The simplest form of a traveling electromagnetic wave is a plane wave. For a wave traveling in the x direction whose electric field is in the y direction, the electric and magnetic fields are given below. This wave is linearly polarized in the y direction.
E= EoSin(kx - ωt)y
B= Bosin(kx- ωt)z
In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0 and B0 are the________ of the electric and magnetic fields.
a. maximas
b. wavelenghts
c. amplitudes.
d. velocities
Answer:
amplitudes
Explanation:
In everyday physics we define the amplitude of a wave as the maximum (this can also be called the highest)displacement or distance moved by a point on a given vibrating body or wave as measured from its equilibrium position. The key idea in defining the amplitude of a wave motion is the idea of a 'maximum displacement from the position of equilibrium'.
Given the equations;
E= EoSin(kx - ωt)y
B= Bosin(kx- ωt)z
Both Eo and Bo refer to the maximum displacement of the electric and magnetic field components of the electromagnetic wave. This maximum displacement is known as the amplitude of the electric and magnetic components of the electromagnetic wave.
When a central dark fringe is observed in reflection in a circular interference pattern, waves reflected from the upper and lower surfaces of the medium must have a phase difference, in radians, of
Explanation:
Let the first wave is :
[tex]y_1=A\sin\omega t[/tex]
And another wave is :
[tex]y_2=A\sin (\omega t+\phi)[/tex]
[tex]\phi[/tex] is phase difference between waves
Let y is the resultant of these two waves. So,
[tex]y =y_1+y_2[/tex]
The waves reflected from the upper and lower surfaces of the medium, it means that the resultant to be zero. So,
[tex]\cos(\dfrac{\phi}{2})=0\\\\\cos(\dfrac{\phi}{2})=\cos(\dfrac{\pi}{2})\\\\\phi=\pi[/tex]
So, the phase difference between the two waves is [tex]\pi[/tex].
A narrow copper wire of length L and radius b is attached to a wide copper wire of length L and radius 2b, forming one long wire of length 2L. This long wire is attached to a battery, and a current is flowing through it. If the electric field in the narrow wire is E, the electric field in the wide wire is
Answer:
electric field in the wide wire is
E₂ =[tex]\frac{E}{4}[/tex]
Explanation:
given
length of the copper wire = L
radius of the copper wire r₁ = b
length of the second copper wire = L
radius of the second copper wire r₂ = 2b
electric field in the narrow wire = E₁=E
recall
resistance R = ρL/A
where ρ is resistivity of the copper wire, L is the length, and A is the cross sectional area.
Resistance of narrow wire, R₁
R₁ = ρL/A
where A = πb²
R₁ = ρL/πb²---------- eqn 1
Resistance of wide wire, R₂
R₂ = ρL/A
where A = π(2b)²
R₂ = ρL/π(2b)²
R₂ = ρL/4πb²-------------- eqn 2
R₂ = ¹/₄(ρL/πb²)
comparing eqn 1 and 2
R₁ = 4R₂
calculating the current in the wire,
I = E/(R₁ + R₂)
recall
R₁ = 4R₂
∴ I = E/(4R₂ + R₂)
I = E/5R₂
calculating the potential difference across R₁ & R₂
V₁ = IR₁
I = E/5R₂
∴ V₁ = ER₁/5R₂
R₁ = 4R₂
V₁ = 4ER₂/5R₂
∴V₁ = ⁴/₅E
potential difference for R₂
V₂= IR₂
I = E/5R₂
∴ V₂ = ER₂/5R₂
V₂ = ER₂/5R₂
∴V₂ = ¹/₅E
so, electric field E = V/L
for narrow wire E₁ = V₁/L ----------- eqn 3
for wide wire, E₂ = V₂/L------------ eqn 4
compare eqn 3 and 4
E₂/E₁ = V₂/V₁( L is constant)
E₂/E₁ = ¹/₅E/⁴/₅E
E₂ = E₁/4
note E₁ = E
∴E₂ =[tex]\frac{E}{4}[/tex]
What happens to a bar of metal when it's heated?
A.It gets longer.
B.The effect depends on the density of the bar.
C.It stays the same length.
D.It gets shorter
Hey There!!
Your answer will be A. It get longer.
Because, The kinetic energy of its atoms increase. They vibrate faster. This means that each atom will take up more space due to its movement so the material will expand. Some metals expand more than others due to differences in the forces between the atoms / molecules. . . .
Hope This Helps <3!!
What would the Hall voltage be if a 2.00-T field is applied across a 10-gauge copper wire (2.588 mm in diameter) carrying a 20.0-A current
Answer:
The hall voltage is [tex]\epsilon =1.45 *10^{-6} \ V[/tex]
Explanation:
From the question we are told that
The magnetic field is [tex]B = 2.00 \ T[/tex]
The diameter is [tex]d = 2.588 \ mm = 2.588 *10^{-3} \ m[/tex]
The current is [tex]I = 20 \ A[/tex]
The radius can be evaluated as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{2.588 * 10^{-3}}{2}[/tex]
[tex]r = 1.294 *10^{-3} \ m[/tex]
The hall voltage is mathematically represented as
[tex]\episilon = B * d * v_d[/tex]
where[tex]v_d[/tex] is the drift velocity of the electrons on the current carrying conductor which is mathematically evaluated as
[tex]v_d = \frac{I}{n * A * q }[/tex]
Where n is the number of electron per cubic meter which for copper is
[tex]n = 8.5*10^{28} \ electrons[/tex]
A is the cross - area of the wire which is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * [ 1.294 *10^{-3}]^2[/tex]
[tex]A = 5.2611 *10^{-6} \ m^2[/tex]
so the drift velocity is
[tex]v_d = \frac{20 }{ 8.5*10^{28} * 5.26 *10^{-6} * 1.60 *10^{-19} }[/tex]
[tex]v_d = 2.7 *10^{-4 } \ m/s[/tex]
Thus the hall voltage is
[tex]\epsilon = 2.0 * 2.588*10^{-3} * 2.8 *10^{-4}[/tex]
[tex]\epsilon =1.45 *10^{-6} \ V[/tex]
Which of the following explains why metallic bonding only occurs between
metallic atoms?
A. Metallic atoms are less likely to give their electrons to nonmetallic
atoms
B. Electrical conductivity is higher in metallic atoms, which means
they are more likely to attract free electrons.
C. Metallic atoms are highly reactive and do not tend to form bonds
with other atoms.
D. Metallic atoms have valence shells that are mostly empty, which
means these atoms are more likely to give up electrons and allow
them to move freely.
Answer:
D. Metallic atoms have valence shells that are mostly empty, which
means these atoms are more likely to give up electrons and allow
them to move freely.
Explanation:
Metals usually contain very few electrons in their valence shells hence they easily give up these few valence electrons to yield metal cations.
In the metallic bond, metal cations are held together by electrostatic attraction between the metal ions and a sea of mobile electrons.
Since metals give up their electrons easily, it is very easy for them to participate in metallic bonding. They give up their electrons easily because their valence shells are mostly empty, metal valence shells usually contain only a few electrons.
You are fixing a transformer for a toy truck that uses an 8.0-V emf to run it. The primary coil of the transformer is broken; the secondary coil has 40 turns. The primary coil is connected to a 120-V wall outlet.
(a) How many turns should you have in the primary coil?
(b) If you then connect this primary coil to a 240-V source, what emf would be across the secondary coil?
Comments: The relevant equation is N1/N2 = V1/V2 where N is the number of turns and V is the voltage. I'm just not sure how to get the voltage of the secondary coil using emf.
Answer:
a. The primary turns is 60 turns
b. The secondary voltage will be 360 volts.
Explanation:
Given data
secondary turns N2= 40 turns
primary turns N1= ?
primary voltage V1= 120 volts
secondary voltage V2= 8 volts
Applying the transformer formula which is
[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]
we can solve for N1 by substituting into the equation above
[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]
the primary turns is 60 turns
If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)
[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]
the secondary voltage will be 360 volts.
(a) In the primary coil, you have "60 turns".
(b) The emf across the secondary coil would be "360 volts".
Transformer and VoltageAccording to the question,
Primary voltage, V₁ = 120 volts
Secondary voltage, V₂ = 8 volts
Secondary turns, N₂ = 40 turns
(a) By applying transformer formula,
→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]
or,
N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]
By substituting the values,
= [tex]\frac{40\times 120}{8}[/tex]
= [tex]\frac{4800}{8}[/tex]
= 60
(2) Again by using the above formula,
→ V₂ = [tex]\frac{60\times 240}{40}[/tex]
= [tex]\frac{14400}{40}[/tex]
= 360 volts.
Thus the above approach is correct.
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Part (a) Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing through the aorta (in grams of blood per second)? 94.2 Attempts Remain 33%
Part (b) If all the blood that flows through the aorta then branches into the major arteries, what is the velocity of blood in the major arteries? Give your answer in cm/s X Attempts Remain v4.71 A 33%
Part (c) The blood flowing in the major arteries then branches into the capillaries. If the velocity of blood in the capillaries is measured to be 0.04 cm/s, what is the cross sectional area of the capillary system in cm2? Grade Summary Deductions Potential cm2 A = 0% 100% cos( tan acos Submissions sin ( 7 8 9 НОME л Attempts remaining: 10 (4% per attempt) detailed view cotan0 asin acotan A E 4 5 6 atan cosh sinh0 1 2 3 tanh) ODegrees cotanh 0 + END Radians BACKSPACE CLEAR DEL
The aorta (the main blood vessel coming out of the heart) has a radius of about 1.0 cm and the total cross section of the major arteries is about 20 cm2. The density of blood is about the same as water, 1 g/cm3.
Answer:
a) 94.26 g/s
b) 4.713 cm/s
c) 2356.5 cm^2
Explanation:
a) velocity of blood through the aorta = 30 cm/s
radius of aorta = 1 cm
density of blood = 1 g/cm^3
Area of the aorta = [tex]\pi r^{2}[/tex] = 3.142 x [tex]1^{2}[/tex] = 3.142 cm^2
Flow rate through the aorta Q = AV
where A is the area of aorta
V is the velocity of blood through the aorta
Q = 3.142 x 30 = 94.26 cm^3/s
Current of blood through aorta [tex]I[/tex] = Qρ
where ρ is the density of blood
[tex]I[/tex] = 94.26 x 1 = 94.26 g/s
b) Velocity of blood in the major aorta = 30 cm/s
Area of the aorta = 3.142 cm^2
Velocity of blood in the major arteries = ?
Area of major arteries = 20 cm^2
From continuity equation
[tex]A_{ao} V_{ao} = A_{ar} V_{ar}[/tex]
where
[tex]V_{ao}[/tex] = velocity of blood in the major arteries
[tex]A_{ao}[/tex] = Area of the aorta
[tex]V_{ar}[/tex] = velocity of blood in the major arteries
[tex]A_{ar}[/tex] = Area of major arteries
substituting values, we have
3.142 x 30 = 20[tex]V_{ar}[/tex]
94.26 = 20[tex]V_{ar}[/tex]
[tex]V_{ar}[/tex] = 94.26/20 = 4.713 cm/s
c) From continuity equation
[tex]A_{ar} V_{ar} = A_{c} V_{c}[/tex]
where
[tex]A_{ar}[/tex] = Area of major arteries = 20 cm/s
[tex]V_{ar}[/tex] = velocity of blood in the major arteries = 4.713 cm/s
[tex]A_{c}[/tex] = Area of the capillary system = ?
[tex]V_{c}[/tex] = velocity of blood in the capillary system = 0.04 cm/s
substituting values, we have
20 x 4.713 = [tex]A_{c}[/tex] x 0.04
94.26 = 0.04[tex]A_{c}[/tex]
[tex]A_{c}[/tex] = 94.26/0.04 = 2356.5 cm^2
This question involves the concepts of volumetric flow rate, continuity equation, and flow velocity.
a) Total current of the blood passing through the aorta is "94.2 g/s".
b) The velocity of blood in major arteries is "4.71 cm/s".
c) The cross-sectional area of the capillary system is "2356.2 cm²".
a)
First, we will find the volumetric flow rate of the blood, using the continuity equation's formula:
[tex]Q=Av[/tex]
where,
Q = volumetric flow rate = ?
A = cross-sectional area of aorta
A = [tex]\pi(r)^2=\pi(1\ cm)^2= 3.14\ cm^2[/tex]
v = flow velocity = 30 cm/s
Therefore,
[tex]Q=(3.14\ cm^2)(30\ cm/s)[/tex]
Q = 94.25 cm³/s
Now, the blood current will be given as:
I = Qρ
where,
I = current = ?
ρ = blood density = 1 g/cm³
Therefore,
I = (94.2 cm³/s)(1 g/cm³)
I = 94.2 g/s
b)
Now, this volumetric flow rate will be constant in major arteries:
[tex]Q = A_r v_r\\\\v_r=\frac{Q}{A_r}[/tex]
where,
Ar = cross-section area of major arteries = 20 cm²
vr = flow velocity of blood in major arteries = ?
Therefore,
[tex]v_r=\frac{94.25\ cm^3/s}{20\ cm^2}[/tex]
vr = 4.71 cm/s
c)
Now, this volumetric flow rate will be constant in capillaries:
[tex]Q = A_c v_c\\\\A_c=\frac{Q}{v_c}[/tex]
where,
Ac = cross-section area of capillaries = ?
vc = flow velocity of blood in capillaries = 0.04 cm/s
Therefore,
[tex]A_c=\frac{94.25\ cm^3/s}{0.04\ cm/s}[/tex]
Ac = 2356.2 cm²
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9. You are given a number of 10 Ω resistors, each capable of dissipating only 1.0 W without being destroyed. What is the minimum number of such resistors that you need to combine in series or in parallel to make a 10 Ω resistance that is capable of dissipating at least 5.0 W?
Answer:
here are 9 resistors, forming a group of 3 resistors in parallel and each group in series with the other.
Explanation:
Let's work carefully this exercise, they indicate that the total resistance 10 ohm and dissipates 5W, so we can use the power equation to find the circuit current
P = Vi = i² R
i = √ P / R
i = √ (5/10)
i = 0.707 A
This is the current that must circulate in the circuit.
Let's build a circuit with three resistors in series and each resistor in series has three resistors in parallel
The equivalent resistance is
1 /[tex]R_{equi}[/tex] = 1/10 + 1/10 + 1/10 = 3/10
Requi = 10/3
Requi = 3.3 Ω
The current in the three series resistors is I = 0.707 A, and this is divided into three equal parts for the parallel resistors
current in each residence in parallel
i_P = 0.707 / 3
I_p = 0.2357 A
now let's look at the power dissipated in each resistor
P = R i²
P = 10 0.2357²
P = 0.56 W
the power dissipated by each resistance is within the range of 1 A, let's see the total power that the 9 resistors dissipate
P_total = 9 P
P = total = 9 0.56
P_total = 5 W
we see that this combination meets the specifications of the problem.
Therefore there are 9 resistors, forming a group of 3 resistors in parallel and each group in series with the other.
collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitude of their velocity
Answer:
metre per seconds
Explanation:
because velocity = distance ÷ time
13. A base is also known as
A) A proton donor
B) An electron donor
C) proton acceptor
D)An electron acceptor
Answer:
C) proton acceptor
Explanation:
A proton acceptor is another name for a base,
Suppose you sit on a rotating piano stool and hold a 2kg mass in each outstreched hand. If without your arms relative to your body you now drop these masses,
a) your angular velocity remains unchanged
b) your angular velocity increases
c) your angular velocity decreases but your kinetic energy increases.
d) your kinetic energy and angular velocity increases.
Answer:
C is the best answer for this question
You are using a hydrogen discharge tube and high quality red and blue light filters as the light source for a Michelson interferometer. The hydrogen discharge tube provides light of several different wavelengths (colors) in the visible range. The red light in the hydrogen spectrum has a wavelength of 656.3 nm and the blue light has a wavelength of 434.0 nm. When using the discharge tube and the red filter as the light source, you view a bright red spot in the viewing area of the interferometer. You now move the movable mirror away from the beam splitter and observe 158 bright spots. You replace the red filter with the blue filter and observe a bright blue spot in the interferometer. You now move the movable mirror towards the beam splitter and observe 114 bright spots. Determine the final displacement (include sign) of the moveable mirror. (Assume the positive direction is away from the beam splitter.)
Answer:
final displacement = +24484.5 nm
Explanation:
The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;
Δr = 2d2 - 2d1 = 150λ1
So, 2d2 - 2d1 = 150λ1
Dividing both sides by 2 to get;
d2 - d1 = 75λ1 - - - - eq1
Where;
d1 = distance between the fixed mirror and the beam splitter
d2 = position of moveable mirror from splitter when 158 bright spots are observed
Now, the path difference between the two waves when 114 bright spots were observed is;
Δr = 2d'2 - 2d1 = 114λ1
2d'2 - 2d1 = 114λ1
Divide both sides by 2 to get;
d'2 - d1 = 57λ1
Where;
d'2 is the new position of the movable mirror from the splitter
Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.
(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2
d2 - d1 - d'2 + d1 = 75λ1 - 57λ2
d2 - d'2 = 75λ1 - 57λ2
We are given;
(λ1 = 656.3 nm) and λ2 = 434.0 nm.
Thus;
d2 - d'2 = 75(656.3) - 57(434)
d2 - d'2 = +24484.5 nm
. A 24-V battery is attached to a 3.0-mF capacitor and a 100-ohm resistor. If the capacitor is initially uncharged, what is the voltage across the capacitor 0.16 seconds after the circuit is connected to the battery
Answer:
The voltage is [tex]V_c = 9.92 \ V[/tex]
Explanation:
From the question we are told that
The voltage of the battery is [tex]V_b = 24 \ V[/tex]
The capacitance of the capacitor is [tex]C = 3.0 mF = 3.0 *10^{-3} \ F[/tex]
The resistance of the resistor is [tex]R = 100\ \Omega[/tex]
The time taken is [tex]t = 0.16 \ s[/tex]
Generally the voltage of a charging charging capacitor after time t is mathematically represented as
[tex]V_c = V_o (1 - e^{- \frac{t}{RC} })[/tex]
Here [tex]V_o[/tex] is the voltage of the capacitor when it is fully charged which in the case of this question is equivalent to the voltage of the battery so
[tex]V_c = 24 (1 - e^{- \frac{0.16}{100 * 3.0 *10^{-1}} })[/tex]
[tex]V_c = 9.92 \ V[/tex]
A screen is placed 43 cm from a single slit which is illuminated with 636 nm light. If the distance from the central maximum to the first minimum of the diffraction pattern is 3.8 mm, how wide is the slit in micrometer
Answer: The width is 1.25692 μm
Explanation:
The data that we have here is:
Distance between the single slit to the screen = L = 43cm
λ = 636 nm
Distance from the central maximum to the first minimum = Z = 3.8mm
We know that the angle for the destructive diffraction is:
θ = pλ/a
where p is the order of the minimum, for the first minimum we have p = 1, and a is the width of the slit,
then we have:
θ = (636nm/a)
And we also know that we can construct a triangle rectangle, where the adjacent cathetus to this angle is the distance between the slit and the screen, and the opposite cathetus is the distance between the first maximum and the first minimum:
Tg(θ) = Z/L
Tan(636nm/a) = 3.8cm/43cm
First, we need to use the same units in the right side:
3.8mm = 0.38cm
Tg(636nm/a) = 0.38cm/43cm
636nm/a = Atg( 0.38cm/43cm ) = 0.506
a = 636nm/0.506 = 1,256.92 nm
1 μm = 1000nm
then:
a = 1,256.92 nm = (1,256.92/1000) μm = 1.25692 μm
In a wire with a 1.05 mm2 cross-sectional area, 7.93×1020 electrons flow past any point during 3.97 s. What is the current ???? in the wire?
Answer:
The current in the wire is 31.96 A.
Explanation:
The current in the wire can be calculated as follows:
[tex] I = \frac{q}{t} [/tex]
Where:
q: is the electric charge transferred through the surface
t: is the time
The charge, q, is:
[tex] q = n*e [/tex]
Where:
n: is the number of electrons = 7.93x10²⁰
e: is the electron's charge = 1.6x10⁻¹⁹ C
[tex] q = n*e = 7.93 \cdot 10^{20}*1.6 \cdot 10^{-19} C = 126.88 C [/tex]
Hence, the current in the wire is:
[tex] I = \frac{126.88 C}{3.97 s} = 31.96 A [/tex]
Therefore, the current in the wire is 31.96 A.
I hope it helps you!
"On a movie set, an alien spacecraft is to be lifted to a height of 32.0 m for use in a scene. The 260.0-kg spacecraft is attached by ropes to a massless pulley on a crane, and four members of the film's construction crew lift the prop at constant speed by delivering 135 W of power each. If 18.0% of the mechanical energy delivered to the pulley is lost to friction, what is the time interval required to lift the spacecraft to the specified height?"
Answer:
The time interval required to lift the spacecraft to this specified height is 123.94 seconds
Explanation:
Height through which the spacecraft is to be lifted = 32.0 m
Mass of the spacecraft = 260.0 kg
Four crew member each pull with a power of 135 W
18.0% of the mechanical energy is lost to friction.
work done in this situation is proportional to the mechanical energy used to move the spacecraft up
work done = (weight of spacecraft) x (the height through which it is lifted)
but the weight of spacecraft = mg
where m is the mass,
and g is acceleration due to gravity 9.81 m/s
weight of spacecraft = 260 x 9.81 = 2550.6 N
work done on the space craft = weight x height
==> work = 2550.6 x 32 = 81619.2 J
this is equal to the mechanical energy delivered to the system
18.0% of this mechanical energy delivered to the pulley is lost to friction.
this means that
0.18 x 81619.2 = 14691.456 J is lost to friction.
Total useful mechanical energy = 81619.2 J - 14691.456 J = 66927.74 J
Total power delivered by the crew to do this work = 135 x 4 = 540 W
But we know tat power is the rate at which work is done i.e
[tex]P = \frac{w}{t}[/tex]
where p is the power
where w is the useful work done
t is the time taken to do this work
imputing values, we'll have
540 = 66927.74/t
t = 66927.74/540
time taken t = 123.94 seconds
A typical home uses approximately 1600 kWh of energy per month. If the energy came from a nuclear reaction, what mass would have to be converted to energy per year to meet the energy needs of the home
Answer:
7.68×10^25kg
Explanation:
The formula for energy used per year is calculated as
Energy used per year =12 x Energy used per month
By substituting Energy used per month in the above formula, we get
Energy used per year =12 x 1600kWh
= 19200kWh
Conversion:
From kWh to J:
1 kWh=3.6 x 10^6 J
Therefore, it is converted to J as
19200 kWh =19200 x 3.6 x 10^6 J
= 6.912×10^10 J
Hence, energy used per year is 6.912×10^10 J
To find the mass that is converted to energy per year.
E = MC^2 ............1
E is the energy used per year
C is the speed of light = 3.0× 10^8m/s
Where E= 6.912×10^10 J
Substituting the values into equation 1
6.912×10^10 J = M × 3.0× 10^8m/s
M = 6.912×10^10 J / (3.0× 10^8m/s)^2
M = 6.912×10^10 J/9×10^16
M = 7.68×10^25kg
Hence the mass to be converted is
7.68×10^25kg
A metal rod of length 2.0 m is moved at 6.0 m/s in a direction perpendicular to its length. A 5.0 mT magnetic field is perpendicular to both the rod and its velocity. What is the potential difference between the ends of the rod? 30 mV 15 mV 0 mV 12 mV 60 mV
Answer:
The potential difference is 60mVExplanation:
This problem bothers on application of the expression for motion emf which is expressed as
[tex]E= Blv[/tex]
where B= magnetic field in Tesla
l= length of the conductor
v= speed of conductor
Given data
l= 2 meters
v= 6 m/s
B= 5 Tesla
Applying the formula we have
[tex]E=5*2*6= 60mV[/tex]
Before you start taking measurements though, we’ll first make sure you understand the underlying concepts involved. By what method is each of the spheres charged?
Answer:
If they are metallic spheres they are connected to earth and a charged body approaches
non- metallic (insulating) spheres in this case are charged by rubbing
Explanation:
For fillers, there are two fundamental methods, depending on the type of material.
If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.
If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.
A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite from the planet is 6600 N. What is the kinetic energy of the satellite
Answer:
The kinetic energy is [tex]KE = 7.59 *10^{10} \ J[/tex]
Explanation:
From the question we are told that
The radius of the orbit is [tex]r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m[/tex]
The gravitational force is [tex]F_g = 6600 \ N[/tex]
The kinetic energy of the satellite is mathematically represented as
[tex]KE = \frac{1}{2} * mv^2[/tex]
where v is the speed of the satellite which is mathematically represented as
[tex]v = \sqrt{\frac{G M}{r^2} }[/tex]
=> [tex]v^2 = \frac{GM }{r}[/tex]
substituting this into the equation
[tex]KE = \frac{ 1}{2} *\frac{GMm}{r}[/tex]
Now the gravitational force of the planet is mathematically represented as
[tex]F_g = \frac{GMm}{r^2}[/tex]
Where M is the mass of the planet and m is the mass of the satellite
Now looking at the formula for KE we see that we can represent it as
[tex]KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r[/tex]
=> [tex]KE = \frac{ 1}{2} *F_g * r[/tex]
substituting values
[tex]KE = \frac{ 1}{2} *6600 * 2.3*10^{7}[/tex]
[tex]KE = 7.59 *10^{10} \ J[/tex]
An object on a rope is lowered steadily decreasing speed. Which is true?
A) The tope tensions is greater than the objects weight
B) the rope tension equals the objects weight
C)the rope tension is less than the objects weight
D) the rope tension can’t be compared to the objects weight
Answer:
C) the rope tension is less than the objects weight
Explanation:
According to Newton's Second Law, when an unbalanced or net force is applied to a body, it produces an acceleration in the body in the direction of the net force itself.
In this scenario, we have two forces acting on the object. First is the weight of object acting downward. Second is the tension in the rope acting upwards.
Since, the object is being lowered in the direction of weight. Therefore, weight of the object must be greater than the tension in the rope. So, the net force has the downward direction and the object is lowered. Hence, the correct option is:
C) the rope tension is less than the objects weight
Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.24 x 10-3 rad/s2 for 2.80 x 103 s. For the next 1.57 x 103 s the propeller rotates at a constant angular speed. Then it decelerates at 2.01 x 10-3 rad/s2 until it slows (without reversing direction) to an angular speed of 2.99 rad/s. Find the total angular displacement of the propeller.
Answer:
θ = 26.19 x 10³ radians
Explanation:
FOR ACCELERATED MOTION:
we use 2nd equation of motion for accelerated motion:
θ₁ = ωi t + (1/2)αt²
Where,
θ₁ = Angular Displacement covered during accelerated motion = ?
ωi = Initial Angular Speed = 0 rad/s
t = Time Taken = 2.8 x 10³ s
α = Angular Acceleration = 2.24 x 10⁻³ rad/s²
Therefore,
θ₁ = (0 rad/s)(2.8 x 10³ s) + (1/2)(2.24 x 10⁻³ rad/s²)(2.8 x 10³ s)²
θ₁ = 8.78 x 10³ radians
Now we find final angular velocity (ωf) by using 1st equation of motion:
ωf = ωi + αt
ωf = 0 rad/s + (2.24 x 10⁻³ rad/s²)(2.8 x 10³ s)
ωf = 6.272 rad/s
FOR UNIFORM ANGULAR SPEED:
For uniform angular speed we use following equation:
θ₂ = ωt
where,
θ₂ = Angular Displacement during uniform motion = ?
ω = Uniform Angular Speed = ωf = 6.272 rad/s
t = Time Taken = 1.57 x 10³ s
Therefore,
θ₂ = (6.272 rad/s)(1.57 x 10³ s)
θ₂ = 9.85 x 10³ radians
FOR DECELERATED MOTION:
Now, we use 3rd equation of motion for decelerated motion:
2αθ₃ = ωf² - ωi²
where,
α = Angular deceleration = - 2.01 x 10⁻³ rad/s²
θ₃ = Angular Displacement during decelerated motion = ?
ωf = Final Angular Speed = 2.99 rad/s
ωi = Initial Angular Speed = 6.272 rad/s
Therefore,
2(-2.01 x 10⁻³ rad/s²)θ₃ = (2.99 rad/s)² - (6.272 rad/s)²
θ₃ = (- 30.4 rad²/s²)/(-4.02 x 10⁻³ rad/s²)
θ₃ = 7.56 x 10³ radians
FOR TOTAL ANGULAR DISPLACEMENT:
Total Angular Displacement = θ = θ₁ + θ₂ + θ₃
θ = 8.78 x 10³ radians + 9.85 x 10³ radians + 7.56 x 10³ radians
θ = 26.19 x 10³ radians
A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 20.5 meters?
Answer:
The final speed of the crate after being pulled these 20.5 meters is 13.82 m/s
Explanation:
I'll assume that the correct question is
A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 22.5 meters?
mass of box = 51 kg
for the first 12 m, it is pulled with a constant force of 240 N
The acceleration of the box for this first 12 m will be
from F = ma
a = F/m
where F is the pulling force
m is the mass of the box
a is the acceleration of the box
a = 240/51 = 4.71 m/s^2
Since the body started from rest, the initial velocity u = 0
applying Newton's equation of motion to find the final velocity at the end of the first 12 m, we have
[tex]v^{2}= u^{2}+2as[/tex]
where v is the final velocity
u is the initial velocity which is zero
a is the acceleration of 4.71 m/s^2
s is the distance covered which is 12 m
substituting value, we have
[tex]v^{2}[/tex] = 0 + 2(4.71 x 12)
[tex]v^{2}[/tex] = 113.04
[tex]v = \sqrt{113.04}[/tex] = 10.63 m/s
For the final 10.5 m, coefficient of friction is 0.21
from f = μF
where f is the frictional force,
μ is the coefficient of friction = 0.21
and F is the pulling force of the box 240 N
f = 0.21 x 240 = 50.4 N
Net force on the box = 240 - 50.4 = 189.6 N
acceleration = F/m = 189.6/51 = 3.72 m/s^2
Applying newton's equation of motion
[tex]v^{2}= u^{2}+2as[/tex]
u is initial velocity, which in this case = 10.63 m/s
a = 3.72 m/s^2
s = 10.5 m
v = ?
substituting values, we have
[tex]v^{2}[/tex] = [tex]10.63^{2}[/tex] + 2(3.72 x 10.5)
[tex]v^{2}[/tex] = 112.9 + 78.12
v = [tex]\sqrt{191.02}[/tex] = 13.82 m/s
What sentence best supports the statement that hormones are involved in the regulation of homeostasis? A. The hormone cortisol suppresses the immune system and is produced when the body is under stress. B. The hormone erythropoeitin increases the production of red blood cells when oxygen levels are low. C. The hormone melatonin induces sleep and its production is slowed by exposure to light. D. The hormone oxytocin promotes labor contractions of the uterus during childbirth.
B. The hormone erythropoeitin increases the production of red blood cells when oxygen levels are low.
A 50 g ball of clay traveling at speed v0 hits and sticks to a 1.0 kg brick sitting at rest on a frictionless surface.
Required:
a What is the speed of the block after the collision?
b. What percentage of the ball's initial energy is "lost"?
Answer:
(a) The speed of the block after the collision is 0.0476v0.
(b) The percentage of the ball's initial energy lost, is 0 % (energy is conserved)
Explanation:
Given;
mass of ball of clay, m₁ = 50 g = 0.05 kg
mass of brick, m₂ = 1 kg
initial velocity of the ball of clay, u₁ = v0
initial velocity of the brick, u₂ = 0
Since the clay ball sticks with the brick after collision, it is inelastic collision.
Therefore, let their final velocity = v
(a) What is the speed of the block after the collision?
Apply the principle of conservation linear momentum
m₁u₁ + m₂u₂ = v (m₁ + m₂)
0.05v₀ + 1(0) = v( 0.05 + 1)
0.05v₀ = 1.05v
v = 0.05v₀ / 1.05
v = 0.0476v₀
Thus, the speed of the block after the collision is 0.0476 of its initial velocity.
(b). What percentage of the ball's initial energy is "lost"?
Initial kinetic energy of the ball = ¹/₂mv₀²
= ¹/₂ x 0.05 x v₀²
= 0.025v₀²
Final kinetic energy of the ball, = ¹/₂(m₁ + m₂)v²
= ¹/₂ x 1.05 x 0.0476v₀²
= 0.025v₀²
Change in kinetic energy = 0.025v₀² - 0.025v₀²
= 0
percentage change in the initial kinetic energy of the ball;
= (0 / 0.025v₀²) x 100%
= 0 x 100%
= 0 %
Therefore, the percentage of the ball's initial energy lost, is 0 % (energy is conserved)
A person with normal vision can focus on objects as close as a few centimeters from the eye up to objects infinitely far away. There exist, however, certain conditions under which the range of vision is not so extended. For example, a nearsighted person cannot focus on objects farther than a certain point (the far point), while a farsighted person cannot focus on objects closer than a certain point (the near point). Note that even though the presence of a near point is common to everyone, a farsighted person has a near point that is much farther from the eye than the near point of a person with normal vision.
Both nearsightedness and farsightedness can be corrected with the use of glasses or contact lenses. In this case, the eye converges the light coming from the image formed by the corrective lens rather than from the object itself.
Required:
a. If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that the person would need to see an object at infinity clearly?
b. If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350 m from the person's eyes?
Answer:
a) f₁ = 3.50 m , b) f₂ = 0.84 m
Explanation:
For this exercise we must use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p is the distance to the object and q is the distance to the image
a) the distance where the image should be placed is q = 3.50 m and the object is located at infinity p = ∞
1 / f₁ = 1 /∞ + 1 / 3.50
f₁ = 3.50 m
b) in this case the image is at q = -0.600 m and the object p = 0.350 m
1 / f₂ = 1 / 0.350 -1 / 0.600
the negative sign, is because the image is in front of the object
1 / f₂ = 1,1905
f₂ = 1 / 1,1905
f₂ = 0.84 m
A cylinder is closed by a piston connected to a spring of constant 2.20 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. The piston has a cross sectional area of 0.0100 m^2 and negligible mass. What is the pressure of the gas at 250 °C?
Answer:
1.3515x10^5pa
Explanation:
Plss see attached file
The length of your eye decreases slightly as you age, making the lens a bit closer to the retina. Suppose a man had his vision surgically corrected at age 30. At age 70, once his eyes had decreased slightly in length, what condition would he have?
A. Nearsightedness
B. Farsightedness
C. Neither nearsightedness nor farsightedness
Answer:
A. NearsightednessExplanation:
A nearsightedness is an eye defect that occurs when someone is only able to see close ranged object but not far distance object. According to the question, if the length of my eye decreases slightly as I age, this means there is a possibility that I will find it difficult to view a far distance object as I age.
At 70, once my eyes had decreased slightly in length, this means I will only be able to see close ranged object but not far distant object, showing that I am now suffering from nearsightedness according to its definition above.
What is the minimum magnitude of an electric field that balances the weight of a plasticsphere of mass 5.4 g that has been charged to -3.0 nC
Answer:
E = 17.64 x 10⁶ N/C = 17.64 MN/C
Explanation:
The electric field is given by the following formula:
E = F/q
E= W/q
E = mg/q
where,
E = magnitude of electric field = ?
m = mass of plastic sphere = 5.4 g = 5.4 x 10⁻³ kg
g = acceleration due to gravity = 9.8 m/s²
= charge = 3 nC = 3 x 10⁻⁹ C
Therefore,
E = (5.4 x 10⁻³ kg)(9.8 m/s²)/(3 x 10⁻⁹ C)
E = 17.64 x 10⁶ N/C = 17.64 MN/C