Answer:
The distance is 2.94 mm.
Explanation:
Magnetic field, B = 1.7 mT
Current, I = 25 A
Let the distance is d.
The magnetic field is given by
[tex]B = \frac{\mu o}{4\pi}\times \frac{2I}{r}\\\\1.7\times 10^{-3} = 10^{-7}\times \frac{2\times 25}{r}\\\\r = 2.94\times 10^{-3} m \\\\ r = 2.94 mm[/tex]
Two divers, G and H, are at depths 20 m and 40 m respectively
below the water surface in lake. The pressure on G is P, while
the pressure on H is P2 if the atmospheric pressure is equivalent
to 10 m of water, then the value of P2/P1 is.
A. 1.67.
B. 2.00.
C. 0.50.
D. 0.60.
Answer:
B
Explanation:
P1/P1 = 40/20
=2
Two circular coils are concentric and lie in the same plane.The inner coil contains 120 turns of wire, has a radius of 0.012m,and carries a current of 6.0A. The outer coil contains 150turns and has a radius of 0.017 m. What must be the magnitudeand direction (relative to the current in the inner coil) ofthe current in the outer coil, such that the net magnetic field atthe common center of the two coils is zero?
Answer:
[tex]I_2=6.8A[/tex]
Explanation:
From the question we are told that:
Turns of inner coil [tex]N_1=120[/tex]
Radius of inner coil [tex]r_1=0.012m[/tex]
Current of inner coil [tex]I_1=6.0A[/tex]
Turns of Outer coil [tex]N_2=150[/tex]
Radius of Outer coil [tex]r_2=0.017m[/tex]
Generally the equation for Magnetic Field is mathematically given by
[tex]B =\frac{ \mu N I}{2R}[/tex]
Therefore
Condition for the net Magnetic field to be zero
[tex]\frac{N_1* I_1}{( 2 * r_1 )}=\frac{N_2 * I_2}{2 * r_2}[/tex]
[tex]I_2=\frac{(N_1* I_1)*(( 2 * r_2)}{( 2 * r_1)*N_2}[/tex]
[tex]I_2=\frac{(120*6.0)*(( 2 * 0.017)}{( 2 * 0.012)*150}[/tex]
[tex]I_2=6.8A[/tex]
A roller coaster has a total track length of 500 yards. A complete ride on the roller coaster is considered two times around the track. The start and stop places for the ride are virtually the same. What are the distance and displacement for a ride on the roller coaster? Explain your answers.
Answer:
Explanation:
its 1000 yards if its going around the track 2 times and that if one whole around the track is 500 its 500 x 2
A proposed communication satellite would revolve around the earth in a circular orbit in the equatorial plane at a height of 35880Km above the earth surface. Find the period of revolution of the satellite. (Take the mass of earth =5.98×10²⁴kg, the radius of the earth 6370km and G=6.6×10–¹¹Nm²/kg2)
Answer:
Period is 86811.5 seconds.
Explanation:
[tex]{ \boxed{ \bf{T {}^{2} = (\frac{4 {\pi}^{2} }{GM}) {r}^{3} }}}[/tex]
[tex]{ \tt{T {}^{2} = \frac{4 {(3.14)}^{2} }{(6.6 \times {10}^{ - 11} ) \times (5.98 \times {10}^{24} )} \times {((35880\times {10}^{3}) } + (6370 \times {10}^{3} )) {}^{3} }} \\ \\ { \tt{T {}^{2} = 7.54 \times {10}^{9} }} \\ { \tt{T = \sqrt{7.54 \times {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}[/tex]
What is science?Give two examples of living beings?
Answer:
the study of the past
Explanation:
dogs and cats
Earth’s Moon has a diameter of 3,474 km and orbits at an average distance of 384,000 km. At that distance it subtends and angle just slightly larger than half a degree in Earth’s sky. Pluto’s moon Charon has a diameter of 1,186 km and orbits at a distance of 19,600 km from the dwarf planet. Compare the appearance of Charon in Pluto’s skies with the Moon in Earth’s skies. Describe where in the sky Charon would appear as seen from various locations on Pluto.
The reason for arriving at the above solutions is as follows:
The given dimensions and distance from the Earth of the Moon are;
The diameter of the Moon, d = 3,474 km
The average distance of the Moon from the Earth, R = 384,000 km
Required:
The comparison between Charon's appearance in Pluto and the Moon's appearance on Earth Earth
Solution:
The distance of the Moon's travels in an orbit, C = 2·π·R
∴ C = 2 × π × 384,000 km
The angle subtended by the Moon, θ = d/C × 360°
∴ θ = 3,474/(2 × π × 384,000) × 360° ≈ 0.518°
Pluto's moon Charon, has the following parameters;
The diameter of the Charon, d₂ = 1,186 km
The average distance of the Charon from Pluto, R₂ = 19,600 km
Therefore, the distance of the Moon's travels in an orbit, C₂ = 2·π·R₂
∴ C₂ = 2 × π × 19,600 km
The angle subtended by the Moon, θ₂ = d₂/C₂ × 360°
∴ θ₂ = 1,186/(2 × π × 19,900) × 360° ≈ 3.415°
The angle subtended by Charon in Pluto's sky ≈ 3.415°
Charon therefore, appears 7 times larger in Pluto's skies than the Moon's appearance in Earth's skies
Required:
The appearance of Charon as seen from different locations on Pluto
Solution:
Charon is gravitationally locked to Pluto, therefore, the same side of Pluto is faced with the same side of Charon
Therefore;
Charon appears constantly overhead from the side of Pluto locked to CharonCharon appears constantly at the horizon from the poles on either side of the axis of rotation of Pluto and CharonLearn more about Pluto's moon Charon here:
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A mountain biker takes a jump in a race and goes airborne. The mountain bike is travelling at 10.0 m/s before it goes airborne. If the mass of the front wheel on the bike is 750 g and has radius 35 cm, what is the angular momentum of the spinning wheel in the air the moment the bike leaves the ground?
Answer:
Explanation:
The formula for angular momentum is
L = mvr where L is the angular momentum, m is the mass of the object, v is the velocity of the object, and r is the radius of the object. The problem we have that prevents us from just throwing those numbers in there is that mass has to be in kg and it's not, and radius has to be in meters and it's not.
Changing the mass to kg:
750 g = .750 kg
Changing the radius to m:
35 cm = .35 m
Now we can fill in the variables with their respective values:
L = .750(10.0)(.35) gives us
[tex]L=2.625\frac{kg*m^2}{s}[/tex]
Find out other examples of bodies showing more than one type of motion Tabulate your findings.
Answer:
down below
Explanation:
Image 1- wheels of train showing both translatory motion as well as rotatory motion.
Image 2- rotation of ball shows both rotatory motion as well as translatory motion.
Image 3- the earth rotates about its axis, same time it revolves around the sun thus showing both rotatory motion and curvilinear motion in a fixed time. (perodic motion)
Image 4- while cutting wood, the
carpenter's saw has both
translatory motion and oscillatory
motion, as it moves down while
oscillating.
A cylindrical wire made of an unknown alloy hangs from a support in the ceiling. You measure the relaxed length of the wire to be 16 m long; and the radius of the wire to be 3.5 m. When hang a 5 kg mass from the wire, you measure that it stretches a distance of 4 x 10 m The average bond length between atoms is 2.3 x 10^0 m for th alloy.
Required:
What is the stiffness of a typical interatomic bond in the alloy
Answer: hello some of your values are wrongly written hence I will resolve your question using the right values
answer:
stiffness = 1.09 * 10^-6 N/m
Explanation:
Given data:
Length ( l ) = 16 m
radius of wire ( r ) = 3.5 m
mass ( m ) = 5kg
Distance stretched ( Δl ) = 4 * 10^-3 m ( right value )
average bond length ( between atoms ) = 2.3 * 10^-10 m ( right value)
first step : calculate the area
area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2
γ = MgL / A Δl
= [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]
= 784.8 / 0.165 = 4756.36 N/m^2
hence : stiffness = γ * bond length
= 4756.36 * 2.3 * 10^-10 = 1.09 * 10^-6 N/m
g Light that is incident upon the eye is refracted several times before it reaches the retina. As light passes through the eye, at which boundary does most of the overall refraction occur?
Answer
Explanation
:giác mạc
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 11.0 ft/s at point A and 18.0 ft/s at point C. The cart takes 5.00 s to go from point A to point C, and the cart takes 1.30 s to go from point B to point C. What is the cart's speed at point B
Answer:
The speed at B is 16.18 ft/s .
Explanation:
Speed at A, u = 11 ft/s
Speed at C, v' = 18 ft/s
Time from A to C = 5 s
Time from B to C = 1.3 s
Let the speed of car at B is v.
Let the acceleration is a.
From A to B
Use first equation of motion
v = u + a t
18 = 11 + a x 5
a = 1.4 ft/s^2
Let the time from A to B is t' .
t' = 5 - 1.3 = 3.7 s
Use first equation of motion from A to B
v = 11 + 1.4 x 3.7 = 16.18 ft/s
Two long straight wires are suspended vertically. The wires are connected in series, and a current from a battery is maintained in them. What happens to the wires? What happens if the battery is replaced by an a-c source?
Answer:
(i) When a battery is connected inseries to two long parallel wires, the currents in the two wires will be in opposite directions. Due to which a force of repulsion will be acting between them and they are moving further apart.
(ii) When a battery is connected in parallel to two long parallel wires, the currents in the two wires will be in same direction. Due to it, a force of attraction will be acting between them and they are coming closer to each other.
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Explanation:
Hope it will helps you lot!
Consider a 200-ft-high, 1200-ft-wide dam filled to capacity. Determine (a) the hydrostatic force on the dam and (b) the force per unit area of the dam near the top and near the bottom. Note: we will see that the resultant hydrostatic force will be
Answer:
a) [tex]F_g=1.5*10^9Ibf[/tex]
b) [tex]F_t=12490Ibf/ft^2[/tex]
[tex]F_b=0[/tex]
Explanation:
From the question we are told that:
Height [tex]h=200ft[/tex]
Width [tex]w=1200ft[/tex]
a)
Generally the equation for Dam's Hydro static force is mathematically given by
[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]
Where
[tex]\rho=Density\ of\ water[/tex]
[tex]\rho=62.4Ibm/ft^3[/tex]
Therefore
[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]
[tex]F_g=1.5*10^9Ibf[/tex]
b)
Generally the equation for Dam's Force per unit area is mathematically given by
[tex]F=\rho*g*h[/tex]
For Top
[tex]F_t=\rho*g*h[/tex]
[tex]F_t=62.4*32.2*200[/tex]
[tex]F_t=12490Ibf/ft^2[/tex]
For bottom
[tex]Here \\H=0 zero[/tex]
Therefore
[tex]F_b=0[/tex]
The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].
The force per unit area near the top is 86.74 psi.
The force per unit area near the bottom is zero.
Hydrostatic force
The hydrostatic force on the dam is the force exerted on the dam by the column of the water.
[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]
Force per unit area near the topThe force per unit area is the pressure exerted near the top of the dam.
[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]
where;
P is pressure in PSI
ρ is density of water in lb/gal
h is the vertical height in ft
[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]
The pressure near the bottom is zero, become the vertical height is zero.
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A rod of length L and electrical resistance R moves through a constant uniform magnetic field ; both the magnetic field and the direction of motion are parallel to the rod. The force that must be applied by a person to keep the rod moving with constant velocity is:
Answer:
don't know what class are you you are using which mobile or laptop
What is the order of magnitude of the distance of Sun to nearest star in meters?
Answer:
Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.
Explanation:
Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.
Let's model the Milky Way galaxy as a disk.
The volume of a disk is:
V
=
π
⋅
r
2
⋅
h
Plugging in our numbers (and assuming that
π
≈
3
)
V
=
π
⋅
(
10
21
m
)
2
⋅
(
10
19
m
)
V
=
3
×
10
61
m
3
Is the approximate volume of the Milky Way.
Now, all we need to do is find how many stars per cubic meter (
ρ
) are in the Milky Way and we can find the total number of stars.
Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of
4
×
10
16
m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.
ρ
=
n
V
Using the volume of a sphere
V
=
4
3
π
r
3
ρ
=
1
4
3
π
(
4
×
10
16
m
)
3
ρ
=
1
256
10
−
48
stars /
m
3
Going back to the density equation:
ρ
=
n
V
n
=
ρ
V
Plugging in the density of the solar neighborhood and the volume of the Milky Way:
n
=
(
1
256
10
−
48
m
−
3
)
⋅
(
3
×
10
61
m
3
)
n
=
3
256
10
13
n
=
1
×
10
11
stars (or 100 billion stars)
Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.
Answer:
Mercury, 46,001,272 km from the sun at the nearest point.
Explanation:
elastic wire extend by 1.ocm when a load on 20g range from It, what additional load will it be required Cause the futher extension of 2.0cm
Answer:
40g
Explanation:
20g range > 1.0cm
Therefore,
40g range > 2.0cm
What is hydroelectric power ?
Answer quickly..!
Answer:
It's electricity produced from hydropower. It's also a form of energy that controls the power of water motion.
Explanation:
One pro about hydroelectric power is that it's renewable energy. But one con about hydroelectric power is that it can impact the environment in a negative way.
A Geiger counter registers a count rate of 8,000 counts per minute from a sample of a radioisotope. The count rate 24 minutes later is 1,000 counts per minute. What is the half-life of the radioisotope?
11.54 minutes
Explanation:
The decay rate equation is given by
[tex]N = N_0e^{-\frac{t}{\lambda}}[/tex]
where [tex]\lambda[/tex] is the half-life. We can rewrite this as
[tex]\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}[/tex]
Taking the natural logarithm of both sides, we get
[tex]\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)[/tex]
Solving for [tex]\lambda[/tex],
[tex]\lambda = -\dfrac{t}{\ln \left(\frac{N}{N_0}\right)}[/tex]
[tex]\:\:\:\:= -\dfrac{(24\:\text{minutes})}{\ln \left(\frac{1000\:\text{counts/min}}{8000\:\text{counts/min}}\right)}[/tex]
[tex]\:\:\:\:=11.54\:\text{minutes}[/tex]
Derive the dimension of coefficient of linear expansivity
Answer:
The SI unit of coefficient of linear expansion can be expressed as °C-1 or °K-1. ... The dimension of coefficient of linear expansion will be M0L0T0K−1.
Electron A is fired horizontally with speed 1.00 Mm/s into a region where a vertical magnetic field exists. Electron B is fired along the same path with speed 2.00 Mm/s. (i) Which electron has a larger magnetic force exerted on it
B will have the greater force
Fc=MV2 /R=Fm
The A particle has less centipetal force and larger radius so larger curve
A force of 200 N, acting at 60° to the horizontal, accelerates a block of mass 50 kg along a horizontal plane. Calculate the component of the 200N force that accelerates the block horizontally
Answer:
Explanation:
a) Fx = F cos (θ)
= (200) cos(60)
= 100 N
b) FR = ma
Fx + Ff = ma
100 + Ff = (50)(1,5)
Ff = 75 - 100
= -25 N
c) Fy = F sin θ
= (200) sin(60)
= 173,2 N
A singly charged 7Li ion has a mass of 1.16 10-26 kg. It is accelerated through a potential difference of 523 V and subsequently enters a uniform magnetic field of magnitude 0.370 T perpendicular to the ion's velocity. Find the radius of its path.
Answer:
[tex]R=0.023m[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=1.16*10^{-26}[/tex]
Potential difference [tex]V=523V[/tex]
Magnitude [tex]m=0.370 T[/tex]
Generally the equation for Velocity is mathematically given by
[tex]\frac{1}{2}mv^2=ev[/tex]
[tex]v=\frac{2ev}{m}[/tex]
[tex]v=\frac{2*1.6*10^{-19}*542}{1.16*10^{-26}}[/tex]
[tex]v=12.22*10^4m/s[/tex]
Generally the equation for Force is mathematically given by
[tex]F=qvBsin \theta[/tex]
Where
[tex]qVB=m\frac{v^2}{R}[/tex]
[tex]F=m\frac{v^2}{R}sin\theta[/tex]
Therefore
[tex]R=\frac{mv}{qB sin \theta}[/tex]
[tex]R=\frac{1.6*10^{-26}*12.2*10^{4}}{1.60*10^{-19}*0.394 sin 90}[/tex]
[tex]R=0.023m[/tex]
Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes, subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is Q=ΔV/Δt what is the flow rate in pipe A? Viscosity: Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes, subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is what is the flow rate in pipe A?
a) Q√2
b) 16Q
c) 2Q
d) 4Q
e) 8Q
Answer:
c) 2Q
Explanation:
From the given information:
The pressure inside a pipe can be expressed by using the formula:
[tex]\Delta P = \dfrac{128 \mu L Q}{\pi D^4}[/tex]
Since the diameter in both pipes is the same, we can say:
[tex]D = D_A = D_B[/tex]
where;
length of the first pipe A [tex]L_A = L[/tex] and the length of the second pipe B [tex]L_B = 2L[/tex]
Since the difference in pressure is equivalent in both pipes:
Then:
[tex]\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}[/tex]
[tex]\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}[/tex]
[tex]\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}[/tex]
[tex]\mathbf{Q_1 = 2Q}[/tex]
The flow rate in pipe B is 2Q of the flow rate of the pipe A
What is flow rate?
The flow rate is defined as the flow of the fluid across the cross section in per unit time.
From the given information:
The pressure inside a pipe can be expressed by using the formula:
[tex]\Delta p=\dfrac{128\mu LQ}{\pi D^4}[/tex]
Since the diameter in both pipes is the same, we can say:
[tex]D=D_A=D_B[/tex]
where;
length of the first pipe A [tex]L_A=L[/tex] and the length of the second pipe B
[tex]L_B=2L[/tex]
Since the difference in pressure is equivalent in both pipes:
Then:
[tex]\dfrac{128\mu L_1Q_1}{\pi D_1^4}=\dfrac{128\mu L_2Q_2}{\pi D_2^4}[/tex]
[tex]\dfrac{L_1Q_1}{D_1^4}=\dfrac{L_2Q_2}{D_2^4}[/tex]
[tex]\dfrac{LQ_1}{D_1^4}=\dfrac{2LQ}{D_2^4}[/tex]
[tex]Q_1=2Q[/tex]
Hence the flow rate in pipe B is 2Q of the flow rate of the pipe A
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Calcula la resistencia total del siguiente circuito eléctrico.
In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0E0E_0 and B0B0B_0 are the __________ of the electric and magnetic fields. Choose the best answer to fill in the blank.
Light of a given wavelength is used to illuminate the surface of a metal, however, no photoelectrons are emitted. In order to cause electrons to be ejected from the surface of this metal you should: ___________
a. use light of the same wavelength but increase its intensity.
b. use light of a shorter wavelength.
c. use light of the same wavelength but decrease its intensity.
d. use light of a longer wavelength.
Answer:
use light of the same wavelength but decrease it's intensity
a standard bathroom scale is placed on an elevator. A 34 kg boy enters the elevator on the first floor and steps on the scale. What will the scale read (in newtons) when the elevator begins to accelerate upward at 0.4 m/s2
Answer:F = 255 N
Explanation:
It is given that,
Mass of the boy, m = 25 kg
Acceleration of the elevator,
The elevator is accelerating in upward direction. The net force acting on the boy is given by :
g is the acceleration due to gravity
F = 255 N
The scale reading is 255 N as it begins to accelerate upward. hence, this is the required solution.
Question 3 of 10
What has the same value no matter where it is located in the universe?
A. Volume
B. Weight
C. Mass
D. Density
Reset Selection
Answer:
C. Mass
Explanation:
In the diagram, the crest of the wave is show by:
A
B
C
D
Answer:
D.
Explanation:
The crest of a wave refers to the highest point of a wave. This is illustrated by D.
A large metal sphere has three times the diameter of a smaller sphere and carries three times the charge. Both spheres are isolated, so their surface charge densities are uniform. Compare (a) the potentials (relative to infinity) and (b) the electric field strengths at their surfaces.
Answer:
A. Equals to that of the smaller sphere
B. 3 times less than that of the smaller sphere
Explanation:
(a) Equals to that of the smaller sphere
The potential of an isolated metal sphere, with charge Q and radius R, is kQ=R, so a sphere with charge 3Q and radius 3R has the same potential
b) 3 times less than that of the smaller sphere
However, the electric field at the surface of the smaller sphere is ?=? 0 = kQ=R2 , so tripling Q and R reduces the surface field by a factor of 1/3
