Answer:
Part a)
[tex]m_2 = 4.9 \times 10^7 kg[/tex]
Part b)
[tex]q_1 = q_2 = 5312.6 C[/tex]
Explanation:
Part a)
As we know that both charge particles will exert equal and opposite force on each other
so here the force on both the charges will be equal in magnitude
so we will have
[tex]F = m_1a_1 = m_2a_2[/tex]
here we have
[tex]6.3 \times 10^7(7) = m_2(9)[/tex]
now we have
[tex]m_2 = 4.9 \times 10^7 kg[/tex]
Part b)
Now for the force between two charges we can say
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
now we have
[tex](6.3 \times 10^7)(7) = \frac{(9\times 10^9)q^2}{(24\times 10^3)^2}[/tex]
now we have
[tex]q_1 = q_2 = 5312.6 C[/tex]
Assume that you can heat water with perfect insulation (all the heat from combustion of ethanol is transferred to water). What is the volume of ethanol required to heat 200 mL of water by 10 °C? The density of ethanol is 0.78 g/mL, the specific heat capacity of water is 4.184 J∙g−1∙°C−1and the heat of combustion of ethanol is -1368 kJ/mol. Show your work.
Answer:
volumme =0.36 ml
Explanation:
total heat required can be obtained by using following formula
[tex]q= mC \Delta T[/tex].......(1)
where,
m - mass of water,
C - specific heat capacity of water and = 4.184 j g^{-1} °C
[tex]\Delta T[/tex] - total change in temperature. = 10°C
The density of water is 1 g/cc. hence, 200 mL of water is equal to 200 g
putting all value in the above equation (1)
q = 200*4.184* 10 ° = 8368 J.
Therefore total number of moles of ethanol required to supply 8368 J of heat is
[tex]\frac {8368}{1368000} = 0.006117 moles.[/tex]
The molar mass of ethanol is 46 g/mol.
The mass of ethanol required is 46* 0.006117 = 0.28138 g
The density of ethanol is 0.78 g/ml.
The volume of ethanol required is
[tex]\frac {0.28138}{0.78} = 0.36 ml[/tex]
To heat 200 mL of water by 10 °C, you would need approximately 0.00785 mL of ethanol. This is calculated by converting the volume of water to its mass, calculating the heat energy using the specific heat capacity of water, and then converting that energy to the amount of ethanol required using the heat of combustion of ethanol.
Explanation:To calculate the amount of ethanol required to heat 200 mL of water by 10 °C, we need to use the formula: Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, we need to convert the volume of water to its mass. Since the density of water is 1 g/mL, the mass of 200 mL of water is 200 g.
Next, we can calculate the heat energy using the formula: Q = mcΔT. The specific heat capacity of water is 4.184 J/g °C, the mass of water is 200 g, and the change in temperature is 10 °C. Plugging in these values, we get: Q = (200 g)(4.184 J/g °C)(10 °C) = 8376 J.
Finally, we can convert this heat energy to the amount of ethanol required using the heat of combustion of ethanol. The heat of combustion of ethanol is -1368 kJ/mol. To convert from J to kJ, we divide the heat energy by 1000 to get: 8376 J / 1000 = 8.376 kJ.
Now we can calculate the volume of ethanol. Since the density of ethanol is 0.78 g/mL, we can use the formula: V = m/ρ, where V is the volume, m is the mass, and ρ is the density. The mass of ethanol can be calculated by rearranging the formula: m = Q/ΔH, where Q is the heat energy and ΔH is the heat of combustion. Plugging in the values, we get: m = 8.376 kJ / -1368 kJ/mol = -0.00611 mol.
Now we can calculate the volume of ethanol using the formula: V = m/ρ. The mass of ethanol is -0.00611 mol and the density of ethanol is 0.78 g/mL. Plugging in these values, we get: V = -0.00611 mol / (0.78 g/mL) = -0.00785 mL.
Since volume cannot be negative, the volume of ethanol required to heat 200 mL of water by 10 °C is 0.00785 mL.
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Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of (c) a 2300 kg rhinoceros; (d) a 22 g song sparrow.
Explanation:
1. Force applied on an object is given by :
F = W = mg
(a) A 160 lb human being, F = 160 lb
g = acceleration due to gravity, g = 32 ft/s²
[tex]m=\dfrac{F}{g}[/tex]
[tex]m=\dfrac{160\ lb}{32\ ft/s^2}[/tex]
m = 5 kg
(b) A 1.9 lb cockatoo, F = 1.9 lb
[tex]m=\dfrac{F}{g}[/tex]
[tex]m=\dfrac{1.9\ lb}{32\ ft/s^2}[/tex]
m = 0.059 kg
2. (a) A 2300 kg rhinoceros, m = 2300 kg
[tex]W=2300\ kg\times 32\ ft/s^2=73600\ lb[/tex]
(b) A 22 g song sparrow, m = 22 g = 0.022 kg
[tex]W=0.022\ kg\times 32\ ft/s^2=0.704\ lb[/tex]
Hence, this is the required solution.
Final answer:
The mass of a 160 lb human being is 72.576 kg, and the mass of a 1.9 lb cockatoo is 0.86184 kg. A 2300 kg rhinoceros weighs 5060 lbs in English units, while a 22 g song sparrow weighs 0.0484 lbs.
Explanation:
To calculate the mass (in SI units) of a 160 lb human being, we use the conversion factor 1 lb = 0.4536 kg. So, the calculation is:160 lb x 0.4536 kg/lb = 72.576 kg
(a) The mass of a 160 lb human being in SI units is 72.576 kg.
For a 1.9 lb cockatoo, the conversion to kilograms is:1.9 lb x 0.4536 kg/lb
= 0.86184 kg
(b) The mass of a 1.9 lb cockatoo in SI units is 0.86184 kg
To convert the mass of a 2300 kg rhinoceros to weight in English units, knowing that 1 kg = 2.2 lbs (where weight in pounds is considered the gravitational force on the mass), the weight is calculated as:
2300 kg x 2.2 lbs/kg= 5060 lbs
(c) The weight of a 2300 kg rhinoceros in English units is 5060 lbs.
Finally, for converting a 22 g song sparrow to weight in English units:
22 g x (1 kg/1000 g) x (2.2 lbs/kg)= 0.0484 lbs
(d) The weight of a 22 g song sparrow in English units is 0.0484 lbs.
A large in-falling fragment could be tracked using radar. Explain how distance, speed, and the direction of motion, of the fragment could be determined. (15 points)
Answer:
speed, distance and direction of motion of the object can be determined by analyzing the radio wave.
Explanation:
We know that radar operates by transmitting radio waves to a destination and these waves are comes back to the receiver station. By Considering this transmission and receiver process, we can measure the distance, velocity and path of an object's movement.
Distance can be assessed by taking following consideration, the velocity of the waves is V. It can help to assess the time made for the waves to be emitted by the radar and felt by the receiver, let the time be t.
Therefore distance can be determine as D= v*t/2,
here 2 signifies that the distance travelled by the wave in either direction ( from transmitter to receiver and vice verse)
Using the source wave frequency, speed can be computed. In a specific frequency, the radar starts sending out the frequencies and the reflected wave will have a distinct frequency. The velocity can be determine by
[tex]v= (\Delta f/f)(c/2),[/tex]
where[tex]\Delta f[/tex]is the change in frequency and
c is the speed of light (the wave).
Direction can be determine by applying above principle. change in frequency is used to determine the direction in the following way:
When the frequency transition is very low, the object moves away from the radar and vice verse.
A 61 kg skier starts from rest at the top of a 1200 m long trail which drops a total of 227 m from top to bottom. At the bottom, the skier is moving 11 m/s. How much energy was dissipated by friction?
Answer:
energy dissipated = 132 kJ
Explanation:
mass = 61 kg
height drop = 227 m
velocity = 11 m/s
potential energy due to height drop from top to bottom
P.E. = m g h
P.E. = 61× 9.8× 227
P.E. = 135,700 J
kinetic energy = [tex]\frac{1}{2}mv^2[/tex]
= [tex]\frac{1}{2}\times 61 \times 11^2[/tex]
= 3690.5 J
energy dissipated = P.E - K.E.
= 135,700 J -3690.5 J
=132,009.5 J = 132 kJ
A batter hits a baseball into the air. The formula y=-16x^2+64x+5models the baseball’s height above the ground, y, in feet, x seconds after it is hit.When does the baseball reach its maximum height? What is that height?
Explanation:
A batter hits a baseball into the air. The formula that models the baseball’s height above the ground, y, in feet, x seconds after it is hit is given by :
[tex]y=-16x^2+64x+5[/tex].........(1)
(a) We need to find the maximum height reached by the baseball. The maximum height reached is calculated by differentiating equation (1) wrt x as :
[tex]\dfrac{dy}{dx}=0[/tex]
[tex]\dfrac{d(-16x^2+64x+5)}{dx}=0[/tex]
[tex]-32x+64=0[/tex]
x = 2 seconds
(b) The height of the ball is given by equation (1) and putting the value of x in it as :
[tex]y=-16(2)^2+64(2)+5[/tex]
y = 69 feets
So, at 2 seconds the baseball reaches its maximum height and its maximum height is 69 feets.
Answer:
wow
Explanation:
the pressure and absolute temperature of an ideal gas are both tripled, the volume is __________. not changed
increased nine times
decreased to one ninth
decreased to one third
Answer:
Not Changed
Explanation:
To know what happened with the volume you need to know the Ideal gas Law
[tex]\frac{P_{1}V_{1}}{T1} =\frac{P_{2}V_{2} }{T2}[/tex]
This law is a combination of the other four laws: Boyles's, Charles's, Avogadro's, and Guy-Lussac's.
The initial state is represented by P1, V1, T1 and the final by P2, V2, T2.
In this case:
[tex]T_{2} =3T_{1} \\P_{2} =3P_{1}[/tex]
Replacing on the equation
[tex]\frac{P_{1}V_{1}}{T1} =\frac{3P_{1}V_{2}}{3T_{1} }[/tex]
If we clear from the equation V2
[tex]\frac{P_{1}V_{1}3T_{1}}{T_{1} 3P_{1}} ={V_{2}}[/tex]
Then cancel both P1 and T1
[tex]\frac{3V_{1}}{3} =V_{2}[/tex]
You will found that
[tex]V_{1} =V_{2}[/tex]
Velocity of a wave is the. (a) Wavelength x frequency (b) Wave number x frequency (c) Time period x phase (d) None
Answer:
option (a)
Explanation:
Wavelength is defined as the distance traveled by the wave in one complete oscillation.
The number of oscillations completed in one second is called frequency.
The relation for the wave velocity is given by
wave velocity = frequency x wavelength
A 2.4-m-diameter merry-go-round with a mass of 270 kg is spinning at 20 rpm. John runs around the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John’s mass is 34 kg . Part A Part complete What is the merry-go-round's angular speed, in rpm, after John jumps on?
Answer:
23.98 rpm
Explanation:
d = diameter of merry-go-round = 2.4 m
r = radius of merry-go-round = (0.5) d = (0.5) (2.4) = 1.2 m
m = mass of merry-go-round = 270 kg
I = moment of inertia of merry-go-round
Moment of inertia of merry-go-round is given as
I = (0.5) m r² = (0.5) (270) (1.2)² = 194.4 kgm²
M = mass of john = 34 kg
Moment of inertia of merry-go-round and john together after jump is given as
I' = (0.5) m r² + M r² = 194.4 + (34) (1.2)² = 243.36 kgm²
w = final angular speed
w₀ = initial angular speed of merry-go-round = 20 rpm = 2.093 rad/s
v = speed of john before jump
using conservation of angular momentum
Mvr + I w₀ = I' w
(34) (5) (1.2) + (194.4) (2.093) = (243.36) w
w = 2.51 rad/s
w = 23.98 rpm
A circular coil of wire of radius 8 cm and 19 turns is placed with its plane perpendicular to a 5.6 T magnetic field, produced by an electromagnet. The current to the electromagnet is then switched off, causing the magnetic field to collapse to zero in a time of 744 ms. What is the emf generated in the coil while the magnetic field is collapsing, in units of volts?
Answer:
2.87 Volt
Explanation:
N = 19
change in magnetic field = 5.6 T
Time, dt = 744 ms = 0.744 s
Induced emf,
e = N × change in flux / time
e = 19 × 3.14 × 0.08 × 0.08 ×5.6/0.744
e = 2.87 Volt
9.96 kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the container. The container is now heated until the pressure is 600 kPa. Determine the temperature and total enthalpy when the heating is completed. Use data from the steam tables.
By using steam tables or similar thermodynamic property charts, you can determine the initial and final temperature and enthalpy of R-134a under specified pressure conditions in a rigid container. As the pressure changes with heating, you can find the corresponding changes in temperature and enthalpy.
Explanation:The subject matter of this question falls under the domain of thermodynamics in physics. It involves an element of heating, changes in pressure and the resultant alterations in temperature. Given the specifics, we can't directly provide a numerical solution in this setting as it requires the usage of refprop or other similar thermodynamic property database or the proper specific enthalpy and temperature charts for R-134a. However, conceptually, it can be handled via using steam tables (or more specifically refrigerant tables) for the substance R-134a. Initially, you lookup the properties using the given pressure and known mass and volume to calculate the initial temperature and enthalpy. Similarly, when the container is heated and the pressure raises to 600 kPa, you can again refer to the steam tables with this new pressure (and the volume which remains unchanged as the container is rigid) to determine the new temperature and enthalpy.
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What is the wavelength of an electron that has a kinetic energy of 0.50 MeV (relativistic)?
Answer:
The wavelength of electron is [tex]6.99\times 10^{-22}\ m[/tex]
Explanation:
The kinetic energy of the electron is, [tex]E=0.5\ MeV=0.5\times 10^6\ eV[/tex]
We need to find the wavelength of this electron. It can be calculated using the concept of DE-broglie wavelength as :
[tex]\lambda=\dfrac{h}{\sqrt{2mE} }[/tex]
h is Plank's constant
m is the mass of electron
[tex]\lambda=\dfrac{6.67\times 10^{-34}\ J-s}{\sqrt{2\times 9.1\times 10^{-31}\ kg\times 0.5\times 10^6\ eV} }[/tex]
[tex]\lambda=6.99\times 10^{-22}\ m[/tex]
So, the wavelength of electron is [tex]6.99\times 10^{-22}\ m[/tex]. Hence, this is the required solution.
The wavelength of an electron with a kinetic energy of 0.50 MeV (relativistic) is approximately 7.28 x 10^-12 m.
Explanation:The wavelength of an electron with a kinetic energy of 0.50 MeV can be calculated using the relativistic de Broglie equation:
λ = h/(m*c)
Where λ is the wavelength, h is Planck's constant (6.63 x 10^-34 Js), m is the mass of the electron (9.11 x 10^-31 kg), and c is the speed of light (3.00 x 10^8 m/s).
Substituting the values:
λ = (6.63 x 10^-34 Js)/((9.11 x 10^-31 kg)*(3.00 x 10^8 m/s))
λ ≈ 7.28 x 10^-12 m
Therefore, the wavelength of the electron is approximately 7.28 x 10^-12 m.
The sound intensity at the ear of passenger in a car with a damaged muffler is 8.0 × 10-3 W/m2. What is the intensity level of this sound in decibels? Use the threshold of hearing (1.0 × 10-12 W/m2) as the refere
Answer:
99 dB
Explanation:
We have given that sound intensity with a damaged muffler =8× [tex]10^{-3}[/tex]
we have to find the intensity level of the this sound in decibels
for calculating in decibels we have to use the formula
β=[tex]10log\frac{I}{I_0}[/tex]
=[tex]10log\frac{.008}{10^{-12}}[/tex]
=10 log8+10 log[tex]10^{9}[/tex]
=10 log8+90 log10
=10×0.9030+90
=99 dB
Final answer:
To calculate the intensity level of a sound in decibels, use the formula β = 10 log10(I/I0). Given the intensity of [tex]8.0 \times 10^{-3[/tex] threshold of hearing of [tex]10^{-12[/tex] the sound intensity level is approximately 99 decibels.
Explanation:
The student asks about calculating the intensity level of sound in decibels, given the sound intensity at the ear of a passenger in a car with a damaged muffler (8.0 × 10-3 W/m2). To find the intensity level in decibels (dB), we use the formula:
β = 10 log10(I/I0)
where:
β is the sound intensity level in decibels (dB)I is the sound intensity in watts per meter squared (W/m2)I0 = 10-12 W/m2 is the reference intensity, which is the threshold of hearingPlugging in the values:
β = 10 log10(8.0 × 10-3 / 10-12)
β = 10 log10(8.0 × 109)
β = 10 (log10(8) + log10(109))
β = 10 (0.903 + 9)
β = 10 × 9.903
β = 99.03 dB
The intensity level of the sound in the car with the damaged muffler is approximately 99 decibels.
An archer pulls her bowstring back 0.376 m by exerting a force that increases uniformly from zero to 251 N. (a) What is the equivalent spring constant of the bow? N/m (b) How much work does the archer do on the string in drawing the bow? J
Answer:
(A) 667.5 N/m
(B)
Explanation:
(A) Let the spring constant be k.
Using the formula F = kx
k = 251 / 0.376
K = 667.5 N/m
(B)
Work done
W = 0.5 × kx^2
W = 0.5 × 667.5 × 0.376 × 0.376
W = 47.2 J
Nitroglycerin flows through a pipe of diameter 3.0 cm at 2.0 m/s. If the diameter narrows to 0.5 cm, what will the velocity be?
Answer:
72 m/s
Explanation:
D1 = 3 cm, v1 = 2 m/s
D2 = 0.5 cm,
Let the velocity at narrow end be v2.
By use of equation of continuity
A1 v1 = A2 v2
3.14 × 3 × 3 × 2 = 3.14 × 0.5 ×0.5 × v2
v2 = 72 m/s
A ball having a mass of 200 g is released from rest at a height of 400 mm above a very large fixed metal surface. If the ball rebounds to a height of 325 mm above the surface, determine the coefficient of restitution between the ball and the surface.
Answer:
0.9
Explanation:
h = 400 mm, h' = 325 mm
Let the coefficient of restitution be e.
h' = e^2 x h
325 = e^2 x 400
e^2 = 0.8125
e = 0.9
The coefficient of restitution between the ball and the surface is 0.9.
The coefficient of restitution (e) between the ball and the surface can be determined using the formula:
[tex]\[ e = \sqrt{\frac{\text{height after collision}}{\text{height before collision}}} \][/tex]
First, we need to convert the heights from millimeters to meters for consistency, since the standard units for height in physics are meters.
The initial height (h_i) before the collision is 400 mm, which is equivalent to 0.4 m (since 1 m = 1000 mm).
The final height (h_f) after the collision is 325 mm, which is equivalent to 0.325 m.
Now, we can plug these values into the formula for the coefficient of restitution:
[tex]\[ e = \sqrt{\frac{0.325 \text{ m}}{0.4 \text{ m}}} \][/tex]
[tex]\[ e = \sqrt{\frac{325}{400}} \][/tex]
[tex]\[ e = \sqrt{0.8125} \][/tex]
[tex]\[ e = 0.9 \][/tex]
Therefore, the coefficient of restitution between the ball and the surface is 0.9. This means that the collision is relatively elastic, with the ball retaining a significant portion of its initial kinetic energy after the rebound.
An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion. At time t = 0.00 s the cart is released at the position x = +0.250 m. What is the position of the cart when t = 29.6 s?
Answer:
0.2447 m
Explanation:
Amplitude, A = 0.25 m, T = 5.67 s, t = 29.6 s, y = ?
The general equation of SHM is given by
y = A Sin wt
y = 0.25 Sin (2 x 3.14 t /5.67)
Put t = 29.6 s
y = 0.25 Sin (2 x 3.14 x 29.6 / 5.67)
y = 0.2447 m
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in parallel with the same battery, the current is 1.60 A. Part A What are the values of the two resistors?
Answer:
If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
Explanation:
R₁ = Resistance of first resistor
R₂ = Resistance of second resistor
V = Voltage of battery = 12 V
I = Current = 0.33 A (series)
I = Current = 1.6 A (parallel)
In series
[tex]\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2[/tex]
In parallel
[tex]\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}[/tex]
[tex]\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0[/tex]
Solving the above quadratic equation
[tex]\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}[/tex]
[tex]\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega[/tex]
∴ If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
The two resistors in question, when wired in a series, have a combined resistance of about 36.36 Ohms. When wired in parallel, they share a resistance of roughly 7.5 Ohms. The two resistor values would then most likely be around 28.86 Ohms and 7.5 Ohms.
Explanation:The nature of your question indicates a focus on the properties and calculations associated with electrical resistors in a circuit. These resistors can either be configured in a series or parallel connection, which will drastically change their behavior and derived readings.
From the context of your question, it is evident that a 12V battery is being used in conjunction with two resistors. In a series connection, the sum of the voltages across each resistor will equal the total voltage, effectively segmenting the 12V battery's output. In a parallel connection, each resistor would experience the full 12V impact, leading to larger current readings. Your values indicate a 0.33A current in a series scenario and a 1.60A current in a parallel situation.
Knowing this, we can apply Ohm's Law, which states Voltage (V) equals Current (I) times Resistance (R). In the series connection, the total resistance can be calculated as 12V divided by 0.33A equals approximately 36.36 Ohms. For the parallel connection, the total resistance would be 12V divided by 1.60A equals approximately 7.5 Ohms. Subtracting these two values, we can find that one of the resistors is around 28.86 Ohms while the other is roughly 7.5 Ohms. This would satisfy both the series and parallel conditions outlined in your question.
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Four objects are situated along the y axis as follows: a 2.00-kg object is at 13.00 m, a 3.00-kg object is at 12.50 m, a 2.50-kg object is at the origin, and a 4.00-kg object is at 20.500 m. Where is the center of mass of these objects?
The center of mass of these objects is located at approximately 12.652 meters on the y-axis.
We can find the center of mass (center of gravity) of these objects by calculating the weighted average of their positions along the y-axis. Here's how:
Consider Each Object:
We have four objects with masses m1, m2, m3, and m4 at positions y1, y2, y3, and y4 respectively.
Center of Mass Formula:
The center of mass coordinate (y_cm) is calculated as:
y_cm = (Σ(mi * yi)) / Σ(mi)
Σ (sigma) represents summation over all objects (i = 1 to 4 in this case).
mi is the mass of the i-th object.
yi is the y-axis position of the i-th object.
Apply the formula to our case:
y_cm = [(2.00 kg * 13.00 m) + (3.00 kg * 12.50 m) + (2.50 kg * 0.00 m) + (4.00 kg * 20.500 m)] / (2.00 kg + 3.00 kg + 2.50 kg + 4.00 kg)
Calculate:
y_cm = [26.00 kgm + 37.50 kgm + 0.00 kgm + 82.00 kgm] / 11.50 kg
y_cm ≈ 145.50 kg*m / 11.50 kg ≈ 12.652 m (rounded to four decimal places)
Therefore, the center of mass of these objects is located at approximately 12.652 meters on the y-axis.
A 1210 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating? (b) What is the resistance of the heating coil? (c) How much thermal energy is produced in 4.07 h?
Answer:
(a) 10.52 A
(b) 11 ohm
(c) 1.77 x 10^7 J
Explanation:
P = 1210 W, V = 115 V,
Let i be the current.
(a) Use the formula of power
P = V i
1210 = 115 x i
i = 10.52 A
(b) Let the resistance of heating coil is R.
Use the formula given below
V = i x R
R = V / i = 115 / 10.52 = 11 ohm
(c) Use the formula for energy
E = V x i x t
E = 115 x 10.52 x 4.07 x 60 x 60 = 1.77 x 10^7 J
a) How fast must a meter stick be moving if its length is observed to shrink to 0.6 m? b) With what speed will a clock have to be moving in order to run at a rate that is one-half the rate of a clock at rest?
Answer:
A)
0.8 c
B)
0.87 c
Explanation:
A)
L₀ = Original length of the meter stick = 1 m
L = Length observed = 0.6 m
[tex]v[/tex] = speed of the meter stick
Using the equation
[tex]L = L_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]0.6 = 1 \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]0.36 = 1 - \left ( \frac{v}{c} \right )^{2}[/tex]
[tex]v[/tex] = 0.8 c
B)
T₀ = Time of the clock at rest = t
T = Time of the clock at motion = (0.5) t
[tex]v[/tex] = speed of the clock
Using the equation
[tex]T = T_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]0.5 t = t \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]0.5 = \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
v = 0.87 c
A 120-V rms voltage at 1000 Hz is applied to an inductor, a 2.00-μF capacitor and a 100-Ω resistor, all in series. If the rms value of the current in this circuit is 0.680 A, what is the inductance of the inductor?
Answer:
The inductance of the inductor is 35.8 mH
Explanation:
Given that,
Voltage = 120-V
Frequency = 1000 Hz
Capacitor [tex]C= 2.00\mu F[/tex]
Current = 0.680 A
We need to calculate the inductance of the inductor
Using formula of current
[tex]I = \dfrac{V}{Z}[/tex]
[tex]Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}[/tex]
Put the value of Z into the formula
[tex]I=\dfrac{V}{\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}}[/tex]
Put the value into the formula
[tex]0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}[/tex]
[tex]L=35.8\ mH[/tex]
Hence, The inductance of the inductor is 35.8 mH
Answer:
Inductance,L:
"The property of the conductor or the solenoid to generate the electromotive force,emf due to the flow of current,I."
Unit: henry,H as it is equivalent to, kg.m².sec⁻².A⁻².
Explanation:
Data:
Voltage,v=120 v-rms,Frequency,f=1000 Hz,Capacitor, C=2.00 μF,Current,I=0.680 A,Solution:
We need to calculate the inductance, L of the solenoid inside a circuit,
I=v/z,Z=√R²+(Lω-1/Cω)²,putting the values I=V/√R²+(Lω-1/Cω)²,0.680=120/√(100)²+(L×2π×1000-1/2×10⁻⁶×2π×1000)²,L=35.8×10⁻³H, or L=35.8 mH.⇒AnswerA steel wire of length 4.7 m and cross section 3 x 103 m2 stretches by the same amount as a copper wire of length 3.5 m and cross section 4 x 10-5 m2 under a given load. What is the ratio of the Young's modulus of steel to that of copper? (a) 3.83 x 103 (b) 1.46 x 10-2 (d) 5.85 x 10-3 (c) 1.79 x 10-2 2.
Answer:
The ratio of the young's modulus of steel and copper is [tex]1.79\times10^{-2}[/tex]
(c) is correct option.
Explanation:
Given that,
Length of steel wire = 4.7 m
Cross section[tex]A = 3\times10^{-3}\ m^2[/tex]
Length of copper wire = 3.5 m
Cross section[tex]A = 4\times10^{-5}\ m^2[/tex]
We need to calculate the ratio of young's modulus of steel and copper
Using formula of young's modulus for steel wire
[tex]Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}[/tex]
[tex]Y_{s}=\dfrac{Fl_{s}}{A_{s}\Delta l}[/tex]....(I)
The young's modulus for copper wire
[tex]Y_{c}=\dfrac{Fl_{c}}{A_{c}\Delta l}[/tex]....(II)
From equation (I) and (II)
The ratio of the young's modulus of steel and copper
[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{\dfrac{Fl_{s}}{A_{s}\Delta l}}{\dfrac{Fl_{c}}{A_{c}\Delta l}}[/tex]
[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{A_{c}\times l_{s}}{A_{s}\times l_{c}}[/tex]
[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{4\times10^{-5}\times4.7}{3\times10^{-3}\times3.5}[/tex]
[tex]\dfrac{Y_{s}}{Y_{c}}=1.79\times10^{-2}[/tex]
Hence, The ratio of the young's modulus of steel and copper is [tex]1.79\times10^{-2}[/tex]
An electron enters a magnetic field of 0.43 T with a velocity perpendicular to the direction of the field. At what frequency does the electron traverse a circular path? ( m el = 9.11 × 10-31 kg, e = 1.60 × 10-19 C)
Answer:
Frequency, [tex]f=1.2\times 10^{10}\ Hz[/tex]
Explanation:
It is given that,
Magnetic field, B = 0.43 T
We need to find the frequency the electron traverse a circular path. It is also known as cyclotron frequency. It is given by :
[tex]f=\dfrac{qB}{2\pi m}[/tex]
[tex]f=\dfrac{1.6\times 10^{-19}\ C\times 0.43\ T}{2\pi \times 9.11\times 10^{-31}\ Kg}[/tex]
[tex]f=1.2\times 10^{10}\ Hz[/tex]
Hence, this is the required solution.
A capacitor has a charge of 4.6 μC. An E-field of 1.8 kV/mm is desired between the plates. There's no dielectric. What must be the area of each plate?
Answer:
[tex]A = 0.2875 m^2[/tex]
Explanation:
As we know that
[tex]Q = 4.6 \mu C[/tex]
E = 1.8 kV/mm
now we know that electric field between the plated of capacitor is given as
[tex]E = \frac{\sigma}{\epsilon_0}[/tex]
now we will have
[tex]1.8 \times 10^6 = \frac{\sigma}{\epsilon_0}[/tex]
[tex]\sigma = (1.8 \times 10^6)(8.85 \times 10^{-12})[/tex]
[tex]\sigma = 1.6 \times 10^{-5} C/m^2[/tex]
now we have
[tex]\sigma = \frac{Q}{A}[/tex]
now we have area of the plates of capacitor
[tex]A = \frac{Q}{\sigma}[/tex]
[tex]A = \frac{4.6 \times 10^{-6}}{1.6 \times 10^{-5}}[/tex]
[tex]A = 0.2875 m^2[/tex]
A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is 4.5 x 10 -3 Wb. What is the flux that passes through the circular loop?
Calculate the force required to double the length of a steel wire of area of cross section 5 x 10-5 m2? (Y=2x 1011N m2) (a) 10-5 N (b) 10-7 N n's D (c) 107 N (d) 105 N
Answer:
Option C is the correct answer.
Explanation:
We equation for elongation
[tex]\Delta L=\frac{PL}{AE}[/tex]
Here we need to find load required,
We need to double the wire, that is ΔL = 2L - L = L
A = 5 x 10⁻⁵ m²
E = 2 x 10¹¹ N/m²
Substituting
[tex]L=\frac{PL}{5\times 10^{-5}\times 2\times 10^{11}}\\\\P=10^7N[/tex]
Option C is the correct answer.
In a head-on collision, an alpha particle (Z = 2) of energy 8.80 MeV bounces straight back from a nucleus of charge 82.0 e. How close were the centers of the objects at closest approach?
The close were the centers of the objects at closest approach 2.7 x 10^-14 m
E =8.8 MeV = 8.8 x 1.6 x 10^-13 J
q = 2 e = 2 x 1.6 x 10^-19 C
Q = 82 e = 82 x 1.6 x 10^-19 C
Let d be the distance of closest approach
E = k Q q / d
Where, K = 9 x 10^9 Nm^2 / C^2
d = k Q q / E
d = (9 x 10^9 x 82 x 1.6 x 10^-19 x 2 x 1.6 x 10^-19) / (8.8 x 1.6 x 0^-13)
d = 2.7 x 10^-14 m
In jumping to block a shot, a volleyball player with a weight of 600 N generates an average vertical ground reaction force of 900 N for 0.37 seconds. What is the net vertical impulse that causes her velocity to increase in the upward direction?
Answer:
111 N
Explanation:
weight (downwards) = 600 N
reaction force (upwards) = 900 N
t = 0.37 s
Net force
F = reaction force - weight = 900 - 600 = 300 N
Impulse = force x time = 300 x 0.37 = 111 N
1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be supplied to the fan. Take the density of air to be 1.18 kg/m3.
Answer:
[tex]\dot{W} = 339.84 W[/tex]
Explanation:
given data:
flow Q = 9 m^{3}/s
velocity = 8 m/s
density of air = 1.18 kg/m^{3}
minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as
[tex]\dot{W} =\dot{m}\frac{V^{2}}{2}[/tex]
here [tex]\dot{m}[/tex]is mass flow rate and given as
[tex]\dot{m} = \rho*Q[/tex]
[tex]\dot{W} =\rho*Q\frac{V^{2}}{2}[/tex]
Putting all value to get minimum power
[tex]\dot{W} =1.18*9*\frac{8^{2}}{2}[/tex]
[tex]\dot{W} = 339.84 W[/tex]
The minimum power required to accelerate air to a velocity of 8 m/s at a rate of 9 m^3/s with a density of 1.18 kg/m^3 is roughly 339.84 Watts.
Explanation:
The subject of this question is physics, specifically the topic of power in mechanical systems. To find the minimum power required, we use the formula Power = 0.5 * density * volume flow rate * velocity^2. Plugging in the given values: Power = 0.5 * 1.18 kg/m^3 * 9 m^3/s * (8 m/s)^2, we get approximately 339.84 W. Therefore, the minimum power that must be supplied to the fan is roughly 339.84 Watts.
Learn more about Power Calculation here:https://brainly.com/question/31734672
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A mass of 100 kg is pulled by a 392 N force in the +X direction along a rough surface (uk=0.4) with uniform velocity v=20 m/s. What is the TOTAL work done by ALL forces on the object after s=10 m?
Answer:
The total work done will be zero.
Explanation:
Given that,
Mass = 100 kg
Force = 392 N
Velocity = 20 m/s
Distance s= 10 m
We need to calculate the work done
Using balance equation
The net force will be
[tex]F'=F-\mu mg[/tex]
[tex]F'=392-0.4\times100\times9.8[/tex]
[tex]F'=0[/tex]
The net force is zero.
Hence, The total work done will be zero by all forces on the object.