Answer:
10.8rev
Explanation:
Using
Wf²-wf = 2 alpha x theta
0²- 56.36x56.36/ 2(-20.13) x theta
Theta = 68.09 rad
But 68.09/2π
>= 10.8 revolutions
Explanation:
Calculate the range of wavelengths (in m) for AM radio given its frequency range is 540 to 1,600 kHz. smaller value m larger value m (b) Do the same for the ultraviolet frequency range of 760 to 30,000 THz. smaller value m larger value m
Answer:
a) λ = 555.5 m, λ = 187.5 nm
Explanation:
The velocity of a wave is given by the relation
c = λ f
λ = c /f
a) length of the radii AM
λ = 3 10⁸/540 10⁻⁷
λ= 5.555 102 m
λ = 555.5 m
f = 1600 kHz
λ = 3 108/1600 103
λ = 1.875 102 m
lam = 187.5 nm
b) light = 760 Thz = 760 10-12 Hz
λ= c/f
λ = 3 108/760 10-12
λ = 3.947 10-9
λ= 30000 Thz
λ= c/f
λ = 3 10⁸/ 30000 10-12
λ = 1 m
CAN SOMEONE HELP ME PLEASE ITS INTEGRATED SCIENCE AND I AM STUCK
Answer:
[tex]\huge \boxed{\mathrm{Option \ D}}[/tex]
Explanation:
Two forces are acting on the object.
Subtracting 2 N from both forces.
2 N → Object ← 5 N
- 2 N - 2N
0 N → Object ← 3 N
The force 3 N is pushing the object to the left side.
The mass of the object is 10 kg.
Applying formula for acceleration (Newton’s Second Law of Motion).
a = F/m
a = 3/10
a = 0.3
Please help!
Much appreciated!
Answer:
your question answer is 22°
Which examination technique is the visualization of body parts in motion by projecting x-ray images on a luminous fluorescent screen?
Answer:
Fluoroscopy
Explanation:
A Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of an object. In its primary application of medical imaging, a fluoroscope allows a physician to see the internal structure and function of a patient, so that the pumping action of the heart or the motion of swallowing, for example, can be watched.
an alternating voltage of 100V, 50HZ Is Applied across an impedance of (20-j30) calculate the resistance, the capacitance, current, the phase angle between current and voltage
The resistance R = 20 Ω
The capacitance C = 106.1 μF
The current, I is 2.773 A at 56.31°.
The phase angle of the between the current and the voltage is 56.31° leading.
Since the impedance Z = 20 - j30 Ω, the resistance, R is the real part of the impedance. So R = ReZ = 20 Ω
So, the resistance R = 20 Ω
To find the capacitance, we need first to find the reactance of the capacitor X. Since the impedance Z = 20 - j30, the reactance of the capacitor X. is the imaginary part of the impedance. So X = ImZ = 30 Ω.
Now the reactance of the capacitor X = 1/ωC where ω = angular frequency of the circuit = 2πf where f = frequency of the circuit = 50 Hz and C = capacitance
So, C = 1/ωX = 1/2πfX
Substituting the values of the variables into the equation, we have
C = 1/2πfX
C = 1/(2π × 50 Hz × 30 Ω)
C = 1/3000π
C = 1/9424.778
C = 1.061 × 10⁻⁴ F
C = 106.1 × 10⁻⁶ F
C = 106.1 μF
So, the capacitance is 106.1 μF
The current I = V/Z where V = voltage = 100 V at 0° and Z = impedance.
The magnitude of Z = √(20² + (-30)²)
= √(400 + 900)
= √1300
= 36.06 Ω
and its angle Φ = tan⁻¹(ImZ/ReZ)
= tan⁻¹(-30/20)
= tan⁻¹(-1.5) = -56.31°
So, V = 100 ∠ 0° and Z = 36.06 ∠ -56.31°
So, the current, I = V/Z = (100 ∠ 0°)/36.06 ∠ -56.31°
= 100/36.06 ∠(0° - (-56.31° ))
= 2.773 ∠ 56.31° A
So, the current is 2.773 A at 56.31°.
Since the current is 2.773 A at 56.31°, the phase angle of the between the current and the voltage is 56.31° leading.
So, the phase angle of the between the current and the voltage is 56.31° leading.
Learn more about alternating voltage here:
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A hydraulic lift raises a 2 000-kg automobile when a 500-N force is applied to the smaller piston. If the smaller piston has an area of 10 cm2, what is the cross-sectional area of the larger piston
Answer:
The cross-sectional area of the larger piston is 392 cm²
Explanation:
Given;
output mass of the piston, m₀ = 2000 kg
input force of the piston, F₁ = 500 N
input area of the piston, A₁ = 10 cm² = 0.001 m²
The output force is given by;
F₀ = m₀g
F₀ = 2000 x 9.8
F₀ = 19600 N
The cross-sectional area of the larger piston or output area of the piston will be calculated by applying the following equations;
[tex]\frac{F_i}{A_i} = \frac{F_o}{A_o} \\\\A_o= \frac{F_o A_i}{F_i} \\\\A_o = \frac{19600*0.001}{500} \\\\A_o = 0.0392 \ m^2\\\\A_o = 392 \ cm^2[/tex]
Therefore, the cross-sectional area of the larger piston is 392 cm²
Can a car moving with a negative velocity moves faster than a car moving with a positive velocity? explain.
Answer:
Yes.
Explanation:
This is because "negative velocity" just means it is in the negative in relation to the point of 0. Negative velocity doesn't equal a decrease in velocity. For example lets say you were parked next to a cone (this cone represents zero) if you accelerate forwards then that would be positive acceleration. If you were to accelerate backwards, this would be in the negative direction, aka negative velocity.
SUMMARY:
A negative velocity means that the object which has the negative velocity is moving in the opposite direction of an object moving at a positive velocity. This is a question of frame of reference. The possibility for the velocity is what makes it different to the speed. Speed is only positive.
Grocery store managers contend that there is less total energy consumption in the summer if the store is kept at a low temperature. Make arguments to support or refute this claim, taking into account that there are numerous refrigerators and freezers in the store.
Answer:
Argument in favor of less total energy consumption if the store is kept at a low temperature
Explanation:
Have in mind that if the store has numerous refrigerators and freezers, the energy consumption of those machines have to be included into the analysis.
Recall that the efficiency (or Coefficient Of Performance - COP) of a frezzer or refrigerator is inversely proportional to the temperature difference between the inside of th machine and the environment where it is operation, therefore the smaller the difference, the highest their efficiency. Therefore, the cooler the environment (the temperature at which the store is kept) the better performance of the running refrigerators and freezers.
of
The radii a wheel are 25 cm
and 5cm respectively, it is found
that an effort of 40N is required
to raise slowly a load 16ON
160 N. Find the Mechanical
Adventage and Effeciency,
Answer:
Explanation:
Given that
Effort = 40N
Load = 16ON
M.A = load/effort
M.A= 160N/40N
M.A = 4
Velocity ratio = V.R =radius of the wheel/radius of the axel
= 25cm/5cm
= 5
Efficiency = mechanical advantage/velocity ratio × 100/1
= 4/5 × 100/1
= 0.8×100/1
= 80%
Hence, the mechanical advantage of the machine is 4 while the efficiency is 80%.
A rigid uniform bar of length L and mass m is suspended by a massless wire AC and a rigid massless link BC. Determine the tension in BC immediately after AC breaks.
Answer:
hello the needed diagram is missing attached below is the diagram and the detailed solution
The tension in BC = [tex]\frac{\sqrt{2} }{4} mg[/tex]
Explanation:
ATTACHED BELOW IS THE DETAILED SOLUTION T THE GIVEN PROBLEM
Ma = mg - T/ [tex]\sqrt{2}[/tex] equation 1
Ma = 3T / [tex]\sqrt{2}[/tex] equation 2
equate both equations to determine the tension on BC
One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious space camper and you launch a serious rocket: it reaches an altitude of 211 km . What gain Δ???? in gravitational potential energy does the launch accomplish? The mass and radius of the Moon are 7.36×1022 kg and 1740 km, respectively.
Answer:
ΔP.E = 6.48 x 10⁸ J
Explanation:
First we need to calculate the acceleration due to gravity on the surface of moon:
g = GM/R²
where,
g = acceleration due to gravity on the surface of moon = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of moon = 7.36 x 10²² kg
R = Radius of Moon = 1740 km = 1.74 x 10⁶ m
Therefore,
g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²
g = 2.82 m/s²
now the change in gravitational potential energy of rocket is calculated by:
ΔP.E = mgΔh
where,
ΔP.E = Change in Gravitational Potential Energy = ?
m = mass of rocket = 1090 kg
Δh = altitude = 211 km = 2.11 x 10⁵ m
Therefore,
ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)
ΔP.E = 6.48 x 10⁸ J
Calculate the time it would take a cell phone signal to travel from a point on the equator to the satellite and back.
Ibrah open a bottle of perfume infront of the room. After few minutes the smell of perfume reach the whole room. Explain why this happens
Imagine two pairs of books. In the first pair, two books of mass m_1m 1 m, start subscript, 1, end subscript and m_2m 2 m, start subscript, 2, end subscript are separated by the distance rrr, resulting in a gravitational force F_1F 1 F, start subscript, 1, end subscript. In the second pair, two books of mass 2m_12m 1 2, m, start subscript, 1, end subscript and 4m_24m 2 4, m, start subscript, 2, end subscript are separated by the distance 2r2r2, r, resulting in a gravitational force of F_2F 2 F, start subscript, 2, end subscript. The relationship between these two forces can be written as F_2 = nF_1F 2 =nF 1 F, start subscript, 2, end subscript, equals, n, F, start subscript, 1, end subscript. In the expression F_2 = nF_1F 2 =nF 1 F, start subscript, 2, end subscript, equals, n, F, start subscript, 1, end subscript, what is the value of nnn?
The value of n in the relationship between F₁ and F₂, (F₂ = n·F₁), is n = 2
The reason for the above value is as follows;
The given parameters of the first pair of books are;
The masses of the books = m₁, and m₂
The distance between the two m₁ and m₂ = r
The gravitational force between the masses = F₁
The given parameters of the second pair of books are;
The masses of the books second pair = 2·m₁, and 4·m₂
The distance between the two masses in second pair = 2·r
The gravitational force between the masses in second pair = F₂
The relationship between the two forces is F₂ = n·F₁
The required parameter;
The value of n in F₂ = n·F₁
According to Newton's law of universal gravitation, we have;
[tex]\mathbf{F = G \times \dfrac{m_1 \times m_2}{r^2}}[/tex]
Therefore, we get;
[tex]F_1 = G \times \dfrac{m_1 \times m_2}{r^2}[/tex]
[tex]F_2 = G \times \dfrac{2\cdot m_1 \times 4 \cdot m_2}{(2 \cdot r)^2} = G \times \dfrac{8 \times m_1 \times m_2}{4 \times r^2} = 2 \times G \times \dfrac{ m_1 \times m_2}{r^2} = 2 \times F_1[/tex]
Therefore;
F₂ = 2·F₁
The value of n in F₂ = n·F₁ is n = 2
Learn more about gravitational force between masses here;
https://brainly.com/question/11359658
A mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 6 feet below the equilibrium position, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 1 2 the instantaneous velocity. Find the equation of motion x(t) if the mass is driven by an external force equal to f(t)
Answer:I don’t know
Explanation:
Give examples of motion in which the directions of the velocity and acceleration vectors are (a) opposite, (b) the same, and (c) mutually perpendicular
Answer:
a) When moving body applies brake then velocity and acceleration would be in opposite direction
b) When body starts to increase velocity then velocity and acceleration would be in same direction
c) When body is circulating then velocity and acceleration would be perpendicular to each other
Explanation:
a) When body applies brake then its velocity starts decreasing, in this case its acceleration would try to stop the moving body. So direction of velocity would be same as direction of motion of body but direction of acceleration would be in opposite direction
b) When body starts to increase velocity, its acceleration would make the body to move faster. So direction of velocity would be the direction of motion of body and acceleration would also be in same direction
c) When body moves in circular path then its acceleration would be towards centre of circle and velocity would try to snap the body out of circle to straight line which in tangent to circle.
Describe how you expect the waveform and the sound you hear changes when you hit the tuning fork harder.
Answer:
In a tuning fork, two basic qualities of sound are considered, they are
1) The pitch of the waveform: This pitch depends on the frequency of the wave generated by hitting the tuning fork.
2) The loudness of the waveform: This loudness depends on the intensity of the wave generated by hitting the tuning fork.
Hitting the tuning fork harder will make it vibrate faster, increasing the number of vibrations per second. The number of vibration per second is proportional to the frequency, so hitting the tuning fork harder increase the frequency. From the explanation on the frequency above, we can say that by increasing the frequency the pitch of the tuning fork also increases.
Also, hitting the tuning fork harder also increases the intensity of the wave generated, since the fork now vibrates faster. This increases the loudness of the tuning fork.
A cart rolls 2 m to the right then rolls back 1 m to the left.
a. What is the total distance rolled by the cart?
Explanation:
It is given that,
Distance covered by the cart to the right is 2 m
Distance covered by the cart to the left is 1 m
We need to find the total distance rolled by the cart. Total distance is equal to the sum of the distances covered by an object. It does depend on the direction.
Total distance = 2 m + 1 m
D = 3 m
The cart rolled to a total distance of 3 m.
A 70 kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above water when his lungs are full.
Required:
a. Calculate the volume of air he inhales - called his lung capacity - in liters.
b. Does this lung volume seem reasonable?
Answer:
Explanation:
A) Vair = 1.3 L
B) Volume is not reasonable
Explanation:
A)
Assume
m to be total mass of the man
mp be the mass of the man that pulled out of the water
m1 be the mass above the water with the empty lung
m2 be the mass above the water with full lung
wp be the weight that the buoyant force opposes as a result of the air.
Va be the volume of air inside man's lungs
Fb be the buoyant force due to the air in the lung
given;
m = 78.5 kg
m1 = 3.2% × 78.5 = 2.5 kg
m2 = 4.85% × 78.5 = 3.8kg
But, mp = m2- m1
mp = 3.8 - 2.5
mp = 1.3kg
So using
Archimedes principle, the relation for formula for buoyant force as;
Fb = (m_displaced water)g = (ρ_water × V_air × g)
Where ρ_water is density of water = 1000 kg/m³
Thus;
Fb = wp = 1.3× 9.81
Fb = 12.7N
But
Fb = (ρ_water × V_air × g)
So
Vair = Fb/(ρ_water × × g)
Vair = 12.7/(1000 × 9.81)
V_air = 1.3 × 10^(-3) m³
convert to litres
1 m³ = 1000 L
Thus;
V_air = 1.3× 10^(-3) × 1000
V_air = 1.3 L
But since the average lung capacity of an adult human being is about 6-7litres of air.
Thus, the calculated lung volume is not reasonable
Explanation:
What is the threshold velocity vthreshold(ethanol) for creating Cherenkov light from a charged particle as it travels through ethanol (which has an index of refraction of n
Explanation:
The velocity of light in a medium of refractive index [tex]n[tex] is given by,
[tex]v=\frac{c}{n}[/tex]
[tex]v \text { is the velocity of light in the medium }[/tex]
[tex]c \text { is speed of light in vacuum }[/tex]
The exact value of speed of light in vacuum is [tex]299792458 \mathrm{m} / \mathrm{s}[/tex].
For Cherenkov radiation to be emitted, the velocity of the charged particle traversing the medium must be greater than this velocity. Thus, the threshold velocity of for creating Cherenkov radiation is,
[tex]v_{\text {Cherenkov }} \geq \frac{c}{n}[/tex]
[tex]v_{\text {threshod }}=\frac{c}{n}[/tex]
For water [tex]n=1.33,[tex] thus the threshold velocity for producing Cherenkov radiation in water is,
[tex]v_{\text {threatold }(\text { water })} &=\frac{299792458 \mathrm{m} / \mathrm{s}}{1.33}[/tex]
[tex]=225407863 \mathrm{m} / \mathrm{s}[/tex]
[tex]=2.254 \times 10^{8} \mathrm{m} / \mathrm{s}[/tex]
For ethanol [tex]n=1.36[tex], thus the threshold velocity for producing Cherenkov radiation in water is,
[tex]v_{\text {threstold }( \text { ettanol) } } &=\frac{299792458 \mathrm{m} / \mathrm{s}}{1.36}[/tex]
[tex]=220435630 \mathrm{m} / \mathrm{s}[/tex]
[tex]=2.204 \times 10^{8} \mathrm{m} / \mathrm{s}[/tex]
Answer:
The answer is "2.2 × [tex]\bold{10^8}[/tex]".
Explanation:
In the given question the value of n is missing which can be defined as follows:
n= 1.36
The velocity value of the threshold(ethanol) for a generation the Cerenkov light from the charged particle by travel through ethanol as:
know we will have to use an equation as follows:
Formula:
(ethanol) or the vthreshold = [tex]\frac{c}{n}[/tex]
[tex]= \frac{3\times 10^8} {1.36} \\\\= 2.2 \times 10^8[/tex]
The water in vthreshold:
[tex]= 2.2 \times 10^8 \ \ \frac{m}{ s} \\\\[/tex]
Express the value in c, that is multiple, so, the value of vthreshold(water) is:
=(0.735) c
In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:_____.
Answer:
In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal: speed of light(c)
Explanation:
Generally the ratio of the E(electric field ) and the B(magnetic field ) is equal to the speed of the electromagnetic wave i.e the speed of light (c) the value is
[tex]c = 3.0 *10^{8} \ m/s[/tex]
on which principle does water pump work ?
Answer:
The working principle of a water pump mainly depends upon the positive displacement principle as well as kinetic energy to push the water.
Explanation:
it mainly depends upon the positive displacement principle and also kinetic energy to push water. hope this hepls!
The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various atmosphericprocesses, including lightning.
What is the excess charge on the surface of the earth? inC
Answer:
[tex]q = -461532.5 \ C[/tex]
Explanation:
From the question we are told that
The electric filed is [tex]E = 102 \ N/C[/tex]
Generally according to Gauss law
=> [tex]E A = \frac{q}{\epsilon_o }[/tex]
Given that the electric field is pointing downward , the equation become
[tex]- E A = \frac{q}{\epsilon_o }[/tex]
Here [tex]q[/tex] is the excess charge on the surface of the earth
[tex]A[/tex] is the surface area of the of the earth which is mathematically represented as
[tex]A = 4\pi r^2[/tex]
Where r is the radius of the earth which has a value [tex]r = 6.3781*10^6 m[/tex]
substituting values
[tex]A = 4 * 3.142 * (6.3781*10^6 \ m)^2[/tex]
[tex]A =5.1128 *10^{14} \ m^2[/tex]
So
[tex]q = -E * A * \epsilon _o[/tex]
Here [tex]\epsilon_o[/tex] s the permitivity of free space with value
[tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}[/tex]
[tex]q = -461532.5 \ C[/tex]
Why does cold water kept in an open container become warm on a hot summer afternoon?
Cold water kept in an open container acquires heat from the warmer surrounding becomes warm like the air around it due to the transfer of thermal energy
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A rope, under a tension of 153 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by . where at one end of the rope, is in meters, and is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation?
Complete question is;
A rope, under a tension of 153 N and fixed at both ends, oscillates in a second harmonic standing wave pattern. The displacement of the rope is given by
y = (0.15 m) sin[πx/3] sin[12π t].
where x = 0 at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c)the mass of the rope? (d) If the rope oscillates in a third - harmonic standing wave pattern, what will be the period of oscillation?
Answer:
A) Length of rope = 4 m
B) v = 24 m/s
C) m = 1.0625 kg
D) T = 0.11 s
Explanation:
We are given;
T = 153 N
y = (0.15 m) sin[πx/3] sin[12πt]
Comparing this displacement equation with general waveform equation, we have;
k = 2π/λ = π/2 rad/m
ω = 2πf = 12π rad/s
Since, 2π/λ = π/2
Thus,wavelength; λ = 4 m
Since, 2πf = 12π
Frequency;f = 6 Hz
A) We are told the rope oscillates in a second-harmonic standing wave pattern. So, we will use the equation;
λ = 2L/n
Since second harmonic, n = 2 and λ = L = 4 m
Length of rope = 4 m
B) speed is given by the equation;
v = fλ = 6 × 4
v = 24 m/s
C) To calculate the mass, we will use;
v = √T/μ
Where μ = mass(m)/4
Thus;
v = √(T/(m/4))
Making m the subject;
m = 4T/v²
m = (4 × 153)/24²
m = 1.0625 kg
D) Now, the rope oscillates in a third harmonic.
So n = 3.
Using the formula f = 1/T = nv/2L
T = 2L/nv
T = (2 × 4)/(3 × 24)
T = 0.11 s
3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing Photo 2.
Answer:
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
Explanation:
I cannot find any attached photo, but we can proceed anyways theoretically.
The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point
i.e
[tex]E=\frac{F}{Q}[/tex]
But the force F
[tex]F= \frac{kQ1Q2}{r^2}[/tex]
But the electric field intensity due to a point charge Q at a distance r meters away is given by
[tex]E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }[/tex]
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
For an object to move, a(n) _______ force must be applied. Question 1 options: Balanced Unbalanced
Answer:
Unbalenced
Explanation:
when balenced forces are applied to an object there is no motion. When you apply unbalenced force the object you are applying the force to will move in the opposite direction of the force.
Answer:
im pretty sure it unbalenced
Explanation:
i just am
Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the concentric spheres. When compared to the number of field lines N1 going through the sphere of radius R, the number of electric field lines N2 going through the sphere of radius 2R is
Answer:
N2 = ¼N1
Explanation:
First of all, let's define the terms;
N1 = number of electric field lines going through the sphere of radius R
N2 = number of electric field lines going through the sphere of radius 2R
Q = the charge enclosed at the centre of concentric spheres
ε_o = a constant known as "permittivity of the free space"
E1 = Electric field in the sphere of radius R.
E2 = Electric field in the sphere of radius 2R.
A1 = Area of sphere of radius R.
A2 = Area of sphere of radius 2R
Now, from Gauss's law, the electric flux through the sphere of radius R is given by;
Φ = Q/ε_o
We also know that;
Φ = EA
Thus;
E1 × A1 = Q/ε_o
E1 = Q/(ε_o × A1)
Where A1 = 4πR²
E1 = Q/(ε_o × 4πR²)
Similarly, for the sphere of radius 2R,we have;
E2 = Q/(ε_o × 4π(2R)²)
Factorizing out to get;
E2 = ¼Q/(ε_o × 4πR²)
Comparing E2 with E1, we arrive at;
E2 = ¼E1
Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;
N2 = ¼N1
A balloon contains 1.21 x 105 L of ideal gas at 265K. The gas is then cooled to 201 K. What is the volume (L) assuming no gas enters or exits the balloon
Answer:
The new volume will be 0.918 x 10^5 L
Explanation:
initial volume [tex]V_{1}[/tex] = 1.21 x 10^5 L
Initial temperature [tex]T_{1}[/tex] = 265 K
Final volume [tex]V_{2}[/tex] = ?
Final temperature [tex]T_{2}[/tex] = 201 K
Th gas is an ideal gas.
For ideal gases, the equation [tex]V_{1}[/tex]/[tex]T_{1}[/tex] = [tex]V_{2}[/tex]/[tex]T_{2}[/tex] = constant
substituting value, we have
(1.21 x 10^5)/265 = [tex]V_{2}[/tex]/201
[tex]V_{2}[/tex] = 24321000/265 = 91777.4 L
= 0.918 x 10^5 L
A flat loop of wire consisting of a single turn of cross-sectional area 7.30 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.50 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 2.60
Answer:
-0.73mA
Explanation:
Using amphere's Law
ε =−dΦB/ dt
=−(2.6T)·(7.30·10−4 m2)/ 1.00 s
=−1.9 mV
Using ohms law
ε=V =IR
I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA