Answer:
Following are the solution to the given points:
Explanation:
For point a:
Following are the expression to the self- inductance and the solenoid:
[tex]L=\frac{\mu_0 N^2 A}{l}[/tex]
[tex]=\frac{\mu_0 (\frac{l}{d})^2 (\pi r^2)}{l}\\\\=\frac{\mu_0 ( \pi(\frac{D}{2})^2)}{d^2}\\\\=\frac{Ld^2}{\mu_0( \pi (\frac{D}{2})^2)}\\\\=\frac{(1.0 \ H)(0.77 \times 10^{-3} \ m)^2}{ 4 \pi \times 10^{-7} \frac{H}{m} (\pi (\frac{13 \times 10^{-2} \ m }{2})^2)}\\\\=35.5 \ m[/tex]
For point b:
Calculting the length of the copper wire:
[tex]L'=N(\pi D)[/tex]
[tex]=\frac{l}{d}(\pi D)\\\\=\frac{35.5 m}{0.77 \times 10^{-3} m}\pi (13\times 10^{-2} m)\\\\=18829.2 \ m(\frac{1 km}{10^3 \ m})\\\\=18.8 \ km[/tex]
For point c:
The resistance of the solenoid is given by,
[tex]R=\frac{\rho L'}{A}[/tex]
[tex]=\frac{\rho L'}{\pi r'^2}\\\\=\frac{\rho L'}{\pi (\frac{d}{2})^2}\\\\=\frac{(1.68\times 10^{-8} \Omega . m) (18829.2 \ m)}{\pi (\frac{0.77 \times 10^{-3} \ m}{2})^2}\\\\=6793 \ \Omega[/tex]
A 70 kg human body typically contains 140 g of potassium. Potassium has a chemical atomic mass of 39.1 u and has three naturally occurring isotopes. One of those isotopes, 40K (potassium), is radioactive with a half-life of 1.3 billion years and a natural abundance of 0.012 %. Each 40K (potassium) decay deposits, on average, 1.0 MeV of energy into the body.
What yearly dose in Gy does the typical person receive from the decay of 40K (potassium) in the body? Express your answer using two significant figures.
Answer:
Gy = 3.14x10⁻⁴ Gy
Explanation:
To get the dose in Gy we need to use the following expression:
Gy = E / m (1)
Where:
Gy: dose
E: energy absorbed per atom
m: mass of the human body.
We don't have the energy per atom, but we can calculate that by following the next procedure.
First, let's determine the number of atoms of potassium in our body. For that we need to determine the moles in the 140 g of potassium, with the molecular mass and then, use the avogadro's number:
moles = m/MM
moles = 140 / 39.1 = 3.58 moles
N° atoms = 3.58 * 6.02x10²³ atoms = 2.16x10²⁴ atoms of K.
The abundance of the ⁴⁰K is 0.012% so the atoms of this isotope would be:
N = 2.16x10²⁴ * (0.012/100) = 2.59x10²⁰ atoms of ⁴⁰K.
With this number, and the half life rate, we can determine the number of decay atoms in a year (λ) using the following expression:
λ = ln2 / t(1/2)
λ = ln2 / 1.3x10⁹ = 5.33x10⁻¹⁰ year⁻¹
This number, multiplied by the number of atoms:
R = 5.33x10⁻¹⁰ * 2.59x10²⁰ = 1.38x10¹¹ atoms/year
Now, each atom of K gives an average energy of 1 MeV, so with the atoms we have:
E = 1.38x10¹¹ * 1x10⁶ eV = 1.38x10¹⁷ eV
This value can be expressed in Joules so:
E = 1.38x10¹⁷ eV * (1 J / 6.24x10¹⁸ eV) = 0.022 J
Finally, we can use (1) to get the dose in Gy:
Gy = 0.022 / 70
Gy = 3.14x10⁻⁴ GyHope this helps
What happens to the force between two magnets as the magnets come closer together? *
Explanation:
So if two magnets are pointing with unlike-poles together (north pole to a south pole), then bringing them closer together decreases the energy stored up in the magnetic field. They will be pushed in the direction that decreases the amount of stored-up energy.
Explanation:
depending on the polarity they'll increasingly repel or attract
What is Ship Handling, an art or a science? Why?
Ship handling is both a science and an art. Science because it requires knowledge of various forces acting on the ship. Art because it requires the skills of an experienced navigator to use these forces in his favour. We may learn the science part from the various ship handling courses.
please give brainliest
A physics student mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them, and they appear concentric when viewed from either end). The lenses are identical, each with a positive (converging) focal length of 14.8 cm. They are separated by a distance of 39.4 cm. Lens 1 is to the left of Lens 2.
Required:
a. What is the final image's distance (in cm) from Lens 2?
b. Where is the final image located?
c. What is the overall magnification of the lens pair, considered as a single optical instrument?
Answer:
A) q₂ = 75.98 cm, B) q₂' = 115.38 cm, C)
Explanation:
A) This is an exercise in geometric optics, as the two lenses are separated by a greater distance than their focal lengths from each lens, they must be worked as independent lenses.
Lens 1. More to the left
let's use the constructor equation
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively,
We must assume a distance to the object to perform the calculation, suppose that the object is 50 cm from lens 1 that is further to the left of the system.
[tex]\frac{1}{q_1} = \frac{1}{f} - \frac{1}{p}[/tex]
[tex]\frac{1}{q_1} = \frac{1}{14.8} - \frac{1}{50}[/tex]
1 / q₁ = 0.04756
q₁ = 21.0227 cm
this image is the object for the second lens that has f₂ = 14.8 cm
the distance must be measured from the second lens
p₂ = 39.4 -q₁
p₂ = 39.4 -21.0227
p₂ = 18.38 cm
let's use the constructor equation
1 / q₂ = 1 / f - 1 / p2
[tex]\frac{1}{q_2} = \frac{1}{14.8} - \frac{1}{18.38}[/tex]
[tex]\frac{1}{q_2}[/tex] = 0.01316
q₂ = 75.98 cm
measured from the second lens
B) the position of the final image with respect to the first lens is
q₂’= q₂ + 39.4
q₂'= 75.98 +39.4
q₂' = 115.38 cm
C) the magnification of a lens is
m = - q / p
in this case the image measured from lens 2 is q2 = 75.98 cm
the distance to the object from the first lens is p1 = 50cm
m = - 75.98 / 50
m = -1.5 X
the negative sign indicates that the image is inverted
The photo shows falling water droplets.
Which quality of a drop of water increases as it gets closer to Earth's surface?
A. Resistance
B. Mass
C. Magnetism
D. Velocity
A block of mass m slides along a frictionless surface with kinetic energy Kiit collides with a block of mass 3m that is initially at rest, and the two blocks stick together and slide with total kinetic energy Kf . What is the ratio Kf:Ki
Answer:
the ratio Kf : Ki is 1 / 4 or 1 : 4
Explanation:
Given the data in the question;
Since this is a perfectly inelastic collision, momentum is conserved;
[tex]P_{initial[/tex] = [tex]P_{final[/tex]
Now for BLOCK 1
mass = M₁ = M
KE = K[tex]_i[/tex]
[tex]\frac{1}{2}[/tex]mv[tex]_i[/tex]₁² = K[tex]_i[/tex]
we solve for v[tex]_i[/tex]₁
mv[tex]_i[/tex]₁² = 2K[tex]_i[/tex]
v[tex]_i[/tex]₁ = √( 2K[tex]_i[/tex] / m )
for BLOCK 2
mass = M₂ = 3m and since its at rest v[tex]_i[/tex]₂ = 0
Now after the collision; Total mass = m + 3m = 4m
KE = K[tex]_f[/tex]
[tex]\frac{1}{2}[/tex]( 4m )v[tex]_f[/tex]² = K[tex]_f[/tex]
(2m)v[tex]_f[/tex]² = K[tex]_f[/tex]
v[tex]_f[/tex] = √(K[tex]_f[/tex] / 2m)
so since [tex]P_{initial[/tex] = [tex]P_{final[/tex]
[m₁ × v[tex]_i[/tex]₁] + [m₂ × v[tex]_i[/tex]₂] = ( m + 3m ) × v[tex]_f[/tex]
so
[ m₁ × √( 2K[tex]_i[/tex] / m ) ] + [ m₂ × 0 ] = ( m + 3m ) × [ √(K[tex]_f[/tex] / 2m) ]
[ m × √( 2K[tex]_i[/tex] / m ) ] = 4m × [ √(K[tex]_f[/tex] / 2m) ]
square both side
m² × 2K[tex]_i[/tex] / m = (4m)² × K[tex]_f[/tex] / 2m
m² × 2K[tex]_i[/tex] / m = 16m² × K[tex]_f[/tex] / 2m
m × 2K[tex]_i[/tex] = 8m × K[tex]_f[/tex]
2K[tex]_i[/tex] = 8K[tex]_f[/tex]
K[tex]_f[/tex] = 2K[tex]_i[/tex] / 8
K[tex]_f[/tex] / K[tex]_i[/tex] = 2 / 8
K[tex]_f[/tex] / K[tex]_i[/tex] = 1 / 4
Therefore, the ratio Kf : Ki is 1 / 4 or 1 : 4
The ratio of the final kinetic energy to initial kinetic energy of the blocks is determined as 1:4.
Final speed of the blocks after collisionThe speed of the two blocks after collision is determined by applying the principle of conservation of linear momentum as follows;
m1u1 + m2u2 = v(m1 + m2)
mu1 + m2(0) = v(m + 3m)
mu1 = v(4m)
u1 = 4v
Ratio of the final kinetic energy to initial kinetic energy[tex]\frac{K.E_f}{K.E_i} = \frac{0.5(4m)v_f^2}{0.5(m)u_1^2} \\\\\frac{K.E_f}{K.E_i} = \frac{4(v_f)^2}{u_1^2} \\\\\frac{K.E_f}{K.E_i} = \frac{4v_f^2}{(4v_f)^2} = \frac{4v_f^2}{16v_f^2} = \frac{1}{4} = 1:4[/tex]
Learn more about kinetic energy here: https://brainly.com/question/25959744
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For which planet is the length of the plants day longer than the planets year
two spheres of radii 5cm and 3cm are given charges on risk volume and 50 calling respectively and then connected by a wire calculate the loss of energy after connection
Answer:
Solution given:
Radius of small sphere[r]=5cm=0.05m
Radius of large sphere[R]=10cm=0.1m
capacitance of small sphere[c]=4πε0r
=[tex]4π*8.85×10^{-12}×0.05=5.56*10^{-12}F[/tex]
Charge for small sphere[Q1]=100C
Charge for small sphere[Q2]=50C
Potential difference [V1]=[tex] \frac{charge}{capacitance}=\frac{100}{5.56*10^{-12}}=1.8×10^{13}[/tex]V
.
again
capacitance of small sphere[C]=4πε0R
=[tex]4π*8.85×10^{-12}×0.1=1.11*10^{-11}F[/tex]
Potential difference [V2]=[tex] \frac{charge}{capacitance}=\frac{50}{1.11*10^{-11}}=4.5×10^{12}[/tex]V
Now
Loss of energy:
[tex] \frac{cC(V1-V2)^{2}}{2(c+C)}[/tex]
=[tex] \frac{5.56*10^{-12}*1.11*10^{-11}(1.8*10^{13}-4.5*10^{12})^{2}}{2(5.56*10^{-12}+1.11*10^{-11})}[/tex]
=25Joule
A sheet of glass having an index of refraction of 1.52 is to be coated with a film of material having a refractiveindex of 1.75 such that green light with a wavelength of 525 nm (measured in vacuum) is preferentially transmitted when white light is incident normally on the film.What is the minimum thickness of the film that will achieve this result
Answer:
The correct answer is "164.06 nm".
Explanation:
Refractive index,
n = 1.75
Wavelength,
λ = 525 nm
Now,
The maximum thickness of the film will be:
⇒ [tex]t=\frac{\lambda}{2n}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{525}{2\times 1.75}[/tex]
⇒ [tex]=\frac{525}{3.5}[/tex]
⇒ [tex]=164.06 \ nm[/tex]
How do tsunamis cause flooding? Please answer in 4-5 sentences.
PLEASE HELP! GIVING BRAINLIEST!
Answer: YES
Explanation: The main difference between flood and tsunami is that the flood is overflow of water that submerges land and tsunami is a series of water waves caused by the displacement by large volume of a volume of water
Answer:
WEEEEEEEEEEEEEEEEEEEEEEEEE
Explanation:
Two electromagnetic waves are traveling through empty space. The wave with the greater wavelength is also characterized by _____.
a. the lower frequency
b. the higher frequency
c. the faster speed
d. the slower speed
A train moving at 30km/h accelerates to a speed of 50km/h in 30s. find the average acceleration in m/s²
Answer:
Explanation:
a= (v-u)/t where v = the final velocity at 50m/s, u= the inital velocity at 30m/s and t= the time taken at 30 seconds
a=(50-30)/30
a= 2/3m/s^-2 or 0.67m/s^-2 (m/s^-2 is the units for accelaration)
It is well known that bullets and other missiles fired at Superman simply bounce off his chest. Suppose that a gangster sprays Superman's chest with 6.4 g bullets at the rate of 92 bullets/min, and the speed of each bullet is 400 m/s. Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest from the stream of bullets
Answer:
Magnitude of the average force = 7.85 N
Explanation:
Data given:
Mass of bullets, m = 6.4 g
Rate of bullets/min, r = 92 bullets/min
Speed of each bullet, v = 400 m/s
Change in momentum here, Δ B = Bf - Bi
where f is the final and i is the initial
Note that change in momentum = force X time
So, Δ B = m(vf - vi)
= 2mv
= 2 X 0.0064 kg X 400 m/s (convert g to kg)
= 5.12 kg.m/s (for one bullet)
so for the 92 bullets = 92 X 5.12 kg.m/s
= 471.04 kg.m/s
The force = Δ B ÷ Δt
where t = time measured in 60 seconds
= 471.04 kg.m/s ÷ 60 seconds
= 7.85 N
A cannon at rest fires a 32.5 kg cannonball forward at 388 m/s. After firing, the cannon recoils at 7.42 m/s. What is the mass of the cannon?
(Unit = kg)
Remember: right is +, left is -
Answer:
1700 kg
Explanation:
Let’s use conservation of momentum
32.5 * 388 = 7.42 * mc
mc = 1699.46
mc = 1700 kg
What is the resistance of a rheostat coil, if 0.05 A of current flows through it when 6 V is applied
across it? 1200
Answer:
i have no idea i came here to find out too :(
Explanation:
True or false in order for light to be emitted electrons must move from high energy state to a lower energy state?
1 point
A.True
B.False
Answer:
B
I think it is false.
It does not move to lower energy state.
I will mark u brainlist
Answer:
DragonFly or a dog
Explanation:
Two children are throwing a ball back-and-forth straight across the back seat of a car. The ball is being thrown 7 mph relative to the car, and the car is travelling 45 mph down the road. If one child doesn't catch the ball and it flies out the window, in what direction does the ball fly (ignoring wind resistance)
Answer:
the ball will fly in AX direction, making angle of 8.84° from the motion of the car
Explanation:
Given the data in the question and as illustrated in the diagram below;
Now, Lets assume line AB represent the movement of the car,
AC is the movement of the ball been thrown back and forth in the back seat
Ax is the motion of the ball it flies off the window
so from the diagram, We can see triangle ABC
where AB is 45 mph and AC = 7 mph
and angle ∠CAB = 90°
using SOH CAH TOA
TOA; tanθ = Opposite / Adjacent
tanθ = Opposite / Adjacent
tan( ∠ ABC ) = AC / AB
we substitute
tan( ∠ ABC ) = 7 / 45
tan( ∠ ABC ) = 0.15555
( ∠ ABC ) = tan⁻¹ 0.15555
( ∠ ABC ) = 8.84°
Therefor, angle ( ∠ ABC ) is 8.84°
Meaning angle ( ∠ XAA' ) is also 8.84°
Therefore, the ball will fly in AX direction, making angle of 8.84° from the motion of the car
A uniform electric field exists everywhere in the x,y plane. The electric field has a magnitude of 3500 N/coil, and is directed in the positive x direction. A point charge of -9.0 x 10-9 coil is placed at the origin. Determine the magnitude of the net electric field at: (a) x
Answer:
5525 N/C
Explanation:
Magnitude of electric field ( E ) = 3500 N/c
Direction of electric field : positive X axis
point charge ( q ) = -9.0 * 10^-9
Calculate the Magnitude of the net electric field at (a) x = -0.20 m
Magnitude = 5525 N/C
Electric field due to q = ( 9 * 10^9 * 9 * 10^-9 ) / ( -0.2 )^2
= 81 / 0.04 = 2025 N/c
Therefore the magnitude of the net electric field
= 2025 + 3500
= 5525 N/C
Si un vector tiene una dirección de 2300 a partir del eje x positivo, ¿Qué signos tendrán sus componentes x y y? Si la la razón de R/R, es negativa. ¿cuáles son los ángulos posibles de R, medidos a partir del eje x positivo?
A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 ms from an initial speed of 5.25 m/s . What is the magnitude of the average contact force exerted on the leg, assuming the total mass of the hand and the forearm to be 1.75 kg
Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F[tex]_{hand[/tex]
using equation for impulse in momentum
F[tex]_{hand[/tex] × t = m( v - v₀ )
we substitute
F[tex]_{hand[/tex] × 0.00265 = 1.75( 0 - 5.25 )
F[tex]_{hand[/tex] × 0.00265 = 1.75( - 5.25 )
F[tex]_{hand[/tex] × 0.00265 = -9.1875
F[tex]_{hand[/tex] = -9.1875 / 0.00265
F[tex]_{hand[/tex] = -3466.98 N
Next we determine force on the leg F[tex]_{leg[/tex]
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F[tex]_{leg[/tex] = - F[tex]_{hand[/tex]
we substitute
F[tex]_{leg[/tex] = - ( -3466.98 N )
F[tex]_{leg[/tex] = 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N
An engine flywheel initially rotates counterclockwise at 5.03 rotations/s. Then, during 23.5 s, its rotation rate changes to 2.63 rotations/s clockwise. Find the flywheel's average angular acceleration (including its sign) in radians per second squared. Define counterclockwise rotation as positive.
Answer:
the flywheel's average angular acceleration is -2.05 rad/s²
Explanation:
Note: counterclockwise is positive
clockwise is negative
Given;
initial angular velocity, [tex]\omega _i[/tex] = 5.03 rev/s = [tex]5.03\frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 31.61 \ rad/s[/tex]
final angular velocity, [tex]\omega_f[/tex]= -2.63 rev/s = [tex]-2.63 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = -16.53 \ rad/s[/tex]
duration of the flywheel rotation, Δt = 23.5 s
The average acceleration of the flywheel is calculated as;
[tex]a_r = \frac{\Delta \omega}{\Delta t} = \frac{\omega_f - \omega _i}{t_2-t_1} = \frac{-16.53 \ - \ 31.61}{23.5} = -2.05 \ rad/s^2[/tex]
Therefore, the flywheel's average angular acceleration is -2.05 rad/s²
Two horizontal strings are attached to a block, resting on a frictionless surface as shown
below.A force of 100N pulls on one string. The block does not move. Find the value o
force, F on the other string
ON
Answer:
100N in the opposite direction
Explanation:
as the block is not moving, we can apply the formula f=ma where the total force is equal to the mass times accelaration, since the block is not moving, the net force is zero meaning that the other force is completely cancelling out the force of 100N, or to put it in a maths equation, 100-force=0 as the block is stationary. we can therefore assume that the force cancelling the 100N out is equal to it and therefore 100N in the opposite direction
How many spoonfuls of water did it take for your sponge to be 100% saturated?
Answer:
19
Explanation:
I legit did this and it took 19.
What is the typical pH of acid rain?
Answer:
5.0-5.5 is the answer to your question
Which conclusion can be made based on the information in the table? Wave speed and wavelengths can vary inversely to produce the same frequency. Frequency and wave speed can vary directly to produce the same wavelength. Wavelengths and frequency can vary inversely to produce the same wave speed. Frequency and wavelengths can vary directly to produce the same wave speed.
Answer: I think it’s C
Explanation:
On edge.
Answer:
It's (C) Wavelengths and frequency can vary inversely to produce the same wave speed.
An object with a mass of 2.00 kg is placed at the end of a spring, having a spring constant of 180.0 N/m. The spring is then compressed 0.890 m. What is the maximum velocity of the mass?
Answer:
the velocity of the mass is 8.44 m/s
Explanation:
Given;
mass of the object, m = 2 kg
spring constant, k = 180 N/m
extension of the spring, x = 0.89 m
The maximum velocity of the mass is calculated as follows;
By the principle of conservation of energy;
Elastic potential energy = kinetic potential energy
¹/₂kx² = ¹/₂mv²
kx² = mv²
[tex]v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{180 \times 0.89^2}{2} }\\\\v = 8.44 \ m/s[/tex]
Therefore, the velocity of the mass is 8.44 m/s
In any given wave, when the frequency of the wave doubles (f = 2f), which of the following other changes would also take place?
A The wavelength would double (λ = 2λ).
B The velocity would double (v = 2v).
C The wavelength would be half (λ = λ/2).
D The velocity would be half (v = v/2).
Answer:
The correct answer is -
B. The velocity would double (v = 2v).
C. The wavelength would be half (λ = λ/2).
Explanation:
A wave has a speed or velocity that is related to the wavelength of the wave and the frequency of the wave and this relationship can be represented by the following equation-
Wave velocity V = Wavelength (λ) * Frequency (f)
Frequency (f) = Velocity (V) / Wavelength(λ).
The frequency and wavelength are inversely proportional and frequency and velocity are directly proportional to each other.
So, if f = 2f then,
putting value in the formula,
2f = 2v/λ, which means, f = 2v and f = λ/2
when the frequency is doubled, the wavelength will be halved and velocity will be doubled.
What are the functionssss of non-geostationary satellite?
Answer:
A communication system comprising a set of non-geostationary satellites provides communication links between ground stations and terminals, a communication link from a particular station via a satellite terminating at associated terminals in a limited geographical era including the station and providing a.Explanation:
Hopes this helps. Mark as brainlest plz!Geostationary satellites orbit around the Earth at the same rate as the Earth rotates so that the satellites are over the same spot on Earth all the time.
They are called geostationary due to their movement.
Geostationary satellites are a key tool for scientists to monitor and observe the Earth's atmosphere.
A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for small-angle oscillations is 0.940 s. (a) What is the moment of inertia of the wrench about an axis through the pivot
Answer:
[tex]I=0.0987kg.m^2[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=1.80kg[/tex]
Deviation [tex]d=0.250[/tex]
Time [tex]t=0.940s[/tex]
Generally the equation for moment of inertia is mathematically given by
[tex]I=\frac{T}{2\pi}^2(mgd)[/tex]
[tex]I=\frac{0.94}{2.3.142}^2(1.80*9.8*0.250)[/tex]
[tex]I=0.0987kgm^2[/tex]