 27 * 9.8[/tex]
[tex]N = 294.6 \ N[/tex]
Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as
[tex]F_f = \mu_s * N[/tex]
substituting values
[tex]F_f = 0.65 * 264.6[/tex]
[tex]F_f = 172 \ N[/tex]
Now for this block to move the force require is equal to [tex]F_f[/tex] i.e
[tex]F= F_f[/tex]
=> [tex]F = 172 \ N[/tex]
A proton moving at 4.80 106 m/s through a magnetic field of magnitude 1.74 T experiences a magnetic force of magnitude 7.00 10-13 N. What is the angle between the proton's velocity and the field? (Enter both possible answers from smallest to largest. Enter only positive values between 0 and 360.)
Answer:
31.55° and 148.45°
Explanation:
Formula for calculating the force experiences by the proton placed in a magnetic field is as expressed below;
F = qvBsinθ where;
F is the magnetic force experienced by the proton
q is the charge on the proton
v is the velocity of the proton
B is the magnetic field
θ is the angle between the proton's velocity and the field (Required)
Given parameters
F = 7.00 * 10⁻¹³N
q = 1.602*10⁻¹⁹C
v = 4.80 * 10⁶ m/s
B = 1.74 T
θ =?
From the formula F = qvBsinθ;
sinθ = F/qvB
sinθ = 7.00 * 10⁻¹³/1.602*10⁻¹⁹* 4.80 * 10⁶*1.74
sinθ = 7.00 * 10⁻¹³/13.38*10⁻¹³
sinθ = 0.5231689 * 10⁰
sinθ = 0.5231689
θ = sin⁻¹0.5231689
θ = 31.55°
The following are the positive values of the angle between 0° and 360°
Sin is positive in the first and second quadrant. In the second quadrant the angle is equal to 180°-31.55° = 148.45°.
Hence the possible values of the angle from smallest to largest are 31.55° and 148.45°
A spherical capacitor contains a charge of 3.40 nC when connected to a potential difference of 240.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 4.10 cm.
Calculate:
a. The capacitance
b. The radius of the inner sphere.
c. The electric field just outside the surface of the inner sphere.
Answer:
A) 1.4167 × 10^(-11) F
B) r_a = 0.031 m
C) E = 3.181 × 10⁴ N/C
Explanation:
We are given;
Charge;Q = 3.40 nC = 3.4 × 10^(-9) C
Potential difference;V = 240 V
Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m
A) The formula for capacitance is given by;
C = Q/V
C = (3.4 × 10^(-9))/240
C = 1.4167 × 10^(-11) F
B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.
C = (4πε_o)/(1/r_a - 1/r_b)
Rearranging, we have;
(1/r_a - 1/r_b) = (4πε_o)/C
ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m
Plugging in the relevant values, we have;
(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))
(1/r_a) - 24.3902 = 7.8501
1/r_a = 7.8501 + 24.3902
1/r_a = 32.2403
r_a = 1/32.2403
r_a = 0.031 m
C) Formula for Electric field just outside the surface of the inner sphere is given by;
E = kQ/r_a²
Where k is a constant value of 8.99 × 10^(9) Nm²/C²
Thus;
E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²
E = 3.181 × 10⁴ N/C
Suppose you are looking into the end of a long cylindrical tube in which there is a uniform magnetic field pointing away from you. If the magnitude of the field is decreasing with time the direction of the induced magnetic field is
Answer:
If the magnitude of the field is decreasing with time the direction of the induced magnetic field is CLOCKWISE
Explanation
This is because If the magnetic field decreases with time, the electric field will be produced in order to oppose the change in line with lenz law. Thus The right hand rule can be applied to find that the direction of electric field is in the clockwise direction.
Help me with these question and please explainnn
Explanation:
1. Impulse = change in momentum
J = Δp
J = mΔv
In the x direction:
Jₓ = mΔvₓ
Jₓ = (0.40 kg) (30 m/s cos 45° − (-20 m/s))
Jₓ = 16.5 kg m/s
In the y direction:
Jᵧ = mΔvᵧ
Jᵧ = (0.40 kg) (30 m/s sin 45° − 0 m/s)
Jᵧ = 8.49 kg m/s
The magnitude of the impulse is:
J = √(Jₓ² + Jᵧ²)
J = 18.5 kg m/s
The average force is:
FΔt = J
F = J/Δt
F = 1850 N
2. Momentum is conserved.
m₁u₁ + m₂u₂ = (m₁ + m₂) v
In the x direction:
(1000 kg) (0 m/s) + (1500 kg) (-12 m/s) = (1000 kg + 1500 kg) vₓ
vₓ = -7.2 m/s
In the y direction:
(1000 kg) (20 m/s) + (1500 kg) (0 m/s) = (1000 kg + 1500 kg) vᵧ
vᵧ = 8 m/s
The magnitude of the final speed is:
v = √(vₓ² + vᵧ²)
v = 10.8 m/s
3. Momentum is conserved.
m₁u₁ + m₂u₂ = (m₁ + m₂) v
In the x direction:
(0.8 kg) (18 m/s cos 45°) + (0.36 kg) (9.0 m/s) = (0.8 kg + 0.36 kg) vₓ
vₓ = 11.6 m/s
In the y direction:
(0.8 kg) (-18 m/s sin 45°) + (0.36 kg) (0 m/s) = (0.8 kg + 0.36 kg) vᵧ
vᵧ = -8.78 m/s
The magnitude of the final speed is:
v = √(vₓ² + vᵧ²)
v = 14.5 m/s
An unknown additional charge q3q3q_3 is now placed at point B, located at coordinates (0 mm, 15.0 mm ). Find the magnitude and sign of q3q3q_3 needed to make the total electric field at point A equal to zero.
Answer:
0.3nanocouloumb
Explanation:
Pls see attached file
Three identical resistors are connected in series to a battery. If the current of 12 A flows from the battery, how much current flows through any one of the resistors
Answer:
Current that flows through any one of the resistors has a value of 12 amperes.
Explanation:
When a group of resistors are connected in series, the same current flows through each resistor. According to the Ohm's Law, the circuit can be represented as follows:
[tex]V_{batt} = i\cdot (R_{1}+R_{2}+R_{3})[/tex]
[tex]i = \frac{V_{batt}}{R_{1}+R_{2}+R_{3}}[/tex]
Where:
[tex]V_{batt}[/tex] - Voltage of the battery, measured in volts.
[tex]R_{1}[/tex], [tex]R_{2}[/tex], [tex]R_{3}[/tex] - Electric resistance of the first, second and third resistors, measured in ohms.
[tex]i[/tex] - Current, measured in amperes.
If [tex]R_{1} = R_{2} = R_{3} = R[/tex], then:
[tex]i = \frac{V_{batt}}{3\cdot R}[/tex]
Current that flows through any one of the resistors has a value of 12 amperes.
The current flows via any of the resistors should have a value of 12 amperes.
Ohm law:At the time When a group of resistors are linked in series, so there is a similar current flow via each resistor.
Here the circuit should be
vbatt = i.(R1 + R2+ R3)
i = Vbatt/R1 + R2 + R3
here
Vbatt means the voltage of the battery
R1,R2, and R3 means the resistance of the first, second and third resistors
I means the current
So, in the case when
R1 = R2 = R3 = R
So,
i = Vbatt/3.R
Learn more about current here: https://brainly.com/question/14956680
If the speed of a "cheetah" is 150 m / s. How long does it take to cover 800 m?
Answer:
5.33333... seconds
Explanation:
800 divided by 150 is equal to 5.33333... because it is per second that the cheetah moves at 150miles, the answer is 5.3333.....
You set two parallel slits 0.1 mm apart at a distance of 2 m from a screen and illuminate them with light of wavelength 450 nm. The distance between a bright spot in the interference pattern and the dark spot adjacent to it is Group of answer choices
Answer:
Δx = 9 x 10⁻³ m = 9 mm
Explanation:
The formula for fringe spacing in Young's Double Slit Experiment is given as follows:
Δx = λL/d
where,
Δx = fringe spacing = ?
λ = wavelength of light = 450 nm = 450 x 10⁻⁹ m
L = Distance between slits and screen = 2 m
d = distance between slits = 0.1 mm = 0.1 x 10⁻³ m
Therefore,
Δx = (450 x 10⁻⁹ m)(2 m)/(0.1 x 10⁻³ m)
Δx = 9 x 10⁻³ m = 9 mm
"Pipe A has length L and is open at one end and closed at the other. Pipe B is open at both ends and has length 2L. Which harmonic of pipe B matches in frequency the fundamental of pipe A?"
Answer:
Explanation:
length of pipe A is L
frequency of fundamental note of pipe A which is closed organ pipe
= velocity of sound / 4 x length of pipe
= V / 4 x L where V is velocity of sound .
Similarly frequency of fundamental note of pipe B which is open organ pipe
= velocity of sound / 2 x length of pipe
= V / (2 x 2L) = V / 4L , This is also called first harmonic
so fundamental frequency of pipe A will be equal to first harmonic of pipe B .
action and reaction are equal in magnitude and opposite in direction.Then why do they not balance each other.
Explanation:
Newton's third law of motion states that every action has an equal and opposite reaction. This means that forces always act in pairs. Action and reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don't cancel out.
Determine the position in the oscillation where an object in simple harmonic motion: (Be very specific, and give some reasoning to your answer.) has the greatest speed has the greatest acceleration experiences the greatest restoring force experiences zero restoring force g
Answer:
Explanation:
The greatest speed is attained at middle point or equilibrium point or where displacement from equilibrium point is zero .
When the object remains at one of the extreme point it experiences greatest acceleration but at that point velocity is zero . Due to acceleration , its velocity goes on increasing till it come to equilibrium point . At this point acceleration becomes zero . After that its velocity starts decreasing because of negative acceleration . Hence at middle point velocity is maximum .
The greatest acceleration is attained at maximum displacement or at one of the two extreme end .
Greatest restoring force too will be at position where acceleration is maximum because acceleration is produced by restoring force .
Restoring force is proportional to displacement or extension against restoring force . So it will be maximum when displacement is maximum .
Zero restoring force exists at equilibrium position or middle point or at point where displacement is zero . It is so because acceleration at that point is zero .
The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming to rest. The total mass of the boat and trailer is 1 Mg. Determine the constant horizontal force developed in the coupling C, and the friction force developed between the tires of the truck and the road during this time.
Answer:
constant horizontal force developed in the coupling C = 11.25KN
the friction force developed between the tires of the truck and the road during this time is 33.75KN
Explanation:
See attached file
The friction force between the tires of the truck and the road is 22500 N.
Calculating the friction force:It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.
Distance traveled before coming to rest, s = 10m
The final velocity of the truck will be zero, v = 0
When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.
The frictional force is given by:
f = -ma
the acceleration of the truck = -a
The negative sign indicates that the acceleration is opposite to the motion.
Applying the third equation of motion we get:
v² = u² -2as
0 = 15² - 2×a×10
225 = 20a
a = 11.25 m/s²
So the magnitude of frictional force is:
f = ma = 2000 × 11.25 N
f = 22500 N
Learn more about friction force:
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