Answer:
Magnetic field, B = 0.11 i (in k direction)
Explanation:
It is given that,
Velocity of the proton, [tex]v=4.5\times 10^6\ m/s[/tex]
Magnetic force, [tex]F=8\times 10^{-14}\ N[/tex] (due south)
The magnetic force acting on the electron is given by :
[tex]F=qvB[/tex]
(-j) = (i) × (B)
[tex]B=\dfrac{F}{qv}[/tex]
q is the charge on proton
[tex]B=\dfrac{8\times 10^{-14}\ N}{1.6\times 10^{-19}\ C\times 4.5\times 10^6\ m/s}[/tex]
B = 0.11
So, the magnitude of 0.11 T is acting on the proton in and is acting directed upward above the plane. Hence, this is the required solution.
Final answer:
To calculate the magnitude of the magnetic field that causes a force on a moving proton, we use the rearranged Lorentz force law, with inputs of proton charge, velocity, and the force experienced. The direction of the magnetic field is determined by the right-hand rule, given the directions of the force and proton's velocity.
Explanation:
The student's question involves the effect of a magnetic field on a moving charged particle, specifically a proton, which falls under the subject of Physics. The Lorentz force law states that the force (F) on a charged particle is directly proportional to the charge (q), the velocity (v), and the magnetic field (B), and is given by the equation F = q(v x B), where v x B denotes the cross product of the velocity vector and the magnetic field vector. The force is perpendicular to both the velocity and the magnetic field.
To find the magnitude of the magnetic field, we can rearrange this equation to B = F / (q * v * sin(θ)), where θ is the angle between the velocity of the proton and the direction of the magnetic field. In the case where force is maximum, the angle is 90 degrees, and the sine of 90 degrees is 1. Given the maximum magnetic force (8.0 × 10⁻¹⁴ N), the charge of a proton (approximately 1.6 × 10⁻¹⁹ C), and the velocity of the proton (4.5 × 10⁶ m/s), we can calculate the magnitude of the magnetic field. As the force experienced by the proton is directed due south and the velocity is due east, the magnetic field must be pointing downwards to produce a force in that direction according to the right-hand rule.
You hold a 50-g sphere of copper (c = 0.4J/(g*C)) in one hand and a 25-g sphere of aluminum ( = 0.9 J/(g*C)) in the other hand. If both absorb energy at the same rate, which will come to your body temperature first and why?
Answer:
rate of change in temperature of copper is more than the rate of change in temperature of aluminium.
so here copper will reach to our body temperature first
Explanation:
As we know that rate of energy absorb by the two sphere is same
so here we will have
[tex]\frac{dQ}{dt} = ms\frac{\Delta T}{\Delta t}[/tex]
now for copper sphere we will have
[tex]\frac{dQ}{dt} = 50(0.4)\frac{\Delta T}{\Delta t}[/tex]
[tex]\frac{\Delta T}{\Delta t} = \frac{1}{20}\frac{dQ}{dt}[/tex]
now for Aluminium sphere we will have
[tex]\frac{dQ}{dt} = 25(0.9)\frac{\Delta T}{\Delta t}[/tex]
[tex]\frac{\Delta T}{\Delta t} = \frac{1}{22.5}\frac{dQ}{dt}[/tex]
So rate of change in temperature of copper is more than the rate of change in temperature of aluminium.
so here copper will reach to our body temperature first
A railroad car moving at a speed of 3.41 m/s overtakes, collides, and couples with two coupled railroad cars moving in the same direction at 1.40 m/s. All cars have a mass of mass 1.07 x 10^5 kg. Determine the following. (a) speed of the three coupled cars after the collision (Give your answer to at least two decimal places.) (b) kinetic energy lost in the collision
Answer:
2.07 m/s
Explanation:
m = 1.07 x 10^5 kg, u1 = 3.41 m/s, u2 = 1.4 m/s
Let the speed of three coupled car after collision is v
Use conservation of momentum
m x u1 + 2 m x u2 = 3 m x v
u1 + 2 u2 = 3 v
3.41 + 2 x 1.4 = 3 v
v = 2.07 m/s
A closed container is filled with oxygen. The pressure in the container is 245 kPa . What is the pressure in millimeters of mercury? Express the pressure numerically in millimeters
Answer:
Answer to the question is: 1837.65 millimeters of mercury are equal to 245 kPa.
Explanation:
1 kPa are equal to 7.50062 millimeters of mercury.
Final answer:
To convert the pressure from 245 kPa to mmHg, first convert kPa to atm, then multiply by the conversion factor from atm to mmHg. The pressure is 1837.68 mmHg.
Explanation:
To convert the pressure in a container from kilopascals (kPa) to millimeters of mercury (mmHg), we use the conversion factor that 1 atmosphere (atm) is equivalent to 760 mmHg. First, we convert the given pressure in kilopascals to atmospheres:
1 atm = 101.325 kPa
So, to convert 245 kPa to atm, we divide 245 kPa by 101.325 kPa/atm:
245 kPa / 101.325 kPa/atm = 2.418 atm
Next, we convert atmospheres to millimeters of mercury (mmHg) using the conversion factor:
2.418 atm x760 mmHg/atm = 1837.68 mmHg
Therefore, the pressure in the container is 1837.68 mmHg.
A solid uniform cylinder of mass 4.1 kg and radius 0.057 m rolls without slipping at a speed of 0.79 m/s. What is the cylinder’s total kinetic energy?
Answer:
The cylinder’s total kinetic energy is 1.918 J.
Explanation:
Given that,
Mass = 4.1 kg
Radius = 0.057 m
Speed = 0.79 m/s
We need to calculate the linear kinetic energy
Using formula of linear kinetic energy
[tex]K.E_{l}=\dfrac{1}{2}mv^2[/tex]
[tex]K.E_{l}=\dfrac{1}{2}\times4.1\times(0.79)^2[/tex]
[tex]K.E_{l}=1.279\ J[/tex]
We need to calculate the rotational kinetic energy
[tex]K.E_{r}=\dfrac{1}{2}\times I\omega^2[/tex]
[tex]K.E_{r}=\dfrac{1}{2}\times\dfrac{1}{2}\times mr^2\times(\dfrac{v}{r})^2[/tex]
[tex]K.E_{r}=\dfrac{1}{4}\times m\times v^2[/tex]
[tex]K.E_{r}=\dfrac{1}{4}\times4.1\times(0.79)^2[/tex]
[tex]K.E_{r}=0.639\ J[/tex]
The total kinetic energy is given by
[tex]K.E=K.E_{l}+K.E_{r}[/tex]
[tex]K.E=1.279+0.639[/tex]
[tex]K.E=1.918\ J[/tex]
Hence, The cylinder’s total kinetic energy is 1.918 J.
A raindrop of mass 3.26 10-5 kg falls vertically at constant speed under the influence of gravity and air resistance. Model the drop as a particle.(a) As it falls 115 m, what is the work done on the raindrop by the gravitational force?
Answer:
Work done by the gravitational force is 37 mJ.
Explanation:
It is given that,
Mass of the raindrop, [tex]m=3.26\times 10^{-5}\ kg[/tex]
It falls from a height of, h = 115 m
It falls vertically at constant speed under the influence of gravity and air resistance. We need to find the work done on the raindrop by the gravitational force. It is given by :
[tex]W=mgh[/tex]
[tex]W=3.26\times 10^{-5}\ kg\times 9.8\ m/s^2\times 115\ m[/tex]
W = 0.0367 J
or
W = 0.037 J = 37 mJ
So, the work done on the raindrop by the gravitational force is 37 mJ. Hence, this is the required solution.
Final answer:
The work done on a raindrop of mass 3.26 times [tex]10^{-5}[/tex] kg by the gravitational force.
Explanation:
To calculate the work done on a raindrop by the gravitational force as it falls, we use the formula for work: W = mgh, where W is the work done, m is the mass of the object, g is the acceleration due to gravity (9.8 m/s2), and h is the height the object falls through.
In this case, the mass m of the raindrop is 3.26 times 10-5 kg, and the height h is 115 m. So:
W = (3.26 times 10-5 kg)(9.8 [tex]m/s^2[/tex])(115 m) = 0.0368134 J
Therefore, the work done on the raindrop by the gravitational force as it falls 115 m is approximately 0.0368134 joules.
The gas within a cylinder of an engine undergoes a net change in volume of 1.50 × 10-3 m3 when it does work at a constant pressure of 3.27 x 105 Pa If the efficiency of the engine is 0.225, how much work must the engine give up as heat to the low-temperature reservoir?
Answer:
work =p×v =3.27×10^5×1.5×10^-3 =490.5 joule
efficiency =w/q in
:. qin= w/efficiency =490.5/0.225=2180 joule
qout =q in - work =1689.5 joule
q out is work given as heat
The engine must give up 1689.5 J of heat to the low-temperature reservoir after calculating the total work done by the gas and accounting for the engine's efficiency.
To find the amount of work the engine gives up as heat, we first calculate the total work done by the gas using the formula W = PΔV, where W is work, P is pressure, and ΔV is the change in volume. Given a constant pressure of 3.27 x 105 Pa and a change in volume of 1.50 x 10-3 m3, the work done is:
W = PΔV = 3.27 x 105 Pa x 1.50 x 10-3 m3 = 490.5 J.
The efficiency of the engine is the ratio of the useful work output to the total work input, given by ext_eta = useful work / total work. The equation that relates efficiency, work done (W), and heat given up (Q) is ext_eta = W / (Q + W). We rearrange the equation to solve for Q:
Q = W / ext_eta - W
Substituting the known values:
Q = 490.5 J / 0.225 - 490.5 J = 2180 J - 490.5 J = 1689.5 J.
Therefore, the engine must give up 1689.5 J of heat to the low-temperature reservoir.
A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be if you let out an amount of air that has a volume of 100cm3 at atmospheric pressure? Assume tire temperature and volume remain constant.
Answer:
[tex]p_2 = 664081 N/m^{2}[/tex]
Explanation:
from the ideal gas law we have
PV = mRT
[tex]P = \rho RT[/tex]
[tex]\rho = \frac{P}{RT}[/tex]
HERE R is gas constant for dry air = 287 J K^{-1} kg^{-1}
[tex]\rho = \frac{7.00 10^{5}}{287(18+273)}[/tex]
[tex]\rho = 8.38 kg/m^{3}[/tex]
We know by ideal gas law
[tex]\rho = \frac{m_1}{V_1}[/tex]
[tex]m_1 = \rho V_1 = 8.38 *2*10^{-3}[/tex]
[tex]m_1 = 0.0167 kg[/tex]
for m_2
[tex]m_2 = \rho V_i - V_removed[/tex]
[tex]m_2 = 8.38*(.002 - 10^{-4})[/tex]
[tex]m_2 = 0.0159 kg[/tex]
WE KNOW
PV = mRT
V, R and T are constant therefore we have
P is directly proportional to mass
[tex]\frac{p_2}{p_1}=\frac{m_2}{m_1}[/tex]
[tex]p_2 = p_1 * \frac{m_2}{m_1}[/tex]
[tex]p_2 =7*10^{5} * \frac {.0159}{0.0167}[/tex]
[tex]p_2 = 664081 N/m^{2}[/tex]
This problem can be solved using Boyle's Law, which relates the pressure and volume of a gas at a constant temperature. The question asks for the new pressure of a bicycle tire after letting out a certain volume of air. The answer is approximately 7.37 x 10⁵ Pa.
Explanation:The subject of this question is gas laws, specifically Boyle's Law which states that the pressure and volume of a gas have an inverse relationship when the temperature is kept constant. Assuming the temperature and volume of the tire remain constant before and after you let out the air, when a volume of 100 cm³ (which we will convert to 0.1 L for consistency) of air is let out, the new total volume of the gas is 1.9 L.
According to Boyle's Law, P1*V1 = P2*V2, where P1 and V1 represent the initial pressure and volume, and P2 and V2 represent the final pressure and volume. Plugging the values into this equation, we get:
(7.00 x 10⁵ Pa)(2.00 L) = P2 * (1.9 L)
Which gives us:
P2 = (7.00 x 10⁵ Pa * 2.00 L) / 1.9 L
Therefore, the pressure in the bike tire after letting out 100 cm³ of gas is approximately 7.37 x 10⁵ Pa.
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A cylindrical blood vessel is partially blocked by the buildup of plaque. At one point, the plaque decreases the diameter of the vessel by 59.0%. The blood approaching the blocked portion has speed V0. Just as the blood enters the blocked portion of the vessel, what is its speed V, expressed as a multiple of V0?
Using the principle of continuity for incompressible fluids, if the diameter of a blood vessel is reduced by 59.0% due to plaque, the speed of the blood just as it enters this section will be approximately 2.44 times its initial speed.
Explanation:The subject of the question falls under the topic of fluid flow in physics, specifically concerning the principle of continuity for incompressible fluids. This principle, often applied in fluid dynamics, suggests that in an area of steadily flowing fluid, the mass passing through one cross-section in a unit of time equals the mass passing through other sections.
Given this principle, if the cross-sectional area of the blood vessel decreases due to plaque buildup, the speed of the blood flow must increase accordingly to maintain a steady flow rate. If the diameter of the vessel decreases by 59.0%, the cross-sectional area A, which is proportional to the square of the diameter (A ~ D²), will be reduced to 0.41 of its original value (because (1 - 59/100)² = 0.41). Therefore, the speed V would be 1/0.41, or approximately 2.44 times the original speed V0.
So, if the blood vessel's diameter is reduced by 59.0%, then just as the blood enters the blocked portion of the vessel, its speed V will be 2.44 times the initial speed V0.
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A car weighing 11.1 kN and traveling at 13.4 m/s without negative lift attempts to round an unbanked curve with a radius of 61.0 m. (a) What magnitude of the frictional force on the tires required to keep the car f static friction between the tire or not ("yes" or "no")? on its circular If the coefficient ro s 0.35, is the attempt at taking the curve successf (a) Number Units (b)
Answer:
Well..
Explanation:
That's impossible. I know because I once weighed 11.1 kN, and I was temporarily immobile. It's probably the same for a car, and therefore it can not be "traveling" anywhere at all.. unless you put the car on an airplane or a boat or something.
A capacitor is being charged from a battery and through a resistor of 10 kΩ. It is observed that the voltage on the capacitor rises to 80% of its maximal value in 4 seconds. Calculate the capacitor's capacitance.
Answer:
[tex]C = 2.48 \times 10^{-4} Farad[/tex]
Explanation:
As per the equation of voltage on capacitor we know that
[tex]V = V_{max}(1 - e^{-\frac{t}{\tau}})[/tex]
now we know that voltage reached to its 80% of maximum value in 4 second time
so we will have
[tex]0.80 V_{max} = V_{max}(1 - e^{-\frac{4}{\tau}})[/tex]
[tex]0.20 = e^{-\frac{4}{\tau}}[/tex]
[tex]-\frac{4}{\tau} = ln(0.20)[/tex]
[tex]-\frac{4}{\tau} = -1.61[/tex]
[tex]\tau = 2.48[/tex]
as we know that
[tex]\tau = RC[/tex]
[tex](10 k ohm)(C) = 2.48[/tex]
[tex]C = 2.48 \times 10^{-4} Farad[/tex]
An empty, free-moving box car with a mass of 22,509 kg is coasting along at 4.21 m/s, when it runs into a second, stationary loaded box car with a mass of 31,647 kg. What is the speed of the two cars after they collide and attach?
Answer:
Final velocity, v = 1.74 m/s
Explanation:
Given that,
Mass of car 1, m₁ = 22509 kg
Velocity of car 1, v₁ = 4.21 m/s
Mass of car 2, m₂ = 31647 kg
It is stationary, v₂ = 0
Let v be the velocity of the two cars after they collide and attach. It can be calculated using law of conservation of momentum as :
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
[tex]v=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}[/tex]
[tex]v=\dfrac{22509\ kg\times 4.21\ m/s+0}{22509\ kg+31647\ kg}[/tex]
v = 1.74 m/s
So, the velocity of two cars after the collision is 1.74 m/s. Hence, this is the required solution.
(d) If η = 40% and TH = 427°C, what is TC, in °C?
Answer:
[tex]T_C=256.2^{\circ}C[/tex]
Explanation:
Given that,
Efficiency of heat engine, [tex]\eta=40\%=0.4[/tex]
Temperature of hot source, [tex]T_H=427^{\circ}C[/tex]
We need to find the temperature of cold sink i.e. [tex]T_C[/tex]. The efficiency of heat engine is given by :
[tex]\eta=1-\dfrac{T_C}{T_H}[/tex]
[tex]T_C=(1-\eta)T_H[/tex]
[tex]T_C=(1-0.4)\times 427[/tex]
[tex]T_C=256.2^{\circ}C[/tex]
So, the temperature of the cold sink is 256.2°C. Hence, this is the required solution.
You are looking up at the top of a building at an angle of 30.6 degrees from the horizontal. If the building is 42.0m tall, how far are you from the building? Assume that you are 1.50m tall.
Answer:
The distance between the person and the building is 68.48 meters.
Explanation:
It is given that,
Angle of elevation, θ = 30.6 degrees
Height of building, MP = 42 m
Height of person, AB = 1.5 m
We need to find the distance between person and building. It is given by BP.
Since, MN + NP = 42
So, MN = 40.5 m
Using trigonometric equation as :
[tex]tan\theta=\dfrac{MN}{AN}[/tex]
[tex]tan(30.6)=\dfrac{40.5}{AN}[/tex]
AN = 68.48 meters.
So, the distance between the person and the building is 68.48 meters. Hence, this is the required solution.
To determine the distance from a building, we use trigonometry and the formula adjacent = opposite / tangent(angle), taking into account the height of the building minus your height. The distance is calculated to be approximately 68.88 meters.
Explanation:To find out how far you are from the building, we need to calculate the horizontal distance from the building's base to the point where you are standing. To do this, we can use trigonometry, specifically the tangent function which relates the angle of elevation to the opposite side and the adjacent side of a right-angle triangle. We need to consider the height of the building minus your height to find the correct opposite side.
Since the building is 42.0 meters tall, and you are 1.50 meters tall, the effective height we are looking at is 42.0 m - 1.50 m = 40.5 m. The angle of elevation you are looking at is 30.6 degrees. By using the formula tangent (angle) = opposite / adjacent, we can rearrange this to find the adjacent side (the distance from you to the building): adjacent = opposite / tangent (angle).
Therefore, the distance from you to the building is approximately adjacent = 40.5 m / tan(30.6°). Plugging in the values, we get:
Distance = 40.5 m / tan(30.6°) ≈ 40.5 m / 0.588 ≈ 68.88 m.
So, you are approximately 68.88 meters away from the building.
You have a double slit experiment, with the distance between the two slits to be 0.025 cm. A screern is 120 cm behind the double slits. The distance between the central maximum and the Sh maximum is 1.52 cm. Please calculate the wavelength of the light used in the experiment
Answer:
The wavelength of the light is 633 nm.
Explanation:
Given that,
Distance between the two slits d= 0.025 cm
Distance between the screen and slits D = 120 cm
Distance between the slits y= 1.52 cm
We need to calculate the angle
Using formula of double slit
[tex]\tan\theta=\dfrac{y}{D}[/tex]
Where, y = Distance between the slits
D = Distance between the screen and slits
Put the value into the formula
[tex]\tan\theta=\dfrac{1.52}{120}[/tex]
[tex]\theta=\tan^{-1}\dfrac{1.52}{120}[/tex]
[tex]\theta=0.725[/tex]
We need to calculate the wavelength
Using formula of wavelength
[tex]d\sin\theta=n\lambda[/tex]
Put the value into the formula
[tex]0.025\times\sin0.725=5\times\lambda[/tex]
[tex]\lambda=\dfrac{0.025\times10^{-2}\times\sin0.725}{5}[/tex]
[tex]\lambda=6.326\times10^{-7}\ m[/tex]
[tex]\lambda=633\ nm[/tex]
Hence, The wavelength of the light is 633 nm.
with what speed will water emerge from a 5 cm diameter nozzle 10 m above the height of the pump? O A. 8600 m/s O B. 7100 m/s C. 17 ms D. 14 m/s
Answer:
d
Explanation:
HOPE THIS HELPS!!
A pendulum of length L is suspended from the ceiling of an elevator. When the elevator is at rest the period of the pendulum is T. How would the period of the pendulum change if the supporting chain were to break, putting the elevator into freefall?
Answer:
Explanation:
When the pendulum falls freely the net acceleration due to gravity is zero.
As we know that the time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity, thus the time period becomes infinity.
In freefall, the pendulum's effective acceleration due to gravity becomes zero, causing the pendulum to not swing, and its period becomes theoretically infinite and immeasurable.
Explanation:Effect of Freefall on a Pendulum's Period
When considering simple pendulum motion in an elevator under normal conditions, we can determine its periodic time (T) using the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. This equation illustrates that the period of the pendulum (T) is affected by two variables: the length of the pendulum (L) and the acceleration due to gravity (g).
When the elevator is in free fall, the effective acceleration g becomes zero because the elevator and the pendulum are both in a state of free fall with the same acceleration due to gravity. Therefore, in this scenario, the pendulum would experience weightlessness and would not oscillate, resulting in an infinite theoretical oscillation period, making the concept of a period inapplicable.
The period is normally independent of mass or amplitude for small angles, but since freefall changes the acceleration experienced by the pendulum to zero, it significantly affects the pendulum's oscillation, negating the normal conditions for calculating a pendulum's period.
While on a moving elevator during a certain perfod or time, Frank's apparent weight is 620 N. If Frank's mass is 70 kg, what is the magnitude and direction of Frank's acceleration?
Answer:
0.94 m/s^2 downwards
Explanation:
m = 70 kg, m g = 70 x 9.8 = 686 N
R = 620 N
Let the acceleration be a, as the apparent weight decreases so the elevator is moving downwards with an acceleration a.
mg - R = ma
686 - 620 = 70 x a
a = 0.94 m/s^2
The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in Figure below.. Determine the gage pressure of air in the tank if hl -0.2 m, h2 = 0.3 m, and h3 = 0.4 m. Take the densities of water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively.
Answer:
Spongebob: Bye Mr. Krabs! Bye Squidward! BYE SQUIDWARD!
Patrick: (clearly triggered) Why'd you say "bye squidward" twice?
Spongebob: I LiKe SqUiDwArD
What is the current produced by the solar cells of a pocket calculator through which 4.00 C of charge passes in 4.00 hr? Give your answer in mA.
The average current passing through a device is given by:
I = Q/Δt
I is the average current
Q is the amount of charge that has passed through the device
Δt is the amount of elapsed time
Given values:
Q = 4.00C
Δt = 4.00hr = 14400s
Plug in the values and solve for I:
I = 4.00/14400
I = 0.000277777778A
I = 0.278mA
Final answer:
The current produced by the solar cells of a pocket calculator through which 4.00 C of charge passes in 4.00 hours is 0.278 milliamperes.
Explanation:
The current produced by the solar cells of a pocket calculator when 4.00 C of charge passes through it in 4.00 hours can be calculated using the formula for electric current I = Q / t, where I is the current in amperes, Q is the charge in coulombs, and t is the time in seconds.
To find the current in milliamperes (mA), first convert the time to seconds:
4.00 hours × 3600 seconds/hour = 14400 seconds.
Next, use the formula to calculate current:
I = 4.00 C / 14400 s = 0.00027778 A,
which is equivalent to 0.278 mA
An amplifier has a power output of 100 mW when the input power is 0.1 mW. The amplifier gain is_________ dB.
a. 10
b. 20
c. 30
d. 40
Answer:
The amplifier gain is 30 dB.
(c) is correct option.
Explanation:
Given that,
Output power = 100 mW
Input power = 0.1 mW
We need to calculate the amplifier gain in dB
Using formula of power gain
[tex]a_{p}=10\ log_{10}(A_{p})[/tex]....(I)
We calculate the [tex]A_{p}[/tex]
[tex]A_{p}=\dfrac{P_{o}}{P_{i}}[/tex]
[tex]A_{p}=\dfrac{100}{0.1}[/tex]
[tex]A_{p}=1000[/tex]
Now, put the value of [tex]A_{p}[/tex] in equation (I)
[tex]a_{p}=10\ log_{10}(1000)[/tex]
[tex]a_{p}=10\times log_{10}10^{3}[/tex]
[tex]a_{p}=10\times 3log_{10}10[/tex]
[tex]a_{p}=30\ dB[/tex]
Hence, The amplifier gain is 30 dB.
The gain of an amplifier, given a power output of 100 mW and an input power of 0.1 mW, can be calculated using the gain formula in decibels, which results in a gain of 30 dB.
Explanation:In this context, the gain of the amplifier can be calculated using the formula for Gain in decibels (dB), which is 10 times the log base 10 of the output power divided by the input power. Therefore, we first divide 100 mW by 0.1 mW, which gives us 1000. Taking the log base 10 of 1000 returns 3, and multiplying 3 by 10 gives us a gain of 30 dB.
So the correct answer to your question: 'An amplifier has a power output of 100 mW when the input power is 0.1 mW, what is the amplifier gain?' is option c which states that the gain is 30 dB.
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Calculate the Reynolds number for a person swimming through maple syrup. The density of syrup is about 1400 kg/m^3 and the viscosity is about 0.5 Pa's. A person is about 2m in length and can swim about 1 m/s.
Answer:
The Reynolds number is 5600.
Explanation:
Given that,
Density = 1400 kg/m³
Viscosity = 0.5 Pa's
Length = 2 m
Speed = 1 m/s
We need to calculate the Reynolds number
Using formula of Reynolds number
[tex]R_{e}=\dfrac{\rho V\times L}{\mu}[/tex]
Where, [tex]\rho[/tex] = density of fluid
v = speed of syrup
l = length of a person
[tex]\mu[/tex]=Viscosity
Put the all value into the formula
[tex]R_{e}=\dfrac{1400\times1\times2}{0.5}[/tex]
[tex]R_{e}=5600[/tex]
Hence, The Reynolds number is 5600.
(a) Find the voltage near a 10.0 cm diameter metal sphere that has 8.00 C of excess positive charge on it. (b) What is unreasonable about this result? (c) Which assumptions are responsible?
Answer:
Part a)
[tex]V = 7.2 \times 10^{11} Volts[/tex]
Part b)
this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.
Part C)
here we can assume the sphere is placed at vacuum so that there is no break down of air.
Explanation:
Part a)
As we know that the potential near the surface of metal sphere is given by the equation
[tex]V = \frac{kQ}{R}[/tex]
here we have
Q = 8 C
R = 10.0 cm
now we have
[tex]V = \frac{(9\times 10^9)(8 C)}{0.10}[/tex]
[tex]V = 7.2 \times 10^{11} Volts[/tex]
Part b)
this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.
Part C)
here we can assume the sphere is placed at vacuum so that there is no break down of air.
Final answer:
The voltage near a 10.0 cm diameter metal sphere with 8.00 C of excess charge is calculated to be 1.438 x 10^12 V, which is unreasonable due to the high value leading to inevitable discharge. The assumption of an 8.00 C charge on such a small sphere is responsible for this unrealistic result.
Explanation:
Calculating the Voltage near a Charged Sphere
To find the voltage near a 10.0 cm diameter metal sphere with an excess positive charge of 8.00 C, we use the formula V = kQ/r, where V is the voltage, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), Q is the charge, and r is the radius of the sphere. For a diameter of 10.0 cm, the radius (r) is 0.05 m. Thus, V = (8.99 x 10^9 N m^2/C^2 * 8.00 C) / 0.05 m = 1.438 x 10^12 V.
Unreasonable Voltage
This voltage is extremely high and unreasonable because a metal sphere of that size could not sustain such a high voltage without discharging. The consequence of such a high voltage would include electric breakdown of the air around the sphere, leading to sparks or lightning-like discharges.
Erroneous Assumptions
The assumption responsible for this unreasonable result is the magnitude of charge being considered. An 8.00 C charge on a small metal sphere is significantly larger than what could realistically accumulate on the surface, given the limits of charge density and material breakdown thresholds.
Oil is poured into the open side of an open-tube manometer containing mercury. What is the density of the oil if a column of mercury 5.50 cm high supports a column of oil 85.0 cm high? (The density of mercury is 13,600 kg/m³)
Answer:
880 kg / m^3
Explanation:
height of column of oil = 85 cm = 0.85 m
height of column of mercury = 5.5 cm = 0.055 m
Density of mercury = 13600 kg/m^2
Let teh density of oil is d.
A the height of mercury column is balanced by the height of oil column
So, the pressure due to the mercury column = pressure by teh oil column
height of mercury column x density of mercury x g = height of oil column
x density of oil x g
0.055 x 13600 x g = 0.85 x d x g
748 = 0.85 d
d = 880 kg / m^3
You have just landed on Planet X. You take out a 100-g ball, release it from rest from a height of 10.0 m, and measure that it takes 2.2 s to reach the ground. You can ignore any force on the ball from the atmosphere of the planet. How much does the 100-g ball weigh on the surface of Planet X?
Answer:
Weight in planet X = 0.413 N
Explanation:
Weight = Mass x Acceleration due to gravity.
W = mg
Mass, m = 100 g = 0.1 kg
We have equation of motion s = ut + 0.5 at²
Displacement, s = 10 m
Initial velocity, u = 0 m/s
Time, t = 2.2 s
Substituting
s = ut + 0.5 at²
10 = 0 x 2.2 + 0.5 x a x 2.2²
a = 4.13 m/s²
Acceleration due to gravity, a = 4.13 m/s²
W = mg = 0.1 x 4.13 = 0.413 N
Weight in planet X = 0.413 N
In Planet X, a 100-g ball is released from rest from a height of 10.0 m and it takes 2.2 s for it to reach the ground. The weight of the ball on the surface of Planet X is 0.41 N.
What is the gravitational acceleration (g)?In physics, gravitational acceleration is the acceleration of an object in free fall within a vacuum (and thus without experiencing drag).
Step 1. Calculate the gravitational acceleration of Planet X.A 100-g (m) ball is released from rest from a height of 10.0 m (s) and it takes 2.2 s (t) for it to reach the ground. We can calculate the gravitational acceleration using the following kinematic equation.
s = 1/2 × g × t²
g = 2 s / t² = 2 (10.0 m) / (2.2 s)² = 4.1 m/s²
Step 2. Calculate the weight (w) of the ball on the surface of Planet X.We will use Newton´s second law of motion.
w = m × g = 0.100 kg × 4.1 m/s² = 0.41 N
In Planet X, a 100-g ball is released from rest from a height of 10.0 m and it takes 2.2 s for it to reach the ground. The weight of the ball on the surface of Planet X is 0.41 N.
Learn more about gravity here: https://brainly.com/question/557206
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of an 80.0-kg skydiver falling in a pike (headfirst) position with a surface area of 0.140 m2
Answer:
115 m/s, 414 km/hr
Explanation:
There are two forces acting on a skydiver: gravity and air resistance (drag). At terminal velocity, the two forces are equal and opposite.
∑F = ma
D − mg = 0
D = mg
Drag force is defined as:
D = ½ ρ v² C A
where ρ is the fluid density,
v is the velocity,
C is the drag coefficient,
and A is the cross sectional surface area.
Substituting and solving for v:
½ ρ v² C A = mg
v² = 2mg / (ρCA)
v = √(2mg / (ρCA))
We're given values for m and A, and we know the value of g. We need to look up ρ and C.
Density of air depends on pressure and temperature (which vary with elevation), but we can estimate ρ ≈ 1.21 kg/m³.
For a skydiver falling headfirst, C ≈ 0.7.
Substituting all values:
v = √(2 × 80.0 kg × 9.8 m/s² / (1.21 kg/m³ × 0.7 × 0.140 m²))
v = 115 m/s
v = 115 m/s × (1 km / 1000 m) × (3600 s / hr)
v = 414 km/hr
The terminal velocity of the skydiver in m/s and km/h is; 115m/s and 414 km/h
Using Given data :
mass of skydiver ( M ) = 80 kg
Cross sectional surface area ( A ) = 0.14 m^2
p ( fluid density ) ≈ 1.21 kg/m³.
C ( drag coefficient ) = 0.7
Determine the terminal velocity of the skydiver
At terminal velocity drag force and gravity is equal and opposite therefore canceling out each other
∑ F = ma
Drag force - Mg = 0
therefore; D = Mg ----- ( 1 )
where D ( drag force ) = 1/2 pv² C A ---- ( 2 )
p = fluid density , C = drag coefficient , A = cross sectional area
v = velocity
Back to equations 1 and 2 ( equating them )
1/2 pv² CA = Mg ---- ( 3 )
v² = 2mg / ( p C A )
∴ V = √ ( 2mg / (p C A ))
V = √ ( 2 * 80 * 9.8 ) / ( 1.21 * 0.7 * 0.140 ))
= 115 m/s
also V = 414 km/h
Hence we can conclude that the terminal velocity of the skydiver is in m/s and km/h are 115m/s and 414 km/h
Learn more ; https://brainly.com/question/3049973
An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the planet's surface. The period of revolution of the satellite around the planet is T 1.15 hours. What is the average density of the planet?
Answer:
[tex]\rho = 12580.7 kg/m^3[/tex]
Explanation:
As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet
So here we will have
[tex]F = \frac{GMm}{(r + h)^2}[/tex]
here we have
[tex]F =\frac {mv^2}{(r+ h)}[/tex]
[tex]\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}[/tex]
here we have
[tex]v = \sqrt{\frac{GM}{(r + h)}}[/tex]
now we can find time period as
[tex]T = \frac{2\pi (r + h)}{v}[/tex]
[tex]T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}[/tex]
[tex]1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}[/tex]
[tex]M = 4.54 \times 10^{23} kg[/tex]
Now the density is given as
[tex]\rho = \frac{M}{\frac{4}{3}\pi r^3}[/tex]
[tex]\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}[/tex]
[tex]\rho = 12580.7 kg/m^3[/tex]
An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 1.98 mT. If the speed of the electron is 1.53 107 m/s, determine the following.(a) the radius of the circular path
Answer:
4.4 cm
Explanation:
B = 1.98 mT = 1.98 x 10^-3 T, v = 1.53 x 10^7 m/s, m = 9.1 x 10^-31 kg
q = 1.6 x 10^-19 C
(a) The force due to the magnetic field is balanced by the centrpetal force
mv^2 / r = q v B
r = m v / q B
r = (9,1 x 10^-31 x 1.53 x 10^7) / (1.98 x 10^-3 x 1.6 x 10^-19)
r = 0.044 m = 4.4 cm
Calculate the power output in watts and horsepower of a shot-putter who takes 1.30 s to accelerate the 7.27-kg shot from rest to 16.0 m/s, while raising it 0.900 m. (Do not include the power produced to accelerate his body.) Shot putter at the Dornoch Highland Gathering in 2007.
Explanation:
It is given that,
Mass of the shot, m = 7.27 kg
Time taken to accelerate, t = 1.3 s
It is shot from rest to 16 m/s and it raises to a height of 0.9 m. We need to find the power output of the shot-putter. It is given by :
[tex]P=\dfrac{energy}{time}[/tex]
Energy = kinetic energy + potential energy
[tex]E=\dfrac{1}{2}\times 7.27\ kg\times (16\ m/s)^2+7.27\ kg\times 9.8\ m/s^2\times 0.9\ m[/tex]
E = 994.68 J
Power, [tex]P=\dfrac{994.68\ J}{1.3\ s}[/tex]
P = 765.13 Watts
We know that, 1 horse power = 745.7 watts
Or P = 1.02 horse power
Hence, this is the required solution.
A proton initially at rest is accelerated by a uniform electric field. The proton moves 5.62 cm in 1.15 x 10^-6 s. Find the voltage drop through which the proton moves. (Answer should be positive)
Answer:
49.85 V
Explanation:
u = 0, s = 5.62 cm, t = 1.15 x 10^-6 s
Let the electric field is E and voltage is V.
Use second equation of motion
s = ut + 1/2 a t^2
5.62 x 10^-2 = 0 + 0.5 a x (1.15 x 10^-6)^2
a = 8.5 x 10^10 m/s^2
m x a = q x E
E = m x a / q
E = (1.67 x 10^-27 x 8.5 x 10^10) / (1.6 x 10^-19)
E = 887.19 V/m
V = E x s
V = 887.19 x 5.62 x 10^-2 = 49.85 V
The International Space Station operates at an altitude of 350 km. When final construction is completed, it will have a weight (measured at the Earth’s surface) of 4.22 x 106 N. What is its weight when in orbit?
The International Space Station's operational weight in orbit is effectively zero due to its state of continuous free-fall around Earth, even though the gravitational force at its altitude is not significantly less than on Earth's surface.
Explanation:The question relates to the weight of the International Space Station (ISS) when in orbit. Weight in physics is defined as the force exerted on an object due to gravity, calculated as the product of mass and gravitational acceleration (g). On Earth's surface, g is approximately 9.81 m/s2, but this value decreases with altitude due to the equation g = GME / r2, where G is the gravitational constant, ME is Earth's mass, and r is the distance from Earth's center. At the ISS's altitude (> 350 km), g is about 8.75 m/s2. However, it's crucial to understand that while the ISS has a significant mass, leading to a large weight calculation on Earth, its apparent weight in orbit is effectively zero due to it being in a continuous free-fall state around Earth, experiencing microgravity. This explains why astronauts appear weightless, even though the actual gravitational force at that altitude is not much less than on Earth's surface. Therefore, while the ISS has a calculable weight based on its mass and Earth's gravitational pull at its altitude, its operational weight in orbit, in terms of the experience within it, is zero.