I think any pendulum would swing slower on the moon than it would on Earth.
The time it takes a pendulum to go through a complete back and forth swing is:
Time period = 2 π √(length/gravity)
You can see that 'gravity' is in the denominator of the fraction, so the smaller gravity gets, the longer the period gets.
To be a little bit more technical, the period is inversely proportional to the square root of gravity.
So the period for a complete swing on the moon would be √(9.8/1.6) times as long as the complete swing of the same pendulum on Earth.
That number is roughly 2.47 .
So, for every 1 second that a pendulum takes to swing back and forth once on Earth, the same pendulum would take 2.47 seconds to do it on the moon.
Answer:
based on my opinion....
as we know that gravity in moon are less than gravity in earth.. since the force of gravity is less on the moon, the pendulum would swing slower at the same length, angle
and the frequency would be less.
I hope this helps
The magnetic field perpendicular to a single 13.2-cm diameter circular loop of copper wire decreases uniformly from 0.670 T to zero. If the wire is 2.25 mm in diameter, how much charge moves past a point in the coil during this operation? (rhoCu = 1.68 x 10-8 Ω.m)
Answer:
5.23 C
Explanation:
The current in the wire is given by I = ε/R where ε = induced emf in the wire and R = resistance of wire.
Now, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = AΔB and A = area of loop and ΔB = change in magnetic field intensity = B₂ - B₁
B₁ = 0.670 T and B₂ = 0 T
ΔB = B₂ - B₁ = 0 - 0.670 T = - 0.670 T
A = πD²/4 where D = diameter of circular loop = 13.2 cm = 0.132 m
A = π(0.132 m)²/4 = 0.01368 m² = `1.368 × 10⁻² m²
ε = -ΔΦ/Δt = -AΔB/Δt = -1.368 × 10⁻² m² × (-0.670 T)/Δt= 0.9166 × 10⁻² Tm²/Δt
Now, the resistance R of the circular wire R = ρl/A' where ρ = resistivity of copper wire = 1.68 x 10⁻⁸ Ω.m, l = length of wire = πD and A' = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m
R = ρl/A' = 1.68 x 10⁻⁸ Ω.m × π × 0.132 m÷π(2.25 × 10⁻³ m)²/4 = 0.88704/5.0625 = 0.1752 × 10⁻² Ω = 1.752 × 10⁻³ Ω
So, I = ε/R = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω
IΔt = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω = 0.5232 × 10 C
Since ΔQ = It = 5.232 C ≅ 5.23 C
So the charge is 5.23 C
The induced emf through wire depends on the current flow (indirectly with charge flow as well).
The value of charge moving past a point in the coil during its operations is 5.23 C.
What is the magnetic field?The region in a space where a particle experiences some magnetic force, is known as the magnetic field.
Given data -
The diameter of the circular loop is, d = 13.2 cm = 0.132 m.
The change in magnetic field strength is, ΔB = 0.670 T.
The new diameter of the wire is, d' = 2.25 mm = 2.25 × 10³ m.
The resistivity of the wire is, [tex]\rho = 1.68 \times 10^{-8} \;\rm \Omega.m[/tex].
The current in the wire is given by the following expression,
I = ε/R
Here,
ε is the induced emf in the wire.
R is the resistance of the wire.
And the expression for the induced emf of the wire is given as,
ε = -ΔΦ/Δt
Here,
ΔΦ is the change in magnetic flux. And its expression is,
ε = A × ΔB
Here,
A is the area of the loop. And its value is, A = πd²/4.
Solving as,
A = π(0.132 m)²/4
A = 0.01368 m²
A = `1.368 × 10⁻² m²
Now, calculating the induced emf as,
ε = ΔΦ/Δt
ε = A × ΔB/Δt
ε = 1.368 × 10⁻² m² × (0.670 T)/Δt
ε = 0.9166 × 10⁻² Tm²/Δt
Now, the resistance R of the circular wire is,
R = ρ × L/A'
Here,
L is the length of the wire and its value is. L = πd .
And A' is the new cross-sectional area of wire,
A' = πd'²/4
So, the resistance is,
R = ρ × L/A'
R = 1.68 x 10⁻⁸ ×( π × 0.132 m) ÷ π(2.25 × 10⁻³ m)²/4 =
R = 0.88704/5.0625
R = 0.1752 × 10⁻² Ω
R = 1.752 × 10⁻³ Ω
Now, the current flow (I) in the wire is given as,
I = ε/R
I = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω
And obtaining the value of charge from the expression of current as,
Q = IΔt
Q = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω
Q = 0.5232 × 10 C
Q = 5.23 C
Thus, we can conclude that the value of charge moving past a point in the coil during its operations is 5.23 C.
Learn more about the resistance of a wire here:
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A car is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of -0.5 m/s2(metre per sec.Square). Find how far the car will go before it is brought to rest?
The only thing that mkes this question inconvenient is that it uses a mixture of units ... speed in km/hr, and acceleration in m/s². You can't directly mash those together.
What's the speed when we express it in m/s ?
Speed = (90 km/hr) · (1,000 m/km) · (1 hr/3,600 sec)
Speed = (90 · 1,000 · 1 / 3,600) · (km-m-hr / hr-km-sec)
Speed = 25 m/s
OK great !
-- The car is traveling at 25 m/s when the brakes are applied.
-- The brakes slow it down by 0.5 m/s every second.
-- So it takes (25/0.5) = 50 seconds to stop the car.
-- During that time, the car's average speed is (1/2)·(25m/s + 0) = 12.5 m/s .
-- Moving at an average speed of 12.5 m/s for 50 sec, the car travels
(12.5 m/s) · (50 s) = 625 meters
The angle of incidence of a ray of light striking an equilateral triangular prisms ABC of refracting angle 60o is 40o. Calculate:
(i)the angle of refraction at the first face
(ii)the angle of emergence
Answer:
1: the refracted angle in the first face is equal to the incident angle that is 60degrees
2. Emergence Angle is 42degrees
Explanation:
Pls see attached file
A spherical mirror gives an image magnified 5 times at a distance 5 m. determine whether the mirror is convex or concave? How much will be the focal length of the mirror?
Answer:
1. Concave mirror.
2. 4.17 m or 417 cm.
Explanation:
The following data were obtained from the question:
Object distance (u) = 5 m
Magnification (M) = 5
Focal length (f) =..?
1. Identification of the mirror.
To determine whether or not the mirror is concave or convex, we must first of all calculate the image distance.
This can be obtained as follow:
Object distance (u) = 5 m
Magnification (M) = 5
Image distance (v) =.?
Magnification (M) = image distance (v) /object distance (u).
M = v/u
5 = v/5
Cross multiply
v = 5 x 5
v = 25 m
Since the image distance obtained is positive, the mirror is said to be a concave mirror.
2. Determination of the focal length of the mirro.
This can be obtained as follow:
Object distance (u) = 5 m
Image distance (v) = 25 m
Focal length (f) =...?
1/f = 1/v + 1/u
1/f = 1/25 + 1/5
1/f = 0.04 + 0.2
1/f = 0.24
Cross multiply
f x 0.24 = 1
Divide both side by 0.24
f = 1/0.24
f = 4.17 m
Converting the focal length of cm, we have:
1 m = 100 cm
Therefore, 4.17 m = 4.17 x 100 = 417 cm
Therefore, the focal length of the mirror is 4.17 or 417 cm.
Fill in the blanks in the following statements:
1. A parsec is defined as the distance from the Sun which would result in a parallax of_____ arcsecond as seen from Earth when observed 6 months apart.
2. One parsecs is about_____light years.
3. The more distant a star, the_____its parallax.
4. The statement "we can measure stellar parallax for most stars in our galaxy" is_____.
5. The closest stars have parallaxes smaller than______arcsecond.
6. The first successful measurements of stellar parallax were made by______in year______for the star named.
7. Star Fred has parallax four times greater than star Bob. Star Fred is______times______then star Bob.
Answer:
Explanation:
1. A parsec is defined as the distance from the Sun which would result in a parallax of_TWO____ arcsecond as seen from Earth when observed 6 months apart.
2. One parsecs is about_3.3____light years
3. The more distant a star, the_LESS ____its parallax.
4. The statement "we can measure stellar parallax for most stars in our galaxy" is__TRUE .___.
5. The closest stars have parallaxes smaller than_.002 _____arcsecond.
6. The first successful measurements of stellar parallax were made by_Friedrich Bessel_____in year_1838_____for the star named.
7. Star Fred has parallax four times greater than star Bob. Star Fred is__4____times_nearer_____than star Bob.
Given a volume of 1000. сm of an ideal gas at 300. K, what volume would it occupy at a temperature of 600. K?
Answer:
2000 cm³
Explanation:
Assuming the pressure is constant:
V / T = V / T
1000 cm³ / 300 K = V / 600 K
V = 2000 cm³
A nonrenewable resource
Answer:
It is a finite resource.
Explanation:
A nonrenewable resource is a natural substance that is not replenished with the speed at which it is consumed. Fossil fuels such as oil, natural gas, and coal are examples of nonrenewable resources.
Answer:
Coal, oil, natural gas.
Explanation:
Non renewable resource -A non-renewable resource (also called a finite resource) is a natural resource that cannot be readily replaced by natural means at a quick enough pace to keep up with consumption.
Coal, oil, and natural gas are all examples of non-renewable resources and cannot be readily replaced to keep up with consumption.
PLEASEEE HELP, thank you :)
Answer:
The answer is B.
Explanation:
Given that the current (Ampere) in a series circuit is same so we can ignore it. We can assume that the total voltage is 60V and all the 3 resistance are different, 20Ω, 40Ω and 60Ω. So first, we have to find the total resistance by adding :
Total resistance = 20Ω + 40Ω + 60Ω
= 120Ω
Next, we have to find out that 1Ω is equal to how many voltage by dividing :
120Ω = 60V
1Ω = 60V ÷ 120
1Ω = 0.5V
Lastly, we have to calculate the voltage at R1 so we have to multiply by 20 (R1) :
1Ω = 0.5V
20Ω = 0.5V × 20
20Ω = 10V
Increase in Space Suit Pressure 0.0/3.0 points (graded) If the pressure in a space suit increases, how will each of the following be affected? Flexilibity will: Increase Decrease Stay the same unanswered The required pre-breathe time will: Increase Decrease Stay the same unanswered The mass of the suit will: Increase Decrease Stay the same
Answer:
Flexibility Increases
Pre-breathe time decreases
Mass of suit decreases.
Explanation:
Spacesuits are designed for space shuttles when a person goes to explore the galaxy. The spacesuits shuttle era are pressurized at 4.3 pounds per inch. The gas in the suit is 100% of oxygen and there is more oxygen to breathe when the altitude of 10,000 is reached. This will decrease the breathing time and mass of suit.
what nuclear fission and nuclear fusion have in common? A. They’re both used in power plants B. they have less mass than their reactants C. they start with the same reactants D. their products have more mass than their reactants
Answer:
B. they have less mass than their reactants
Explanation:
In both nuclear fission and nuclear fusion, the products have less mass than their reactants. So, option B is correct.
What is meant by nuclear reaction ?Nuclear reaction is defined as the reaction that involves the collision between one or more atomic nuclei and a highly energetic subatomic particle to produce one or more nuclei along with the emission of a large amount of energy.
Here,
Nuclear fission and fusion are two types of nuclear reactions that involves the production of large amount of energy from atomic nuclei.
Nuclear fission is the nuclear reaction in which a heavier nucleus is split into two or more lighter nuclei, under radioactive condition, with the release of a massive amount of energy.
Nuclear fusion is the nuclear reaction in which two or more lighter nuclei nuclei fuse together under radioactive condition and forms a heavier nuclei and by releasing a large amount of energy.
In both nuclear fission and nuclear fusion, there occurs a release of large amount of energy. This is because most of the mass of the reactants is converted into energy which is known as the mass-energy. So the products of these nuclear reactions will have lesser mass than their reactants.
Hence,
In both nuclear fission and nuclear fusion, the products have less mass than their reactants.
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An electron of kinetic energy 1.39 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 20.8 cm. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.
Answer:
a) v = 2.21 10⁷ m / s, b) B = 6.04 10⁻⁴ T , c) f = 1.69 10⁷ Hz , d) T = 5.9 10⁻⁸ s
Explanation:
a) For this exercise, let's start by using the concept of kinetic energy
K = ½ m v²
v = √ 2K / m
Let's reduce the magnitudes to SI units
E = K = 1.39 kev (1000 eV / 1 keV) (1.6 10⁻¹⁹ J / 1 eV) = 2,224 10⁻¹⁶ J
r = 20.8 cm (1 m / 100 cm) = 0.208 m
let's calculate
v = √ (2 2,224 10⁻¹⁶ / 9.1 10⁻³¹)
v = √ (4.8879 10¹⁴)
v = 2.21 10⁷ m / s
b) let's use Newton's second law where the force is magnetic
F = m a
where the acceleration is centripetal
a = v² / r
the magnetic force is
Fm = q v x B = q v B sin θ
since the circle is perpendicular to the magnetic field, the angle is 90º and the sine is equal to one, let's substitute
qv B = m v² / r
B = m v / rq
let's calculate
B = 9.1 10⁻³¹ 2.21 10⁷ / (0.208 1.6 10⁻¹⁹)
B = 6.04 10⁻⁴ T
c) Linear and angular variables are related
v = w r
w = v / r
w = 2.21 10⁷ / 0.208
w = 1.0625 10⁸ ras / s
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 1.0625 10⁸ / (2π)
f = 1.69 10⁷ Hz
d) the frequency is the inverse of the period
f = 1 / T
T = 1 / 1.69 10⁷
T = 5.9 10⁻⁸ s
Your car's 32.5 W headlight and 2.00 kW starter are ordinarily connected in parallel in a 12.0 V system. What power (in W) would one headlight and the starter consume if connected in series to a 12.0 V battery
Answer:
Explanation:
the resistance of a electrical device
R = V² / P where V is volt and P is power .
The devices are in parallel so same volt will apply on them
So R₁ = 12² / 32.5 = 4.431 ohm
R₂ = 12² / 2 x 10³ = .072 ohm
when they are in series
Common Current in them = 12 / 4.431 + .072
= 2.6649 A
power consumed by first device when they are in series
= current² x resistance
= 2.6649² x 4.431 = 31.46 W
power consumed by other
= 2.6649² x .072 = .511 W
a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on the earth surface.
Answer:
The object would weight 63 N on the Earth surface
Explanation:
We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:
[tex]F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}[/tex]
Now, if the body is on the surface of the Earth, its weight (w) would be:
[tex]F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}[/tex]
Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:
[tex]\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\[/tex]
Complete all four parts. 15 points. Will give brainliest! Show work!
Answer:
A. 5.08 secs.
B. 10.16 secs.
C. 126.50 m.
D. 373.36 m
Explanation:
Data obtained from the question include the following:
Initial velocity (u) = 65 m/s
Angle of projection θ = 50°
A. Determination of the time taken to reach the peak.
Initial velocity (u) = 65 m/s
Angle of projection θ = 50°
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =.?
t = u•Sine θ/g
t = (65 × Sine 50) /9.8
t = 5.08 secs.
B. Determination of the total time spent by the ball in air.
Time (t) taken to reach the peak = 5.08 secs.
Total time (T) spent by the ball in air =?
T = 2t
T = 2 × 5.08
T = 10.16 secs
Therefore, the total time spent by the ball in air is 10.16 secs.
C. Determination of the maximum height.
Initial velocity (u) = 65 m/s
Angle of projection θ = 50°
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (H) =..?
H = u²•Sine² θ / 2g
H = 65² × (Sine 50)² / 2 × 9.8
H = 4225 × (Sine 50)² /19.6
H = 126.50 m
Therefore, the maximum height reached by the ball is 126.50 m.
D. Determination of the horizontal distance travelled by the ball.
Initial velocity (u) = 65 m/s
Angle of projection θ = 50°
Acceleration due to gravity (g) = 9.8 m/s²
Horizontal distance (R) =..?
R = u²•Sine 2θ / g
R = 65² × Sine (2×30) / 9.8
R = (4225 × Sine 60) / 9.8
R = 373.36 m
Therefore, the horizontal distance travelled by the ball is 373.36 m
If the absolute pressure inside the bottom of a container open to the atmosphere and filled with an unknown substance is 300 Pa. And the height of the container is 800.0 cm. What is the density of the substance?
Answer:
2.5 kg/m³
Explanation:
Absolute pressure = gauge pressure + atmospheric pressure
P = Pg + Pa
The gauge pressure caused by the weight of a fluid is called the static pressure. It is equal to the density of the fluid × acceleration due to gravity × depth of the fluid,
Pg = ρgh
Therefore:
P = ρgh + Pa
300 Pa = ρ (10 m/s²) (8.00 m) + 100 Pa
ρ = 2.5 kg/m³
When we double the distance between a source of light and the
surface on which it falls, the amount of illumination on the surface
decreases to
(what fraction) of the
original illumination.
The amount of lighting on a surface drops to (1/4) of the initial illumination when the distance between a light source and the surface it falls on is doubled.
What is illumination?The amount of light or luminous flux that falls on a surface is known as illumination. It is expressed as lux or lumens per square meter.
The square of the distance has an inverse relationship with the light intensity;
[tex]\rm I = \frac{1}{r^2}[/tex]
Where,
I is the light intensity
r is the distance
Let r is the distance and I is the sound intensity for case 2;
r' = 2r
[tex]\rm I' = \frac{1}{(2r)^2} \\\\ I' = \frac{1}{4r^2} \\\\ I' = \frac{I}{4}[/tex]
When we double the distance between a source of light and the surface on which it falls, the amount of illumination on the surface decreases to(1/2) of the original illumination.
Hence the value of the fraction is 1/4.
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Please answer this question
Explanation:
m = kg. v=m/s. g=m/s^2. h= m
>>1/2mv^2=mgh
>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m
>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m>>kg m^2/s^2=kg m^2/s^2 the fraction 1/2 won't be able to make any changes to to the dimensional expression of energy i.e half of energy is still energy therefore you can neglect the number .
>>kg m^2/s^2=kg m^2/s^2
>>J= J
A machine takes 0.5 seconds to move a brick 1 meter and put 100 Joules of energy into it. (hint: Power is the amount of energy transferred or converted per unit time. Or we say power is work done per unit time. LaTeX: P\:=\:\frac{W}{t}P = W t) This machine would expend more power in this action if it:
Answer:
200 Watts.
Explanation:
Power is defines as the amount of work expended per unit time. Mathematically, it is expressed as Power = Workdone/Time
Given parameters
Energy used up 100Joules
Distance moved by brick = 1 meters
Time taken by the machine = 0.5 secs
Power can also be written as Energy/Time
Required
We need to calculate the amount of power used up.
Power = 100J/0.5s
Power = 100/(1/2)
Power = 100 * 2/1
Power = 200Watts.
This shows that the machine would expend 200Watts of power
define motion also justify that rest and motion are related terms
Answer;
Motion: A body is said to be in motion if it changes its position with respect to its surroundings.
Explanation:
Rest and motion are the relative terms because they depend on the observer's frame of reference. So if two different observers are not at rest with respect to each other, then they too get different results when they observe the motion or rest of a body .
one example for each. Rest: If a body does not change its position with respect to its surroundings, the body is said to be at rest. ... Motion: A body is said to be in motion if it changes its position with respect to its surroundings.
I REALLY NEED HELP WITH PHYSICS!! HELPPP!! I’m giving 50 points :)))
Answer:
59%
Explanation:
Data obtained from the question include the following:
Input temperature = 495 K
Output temperature = 293 K
Efficiency =?
The efficiency of the car engine can be obtained by multiplying the ratio of the output temperature to the input temperature by 100 as shown below:
Efficiency = output /input x 100
Efficiency = 293/495 x 100
Efficiency = 59.1 ≈ 59%
Therefore, the efficiency of the car engine is approximately 59%
Why does time seem to flow only in one direction?
Answer:
Time seem to flow only in one direction because if it started to go in backward direction that would break the second law of thermodynamics. We do not find time to be moving in any direction because time is not an object that can move nor is it a force that can move any object.
When the current in a toroidal solenoid is changing at a rate of 0.0260 A/s, the magnitude of the induced emf is 12.2 mV. When the current equals 1.40 A, the average flux through each turn of the solenoid is 0.00308 Wb. How many turns does the solenoid have?
Answer:
The solenoid has 213 turns.
Explanation:
The number of the solenoid's turns (N) can be found as follows:
[tex] N = \frac{L*I}{\phi_{B}} [/tex] (1)
Where:
L: is the self-inductance of the solenoid
I: is the current = 1.40 A
[tex]\phi_{B}[/tex]: is the magnetic flux = 0.00308 Wb
The self-inductance of the solenoid (L) is:
[tex] L = \frac{|\epsilon|}{|dI/dt|} [/tex] (2)
Where:
ε: is the induced emf = 12.2x10⁻³ V
dI/dt: is the rate changing of the current = 0.0260 A/s
By entering equation (2) into (1) we can find the number of turns:
[tex] N = \frac{|\epsilon|*I}{\phi_{B}|dI/dt|} = \frac{12.2 \cdot 10^{-3} V*1.40 A}{0.00308 Wb*0.0260 A/s} = 213 [/tex]
Therefore, the solenoid has 213 turns.
I hope it helps you!
we can catch a rolling ball but not a flying bullet?? give reason
Answer:
yeah this statement is tru
Explanation:
it is because the speed of the bullet is more than the speed of rolling ball .so from this reason we cannot catch a bullet.
A bullet will be moving much faster than a rolling ball. Even seeing a bullet in flight requires it to be extremely low velocity.
CAN YOU HELP ME IN THE FOLLOWING EQUATION THE SPEED IS V = A + BT. DETERMINE THE DIMENSIONS OF A AND B
Answer:
A = [m/s]
B = [m/s²]
Explanation:
Assuming that V has SI units of m/s, then A and BT must also have units of m/s.
A = [m/s]
BT = [m/s]
Since T has SI units of s:
B [s] = [m/s]
B = [m/s²]
b. i. In a hydraulic press, a force of 200N is applied to master piston of area 25cm. If
the press is designed to produce a force of 5000N, determine the area of the slave
piston.
(4marks)
Answer:
625 cm²
Explanation:
The pressure is the same on both pistons.
F/A = F/A
200 N / 25 cm² = 5000 N / A
A = 625 cm²
you (60 kg) are standing in a (500 kg) elevator that is moving upwards from a ground floor on a building what is the power rating of the motor that can lift this elevator if the elevator travels a distance of 20m upwards in 15 s
Explanation:
Power = work / time
Power = force × distance / time
P = (650 kg) (10 m/s²) (20 m) / (15 s)
P = 8667 W
Two long, straight wires are parallel and 10 cm apart. One carries a current of 2.0 A, the other a current of 5.0 A. If the two currents flow in opposite directions, what is the magnitude and direction of the force per unit length of one wire on the other
Answer:
The magnitude and direction of the force per unit length of one wire on the other is 2 x 10⁻⁵ N/m, attractive force.
Explanation:
Given;
distance between the two parallel wires, r = 10 cm = 0.1 m
current in the first wire, I₁ = 2A
current in the second wire, I₂ = 5 A
The force per unit length on each wire can be calculated as;
[tex]\frac{F}{L} = \frac{\mu_oI_1I_2}{2\pi r}[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
[tex]\frac{F}{L} = \frac{\mu_oI_1I_2}{2\pi r} \\\\\frac{F}{L} = \frac{4\pi*10^{-7}*2*5}{2\pi *0.1} \\\\\frac{F}{L} = 2 *10^{-5} \ N[/tex]/ m
The direction of the force between the two wires is attractive since the current in the two wires are in opposite direction.
Therefore, the magnitude and direction of the force per unit length of one wire on the other is 2 x 10⁻⁵ N/m, attractive force.
A diver releases an air bubble of volume 3.0 cm3 from a depth of 18 m below the surface of a lake, where the temperature is 9.0 ∘C. Part A What is the volume of the bubble when it reaches just below the surface of the lake, where the temperature is 22 ∘C?
Answer:
Explanation:
We shall apply gas law formula to get the solution .
P₁ V₁ / T₁ = P₂ V₂ / T₂
P₁ = 10⁵ + 18 x 10³ x 9.8 where 10⁵ Pa is atmospheric pressure at surface
= 1.764 x 10⁵ + 10⁵
= 2.764 x 10⁵
2.764 x 10⁵ x 3 / ( 273 + 9 ) = 10⁵ x V₂ / ( 273 + 22 )
V₂ = 8.67 cm³
1. A note has a wavelength of 0.77955 m. If the speed of sound is 343.00 m/s, what pitch is this note? 2. A note has a wavelength of 0.52028 m. If the speed of sound is 343.00 m/s, what pitch is this note? please no funny answers
Answer:
1.) 440 Hz
2.) 659.3 Hz
Explanation:
1.) Given parameters are:
wavelength = 0.77955 m.
speed of sound = 343.00 m/s
Frequency = speed/ wavelength
Substitute speed and wavelength into the formula
Frequency = 343/ 0.77955
Frequency = 439.99
Frequency = 440 Hz approximately
2.) The parameters given are:
wavelength = 0.52028 m.
speed of sound = 343.00 m/s
Using the same formula
Frequency = speed/wavelength
Substitute all the parameters into the formula
Frequency = 343 / 0.52028
Frequency = 659.3 Hz approximately
The pitch of a note depends on the frequency of the sound waves.
The pitch of a sound increases as the frequency of the sound waves increases.
Answer:
1.) 440 Hz
2.) 659.3 Hz
Explanation:
1.) Given parameters are:
wavelength = 0.77955 m.
speed of sound = 343.00 m/s
Frequency = speed/ wavelength
Substitute speed and wavelength into the formula
Frequency = 343/ 0.77955
Frequency = 439.99
Frequency = 440 Hz approximately
2.) The parameters given are:
wavelength = 0.52028 m.
speed of sound = 343.00 m/s
Using the same formula
Frequency = speed/wavelength
Substitute all the parameters into the formula
Frequency = 343 / 0.52028
Frequency = 659.3 Hz approximately
The pitch of a note depends on the frequency of the sound waves.
The pitch of a sound increases as the frequency of the sound waves increases.
Which statement best describes the difference between how lenses and
mirrors interact with light?
A. Lenses bring light to a point; mirrors do not.
B. Lenses form images by reflecting light; mirrors do not.
C. Lenses spread light apart; mirrors do not.
D. Lenses form images by refracting light; mirrors do not.
Answer:D
Explanation:
Apex