Answer:
The kinetic energy of the system after the collision is 9 J.
Explanation:
It is given that,
Mass of object 1, m₁ = 3 kg
Speed of object 1, v₁ = 2 m/s
Mass of object 2, m₂ = 6 kg
Speed of object 2, v₂ = -1 m/s (it is moving in left)
Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,
[tex]E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2[/tex]
[tex]E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2[/tex]
E = 9 J
So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.
A frog jumps for 4.0 seconds at a maximum horizontal distance of 0.8m. what is its velocity along the road?
Answer:
The frog's horizontal velocity is 0.2 m/s.
Explanation:
To solve this problem, we must first remember what velocity is and how we solve for it. Velocity can be solved for using the formula x/t, where x represents horizontal distance and t represents time (in seconds), that it takes to travel this distance. If we plug in the given numbers for these variables and solve, we get the following:
v = x/t
v = 0.8m/4s
v = 0.2 m/s
Therefore, the correct answer is 0.2 m/s. We can verify that these units are correct because the formula calls for distance divided by time, so meters per second is a sensible answer.
Hope this helps!
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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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
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 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.
What is meant by constructive and destructive interference
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 person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force F1 and the other hand holding it up at 50 cm from the end of the plank with force F2. If the plank has a mass of 20 kg and its center of gravity is at the middle of the plank, what are the forces F1 and F2?
Answer:
F1= 196 N
F2= 392 N
Explanation:
Given:
length of the plank = 2 m
mass of the plank = 20 kg
Weight of the plank = 20 x 9.8 =196 N
Torque due to the weight of the plank with respect to the pivoted end (i.e the end held by the hand) Counter clockwise torque = 196 x cog of wood
= 196 x 1 = 196 Nm
Clockwise torque = F2 x 0.5
for the balanced case
F2 x 0.5 = 196
F2 = 196/ 0.5
F2= 392 N
now,
the net force
Net downward force =Net upward force
F1 + weight of plank = F2
F1 + 196 = 392N
F1 = 392 – 196
F1= 196 N
To find the forces F1 and F2 exerted on the plank, we can use the principle of equilibrium. The sum of the forces acting on the plank must be zero, and the sum of the torques must also be zero. By solving the equations derived from these conditions, we can determine the values of F1 and F2.
Explanation:To find the forces F1 and F2, we can use the principle of equilibrium. According to the principle, the sum of the forces acting on an object in equilibrium must be zero, and the sum of the torques must also be zero.
For the vertical forces:
F1 + F2 - mg = 0
Where m is the mass of the plank and g is the acceleration due to gravity.
For the torques:
F1 * 2m = F2 * 0.5m
Solving these equations will give us the values for F1 and F2.
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On July 15, 2004, NASA launched the Aura spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit 705 km above the earth's surface, and we shall assume a circular orbit.
(a) How many hours does it take this satellite to make one orbit?
h
(b) How fast (in km/s) is the Aura spacecraft moving?
km/s
Answer:
a). T = 1.64 hr
b). v = 7.503 km/s
Explanation:
Given :
NASA launched the Aura spacecraft to study the earth's climate and atmosphere.
Height of the satellite orbit from the earth's surface, h = 705 km
= 705000 m
Therefore we know that,
a).Time period of the space craft is
[tex]T = 2\pi \sqrt{\frac{(R+h)^{3}}{G\times M}}[/tex]
where, G = Universal Gravitational constant
= [tex]6.67 \times 10^{-11} N-m^{2}/kg^{2}[/tex]
M = Mass of the earth
= 5.98 x [tex]10^{24}[/tex] kg
R = Radius of the earth
= 6.38 x [tex]10^{6}[/tex] m
∴[tex]T = 2\pi \sqrt{\frac{(R+h)^{3}}{G\times M}}[/tex]
[tex]T = 2\pi \sqrt{\frac{((6.36\times 10^{6})+705000)^{3}}{6.67\times 10^{-11}\times 5398\times 10^{24}}}[/tex]
[tex]T = 5933[/tex] s
= 1.64 hr
Thus, the satellite will take 1.64 hr to make one orbit.
b). We know velocity of the spacecraft is given by
[tex]v=\sqrt{\frac{G\times M}{R+h}}[/tex]
[tex]v=\sqrt{\frac{6.67\times 10^{-11}\times 5.98\times 10^{24}}{(6.38\times 10^{6})+705000}}[/tex]
v = 7503 m/s
= 7.503 km/s
Thus, the Aura satellite is moving with velocity v = 7.503 km/s
A) The number of hours it takes for the satellite to make one orbit ( h ) = 1.64 hours
b) The speed of the Aura spacecraft = 7.5 Km/s
Given data
Height of satellite Orbit ( h ) = 705 km ≈ 705000 m
A) Determine the time taken in hours for the satellite to make a single orbit
T = [tex]2\pi \sqrt{\frac{(R+h)^3}{GM} }[/tex] -------- ( 1 )
where ; G = 6.67 * 10⁻¹¹ N-m²/kg², M ( mass of earth ) = 5.98 * 10²⁴,
R = 6.38 * 10⁶ m , h = 705000 m
Insert the values into equation ( 1 )
T = 5933 secs
= 1.64 hours ( time taken to complete one orbit )
B) Determine how fast Aura spacecraft is moving
V = [tex]\sqrt{\frac{GM}{R+h} }[/tex] -------- ( 2 )
where ; G = 6.67 * 10⁻¹¹ N-m²/kg², M ( mass of earth ) = 5.98 * 10²⁴,
R = 6.38 * 10⁶ m , h = 705000 m
Insert values into equation ( 2 )
∴ V = 7503 m/s
= 7.5 km/s
Hence we can conclude that the number of hours it takes for the satellite to make one orbit ( h ) = 1.64 hours and The speed of the Aura spacecraft = 7.5 Km/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
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
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 sloping surface separating air masses that differ in temperature and moisture content is called a _________.
Answer:
A sloping surface separating air masses that differ in temperature and moisture content is called a front.
A sloping surface separating air masses that differ in temperature and moisture content is called a front.
Explanation:A sloping surface separating air masses that differ in temperature and moisture content is called a front. Fronts occur when warm air and cold air meet, creating a boundary between them. The warm air is forced to rise over the cold air, resulting in changes in weather conditions.
There are several types of fronts, each associated with distinct weather patterns:
Cold Front: A cold front forms when a cold air mass advances and replaces a warmer air mass. As the cold air displaces the warm air, it forces the warm air to rise rapidly, leading to the formation of cumulonimbus clouds and potentially severe weather conditions such as thunderstorms.
Warm Front: In contrast, a warm front occurs when a warm air mass advances and overtakes a retreating cold air mass. As the warm air rises over the colder air, it produces widespread stratiform clouds and precipitation over an extended area.
Stationary Front: When two air masses meet but neither advances, a stationary front is formed. This results in a prolonged period of cloudy and wet weather along the boundary.
Occluded Front: An occluded front develops when a fast-moving cold front overtakes a slow-moving warm front. This complex type of front often leads to a mix of weather conditions, including precipitation and strong winds.
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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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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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easy A solid disk rotates in the horizontal plane at an angular velocity of 0.067 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.10 kg · m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk?
Answer:
[tex]\omega_f = 0.0067\ rad/s[/tex]
Explanation:
Given:
Initial angular velocity of the disc, [tex]\omega_o[/tex] = 0.067 rad/s
Initial moment of inertia of the disc, [tex]I_o[/tex] = 0.10 kg.m²
Distance of sand ring from the axis, r = 0.40m
Mass of the sand ring = 0.50 kg
Now, no external torque is applied to the rotating disk.
Thus, the angular momentum of the system will remain conserved.
Also,from the properties of moment of inertia, the addition of the sand ring will increase the initial moment of inertia by an amount Mr²
thus, we have
Initial Angular Momentum = Final Angular Momentum
or
[tex]I_o\omega_o = I_f\times\omega_f[/tex]
[tex]I_o\omega_o = (I_o + Mr^2)\times\omega_f[/tex]
Where,
[tex]I_f[/tex] = Final moment of inertia
[tex]\omega_f[/tex] = Final angular velocity
substituting the values, we get
[tex]0.10\times0.067 = (0.10 + 0.50\times 0.40^2)\times\omega_f[/tex]
or
[tex]0.0067 = (0.18)\times\omega_f[/tex]
or
[tex]\omega_f = 0.0067\ rad/s[/tex]
The rate of change of angular displacement is defined as angular velocity. After all the sand is in place the angular velocity of the disk will be 0.067 rad/sec.
What is the definition of Angular velocity?The rate of change of angular displacement is defined as angular velocity, and it is stated as follows:
ω = θ t
Where,
θ is the angle of rotation,
t is the time
ω is the angular speed
ω₀ is the initial angular velocity of the disc = 0.067 rad/s
I₀ is the initial moment of inertia of the disc, = 0.10 kg.m²
r is the distance of sand ring from the axis = 0.40m
m is the mass of the sand ring = 0.50 kg
If the net external torque is applied to the rotating disk is zero
According to the angular momentum conservation principle;
Initial Angular Momentum = Final Angular Momentum
[tex]\rm I_0 \omega_0= I_f \omega_f \\\\[/tex]
According to parallel axis theorem ;
[tex]\\\\ I_f = I_0 + mr^2[/tex]
[tex]\rm I_0 \omega_0= (I_0 + mr^2 ) \omega_f[/tex]
[tex]\rm 0.10 \times 0.067= (0.10 + 1020 \times 0.40^2 ) \omega_f \\\\ \rm \omega_f=0.0067\ rad/sec[/tex]
Hence the angular velocity of the disk will be 0.067 rad/sec.
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An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic motion. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s.(a) Calculate the amplitude of the motion._____ m(b) Calculate the maximum velocity attained by the object. [Hint: Use conservation of energy.]______ m/s
Answer:
Part a)
A = 0.066 m
Part b)
maximum speed = 0.58 m/s
Explanation:
As we know that angular frequency of spring block system is given as
[tex]\omega = \sqrt{\frac{k}{m}}[/tex]
here we know
m = 3.5 kg
k = 270 N/m
now we have
[tex]\omega = \sqrt{\frac{270}{3.5}}[/tex]
[tex]\omega = 8.78 rad/s[/tex]
Part a)
Speed of SHM at distance x = 0.020 m from its equilibrium position is given as
[tex]v = \omega \sqrt{A^2 - x^2}[/tex]
[tex]0.55 = 8.78 \sqrt{A^2 - 0.020^2}[/tex]
[tex]A = 0.066 m[/tex]
Part b)
Maximum speed of SHM at its mean position is given as
[tex]v_{max} = A\omega[/tex]
[tex]v_{max} = 0.066(8.78) = 0.58 m/s[/tex]
You recently purchased a large plot of land in the Amazon jungle at an extremely low cost. You are quite pleased with yourself until you arrive there and find that the nearest source of electricity is 1500 miles away, a fact that your brother-in-law, the real estate agent, somehow forgot to mention. Since the local hardware store does not carry 1500-mile-long extension cords, you decide to build a small hydroelectric generator under a 35.0-m high waterfall located nearby. The flow rate of the waterfall is 0.150x10^2 m^3/h, and you anticipate needing 1750 kW h/wk to run you lights, air conditioner, and television.
What is the maximum power theoretically available from the waterfall? Is the power sufficient to meet your needs?
Answer:
The maximum power available from the water fall is 239.568 kWh/week.
Explanation:
Given:
Flow rate = 0.15 × 10² m³/h
now the mass of water flowing per hour will be = flow rate × mass density of water(i.e 1000 kg/m³)
mass of water flowing per hour will be = 0.15 × 10² m³/h × 1000 kg/m³ = 15000 kg/h
The gravitational potential energy of the falling water = mgh = 15000 × 9.8 × 35 = 5145000 J/h
or
5145000/3600 = 1426.166 J/s (as 1 h = 3600 seconds)
or
1426.166 W = 1.426 kW
Now, the number of hours in a week = 7 × 24 = 168 hours
Now the energy produce in a week = 1.426 kW × 168 hr = 239.568 kWh/week.
No it is not sufficient to meet the needs
The maximum hydraulic power generated by the waterfall is around 14.35 kW, which is more than the required energy for weekly consumption (2.28 kW). Therefore, the waterfall can especially serve as a power source.
Explanation:The problem involves the calculation of hydraulic power, which is given by the formula Power = p * g * h * Q, where p is the density of water (around 1000 kg/m³ in standard conditions), g is the gravitational acceleration (approximated as 9.8 m/s²), h is the waterfall height (in meters), and Q is the flow rate (in m³/s). First, we need to convert the flow rate from m³/h to m³/s. We get Q = 0.150 * 10² m³/h = 0.04167 m³/s.
After substituting the figures into the formula, we get: Power = 1000 kg/m³ * 9.8 m/s² * 35.0 m * 0.04167 m³/s = 14350.5 Watts, or about 14.35 kW.
Given the energy requirement per week is 1750 kWh, let's convert it to the energy required per second. That is 1750 kWh/wk = (1750 * 1000) / (7 * 24 * 60 * 60) = 2.28 kW. This means, the maximum power available from the waterfall is more than enough to meet your requirements.
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The sum of potential and kinetic energies in the particles of a substance is called energy.
Explanation:
This sentence is the description of a specific type of energy: the mechanical energy.
To unerstand it better:
The mechanical energy of a body, a system or a substance is that which is obtained from the speed of its movement (kinetic energy) or its specific position (potential energy), in order to produce a mechanical work. This means mechanical energy involves both the kinetic energy and the potential energy (which can be elastic or gravitational, for example).
In addition, it should be noted that mechanical energy is conserved in conservative fields and is a scalar magnitude.
Therefore:
The sum of potential and kinetic energies in the particles of a substance is called Mechanical Energy
Answer:
Explanation:
internal!!!!!!!!
The pressure of a gas in a container is 1.85 atm and occupies a volume of 12.5 L. If the original volume is reduced by half at constant temperature, what would happen to the pressure?
Answer:
3.7 atm i.e., pressure doubles
Explanation:
P₁ = Initial pressure = 1.85 atm
P₂ = Final pressure
V₁ = Initial volume = 12.5 L
V₂ = Final volume = 0.5V₁
T = Temperature is constant
From ideal gas law
P₁V₁ = P₂V₂
[tex]\\\Rightarrow P_2=P_1\frac{V_1}{V_2}\\\Rightarrow P_2=1.85\frac{V_1}{0.5V_1}\\\Rightarrow P_2=1.85\frac{1}{0.5}\\\Rightarrow P_2=1.85\times 2\\\Rightarrow P_2=3.7\ atm[/tex]
∴ Final pressure is 3.7 atm i.e., pressure doubles
Final answer:
Boyle's law describes the relationship between pressure and volume of a gas at constant temperature. When the volume is reduced by half, the pressure of the gas would double.
Explanation:
The pressure of a gas in a container is directly proportional to its volume when the temperature is constant, as described by Boyle's law. In this case, when the original volume is reduced by half, the pressure would double, resulting in a pressure of 3.7 atm.
Suppose the energy transferred to a dead battery during charging is W. The recharged battery is then used until fully discharged again. Is the total energy transferred out of the battery during use also W?
Answer:
No
Explanation:
The amount of energy transferred out of battery will not be the same as given during charging.
A battery has its internal resistance as well which will draw some energy, through this internal resistance energy is lost
P = I² R
where P is energy, I is current passing and R is resistance.
As per law of conservation energy will be saved as it is the sum of drawn energy and energy wasted in internal resistance. But the total energy transferred out will not be same.
______ is the energy released by the sun is caused by thermonuclear fusion.
Answer: Nuclear energy
Explanation:
Nuclear energy (also called atomic energy) is the energy found in the nucleus of an atom.
This energy is released spontaneously (within the stars and the sun) or artificially (in nuclear reactors built by humans) in nuclear reactions, which are divided into two types:
-Fission (separation of the components of the atom nucleous)
-Fusion (joining two light nuclei to form a heavier nucleous)
In the case of the Sun, the nuclear reactions that occur are due fusion, in which the hydrogen is converted into helium.
Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence of dark energy. Rank each model from left to right based on the ratio of its actual mass density to the critical density, from smallest ratio (mass density much smaller than critical density) to largest ratio (mass density much greater than critical density).
(A) coasting universe
(B) critical universe
(C) recollapsing universe
Answer:
1) Recollapsing universe
2) critical universe
3) Coasting universe
Explanation:
According to the smallest ration (ratio actual mass density to current density) to largest ration, rank of models for expansion of universe are
1) Recollapsing universe -in this, metric expansion of space is reverse and universe recollapses.
2) critical universe - in this, expansion of universe is very low.
3) Coasting universe - in this, expansion of universe is steady and uniform
A toy truck has a speed of 4 m/s and kinetic energy of 48 J. What is its mass?
A. 8 kg
B. 4 kg
C. 6 kg
D. 10 kg
48/4= 12
12/2= 6
Answer: C. 6 kg
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]
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
A car travelling 85km/h strikes a tree. The front end compresses and the driver comes to rest after travelling 0.80m. What was the average acceleration of the driver during the collision?
Answer:
The average acceleration of the driver during the collision is [tex]-348.4\frac{m}{s^{2}}[/tex]
Explanation:
Initial speed of car , [tex]u=85\frac{km}{h}=\frac{85\times 5}{18}\frac{m}{s}[/tex]
=>[tex]u=23.61\frac{m}{s}[/tex]
Finally the car comes to rest .
Therefore final speed of the car , [tex]v=0\frac{m}{s}[/tex]
Distance traveled while coming to rest , s = 0.80 m
Using equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
=>[tex]0^{2}=23.61^{2}+2\times 0.80\times a[/tex]
=>[tex]a=-348.4\frac{m}{s^{2}}[/tex]
Thus the average acceleration of the driver during the collision is [tex]-348.4\frac{m}{s^{2}}[/tex]
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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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
Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the centerline does the second bright fringe occur?
Answer: [tex]1.8\°[/tex]
Explanation:
The diffraction angles [tex]\theta_{n}[/tex] when we have a slit divided into [tex]n[/tex] parts are obtained by the following equation:
[tex]dsin\theta_{n}=n\lambda[/tex] (1)
Where:
[tex]d=3.4(10)^{-5}m[/tex] is the width of the slit
[tex]\lambda=540 nm=540(10)^{-9}m[/tex] is the wavelength of the light
[tex]n[/tex] is an integer different from zero.
Now, the second-order diffraction angle is given when [tex]n=2[/tex], hence equation (1) becomes:
[tex]dsin\theta_{2}=2\lambda[/tex] (2)
Now we have to find the value of [tex]\theta_{2}[/tex]:
[tex]sin\theta_{2}=\frac{2\lambda}{d}[/tex] (3)
Then:
[tex]\theta_{2}=arcsin(\frac{2\lambda}{d})[/tex] (4)
[tex]\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})[/tex] (5)
Finally:
[tex]\theta_{2}=1.8\°[/tex] (6)