Answer:
Gravitational potential energy, PE = 2.8 J
Explanation:
It is given that,
Mass of the bluebird, m = 34 g = 0.034 kg
It flies from the ground to the top of an 8.5-m tree, h = 8.5 m
We need to find the change in the bluebird's gravitational potential energy as it flies to the top of the tree. It can be calculated as :
[tex]PE=m\times g\times h[/tex]
[tex]PE=0.034\ kg\times 9.8\ m/s^2\times 8.5\ m[/tex]
PE = 2.83 J
or
PE = 2.8 J
So, the gravitational potential energy as it flies to the top of the tree is 2.8 J. Hence, this is the required solution.
An artificial satellite is in a circular orbit around a planet of radius r = 5.25 Ã 10^3 km at a distance d = 450.0 km from the planet's surface. The period of revolution of the satellite around the planet is T = 2.15 hours. What is the average density of the planet?
Answer:
3020.68 kg/m^3
Explanation:
r = 5.25 x 10^3 km = 5.25 x 10^6 m, d = 450 km = 450 x 10^3 m
T = 2.15 hours = 2.15 x 3600 second = 7740 second
Let the density of the planet is ρ and M be the mass of planet.
The formula for the orbital velocity is
[tex]v = \sqrt{\frac{GM}{r+d}}[/tex]
Time period is given by
[tex]T = {\frac{2\pi (r+d)}{v}}[/tex]
[tex]T = \frac{2\pi (r +d)^{1.5}}{\sqrt{GM}}[/tex]
[tex]7740= \frac{2\pi (5700\times 1000)^{1.5}}{\sqrt{6.67\times 10^{-11}M}}[/tex]
M = 1.83 x 10^24 kg
Density = mass / Volume
ρ = 1.83 x 10^24 / (4/3 x 3.14 x (5.25 x 10^6)^3)
ρ = 3020.68 kg/m^3
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 +3.00); C charge with a mass of 4.00 x 10-3 kg hovers above a horizontal flat insulating surface, neither descending nor ascending, and thus in mechanical equilibrium. Taking g = 9.8 m/s2, what is the magnitude of the electric force on the charge? Hint: keep in mind that the weight force is given by F mg ? A. 0.0281 N O B. 0.0392 N C. 0.0473 N 0 D. 0.0671 N
Answer:
option (B)
Explanation:
q = 3 c, m = 4 x 10^-3 kg, g 9.8 m/s^2,
In the equilibrium condition, the weight of the charge particle is balanced by the electrostatic force.
q E = mg
Electrostatic force = m g = 4 x 10^-3 x 9.8 = 0.0392 N
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
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.
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.
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.
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(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.
A grasshopper floating in water generates waves at a rate of three per second with a wavelength of two centimeters. (a) What is the period of these waves? (b) What is the wave velocity?
Answer:
(a) 0.33 second
(b) 6 cm/s
Explanation:
Frequency, f = 3 waves per second
wavelength, λ = 2 cm = 0.02 m
(a) The period of wave is defined as the time taken by the wave to complete one oscillation. It is the reciprocal of frequency.
T = 1 / f = 1 / 3 = 0.33 second
(b) the relation between wave velocity, frequency and wavelength is given by
v = f x λ
v = 3 x 0.02 = 0.06 m /s
v = 6 cm /s
Final answer:
The period of the waves generated by a grasshopper in water is 0.333 seconds, and the wave velocity is 0.06 m/s.
Explanation:
Calculating the Period and Wave Velocity
When dealing with waves generated by a grasshopper floating in water, two key properties to determine are the period of the waves and the wave velocity.
(a) The Period of the Waves
The period (T) of a wave is the amount of time it takes for one complete wave cycle to pass. It can be calculated as the inverse of the frequency (f), which is the rate at which waves are generated. The formula to find the period is:
T = 1/f
In this case, the grasshopper generates waves at a frequency of 3 waves per second (3 Hz). Therefore, the period is:
T = 1/3 Hz = 0.333 seconds
(b) The Wave Velocity
The velocity (v) of a wave is determined by multiplying the frequency (f) of the wave by its wavelength (λ). The formula for wave velocity is:
v = f × λ
Here, the wavelength given is 2 cm, which we need to convert to meters (since the SI unit for velocity is m/s). Thus:
λ = 2 cm = 0.02 m
The velocity of the waves generated by the grasshopper is then:
v = 3 Hz × 0.02 m = 0.06 m/s
Determine the total impedance of an LRC circuit connected to a 10.0- kHz, 725-V (rms) source if L = 36.00 mL, R = 10.00 kΩ, and C = 5.00 nF.
Answer:
10042.6 ohm
Explanation:
f = 10 kHz = 10000 Hz, L = 36 mH = 0.036 H, R = 10 kilo Ohm = 10000 ohm
C = 5 nF = 5 x 10^-9 F
XL = 2 x π x f x L
XL = 2 x 3.14 x 10000 x 0.036 = 2260.8 ohm
Xc = 1 / ( 2 x π x f x C) = 1 / ( 2 x 3.14 x 10000 x 5 x 10^-9)
Xc = 3184.7 ohm
Total impedance is Z.
Z^2 = R^2 + (XL - Xc)^2
Z^2 = 10000^2 + ( 2260.8 - 3184.7 )^2
Z = 10042.6 ohm
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 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.
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]
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.
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.
An electric field of 8.20 ✕ 105 V/m is desired between two parallel plates, each of area 25.0 cm2 and separated by 2.45 mm. There's no dielectric. What charge must be on each plate?
Answer:
q = 1.815 \times 10^{-8} C
Charge on one plate is positive in nature and on the other plate it is negative in nature.
Explanation:
E = 8.20 x 10^5 V/m, A = 25 cm^2, d = 22.45 mm
According to the Gauss's theorem in electrostatics
The electric field between the two plates
[tex]E = \frac{\sigma }{\varepsilon _{0}}[/tex]
[tex]{\sigma }= E \times {\varepsilon _{0}}[/tex]
[tex]{\sigma }= 8.20 \times 10^{5} \times {8.854 \times 10^{-12}[/tex]
[tex]{\sigma }= 7.26 \times 10^{-6} C/m^{2}[/tex]
Charge, q = surface charge density x area
[tex]q = 7.26 \times 10^{-6} \times 25 \times 10^{-4}[/tex]
q = 1.815 \times 10^{-8} C
An inductor of 299 mH with a resistance of 51 Ω is connected to a power supply with a maximum voltage of 227 V and a frequency of 72 Hz. Find the current in the circuit. Answer in units of A.
Answer:
The answer is 1.1A
Explanation:
See the attached file
Let's begin with the angular acceleration of a compact disk (CD). To play music the CD must rotate at high speed while a laser reads data encoded in a spiral pattern on the disk. The disc has radius R=6.0cm=0.060m; when data are being read, it spins at 7200 rev/min. What is the CD's angular velocity ω in radians per second? How much time is required for it to rotate through 90∘? If it starts from rest and reaches full speed in 4.0 s, what is its average angular acceleration?
The angular velocity of the CD is 754.4 rad/s. The time it takes for the CD to rotate through 90 degrees is 0.0021 seconds. The average angular acceleration of the CD is 188.6 rad/s².
Explanation:To calculate the angular velocity of the CD, we can convert the given 7200 rev/min to radians per second. Since one revolution is equal to 2π radians, we can use the conversion factor to find the angular velocity. Thus, the angular velocity of the CD is 754.4 rad/s.
To calculate the time it takes for the CD to rotate through 90 degrees, we need to find the fraction of the total rotation that corresponds to 90 degrees. Since a full rotation is 360 degrees or 2π radians, 90 degrees is equal to π/2 radians. We can then use the formula Δθ = ωΔt to find the time it takes, where Δθ is the angle in radians, ω is the angular velocity, and Δt is the time. Rearranging the formula, we have Δt = Δθ/ω. Substituting the values, we get Δt = π/2 / 754.4 = 0.0021 seconds.
The average angular acceleration can be found using the formula α = (ωf - ωi) / Δt, where α is the angular acceleration, ωf is the final angular velocity, ωi is the initial angular velocity, and Δt is the time interval. The CD starts from rest and reaches full speed in 4 seconds, so the initial angular velocity is 0. Using the given full speed of 7200 rev/min, we can convert it to radians per second and use it as the final angular velocity. Thus, the average angular acceleration is α = (754.4 rad/s - 0 rad/s) / 4 s = 188.6 rad/s².
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Calculate the speed (in m/s) of an electron and a proton with a kinetic energy of 1.25 electron volt (eV). (The electron and proton masses are me = 9.11 ✕ 10−31 kg and mp = 1.67 ✕ 10−27 kg. Boltzmann's constant is kB = 1.38 ✕ 10−23 J/K.)
The speed of an electron with a kinetic energy of 1.25 eV is approximately 1.57 x 10⁶ m/s, and the speed of a proton with the same kinetic energy is approximately 5.29 x 10⁵ m/s.
To calculate the speed of an electron and a proton with a kinetic energy of 1.25 electron volts (eV), we can use the kinetic energy formula and relate it to the speed of the particles. The kinetic energy (KE) of a particle is given by:
KE = (1/2) * m * v²
Where:
KE = kinetic energy
m = mass of the particle
v = speed of the particle
We are given the kinetic energy in electron volts (eV), but we need to convert it to joules (J) since the mass is given in kilograms (kg). The conversion factor is 1 eV = 1.60219 x 10⁻¹⁹ J.
So, the kinetic energy in joules is:
KE = 1.25 eV * 1.60219 x 10⁻¹⁹ J/eV = 2.0027375 x 10⁻¹⁹ J
Now, we can rearrange the kinetic energy formula to solve for the speed (v):
v = √((2 * KE) / m)
For an electron:
Mass of electron (mₑ) = 9.11 x 10⁻³¹ kg
v(electron) = √((2 * 2.0027375 x 10⁻¹⁹ J) / (9.11 x 10⁻³¹ kg))
Calculating this gives us the speed of the electron.
For a proton:
Mass of proton (m_p) = 1.67 x 10⁻²⁷ kg
v(proton) = √((2 * 2.0027375 x 10⁻¹⁹ J) / (1.67 x 10⁻²⁷ kg))
Calculating this gives us the speed of the proton.
Now, let's calculate these speeds.
After performing the calculations, the speed of the electron is approximately 1.57 x 10⁶ m/s, and the speed of the proton is approximately 5.29 x 10⁵ m/s.
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True False Suppose I have a resistor of some resistance R. If I were to double the length and double the cross-sectional area of the resistor, what is the new resistance?
Explanation:
The resistance of a wire is given by :
[tex]R=\rho\dfrac{l}{A}[/tex]
Where
[tex]\rho[/tex] is the resistivity of the wire
l = initial length of the wire
A = initial area of cross section
If length and the area of cross section of the wire is doubled then new length is l' and A', l' = 2 l and A' = 2 A
So, new resistance of the wire is given by :
[tex]R'=\rho\dfrac{l'}{A'}[/tex]
[tex]R'=\rho\dfrac{l}{A}[/tex]
R' = R
So, the resistance of the wire remains the same on doubling the length and the area of wire.
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
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A sewing machine needle moves in simple harmonic motion with a frequency of 2.5 Hz and an amplitude of 1.27 cm. (a) How long does it take the tip of the needle to move from the highest point to the lowest point in its travel? (b) How long does it take the needle tip to travel a total distance of 11.43 cm?
Answer:
The tip of the needle to move from the highest point to the lowest point in 0.4 sec and the needle tip to travel a total distance in 0.9 sec.
Explanation:
Given that,
Frequency = 2.5 Hz
Amplitude = 1.27 cm
(a). We need to calculate the time
The frequency is the reciprocal of the time.
[tex]f=\dfrac{1}{T}[/tex]
[tex]T=\dfrac{1}{f}[/tex]
[tex]T=\dfrac{1}{2.5}[/tex]
[tex]T=0.4\ sec[/tex]
The time taken from highest point to lowest point
[tex]T'=\dfrac{T}{2}[/tex]
[tex]T'=\dfrac{0.4}{2}[/tex]
[tex]T'=0.2\ sec[/tex]
(b). We need to calculate the time
The time taken in one cycle = 0.4 sec
The distance covered in one sec= 4 times x amplitude
[tex]d=4\times1.27[/tex]
[tex]d=5.08\ m[/tex]
We need to calculate the speed
Using formula of speed
[tex]v=\dfrac{5.08}{0.4}[/tex]
[tex]v=12.7[/tex]
We need to calculate the time
[tex]t=\dfrac{11.43}{12.7}[/tex]
[tex]t= 0.9 sec[/tex]
Hence, The tip of the needle to move from the highest point to the lowest point in 0.4 sec and the needle tip to travel a total distance in 0.9 sec.
The amount of time it took the tip of the needle to move from the highest point to the lowest point in its travel is equal to 0.2 seconds.
Given the following data:
Frequency, F = 2.5 HzAmplitude, A = 1.27 cmHow to calculate the time.Mathematically, the frequency of an object in simple harmonic motion is give by:
[tex]F=\frac{1}{T} \\\\T=\frac{1}{F}\\\\T=\frac{1}{2.5}[/tex]
T = 0.4 seconds.
Now, we can calculate the time it took the needle to move from the highest point to the lowest point:
[tex]t = \frac{T}{2} \\\\t = \frac{0.4}{2}[/tex]
Time, t = 0.2 seconds.
How to calculate the time w.r.t a distance of 11.43 cm.The distance traveled by the needle per seconds is given by:
[tex]d=4A\\\\d=4 \times 1.27\\\\[/tex]
Distance, d = 5.08 cm.
For the speed:
[tex]Speed = \frac{distance}{time} \\\\Speed =\frac{5.08}{0.4}[/tex]
Speed = 12.7 cm/s.
For the time:
[tex]Time = \frac{distance}{speed} \\\\Time =\frac{11.43}{12.7}[/tex]
Time = 0.9 seconds.
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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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An ion source is producing 6Li ions, which have charge +e and mass 9.99 × 10-27 kg. The ions are accelerated by a potential difference of 13 kV and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude B = 1.0 T. Calculate the strength of the smallest electric field, to be set up over the same region, that will allow the 6Li ions to pass through undeflected.
Answer:
6.45 x 10^5 N/C
Explanation:
q = + e = 1.6 x 10^-19 C
m = 9.99 x 10^-27 kg
V = 13 kV = 13000 V
B = 1 T
Let v be the speed and E be the strength of electric field.
1/2 mv^2 = eV
v^2 = 2 e v / m
v^2 = (2 x 1.6 x 10^-19 x 13000) / (9.99 x 10^-27)
v = 6.45 x 10^5 m/s
As the charge particle is undeflected, the force due to magnetic field is counter balanced by the force due to electric field.
q E = q v B
E = v B = 6.45 x 10^5 x 1 = 6.45 x 10^5 N/C
The result will be the magnitude of the smallest electric field (E) required for the 6Li ions to pass through the magnetic field undeflected.
Here's how to calculate the strength of the electric field required for the 6Li ions to pass through the magnetic field undeflected:
Force Balance:
For the ions to move undeflected, the magnetic force acting on them needs to be balanced by the electric force acting in the opposite direction.
Magnetic Force:
The magnetic force (F_magnetic) on a charged particle moving through a magnetic field is given by: F_magnetic = q * v * B
Where:
q is the charge of the particle (q = +e for 6Li ion)
v is the velocity of the particle
B is the magnetic field strength
Electric Force:
The electric force (F_electric) on the charged particle in an electric field (E) is: F_electric = q * E
Balancing Forces:
For undeflected motion: F_magnetic = F_electric
Substitute the expressions from steps 2 and 3: q * v * B = q * E
Solving for Electric Field (E):
Since the charge (q) of the ion appears on both sides, we can cancel it.
E = v * B
Finding Ion Velocity (v):
The ions are accelerated by a potential difference (V) of 13 kV (13000 V).
The potential difference is related to the ion's kinetic energy (KE) by: KE = q * V
Assuming the ion starts from rest, all the potential energy is converted to kinetic energy.
KE = 1/2 * m * v^2 (where m is the mass of the ion)
Solve for v:
Combine equations from steps 6: 1/2 * m * v^2 = q * V
Solve for v: v = sqrt( 2 * q * V / m )
Substitute v in the equation for E (from step 5):
E = sqrt( 2 * q * V / m ) * B
Plug in the values:
q = +e (elementary charge = 1.602 x 10^-19 C)
V = 13000 V
m = 9.99 x 10^-27 kg
B = 1.0 T
Calculate E using the above values and the constant for elementary charge.
A charged particle A exerts a force of 2.45 μN to the right on charged particle B when the particles are 12.2 mm apart. Particle B moves straight away from A to make the distance between them 18.2 mm. What vector force does particle B then exert on A?
Answer:
[tex]F_2 = 1.10 \mu N[/tex]
Explanation:
As we know that the electrostatic force is a based upon inverse square law
so we have
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
now since it depends inverse on the square of the distance so we can say
[tex]\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}[/tex]
now we know that
[tex]r_2 = 18.2 mm[/tex]
[tex]r_1 = 12.2 mm[/tex]
also we know that
[tex]F_1 = 2.45 \mu N[/tex]
now from above equation we have
[tex]F_2 = \frac{r_1^2}{r_2^2} F_1[/tex]
[tex]F_2 = \frac{12.2^2}{18.2^2}(2.45\mu N)[/tex]
[tex]F_2 = 1.10 \mu N[/tex]
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
60. The label on a box of cereal gives the mass of cereal in two units: 978 grams and 34.5 oz. Use this information to find a conversion factor between the English and metric units. How many significant figures can you justify in your conversion factor?
Answer : The mass of cereal in two units are, 34.815 oz and 978.075 grams respectively.
Explanation :
As there are two systems for measuring the units which are English and Metric system.
The Metric System measured the things in meters, grams, liters, etc and adds prefixes like kilo, milli and centi to the count orders of the magnitude.
The English system measured the things in feet, inches, pounds, mile, etc.
As we are given the mass of cereal in 978 grams that means 987 grams is in metric unit. Now we have to convert metric unit into English unit.
Conversion factor used : (1 oz = 28.350 g)
As, 28.350 grams = 1 oz
So, 987 grams = [tex]\frac{987\text{ grams}}{28.350\text{ grams}}\times 1oz[/tex]
= 34.815 oz
As we are given the mass of cereal in 34.5 oz that means 34.5 oz is in metric unit. Now we have to convert metric unit into English unit.
Conversion factor used : (1 oz = 28.350 g)
As, 1 oz = 28.350 grams
So, 34.5 oz = [tex]\frac{34.5\text{ oz}}{1\text{ oz}}\times 28.350\text{ grams}[/tex]
= 978.075 grams
Therefore, the mass of cereal in two units are, 34.815 oz and 978.075 grams respectively.
The conversion factor from ounces to grams, divide 978 grams by 34.5 ounces, resulting in approximately 28.3 g/oz when rounded to three significant figures.
To find a conversion factor between English and metric units using the information that a box of cereal has a mass of 978 grams and 34.5 ounces, you would divide the number of grams by the number of ounces.
Conversion factor = 978 g / 34.5 oz ≈ 28.34783 g/oz
However, we need to consider significant figures in our answer. The number 34.5 has three significant figures, and the number 978 has three significant figures as well. Therefore, we can justify three significant figures in our conversion factor, giving us 28.3 g/oz as the conversion with significant figures properly accounted for.
Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags mutually exert a gravitational attraction F1 on each other. You now take two bricks from one bag and add them to the other bag, causing the bags to attract each other with a force F2. What is the closest expression for F2 in terms of F1?
The closest expression for [tex]\( F_2 \)[/tex] in terms of [tex]\( F_1 \)[/tex] is.
[tex]\[ \\F_2 = \frac{3}{4} F_1} \][/tex]
This means that after transferring 2 bricks from one bag to the other, the gravitational attraction [tex]\( F_2 \)[/tex] between the bags is [tex]\( \frac{3}{4} \)[/tex] of the initial attraction [tex]\( F_1 \).[/tex]
When initially each bag contains 4 bricks of mass [tex]\( M \)[/tex] each, the total mass in each bag is [tex]\( 4M \)[/tex]. The bags exert a gravitational attraction [tex]\( F_1 \)[/tex] on each other.
Let's denote:
-[tex]\( F_1 \):[/tex] Initial gravitational attraction between the bags when each bag has 4 bricks.
- [tex]\( F_2 \):[/tex] Gravitational attraction after transferring 2 bricks from one bag to the other.
Initial Situation Before Transfer
Each bag has 4 bricks so the mass of each bag is [tex]\( 4M \).[/tex]
Gravitational Attraction [tex]\( F_1 \)[/tex]
[tex]\[ F_1 = G \frac{(4M)(4M)}{d^2} \][/tex]
where [tex]\( G \)[/tex] is the gravitational constant and [tex]\( d \)[/tex] is the distance between the centers of the bags.
After Transferring 2 Bricks:
Now, one bag has 6 brick mass 6m and the other bag has 2 bricks mass2m
Gravitational Attraction[tex]\( F_2 \)[/tex]
[tex]\[ F_2 = G \frac{(6M)(2M)}{d^2} \]s[/tex]
Simplifying [tex]\( F_2 \)[/tex]
[tex]\[ F_2 = G \frac{12M^2}{d^2} \][/tex]
Relation between [tex]\( F_2 \)[/tex] and [tex]\( F_1 \)[/tex]
To find the relation between [tex]\( F_2 \)[/tex] and [tex]\( F_1 \)[/tex] we compare them
[tex]\[ \frac{F_2}{F_1} = \frac{G \frac{12M^2}{d^2}}{G \frac{16M^2}{d^2}} \][/tex]
[tex]\[ \frac{F_2}{F_1} = \frac{12M^2}{16M^2} \][/tex]
[tex]\[ \frac{F_2}{F_1} = \frac{3}{4} \][/tex]
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 5.0-kg crate is on an incline that makes an angle 30° with the horizontal. If the coefficient of static friction is 0.5, what is the maximum force that can be applied parallel to the plane without moving the crate?
Answer:
[tex]F_{applied} = 45.8 N[/tex]
Explanation:
When maximum force is applied on the crate along the plane so that it will not move then in that case friction force and component of the weight of the crate is along the plane opposite to the applied force
So here we will have
[tex]mgsin\theta + F_f = F_{applied}[/tex]
now we know that
[tex]F_f = \mu F_n[/tex]
also we know that
[tex]F_n = mg cos\theta[/tex]
so we will have
[tex]F_f = \mu ( mg cos\theta)[/tex]
now we will have
[tex]mg sin\theta + \mu (mg cos\theta) = F_{applied}[/tex]
[tex](5)(9.81)sin30 + (0.5)(5)(9.81)cos30 = F_{applied}[/tex]
so we will have
[tex]F_{applied} = 45.8 N[/tex]
The maximum force that can be applied parallel to the plane without moving the 5.0-kg crate on a 30° incline, considering a static friction coefficient of 0.5, is 45.7N.
Explanation:To determine the maximum force that can be applied without moving the crate, we need to consider the effect of static friction, gravity, and the angle of the incline. The weight of the crate (W) is its mass (m) times acceleration due to gravity (g), which equals 5kg*9.8m/s² = 49N. However, this is the weight vertically down, so the force from gravity parallel to the incline is less, and we should multiply W by sin30⁰, getting 24.5N. The normal force (N) on the incline is W*cos30⁰, equal to 42.4N. Therefore, the maximum static friction force (fs) is coef. of static friction (μs) times N, which equals 0.5*42.4N = 21.2N. The max force applied to keep the crate from moving is the sum of the force of gravity and the static friction forces on the incline, which equals 24.5N + 21.2N = 45.7N.
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