Answer:
[tex]v_1 = 12 volts[/tex]
[tex]v_2 = 6 volts[/tex]
[tex]v_3 = 36 volts[/tex]
Explanation:
As we know that all the batteries are in series
so the net voltage of all three batteries is given as
[tex]V = v_1 + v_2 + v_3[/tex]
now we know that
[tex]v_1 = 2v_2[/tex]
[tex]v_1 = \frac{1}{3}v_3[/tex]
now plug in all the values in it
[tex]54 = v_1 + \frac{v_1}{2} + 3v_1[/tex]
[tex]54 = 4.5 v_1[/tex]
[tex]v_1 = 12 volts[/tex]
now we have
[tex]v_2 = 6 volts[/tex]
[tex]v_3 = 36 volts[/tex]
Final answer:
To find the actual voltage of each battery connected in series to get a total of 54 volts, we can assign variables to represent the voltage of each battery. By using the given information, we can set up and solve an equation to find the values of the variables.
Explanation:
To find the actual voltage of each battery, we'll assign variables to each battery's voltage. Let's say the voltage of the second battery is x volts.
Using the information given, we know that the voltage of the first battery is twice the voltage of the second battery, so it is 2x volts. The voltage of the third battery is 1/3 the voltage of the first battery, so it is (1/3)(2x) = 2x/3 volts.
Since the total voltage when the batteries are connected in series is 54 volts, we can write the equation: 2x + x + 2x/3 = 54. Solving this equation, we find that x = 15 volts. Therefore, the voltage of the first battery is 2x volts = 2(15) = 30 volts, the voltage of the second battery is x volts = 15 volts, and the voltage of the third battery is 2x/3 volts = (2/3)(15) = 10 volts.
A pendulum is made by letting a 2.0-kg object swing at the end of a string that has a length of 1.5 m. The maximum angle the string makes with the vertical as the pendulum swings is 30°. What is the speed of the object at the lowest point in its trajectory?
Answer:
v = 2 m/s
Explanation:
Here we can use energy conservation to find the speed at the lowest point on its trajectory
As we know that by energy conservation
initial total gravitational potential energy = final total kinetic energy
now the height that is moved by the pendulum while it swing down is given as
[tex]h = L(1 - cos30)[/tex]
[tex]h = 1.5(1 - cos30) = 0.200 m[/tex]
now we can use energy conservation as
[tex]mgh = \frac{1}{2}mv^2[/tex]
[tex]v = \sqrt{2gh}[/tex]
[tex]v = \sqrt{2(9.8)(0.200)}[/tex]
[tex]v = 2 m/s[/tex]
Answer:
v = 1.978 m/s
Explanation:
Given that,
Mass of the object, m = 2 kg
Length of the string, l = 1.5 m
The maximum angle the string makes with the vertical as the pendulum swings is 30°, [tex]\theta=30^{\circ}[/tex]
The pendulum have gravitational potential energy when the angle is maximum. The pendulum has only kinetic energy at its lowest point. Let v is the speed of the object at the lowest point in its trajectory. It can be calculated as :
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
h is the height moved by the pendulum.
[tex]h=l(1-cos(30))[/tex]
[tex]h=1.5(1-cos(30))[/tex]
h = 0.2 m
[tex]v=\sqrt{2gh}[/tex]
[tex]v=\sqrt{2\times 9.8\times 0.2}[/tex]
v = 1.978 m/s
So, the speed of the object at the lowest point in its trajectory is 1.978 m/s.
A parallel-plate capacitor has a capacitance of 10 mf and charged with a 20-v power supply. The power supply is then removed and a dielectric material of dielectric constant 4.0 is used to fill the space between the plates. How much energy is now stored by the capacitor?
Answer:
0.5 J
Explanation:
For this capacitor we have:
[tex]C=10 mF = 0.01 F[/tex] is the capacitance
V = 20 V is the potential difference
So the charge stored in the capacitor is
[tex]Q=CV=(0.01 F)(20 V)=0.2 C[/tex]
Later, the power supply is removed, so the charge on the capacitor will remain the same. A dielectric of dielectric constant
k = 4.0
is inserted in the gap between the plates. The capacitance of the capacitor change as follows:
C' = k C = (4.0)(0.01 F) = 0.04 F
The energy stored in the capacitor is given by
[tex]U'=\frac{1}{2}\frac{Q^2}{C'}[/tex]
and using Q = 0.2 C, we find
[tex]U'=\frac{1}{2}\frac{(0.2 C)^2}{(0.04 F)}=0.5 J[/tex]
After placing the dielectrics between the plates of the capacitor, the energy stored in the capacitor becomes 0.5 J.
What is capacitance?Capacitance is a term used to define the amount of energy stored in the form of an electric charge in an electric device.
Given data:
The capacitance of parallel-plate capacitor is, C = 10 mF = 0.01 F.
The potential of power supply is, V' = 20 V.
The dielectric constant of the material is, ∈ = 4.0.
Let us first calculate the charge stored in the capacitor as,
[tex]q = CV'\\\\q = 0.01 \times 20\\\\q = 0.2 \;\rm C[/tex]
After placing the dielectric, the capacitance of the capacitor is,
[tex]C' = \epsilon \times C[/tex]
Solving as,
[tex]C' = 4.0 \times 0.01\\\\C' = 0.04 \;\rm F[/tex]
Now, the expression for the energy stored in the capacitor is,
[tex]U'=\dfrac{q^{2}}{2C'}[/tex]
Solving as,
[tex]U'= \dfrac{0.2^{2}}{2 \times 0.04}\\\\U'= 0.5 \;\rm J[/tex]
Thus, we can conclude that the energy stored in the capacitor is of 0.5 J.
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A resistor R, inductor L, and capacitor C are connected in series to an AC source of rms voltage ΔV and variable frequency. If the operating frequency is twice the resonance frequency, find the energy delivered to the circuit during one period. (Use the following as necessary: the rms voltage ΔV, R, L, and C.)
Answer:E=[tex]\frac{\pi R\left ( \Delta V\right )^2\sqrt{LC}}{\left ( R^2+\frac{9}{4}\left (\frac{L}{C} \right )\right )}[/tex]
Explanation:
We know resonant frequency is given by
[tex]\omega_0=\frac{1}{\sqrt{LC}}[/tex]
and the operating frequency is given by
[tex]\omega =2\omega_0=\frac{2}{\sqrt{LC}}[/tex]
The capacitance reactance is given by
[tex]X_c=\frac{1}{\omega C}=\frac{\sqrt{LC}}{2C}=\frac{1}{2}\sqrt{\frac{L}{C}}[/tex]
inductive reactance is given by
[tex]X_L=\omega L=\left ( \frac{2}{\sqrt{LC}}\right )L=2\sqrt{\frac{L}{C}}[/tex]
Thus impedance is
[tex]Z=\left ( R^2+\left (X_L-X_C \right )^2 \right )^\frac{1}{2}[/tex]
[tex]Z=\left ( R^2+\left (2\sqrt{\frac{L}{C}}-\frac{1}{2}\sqrt{\frac{L}{C}} \right )^2 \right )^\frac{1}{2}[/tex]
[tex]Z=\left ( R^2+\frac{9}{4}\left ( \frac{L}{C} \right ) \right )^\frac{1}{2}[/tex]
The average power delivered is
[tex]P_{avg.}=\frac{\Delta V^2}{Z}cos\phi =\frac{\left ( \Delta V\right )^2}{Z}\left (\frac{R}{Z} \right )[/tex]
[tex]P_{avg.}=\frac{\left (\Delta V \right )^2R}{\left ( R^2+\frac{9}{4}\left (\frac{L}{C} \right )\right )}[/tex]
Energy Delivered in one cycle is given by
[tex]E=P_{avg}T[/tex]
[tex]E=\frac{\left (\Delta V \right )^2R}{\left ( R^2+\frac{9}{4}\left (\frac{L}{C} \right )\right )}\left ( \frac{2\pi }{\frac{2}{\sqrt{LC}}}\right )[/tex]
E=[tex]\frac{\pi R\left ( \Delta V\right )^2\sqrt{LC}}{\left ( R^2+\frac{9}{4}\left (\frac{L}{C} \right )\right )}[/tex]
Two cylindrical resistors are made from same material and have the same length. When connected across the same battery, one dissipates twice as much power as the other. How do their diameters compare ?
Answer:
[tex]\frac{d_2}{d_1} = \sqrt2 = 1.41[/tex]
Explanation:
As resistor is connected to the battery of constant EMF then the power across the resistor is given as
[tex]P = \frac{E^2}{R}[/tex]
now if two resistors are made up of same material and of same length then due to different cross sectional area they both have different resistance
Due to different resistance they both will have different power
Since power is inversely depends on the resistance
So if the power is twice that of the other then the resistance must be half
so we have
[tex]R_1 = \rho \frac{L}{A_1}[/tex]
[tex]R_2 = \rho\frac{L}{A_2}[/tex]
since one resistance is half that of other resistance
So the area of one must be twice that of other
so we have
[tex]\frac{A_2}{A_1} = 2[/tex]
[tex]\frac{\pi d_2^2}{\pi d_1^2} = 2[/tex]
[tex]d_2 = 1.41 d_1[/tex]
A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity is 9.8 m/s 2 . 2 What is the gravitational potential energy associated with the ball relative to the ceiling? Answer in units of J. 012 (part 2 of 3) What is its gravitational potential energy relative to the floor? Answer in units of J. 013 (part 3 of 3) What is its gravitational potential energy relative to a point at the same elevation as the ball? Answer in units of J.
1. -23.2 J
The gravitational potential energy of the ball is given by
[tex]U=mgh[/tex]
where
m = 1.1 kg is the mass of the ball
g = 9.8 m/s^2 is the acceleration of gravity
h is the height of the ball, relative to the reference point chosen
In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is
h = -2.16 m
And the gravitational potential energy is
[tex]U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J[/tex]
2. 41.1 J
Again, the gravitational potential energy of the ball is given by
[tex]U=mgh[/tex]
In this part of the problem, the reference point is the floor.
The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:
h = 5.97 m - 2.16 m = 3.81 m
And so the gravitational potential energy of the ball relative to the floor is
[tex]U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J[/tex]
3. 0 J
As before, the gravitational potential energy of the ball is given by
[tex]U=mgh[/tex]
Here the reference point is a point at the same elevation of the ball.
This means that the heigth of the ball relative to that point is zero:
h = 0 m
And so the gravitational potential energy is
[tex]U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J[/tex]
A balloon is buoyed up with a force equal to the
A) weight of air it displaces.
B) density of surrounding air.
C) atmospheric pressure.
D) weight of the balloon and contents.
Answer:
A. weight of air it displaces.
Explanation:
The force that buoys up a balloon is equal to the weight of the air it displaces, as per Archimedes' Principle.
Explanation:The force that buoys up a balloon is equal to the weight of the air it displaces. This principle is known as Archimedes' principle and it applies to both liquids and gases, like air. According to this principle, the upward buoyant force exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.
In the context of a balloon floating in the air, the balloon and the gas inside it displace a volume of air. The weight of this displaced air pushes upward on the balloon, providing the buoyant force. If the weight of the balloon and the gas inside it are less than the weight of the displaced air, the balloon will rise up into the atmosphere.
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Generation of electricity in coal-burning power plants and nuclear power plants both involve _______.
Answer:
Heating water to produce steam which drives a turbine
Explanation:
Generation of electricity in coal-burning power plants and nuclear power plants both involve heating water to produce steam which drives a turbine.
a beaker containing 200mL of a staurated solution of KI )with some solid KI at the bottom of the beaker) at 25 degrees Celsius is left standing to evaporate water at constant temperature until there is 100 mL of solution in the beaker compared to the original concentration the final concentration of Ki is ____
Answer:
Increased.
Explanation:
Given that in 200 mL of the beaker contains saturated solution of KI.
Whenever we allow the solution to heat or to evaporate, only water present in the solution evaporates because it is volatile while the salt is non volatile and thus do not evaporate.
When we decrease the volume of water in the beaker to 100 mL, the concentration of the salt, KI in the solution increases as the same amount of the salt is present in less amount of volume of the solvent.
To catch a fast-moving softball with your bare hand, you extend your hand forward just before the catch and then let the ball ride backward with your hand. Doing this reduces the catching force because
a. the time of the catch is decreased.
b. relative velocity is more.
c. time of catch is increased.
d. relative velocity is less.
e. none of the above
Answer:
option (c)
Explanation:
If a large force is applied on an object for a small duration of time, it is called impulsive force. for example, a bat's man hit the ball. here the contact time of bat and ball is very small while the force applied by the bat is very large.
Impulse = force x small time
To decrease the impulsive effect, the time should be increased.
Catching a fast-moving softball by extending the hand and reducing the force involves increasing the time of the catch, which decreases acceleration and the force needed to stop the ball.
Explanation:The correct answer to the question is c. time of catch is increased. When catching a fast-moving softball, extending the hand forward just before the catch and allowing the ball to ride backward with the hand effectively increases the time over which the ball is decelerated to a stop. This is due to Newton's second law, which states that the force exerted on an object is equal to the mass of the object times its acceleration (F=ma). By increasing the time of the catch, the acceleration is decreased, and thus the force experienced by the hand is also reduced. This is in contrast to catching the ball abruptly, which would result in a short catch time, higher acceleration, and therefore a larger force needed to stop the ball.
A 10.0-g bullet is fired into, and embeds itself in, a 1.95-kg block attached to a spring with a force constant of 16.6 N/m and whose mass is negligible. How far is the spring compressed if the bullet has a speed of 300 m/s just before it strikes the block and the block slides on a frictionless surface? Note: You must use conservation of momentum in this problem because of the inelastic collision between the bullet and block.
Answer:
Distance the spring is compressed, x = 0.52 m
Explanation:
Given :
Mass of the bullet, m = 10 g = 0.01 kg
Mass of the block, M = 1.95 kg
spring force constant, k = 16.6 N/m
Distance the spring is compressed =x
Speed of the bullet, v = 300 m/s
Speed of the block = V
Therefore we know that according to the law of conservation of momentum,
m.v = ( m+M )V
or [tex]V = \frac{m\times v}{m+M}[/tex]
[tex]V = \frac{0.01\times 300}{0.01+1.95}[/tex]
= 1.53 m/s
Now according to the law of conservation of momentum,
[tex]\frac{1}{2}\times M\times V^{2} =\frac{1}{2}\times k\times x^{2}[/tex]
[tex]x = \sqrt{\frac{\left ( M+m \right ).V^{2}}{k}}[/tex]
[tex]x = \sqrt{\frac{\left ( M+m \right )}{k}}\times V[/tex]
[tex]x = \sqrt{\frac{\left ( 1.95+0.01 \right )}{16.6}}\times 1.53[/tex]
[tex]x = 0.52[/tex] m
Thus, distance the spring is compressed, x = 0.52 m
While many elemental spectral lines are visible, almost all molecular lines lie in the _____ portion of the spectrum, since they are at much lower energy. while many elemental spectral lines are visible, almost all molecular lines lie in the _____ portion of the spectrum, since they are at much lower energy. radio ultraviolet infrared x-ray visible light?
Answer:
Infra and Red
Explanation:
While many elemental spectral lines are visible, almost all molecular lines lie in the infra portion of the spectrum, since they are at much lower energy. while many elemental spectral lines are visible, almost all molecular lines lie in the red portion of the spectrum, since they are at much lower energy.
Most molecular lines lie in the radio and infrared portions of the spectrum due to their lower energy levels. These lines form unique molecular fingerprints that aid scientists in molecular identification.
Explanation:In the electromagnetic spectrum, most molecular lines are found in the infrared and radio portions. This is due to the lower energy levels associated with these wavelengths. Spectral lines are characteristic wavelengths of electromagnetic radiation that are emitted or absorbed by different substances. Atomic spectral lines, such as those observed in elements like hydrogen or iron, are often in the visible part of the spectrum. However, interactions within molecules, specifically vibrations and rotations, create spectral fingerprints in the longer infrared and radio wavelengths, hence, most molecular lines are found in these portions.
It's crucial to note that each molecule has its own unique pattern of spectral lines, creating a molecular fingerprint that scientists use to identify different molecules.
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Two point charges each experience a 1-N electrostatic force when they are 2 cm apart. If they are moved to a new separation of 8 cm, what is the magnitude of the electric force on each of them?
2 N
1/8 N
1/16 N
1/4 N
1/2 N
Electrostatic force between two points in space is defined as,
[tex]F_e=\dfrac{Q_1Q_2}{4\pi r\epsilon_r\epsilon_0}[/tex]
The r is the distance between them.
So if,
[tex]1N=\dfrac{Q_1Q_2}{4\pi 2cm\epsilon_r\epsilon_0}\Rightarrow 2\cdot1N=\dfrac{Q_1Q_2}{4\pi\cdot10^{-2}m\cdot\epsilon_r\epsilon_0}[/tex]
Than,
[tex]\boxed{\dfrac{1}{4}N}=\dfrac{Q_1Q_2}{4\pi\cdot 8cm\cdot\epsilon_r\epsilon_0}\Rightarrow8\dfrac{1}{4}N\Leftrightarrow 2N=\dfrac{Q_1Q_2}{4\pi\cdot10^{-2}m\cdot\epsilon_r\epsilon_0}[/tex]
Hope this helps.
r3t40
Is this statement true or false concerning squall line thunderstorm development? These often form ahead of the advancing front but rarely behind it because lifting of warm, humid air and the generation of a squall line usually occur in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.
Answer: The following statement is true about squall line thunderstorm development: These often form ahead of the advancing front but rarely behind it because lifting of warm, humid air and the generation of a squall line usually occur in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.
An upper-level wave, accountable for the fabrication of a squall line, extend in front of and backside a cold front, the air backside the front is cold, steady and settling while the air ahead of the front is hot and co-seismic.
Final answer:
Squall lines often form ahead of advancing fronts due to warm, humid air lifting, while behind a cold front, the air is cooler and drier.
Explanation:
The statement is true concerning squall line thunderstorm development. Squall lines typically form ahead of an advancing front due to the lifting of warm, humid air in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.
When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of magnitude 172 μC on each plate. While the battery connection is maintained, a dielectric slab is inserted into, and fills, the region between the plates. This results in the accumulation of an additional charge of magnitude 220 μC on each plate. What is the dielectric constant of the dielectric slab?
The dielectric constant or relative permittivity of the dielectric slab inserted into the capacitor is approximately 2.28. This was calculated using the change in charge stored on the capacitor before and after the dielectric was inserted.
Explanation:The question involves understanding the use of a dielectric in a parallel-plate capacitor. The presence of a dielectric alters the capacitance value of the capacitor, allowing it to store more charge for the same applied voltage.
The dielectric constant of a material (also called the 'relative permittivity') is a measure of how much it can increase the capacitance of a capacitor compared to the capacitance when a vacuum is between the plates. The original capacitance C can be calculated as C = Q/V, where Q is the charge stored across the plates, and V is the potential difference across the plates. After the dielectric is inserted, the capacitance C' is calculated as C' = Q'/V, where Q' is the new charge stored.
In this case, you have your original charge (Q) as 172 μC. When the dielectric is inserted, the new charge (Q') is 172 μC + 220 μC = 392 μC. The dielectric constant (k) can be calculated using the equation k = C'/C = Q'/Q = 392/172 ≈ 2.28.
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The time it takes the block to complete one cycle is called the period. Usually, the period is denoted T and is measured in seconds. The frequency, denoted f, is the number of cycles that are completed per unit of time: f=1/T. In SI units, f is measured in inverse seconds, or hertz (Hz).
If the period is doubled, the frequency is ?
A. unchanged.B. doubled.C. halved.
Answer:
option (c)
Explanation:
The time taken by the pendulum to complete one oscillation is called time period.
The number of oscillations completed in one second is called frequency.
The frequency is the reciprocal of time period.
T = 1 / f
If the period is doubled, then the frequency is halved.
Find the angular acceleration produced given the mass lifted is 12 kg at a distance of 27.1 cm from the knee joint, the moment of inertia of the lower leg is 0.959 kg m2 kg m 2 , the muscle force is 1504 1504 N, and its effective perpendicular lever arm is 3.3 3.3 cm.
Answer:
Angular acceleration, α = 26.973 rad/s²
Explanation:
Given data:
Lifted mass, M = 12 kg
Distance of the lifted mass = 27.1 cm = 0.271 m
Effective lever arm, d = 3.3 cm = 0.033 m
Moment of inertia, I = 0.959 kg.m²
Applied force, F = 1504 N
Now,
the torque (T) is given as:
T = F × d
also,
T = I × α
where,
α is the angular acceleration
Now,
Total moment of inertia, I = 0.959 + 12×(0.271)² = 1.840 kg.m²
Now equation both the torque formula and substituting the respective values, we get
1504 × 0.033 = 1.840 × α
⇒ α = 26.973 rad/s²
1. Mars has a mean diameter of 4200 mi (the earth's is 7900 mi). The mass of Mars is .11 times the mass of the earth. What is the value of g on Mars?
2. What is the escape velocity on the moon?
PLEASE HELP ME ASAP
YOU WILL BE MARKED BRAINLIEST!!!!!!
Answer:
1. 3.8 m/s²
2. 2400 m/s
Explanation:
1. Acceleration due to gravity can be found with Newton's law of gravitation:
g = GM / R²
where g is the acceleration at the surface,
G is the universal gravitational constant,
M is the mass of the planet,
and R is the radius of the planet.
If we say M is the mass of the Earth and R is the radius of the Earth, then we know:
9.8 = GM / R²
Mars has a mass of 0.11 M and a radius of (4200/7900) R = 0.53 R. So the acceleration is:
g = G (0.11 M) / (0.53 R)²
g = 0.39 GM / R²
g = 0.39 (9.8)
g = 3.8
So the value of g on Mars is 3.8 m/s².
2. The escape velocity is the velocity at which the kinetic energy of an object at the surface equals its gravitational potential energy.
½ m v² = m (GM / r²) r
½ v² = GM / r
v² = 2GM / r
v = √(2GM / r)
The mass of the moon is 7.35×10²² kg, and the radius of the moon is 1.74×10⁶ m.
v = √(2 (6.67×10⁻¹¹) (7.35×10²²) / (1.74×10⁶))
v = 2400 m/s
Answer: yes that is true
Explanation:
A 6 ft tall person walks away from a 10 ft lamppost at a constant rate of 5 ft/s. What is the rate (in ft/s) that the tip of the shadow moves away from the pole when the person is 15 ft away from the pole?
Answer:
12.5 ft/s
Explanation:
Height of person = 6 ft
height of lamp post = 10 ft
According to the question,
dx / dt = 5 ft/s
Let the rate of tip of the shadow moves away is dy/dt.
According to the diagram
10 / y = 6 / (y - x)
10 y - 10 x = 6 y
y = 2.5 x
Differentiate both sides with respect to t.
dy / dt = 2.5 dx / dt
dy / dt = 2.5 (5) = 12.5 ft /s
When the person is 15 ft away from the pole, the rate is 7.5 ft/s.
Resultant displacementThe resultant displacement is calculated as follows;
[tex]R^2 = 6^2 + 10^2\\\\R = 11.6[/tex]
Rate of change of the displacement10 ft ------ 5 ft
15 ft ------- ?
= (15 x 5)/10
= 7.5 ft/s
Thus, when the person is 15 ft away from the pole, the rate is 7.5 ft/s.
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A car travels from Boston to Hartford in 4 hours. The two cities are 240 kilometers apart. What was the average speed of the car during the trip? A. 4 km/hour B. 960 km/hour C. 60 km/hour
Answer:
Average speed is 60 km/hour
Explanation:
When we need to calculate average speed, we use this equation:
[tex]V = \frac{x_{f} - x_{o}}{t_{f} - t_{o}}[/tex]
Where: [tex]x_{o} = 0 km[/tex] position at the beginning
[tex]x_{f} = 240 km[/tex] at the end
[tex]t_{o} = 0 hours[/tex]
[tex]t_{f} = 4 hours[/tex]
Then: [tex]V = \frac{240 km - 0 km}{4 hours - 0 hours}[/tex]
[tex]V = \frac{240 km}{4 hours}[/tex]
Finally V = 60 km/hour
uniform solid sphere of radius R rotates about a diameter with an angular speed 536 radians/second. The sphere then collapses under the action of internal forces to a final radius R/2. What is the final angular speed of the sphere in radians/second?
Answer:
2144 rad/s
Explanation:
R1 = R
ω1 = 536 rad/s
R2 = R/2
ω2 = ?
Mass is M
By use of angular momentum remains constant if no external force is acting on the body.
I1 ω1 = I2 ω2
The moment of inertia of solid sphere is 12/5 MR^2
So, 2/5 x M R^2 x 536 = 2/5 x M (R/2)^2 x ω2
536 = ω2 / 4
ω2 = 2144 rad/s
Answer:
ω₂ = 2144 rad/s
Explanation:
angular speed = 536 radians/second
as, we all know the moment of inertia of solid sphere
[tex]I_{sphere}= \dfrac{2}{5}MR^2[/tex]
here in the question two radius are given
by using angular momentum conservation
[tex]I_1 \omega_1 = I_2 \omega_2[/tex]
[tex]\dfrac{2}{5}MR_1^2 \omega_1 =\dfrac{2}{5}MR_2^2 \omega_2\\R^2\times 536= \dfrac{R^2}{4}\times \omega_2[/tex]
[tex]\omega_2 = 4 \times 536[/tex]
ω₂ = 2144 rad/s
A proton moving in a region of a space does not experience any acceleration in the direction of motion. Which of the following statements is correct with respect to this situation?A) The electric and magnetic fields must be perpendicular to each other in that region.B)The electric and magnetic fields must be parallel to each other in that region.C) Both electric and magnetic fields must be zero in that regionD) The electric field must be zero but not the magnetic field in that regionE) The magnetic field must be zero but not the electric field in that region
Answer:
A) The electric and magnetic fields must be perpendicular to each other in that region
Explanation:
The proton does not experience any acceleration: this means that the net force acting on it is zero.
Therefore, this also means that the electric force and the magnetic force acting on the proton are balanced.
Let's remind that:
- The electric force acting on a positive charged particle has the same direction as the electric field:
F=qE
where F is the force, q is the charge, E is the electric field
- The magnetic force acting on a positive charged particle in motion is perpendicular to the direction of the magnetic field:
F=qv ∧ B
where q is the charge, v is the velocity of the particle, B is the magnetic field
Therefore, for the two forces to be along the same line (but in opposite directions), we must have that the electric field E and the magnetic field B are perpendicular to each other.
The correct statement is that the magnetic field must be zero but not the electric field for a proton moving without acceleration in a region of space, as only then will it not be deflected from its path.
Explanation:The correct answer to the student's question is E) The magnetic field must be zero but not the electric field in that region. This can be understood by examining the forces experienced by a moving charged particle such as a proton in an electric field and magnetic field. A charged particle moving in a magnetic field experiences a Lorentz force that is always perpendicular to both the particle's velocity and the magnetic field direction.
If the proton is not accelerating in the direction of motion, it implies there's no net force acting on it in that direction. Since a magnetic field's force would always cause an acceleration perpendicular to the velocity, for a proton moving in a straight line with no acceleration, the magnetic field must be zero or parallel to the movement direction.
However, an electric field, if present, could be exerting a force on the proton balanced exactly by other forces (such as another opposite electric field), allowing the proton to have zero net force and continue its motion unaccelerated. Hence, a non-zero electric field could exist, but the magnetic field must be zero or not exerting any influence causing perpendicular acceleration of the proton.
Suppose the room in which you are sitting was magically transported off the Earth, and sent accelerating through the universe at 9.8 m/s2. Assuming your doors and windows are sealed and closed, how could you tell that you'd left the Earth?
Answer:
There is no way by which we can tell weather we are on the earth or we left the earth.
Explanation:
Our body responses to acceleration alone when we accelerate upwards at [tex]9.8m/s^{2}[/tex] a fictitious force equal to our [tex]mass\times acceleration[/tex] acts on our body which in this case of acceleration is same as our weight thus we will still feel the effect of gravity thus cannot say weather we left earth or are still there.
For a line on a graph to represent an object increasing its speed, the line must show
A) an increasingly steeper slope
B) a negative slope
C) a positive slope that is becoming more horizontal
D) a positive slope
Answer:d
Explanation:
Answer: option is A: An increasingly steeper slope
Explanation: Suppose that you are graphing the position of an object with increasing velocity.
in an x vs t graph.
as the velocity increases, the lapse of time needed to travel a fixed distance dx is shorter and shorter, so you will see a positive slope that as the time passes it will become more vertical (never being actually vertical)
Then the correct option is A: An increasingly steeper slope
Imagine that a hypothetical life form is discovered on our moon and transported to Earth. On a hot day, this life form begins to sweat, and it is determined that the heat of vaporization of its sweat is 107 cal/g. The scientist observing the extraterrestrial life form also begins to sweat. The heat of vaporization of the scientist's sweat is 580 cal/g at the skin's surface. If both the extraterrestrial life form and the scientist lose 735 g of sweat over the course of one hour, calculate the heat lost by each body from the vaporization of sweat.
Answer:
The heat of vaporization 580 cal/g times 602g = cal in human and do the same for life form.
Explanation:
Consider a single photon with a wavelength of lambda, a frequency of nu, and an energy of E. What is the wavelength, frequency, and energy of a pulse of light containing 100 of these photons? 0.01 lambda, nu, and 100E 0.01 lambda, 0.01 nu, and 0.01 E 100 lambda, 100 nu, and E 100 lambda, 100 nu, and 100 E lambda, nu, and 100E
Answer: lambda [tex]\lambda[/tex], nu [tex]\nu[/tex], and 100E
Explanation:
The energy [tex]E[/tex] of a photon is given by:
[tex]E=h\nu[/tex] (1)
Where:
[tex]h[/tex] is the Planck constant
[tex]nu[/tex] is the frequency
On the other hand, we have an expression that relates the frequency of the photn with its wavelength [tex]\lambda[/tex]:
[tex]nu=\frac{c}{\lambda}[/tex] (2) where [tex]c[/tex] is the speed of light
Substituting (2) in (1):
[tex]E=h\frac{c}{\lambda}[/tex] (3) This is the energy for a single photon
For 100 photons, the energy is:
[tex]100E=100(h\frac{c}{\lambda})=100h\nu[/tex] (3)
Where the wavelength and the frequency of the light remains constant.
Therefore, the answer is:
[tex]\lambda[/tex], [tex]\nu[/tex], and 100E
The wavelength, frequency, and energy of a pulse of light containing 100 of these photons is a wavelength, a frequency and 100 energy. Therefore the correct answer is E.
1. The wavelength [tex](\(\lambda\))[/tex] and frequency (nu) of a photon are related by the speed of light ([tex]\(c\)[/tex]) in a vacuum: [tex]\(c = \lambda \nu\)[/tex].
2. The energy (E) of a photon is related to its frequency by Planck's equation: [tex]\(E = h \nu\)[/tex], where (h) is Planck's constant.
3. For a pulse of light containing 100 photons, the total energy is the sum of the energies of the individual photons.
4. Therefore, the pulse would have the same wavelength ([tex]\(\lambda\)[/tex]), the same frequency [tex](\(\nu\)),[/tex] and 100 times the energy [tex](\(100E\))[/tex] compared to a single photon.
Thus, the correct choice is option E. [tex]\(\lambda, \nu, \text{ and } 100E\)[/tex].
Complete question:
Consider a single photon with a wavelength [tex]\(\lambda\)[/tex], a frequency (nu), and an energy (E). What is the wavelength, frequency, and energy of a pulse of light containing 100 of these photons?
A. [tex]\(0.01 \lambda, \nu, \text{ and } 100E\)[/tex]
B. [tex]\(0.01 \lambda, 0.01 \nu, \text{ and } 0.01 E\)[/tex]
C. [tex]\(100 \lambda, 100 \nu, \text{ and } E\)[/tex]
D. [tex]\(100 \lambda, 100 \nu, \text{ and } 100 E\)[/tex]
E. [tex]\(\lambda, \nu, \text{ and } 100E\)[/tex]
A car started to move ftom the rest and covers 2 km distance producing an acceleration of 0.5m/s2 find its velocity how long time does it take to cover the distance of 1.6km
(a) 44.7 m/s
We can find the final velocity of the car by using the SUVAT equation:
[tex]v^2-u^2 = 2ad[/tex]
where
v is the final velocity
u = 0 is the initial velocity (the car starts from rest)
[tex]a=0.5 m/s^2[/tex] is the acceleration
[tex]d=2km = 2000 m[/tex] is the displacement
Solving for v,
[tex]v=\sqrt{u^2 +2ad}=\sqrt{0+2(0.5)(2000)}=44.7 m/s[/tex]
(b) 80 s
We can find the time it takes for the car to cover 1.6 km by using the following SUVAT equation:
[tex]d=ut+\frac{1}{2}at^2[/tex]
where in this case we have
d = 1.6 km = 1600 m is the displacement
[tex]a=0.5 m/s^2[/tex] is the acceleration
u = 0
t is the time
Solving for t, we find:
[tex]t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(1600)}{0.5}}=80 s[/tex]
What are the possible values of the angular momentum quantum number l?
The angular momentum quantum number can be any integer from 0 to n-1, describing the shape of the electron's orbital and the level of its angular momentum.
Explanation:The angular momentum quantum number (l) can have any integer value from 0 to n-1, where n is the principal quantum number. This quantum number describes the shape of the electron's orbital, and it essentially tells us about the amount of angular momentum a subshell has. For instance, if the principal quantum number (n) is 3, l can be 0, 1, or 2, corresponding respectively to the s, p, and d orbitals.
Learn more about angular momentum quantum number here:https://brainly.com/question/33722762
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On what factors capacitance of parallel plate capacitor depends?
Answer:
Separation between the plates, area of the plates and dielectric constant
Explanation:
The capacitance of a parallel plate capacitor is given by:
[tex]C=k \epsilon_0 \frac{A}{d}[/tex]
where
k is the dielectric constant
[tex]\epsilon_0[/tex] is the vacuum permittivity (which has a constant value)
A is the area of the plates
d is the separation between the plates
Therefore from the formula we see that the capacitance of a parallel plate capacitor depends on the following factors:
- Separation between the plates
- Area of the plates
- Dielectric constant
Heliox is a helium‑oxygen mixture that may be used in scuba tanks for divers working at great depths. It is also used medically as a breathing treatment. A 7.25 L tank holds helium gas at a pressure of 1906 psi. A second 7.25 L tank holds oxygen at a pressure of 506.7 psi. The two gases are mixed in a 7.25 L tank. If the temperature remains the same throughout the process, what is the pressure of the gas mixture in the tank? Assume ideal gas behavior.
Answer:
2412 psi is the pressure of the gas mixture in the tank.
Explanation:
Pressure of the helium gas = 1,906 psi
Volume of the helium gas = 7.25 L
Pressure of the oxygen gas = 506 psi
Volume of the oxygen gas = 7.25 L
After mixing both gases in a container with volume 7.25 L at with constant temperature.
Since, the temperature and volume remained constant, pressure becomes directly dependent on moles of gases.So, when we mix gases together the moles of gases will also add and along with that pressure of individual gas will also get added to give total pressure of the mixture in a tank.
Total pressure = pressure(Heluim)+pressure(oxygen)
[tex]P_{total}=1,906 psi+506 psi=2,412 psi[/tex]
2412 psi is the pressure of the gas mixture in the tank.
If the speed of a vehicle was reduced from 65 miles per hour to 35 miles per hour, by approximately how many kilometers per hour was the speed reduced? (1 kilometer = 0.62 mile, rounded to the nearest 0.01 mile)
Answer:
reduced speed comes out to be 48.3 km/hr
Explanation:
intial speed = 65 miles per hour
final speed = 35 miles per hour
reduced speed = initial speed - final speed
= 65 - 35
= 30 mile/hr
1 km = 0.62 mile
1 miles = 1.61 km/hr
so reduce speed in (km/hr) = 30 ×1.61 km/hr
= 48.3 km/hr
hence the reduced speed comes out to be 48.3 km/hr