Answer:
8 mm
Explanation:
From the information given:
The Ampere circuital law can be used to estimate the magnetic field strength at two points when the distance is less than the radius and when the distance is greater than the radius.
when the distance is less than the radius ; we have:
[tex]B_1 = \dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2}[/tex]
when the distance is greater than the radius; we have:
[tex]B_2 = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
Equating both equations together ; we have :
[tex]\dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2} = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
[tex]\dfrac{1}{R}= \dfrac{r}{d^2}[/tex]
[tex]R= \dfrac{d^2}{r}[/tex]
where; d = radius of the wire and r = distance;
[tex]R =\dfrac{4^2}{2}[/tex]
[tex]R =\dfrac{16}{2}[/tex]
R = 8 mm
importance of SI system in points
Answer:
SI unit is an international system of measurements that are used universally in technical and scientific research to avoid the confusion with the units. Having a standard unit system is important because it helps the entire world to understand the measurements in one set of unit system.
During World War II, mass spectrometers were used to separate the radioactive uranium isotope U-235 from its far more common isotope, U-238. Estimate the radius of the circle traced out by a singly ionized lead atom moving at the same speed.
Answer:
21.55 m
Explanation:
A 500 kg rocket sled is coasting in reverse at 10 m/s (to the left). It then turns on its rocket engines for 10.0 s, with a thrust of 1500 N (to the right). What is its final velocity? (Remember velocity has magnitude and direction)
Explanation:
F = ma
[tex]a = \frac{f}{m} [/tex]
[tex]a = \frac{1500}{500} = 3[/tex]
[tex]a = \frac{v2 - v1}{t} [/tex]
[tex]3 = \frac{v2 - 10}{10} [/tex]
v2 (final) = 40 m/s to the right direction
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low mass sail and the energy and momentum of sunlight for propulsion.
(a) Should the sail be absorbing or reflective? Why?
(b) How large a sail is necessary to propel a 10000kg
spacecraft against the gravitational force of the sun? Express your result in square kilometers.
(c) Explain why your answer to part (b) is independent of the distance from the sun.
The gravitational constant is G=6.67×10−11m3⋅s−2⋅kg−1.
The mass of the sun is Ms=1.99×1030kg.
Answer:
The complete question is
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is 3.9 x 10^26 W. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.
a) The sail should be reflective because, an incident electromagnetic wave, in this case, light wave, impacts twice the energy density on a reflective sail, and hence twice the force on a totally reflective sail as would be impacted on a sail that is totally absorbing.
For totally reflective, F = (2I/c)A ....1
for totally reflective, F = (I/c)A ....2
where I is the intensity of the light
c is the speed of light = 3 x 10^8 m/s
A is the area the sail
b) The intensity of the light from the sun = power/area
==> I = [tex]\frac{3.9*10^{26}}{4\pi r^{2} }[/tex]
where r is the distance from the sun and the sail
The Force from the sail from equation 1 is therefore
[tex]F[/tex] = [tex]\frac{2*3.9*10^{26}*A}{4\pi r^{2} *3*10^{8}}[/tex] = [tex]2.069*10^{17}\frac{A}{r^{2}}[/tex]
gravitational force between the sail and the sun [tex]F_{g}[/tex] = [tex]\frac{GMm}{r^{2}}[/tex]
where
G is the gravitational constant = 6.67 x 10^−11 m^3⋅s−2⋅kg−1.
m is the mass of the sail = 10000 kg
M is the mass of the sun = 1.99 x 10^30 kg.
==> [tex]F_{g}[/tex] = [tex]\frac{6.67*10^{-11}*1.99*10^{30}*10000}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
Equating the forces, we have
[tex]2.069*10^{17}\frac{A}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
the distance cancels out
A = (1.33 x 10^24)/(2.069 x 10^17) = 6428226.196 m^2
==> 6428.2 km^2
c) The force of the solar radiation is proportional to the intensity of the sun from the light, and the intensity is inversely proportional to the square of the distance from the source. Also, the force of gravitation is inversely proportional to the square of the distance, so they both cancel out.
Which statement about kinetic and static friction is accurate?
Static friction is greater than kinetic friction, and they both act in conjunction with the applied force.
Kinetic friction is greater than static friction, and they both act in conjunction with the applied force.
Kinetic friction is greater than static friction, but they both act opposite the applied force.
Static friction is greater than kinetic friction, but they both act opposite the applied forcr
Answer:
Static friction is greater than kinetic friction, but they both act opposite the applied force.
Explanation:
Newton's 3rd law states that every action has and equal but opposite reaction.
If an object has static friction, that means it stays in one spot, and it takes a great amount of force to get it moving.
Once the object is moving it has kinetic friction, but it's easier to keep it moving unless you are trying to stop it.
The equal but opposite reaction to something moving it is stopping it, and the equal but opposite reaction to stopping something is moving it.
The same amount of force used to move/stop something is used to stop/move it.
Static friction is greater than kinetic friction, but they both act opposite the applied force.
Friction is the force that opposes motion. Frictional force always acts in opposition to the direction of motion.
There are two kinds of friction;
Static frictionDynamic frictionSince more forces tend to act on a body at rest and prevent it from getting into motion than the forces that tend to stop an already moving body, it follows that static friction is greater than kinetic friction. Both act in opposite direction to the applied force.
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The accommodation limits for a nearsighted person's eyes are 20.0 cm and 82.0 cm. When he wears his glasses, he can see faraway objects clearly. At what minimum distance is he able to see objects clearly
Answer;
26.45cm
See attached file for explanation
If 1 kg of each substance in the table absorbs 250 kJ of thermal energy,
which substance will have the smallest change in temperature?
Answer:
Water.
Explanation:
To know which of the substance that will absorbed the greatest amount of thermal energy, we'll simply determine the amount of energy absorbed by each substance.
This is illustrated below:
For Air:
Mass (M) = 1 kg
Change in temperature (ΔT) = 15 °C
Specific heat capacity (C) = 1.01 KJ/Kg°C
Heat Absorbed (Q) =..?
Q = MCΔT
Q = 1 x 1.01 x 15
Q = 15.15 KJ
Therefore, the heat absorbed by air is 15.15 KJ.
For Plastic:
Mass (M) = 1 kg
Change in temperature (ΔT) = 15 °C
Specific heat capacity (C) = 2.60 KJ/Kg°C
Heat Absorbed (Q) =..?
Q = MCΔT
Q = 1 x 2.60 x 15
Q = 39 KJ
Therefore, the heat absorbed by plastic is 39 KJ.
For Water:
Mass (M) = 1 kg
Change in temperature (ΔT) = 15 °C
Specific heat capacity (C) = 4.18 KJ/Kg°C
Heat Absorbed (Q) =..?
Q = MCΔT
Q = 1 x 4.18 x 15
Q = 62.7 KJ
Therefore, the heat absorbed by water is 62.7 KJ.
For Wood:
Mass (M) = 1 kg
Change in temperature (ΔT) = 15 °C
Specific heat capacity (C) = 1.68 KJ/Kg°C
Heat Absorbed (Q) =..?
Q = MCΔT
Q = 1 x 1.68 x 15
Q = 25.2 KJ
Therefore, the heat absorbed by Wood is 25.2 KJ.
Summary:
Substance >>>>> Heat Absorbed
Air >>>>>>>>>>>> 15.15 KJ.
Plastic >>>>>>>>> 39 KJ
Water >>>>>>>>>> 62.7 KJ
Wood >>>>>>>>>> 25.2 KJ.
From the above calculations, we can see that water will absorb the greatest amount of thermal energy.
The substance that will have the smallest change in temperature is water because it has the highest specific heat capacity.
Specific heat capacity
The specific heat capacity of each substance can be used to determine the substance with the smallest change in temperature.
Q = mcΔθ
where;
m is massc is specific heat capacityΔθ is change in temperatureΔθ = Q/mc
For airΔθ = (250)/(1.01)
Δθ = 247.5 ⁰C
For plasticΔθ = (250)/(2.6)
Δθ = 96.15 ⁰C
For waterΔθ = (250)/(4.18)
Δθ = 59.81 ⁰C
For woodΔθ = (250)/(1.68)
Δθ = 148.81 ⁰C
Thus, the substance that will have the smallest change in temperature is water because it has the highest specific heat capacity.
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A fish appears to be 2.00 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish
Answer:
2,66
Explanation:
The refractive index= real depth/ apparent depth
real depth = refractive index * apparent depth
Let's assume index for water is 1.33
real depth = 2*1,33 = 2,66
When you have a straight horizontal line on a velocity time graph, what does this tell you about the object’s motion in terms of velocity and acceleration?
Answer:
It tell you that the velocity is constant, what means that there's no acceleration
A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly above the fish
Answer:
Apparent depth (Da) = 60.15 cm (Approx)
Explanation:
Given:
Distance from fish (D) = 80 cm
Find:
Apparent depth (Da)
Computation:
We know that,
Refractive index of water (n2) = 1.33
So,
Apparent depth (Da) = D(n1/n2)
Apparent depth (Da) = 80 (1/1.33)
Apparent depth (Da) = 60.15 cm (Approx)
The apparent depth of the fish is 60 cm.
To calculate the apparent depth of the fish, we use the formula below.
Formula:
R.F(water) = Real depth(D)/Apparent depth(D')R.F = D/D'.................... Equation 1Where:
R.F = Refractive index of waterMake D' The subject of the equation.
D' = D/R.F................... Equation 2From the question,
Given:
D = 80 cmR.F = 1.333Substitute these values into equation 2
D' = 80/1.33D' = 60.01D' = 60 cmHence, the apparent depth of the fish is 60 cm
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What happens when two polarizers are placed in a straight line, one behind the other? A. They allow light to pass only if they are polarized in exactly the same direction. B. They block all light if they are polarized in exactly the same direction. C. They allow light to pass only if their directions of polarizations are exactly 90° apart. D. They block all light if their directions of polarizations are exactly 90° apart. E. They block all light if their directions of polarizations are either exactly the same or exactly 90° apart.
Answer:
C
They allow light to pass only if their directions of polarizations are exactly 90° apart.
g suppose he used an alpha particle with an energy of 8.3 MeV, what would be the speed of this alpha particle
Answer:
speed of the alpha particle is 2 x 10^7 m/s.
Explanation:
energy of alpha particle = 8.3 Mev
1 Mev = 1.602 x 10^-13 J
8.3 Mev = [tex]x[/tex]
solving, [tex]x[/tex] = 8.3 x 1.602 x 10^-13 = 1.329 x 10^-12 J
mass of a alpha particle = 6.645 x 10^−27 kg
The energy of the alpha particle is the kinetic energy KE of the alpha particle
KE = [tex]\frac{1}{2}mv^{2}[/tex]
where m is the mass of the alpha particle
v is the velocity of the alpha particle
substituting values, we have
1.329 x 10^-12 = [tex]\frac{1}{2}*6.645*10^{-27}*v^{2}[/tex]
[tex]v^{2}[/tex] = 4 x 10^14
[tex]v = \sqrt{4*10^{14} }[/tex] = 2 x 10^7 m/s
Did the kinetic frictional coefficient (for the wood/aluminum and felt/aluminum cases) vary with area of contact
Answer:
Explanation:
Friction is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose motion of one over the other.
While kinetic friction is the force that must be overcome so that a body can move with uniform speed over another.
Hence let consider one of the laws of friction which states that: '' Frictional force is independent of the area of the surfaces in contact.''
The value did not vary with area. This is because when calculating the kinetic fiction, the total contact area is not relevant and only the total weight of the system as well of as the block is put into consideration.
What is the power P of the eye when viewing an object 61.0 cm away? Assume the lens-to-retina distance is 2.00 cm , and express the answer in diopters.
Answer:
The power of the eye is 51.64 diopters
Explanation:
The power of the eye is given by;
[tex]P = \frac{1}{f} = \frac{1}{d_o} +\frac{1}{d_i}[/tex]
where;
P is the power of the eye in diopter
f is the focal length of the eye
[tex]d_o[/tex] is the distance between the eye and the object
[tex]d_i[/tex] is the distance between the eye and the image
Given;
[tex]d_o[/tex] = 61.0 cm = 0.61 m
[tex]d_i[/tex] = 2.0 cm = 0.02 m
[tex]P = \frac{1}{d_o} +\frac{1}{d_i} \\\\P = \frac{1}{0.61} + \frac{1}{0.02} \\\\P = 51.64 \ D[/tex]
Therefore, the power of the eye is 51.64 diopters.
The power P of the eye when viewing an object 61.0 cm away is 51.639D
The power of a lens is a reciprocal of its focal length and it is expressed as:
[tex]P=\frac{1}{f}[/tex]
According to the mirror formula
[tex]\frac{1}{f} =\frac{1}{d_i} +\frac{1}{d_0}[/tex]
where
[tex]d_i[/tex] is the distance from the lens to the image = 61.0cm = 0.61m
[tex]d_0[/tex] is the distance from the lens to the object = 2.00cm = 0.02m
[tex]P=\frac{1}{f} =\frac{1}{0.02} +\frac{1}{0.61}\\P=50+1.639\\P=51.639D[/tex]
Hence the power P of the eye when viewing an object 61.0 cm away is 51.639D
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You are outdoors when you hear the constant chirp of a still cricket. You start walking toward the cricket and at some point you are able to detect that the intensity of the chirp of the cricket has increased by a factor of 4. What of the following statements is true at your new position with respect to the cricket?
a. The power delivered by the sound wave you hear has doubled.
b. The speed of the sound wave emitted by the cricket has decreased by a factor of 4.
c. The distance between you and the cricket has decreased by a factor of 2
Answer:
C
Explanation:
intensity = Power delivered by the sound (Watt)/ Surounding Area (m²)
I = P/A
A = πr²
r = is the distance between you and the cricket.
so in other form we can get
I = P/πr²
let take I(1) as first intensitilynyou heard and I(2) as the increased intensity.
I(1) / I(2) = r(2)² / r(1)²
1/4 = r(2)²/r(1)²
1/2 = r(2) / r(1)
r(2) = ½ r(1)
or r(2) is decreaases by a factor of 2.
In _____ research, a group of people of one age is compared to a group of people who are another age.
Answer:
cross-sectional
Explanation:
The full definition of this is ''a research design in which several different age-groups of participants are studied at one particular point in time.''
A block weighing 400 kg rest on a horizontal surface and supports on top of it another block of weight 100 kg placed on the top of it as shown. The block W2 is attached to a vertical wall by a string 6 m long. Ifthe coefficient of friction between all surfaces is 0.25 and the system is in equilibrium find the magnitude of the horizontal force P applied to the lower block.
The horizontal force applied to the lower block is approximately 1,420.85 Newtons
The known parameters are;
The mass of the block, m₁ = 400 kg, weight, W₁ = 3,924 N
The mass of the block resting on the first block, m₂ = 100 kg, weight, W₂ = 981 N
The length of the string attached to the block, W₂, l = 6 m
The horizontal distance from the point of attachment of the second block to the block W₂, x = 5 m
The coefficient of friction between the surfaces, μ = 0.25
Let T represent the tension in the string
The upward force on W₂ due to the string = T × sin(θ)
The normal force of W₁ on W₂, N₂ = W₂ - T × sin(θ)
The tension in the string, T = N₂ × μ × cos(θ)
∴ T = (W₂ - T × sin(θ)) × μ × cos(θ)
sin(θ) = √(6² - 5²)/6
cos(θ) = 5/6
∴ T = (981 - T × √(6² - 5²)/6) × 0.25 × 5/6
Solving, we get;
T ≈ 183.27 N
The normal reaction on W₂, N₂ = T/(μ × cos(θ))
∴ N₂ = 183.27/(0.25 × 5/6) = 879.7
N₂ ≈ 879.7 N
The friction force, [tex]F_{f2}[/tex] = N₂ × μ
∴ [tex]F_{f2}[/tex] = 879.7 N × 0.25 = 219.925 N
The total normal reaction on the ground, [tex]\mathbf{N_T}[/tex] = W₁ + N₂
[tex]N_T[/tex] = 3,924 N + 879.7 N = 4,803.7 N
The friction force, on the ground [tex]\mathbf{F_T}[/tex] = [tex]\mathbf{N_T}[/tex] × μ
∴ [tex]F_T[/tex] = 4,803.7 N × 0.25 = 1,200.925 N
The horizontal force applied to the lower block, P = [tex]\mathbf{F_T}[/tex] + [tex]\mathbf{F_{f2}}[/tex]
Therefore;
P = 1,200.925 N + 219.925 N = 1,420.85 N
The horizontal force applied to the lower block, P ≈ 1,420.85 N
Given a double slit apparatus with slit distance 2 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 500 nm on the slits
Answer:
The values is [tex]m_{max} = 8001 \ bright \ spots[/tex]
Explanation:
From the question we are told that
The slit distance is [tex]d = 2 \ mm = 2*10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 500 \ nm = 500 *10^{-9} \ m[/tex]
At the first half of the screen from the central maxima
The number of bright spot according to the condition for constructive interference is
[tex]n = \frac{d * sin (\theta )}{\lambda}[/tex]
For maximum number of spot [tex]\theta = 90^o[/tex]
So
[tex]n = \frac{2*10^{-3} * sin (90 )}{500 *10^{-9}}[/tex]
[tex]n =4000[/tex]
Now for the both sides plus the central maxima we have
[tex]m_{max} = 2 * n + 1[/tex]
substituting values
[tex]m_{max} = 2 * 4000 + 1[/tex]
[tex]m_{max} = 8001 \ bright \ spots[/tex]
A radiation worker is subject to a dose of 200 mrad/h of maximum QF neutrons for one 40 h work week. How many times the yearly allowable effective dose did she receive?
Answer:
16 times.
Explanation:
The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr
Time consumed, t = 40 hr
The magnitude of Q.F for the neutrons, Q.F = 2
Thus the effective radiation dose is:
[tex]R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad[/tex]
Thus, the effective dose allowable yearly = 16 times
A student is hammering a nail into a board. Where should he hold the hammer and why?
Answer:
At the end of the handle farthest from the head of the hammer.
Explanation:
The force of the hammer is greatest the longer the radius is on a which would be the length of the handle. Simple mechanical advantage.
Five wheels are connected as shown in the figure. Find the velocity of the block “Q”, if it is known that: RA= 5 [m], RB= 10 [m], RD= 6 [m], RE=12 [m].
Answer:
-5 m/s
Explanation:
The linear velocity of B is equal and opposite the linear velocity of E.
vB = -vE
vB = -ωE rE
10 m/s = -ωE (12 m)
ωE = -0.833 rad/s
The angular velocity of E is the same as the angular velocity of D.
ωE = ωD
ωD = -0.833 rad/s
The linear velocity of Q is the same as the linear velocity of D.
vQ = vD
vQ = ωD rD
vQ = (-0.833 rad/s) (6 m)
vQ = -5 m/s
5. A body falls freely from rest. It covers as much distance in the last second of its
motion as covered in the first three seconds. The body has fallen for a time of:
B) 5s
C) 7s
D) 9s
A) 35
Answer:
B 5s
Explanation:
Because of the Displacement in the nth second of the free fall is
Snth=21g(t12−t22)
Given that (tn−tn−1)=1
Displacement in 3 seconds of the free fall
S=21gt2
S=21×10×32
S=45m
Given that: Snth=45
On solving that we get:
t1=5sec
Two motorcycles are traveling in opposite directions at the same speed when one of the cyclists blasts her horn, which has frequency of 544 Hz. The other cyclist hears the frequency as 563 Hz. If the speed of sound in air is 344 m/s, what is the speed of the motorcycles
Answer:
6ms^-1
Explanation:
Given that the frequency difference is
( 563- 544) = 19
So alsoThe wavelength of each wave is = v/f = 344 /544
and there are 19 of this waves
So it is assumed that each motorcycle has moved 0.5 of this distance
in one second thus the speed of the motorcycles will be
=> 19/2 x 344/544 = 6.0 m/s
Suppose you drop paperclips into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating paper clips, explain whether the momentum and kinetic energy increase, decrease, or stay the same.
Answer:
Stay the same
Explanation:
Since, friction is negligible:
Initial Momentum = Final Momentum
Initial KE = Final KE
m1 * v1 = m2 * v2
When m increases v decreases.
The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.
What is friction?Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.
Given:
The paperclips into an open cart rolling along a straight horizontal track with negligible friction,
Calculate the momentum, Since friction is negligible,
Initial Momentum = Final Momentum
Initial Kinetic Energy = Final Kinetic Energy
m₁ × v₁ = m₁ × v₂
When m increases, v decreases,
Thus, momentum will remain the same.
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An oscillating particle has the equation x = 10cos(8πt +π ) (x in cm, t in s). The number of times the object passes through the equilibrium position in one second is
Answer:
x = A sin (wt + theta) where w = angular frequency - basic SHM equation
w = 8 pi = 2 pi f
f = 4 basic frequency
N = 8 number of times thru origin
Each cycle the particle will pass thru the origin +x and -x twice
A brass rod is 185 cm long and 1.60 cm in diameter. What force must be applied to each end of the rod to prevent it from contract- ing when it is cooled from 120.0°C to 10.0°C?
Answer:
42000N
Explanation:
First you calculate how much it would contract, and secondly you then calculate the force to stretch it by that amount.
1) linear thermal expansion coef brass 19e-6 /K
∆L = αL∆T = (19e-6)(1.85)(110) = 0.00387 meter or 3.87 mm
Second part involves linear elasticity.
for brass, young's modulus is 15e6 psi or 100 GPa
cross-sectional area of rod is π(0.008)² = 0.0002 m²
F = EA∆L/L
F = (100e9)(0.0002)(0.00387) / (1.85)
F = 42000 or 42 kN
the atomic number of a nucleus increases during which nuclear reactions
Answer:
Answer A : Fusion followed by beta decay (electron emission)
Explanation:
Notice that you want the Atomic number to increase, that is the number of protons in a nucleus. So if all four cases given experience the same fusion of nuclei, the only one that net increases the number of protons in the last stage, is the reaction that undergoes a beta decay (with emission of an electron) thus leaving a positive imbalance of positive charge (proton generated in the beta decay of a neutron).
Therefore, answer A is the correct one.
Answer:
A : Fusion followed by beta decay (electron emission)
Explanation:
Ap3x
We observe that a moving charged particle experiences no magnetic force. From this we can definitely conclude that:_______
a. no magnetic field exists in that region of space.
b. the particle must be moving parallel to the magnetic field.
c. the particle is moving at right angles to the magnetic field.
d. either no magnetic field exists or the particle is moving parallel to the magnetic field.
e. either no magnetic field exists or the particle is moving perpendicular to the magnetic field.
Answer:
b. the particle must be moving parallel to the magnetic field.
Explanation:
The magnetic force on a moving charged particle is given by;
F = qvBsinθ
where;
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field
θ is the angle between the magnetic field and velocity of the moving particle.
When is the charge is stationary the magnetic force on the charge is zero.
Also when the charge is moving parallel to the magnetic field, the magnetic force is zero.
Therefore, when a moving charged particle experiences no magnetic force, we can definitely conclude that the particle must be moving parallel to the magnetic field.
b. the particle must be moving parallel to the magnetic field.
what are the property of the image formed by plane mirror
of class 10
Answer:
» The image is laterally inverted.
» The image is upright.
» The image geometry is same as object geometry.
» Image distance is same as object distance.
» Image is not real, it's virtual ( not formed on screen ).
[tex].[/tex]
A 180-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,050 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable
Answer:
27yrs
Explanation:
h= difference in height between the initial position and the bottom position
We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical
h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)
=0.674m
Potential Energy = 28× 9.8×0.674
=184.9J
B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:
E= 0.5mv^2