Answer:
a) 1.94 m/s
b) 120 m
Explanation:
Convert km/h to m/s:
6 km/h = 1.67 m/s
a) The final speed is found with Pythagorean theorem:
v = √((1.67 m/s)² + (1 m/s)²)
v = 1.94 m/s
b) The time it takes the boat to cross the river is:
t = (200 m) / (1.67 m/s)
t = 120 s
The displacement in the direction of the current is:
x = (1 m/s) (120 s)
x = 120 m
An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is 1.38 m/s, and its maximum acceleration is 6.83 m/s2. How much time elapses betwen an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum
Answer:
t = 0.31s
Explanation:
In order to calculate the time that the object takes to travel from the point with its maximum speed to the point with the maximum acceleration, you first use the following formulas, for the maximum speed and the maximum acceleration:
[tex]v_{max}=\omega A\\\\a_{max}=\omega^2A[/tex]
A: amplitude
v_max = 1.38m/s
a_max = 6.83m/s^2
w: angular frequency
From the previous equations you can obtain the angular frequency w.
You divide vmax and amax, and solve for w:
[tex]\frac{v_{max}}{a_{max}}=\frac{\omega A}{\omega^2 A}=\frac{1}{\omega}\\\\\omega=\frac{a_{max}}{v_{max}}=\frac{6.83m/s^2}{1.38m/s^2}=4.94\frac{rad}{s}[/tex]
Next, you take into account that the maximum speed is obtained when the object passes trough the equilibrium point, and the maximum acceleration for the maximum elongation, that is, the amplitude. In such a trajectory the time is T/4 being T the period.
You calculate the period by using the information about the angular frequency:
[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{4.94rad/s}=1.26s[/tex]
Then the required time is:
[tex]t=\frac{T}{4}=\frac{1.26s}{4}=0.31s[/tex]
In a circus act, a uniform board (length 3.00 m, mass 25.0 kg ) is suspended from a bungie-type rope at one end, and the other end rests on a concrete pillar. When a clown (mass 79.0 kg ) steps out halfway onto the board, the board tilts so the rope end is 30∘ from the horizontal and the rope stays vertical. Calculate the force exerted by the rope on the board with the clown on it.
Answer:
Force of Rope = 122.5 N
Force of Rope = 480.2N
Explanation:
given data
length = 3.00 m
mass = 25.0 kg
clown mass = 79.0 kg
angle = 30°
solution
we get here Force of Rope on with and without Clown that is
case (1) Without Clown
pivot would be on the concrete pillar so Force of Rope will be
Force of Rope × 3m = (25kg)×(9.8ms²)×(1.5m)
solve it and we get
Force of Rope = 122.5 N
and
case (2) With Clown
so here pivot is still on concrete pillar and clown is standing on the board middle and above the centre of mass so Force of Rope will be
Force of Rope × 3m = (25kg+73kg)×(9.8ms²)×(1.5m)
solve it and we get
Force of Rope = 480.2N
A rigid tank A of volume 0.6 m3 contains 5 kg air at 320K and the rigid tank B is 0.4 m3 with air at 600 kPa, 360 K. They are connected to a piston cylinder initially empty with closed valves. The pressure in the cylinder should be 800 kPa to float the piston. Now the valves are slowly opened and the entire process is adiabatic. The internal energy of the mixture at final state is:_____.
a. 229 k/kg.
b. 238 kJ/kg
c. 257 kg
d. cannot be determined.
Answer:
the internal energy of the mixture at final state = 238kJ/kg
Explanation:
Given
V= 0.6m³
m=5kg
R=0.287kJ/kg.K
T=320 K
from ideal gas equation
PV = nRT
where P is pressure, V is volume, n is number of mole, R is ideal gas constant , T is the temperature.
Recall, mole = mass/molar mass
attached is calculation of the question.
Find acceleration. Will give brainliest!
Answer:
16200 km/s
270 km/min
4.5 km/h
Explanation:
Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)
Simply plug in our known variables and solve:
a = (45.0 - 0)/10
a = 45.0/10
a = 4.5 km/h
Answer:
[tex]\boxed{\mathrm{4.5 \: kmph/s \: or \: 1.25 \: m/s^2 }}[/tex]
Explanation:
[tex]\displaystyle \mathrm{acceleration = \frac{change \: in \: velocity}{time \: taken}}[/tex]
[tex]\displaystyle \mathrm{a = \frac{v - u}{t}}[/tex]
[tex]\displaystyle \mathrm{v=final \: velocity}\\\displaystyle \mathrm{u=initial \: velocity}[/tex]
[tex]\displaystyle \mathrm{a = \frac{45- 0}{10}}[/tex]
[tex]\displaystyle \mathrm{a = \frac{45}{10}}[/tex]
[tex]\displaystyle \mathrm{a = 4.5}[/tex]
[tex]\mathrm{4.5 \: kmph/s = 1.25 \: m/s^2 }[/tex]
The voltage between the cathode and the screen of a television set is 30 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed (m/s) just before it hits the screen
Answer:
The speed is [tex]v =10.27 *10^{7} \ m/s[/tex]
Explanation:
From the question we are told that
The voltage is [tex]V = 30 kV = 30*10^{3} V[/tex]
The initial velocity of the electron is [tex]u = 0 \ m/s[/tex]
Generally according to the law of energy conservation
Electric potential Energy = Kinetic energy of the electron
So
[tex]PE = KE[/tex]
Where
[tex]KE = \frac{1}{2} * m* v^2[/tex]
Here m is the mass of the electron with a value of [tex]m = 9.11 *10^{-31} \ kg[/tex]
and
[tex]PE = e * V[/tex]
Here e is the charge on the electron with a value [tex]e = 1.60 *10^{-19} \ C[/tex]
=> [tex]e * V = \frac{1}{2} * m * v^2[/tex]
=> [tex]v = \sqrt{ \frac{2 * e * V}{m} }[/tex]
substituting values
[tex]v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }[/tex]
[tex]v =10.27 *10^{7} \ m/s[/tex]
The lower the value of the coefficient of friction, the____the resistance to sliding
Answer: lower
There are a number of factors that can affect the coefficient of friction, including surface conditions.
Values of the coefficient of sliding friction can be a good reference for specific combinations of materials. The frictional force and normal reaction are directly proportion but an increase or decrease in coefficient of friction will cause an increase or decrease in the resistance of sliding respectively
An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answer:
The new voltage between the plates of the capacitor is 18 V
Explanation:
The charge on parallel plate capacitor is calculated as;
q = CV
Where;
V is the battery voltage
C is the capacitance of the capacitor, calculated as;
[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]
[tex]q = \frac{\epsilon _0A V}{d}[/tex]
where;
ε₀ is permittivity of free space
A is the area of the capacitor
d is the space between the parallel plate capacitors
If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;
[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]
Therefore, the new voltage between the plates of the capacitor is 18 V
A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What are the magnitude and direction of the net force on the car
Answer:
9,375
Explanation:
Data provided
The mass of the car m = 1500 Kg.
The diameter of the circular track D = 200 m.
For the computation of magnitude and direction of the net force on the car first we need to find out the radius of the circular path which is shown below:-
The radius of the circular path is
[tex]R = \frac{D}{2}[/tex]
[tex]= \frac{200}{2}[/tex]
= 100 m
after the radius of the circular path we can find the magnitude of the centripetal force with the help of below formula
[tex]Force F = \frac{mv^2}{R}[/tex]
[tex]= \frac{1500\times (25)^2}{100}[/tex]
= 9,375
Therefore for computing the magnitude of the centripetal force we simply applied the above formula.
If a sample of 346 swimmers is taken from a population of 460 swimmers,
the population mean, w, is the mean of how many swimmers' times?
Answer:
It is the mean of 460 swimmers
Explanation:
In this question, we are concerned with knowing the mean of the population w
Now, according to the question at hand, we have a total population of 460 swimmers and we have taken out 346 swimmers for a study
The population mean in this case is simply the mean of the swimming times of the 460 swimmers
There is another related thing here called the sample mean. For the sample mean, we only make a reference to the mean of the 346 swimmers who were taken out from the population to conduct a separate study
So conclusively, the population mean w is simply the mean of the total 460 swimmers
A lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is
The lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is: in the near-ultraviolet.
What is diffraction?The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.
Longer wavelengths of light are diffracted at a greater angle than shorter ones, with the amount of diffraction being dependent on the wavelength of the light. Hence, among the light waves of the visible, near-infrared, and near-ultraviolet range, near-ultraviolet waves have the shortest wavelengths. So, The best resolution of this lens from a diffraction standpoint is in the near-ultraviolet, where diffraction is minimum.
Learn more about diffraction here:
https://brainly.com/question/11176463
#SPJ5
Nerve impulses in a human body travel at a speed of about 100 m/s. Suppose a woman accidentally steps barefoot on a thumbtack. About how much time does it take the nerve impulse to travel from the foot to the brain (in s)
A particle moves along a straight line with the acceleration a = (12t - 3t ^ 1/2) feet / s ^ 2, where t is in seconds. Determine your speed and position as a function of time. When t = 0, v = 0 and s = 15 feet.
Answer:
v = 6t² − 2t^³/₂
s = 2t³ − ⅘t^⁵/₂ + 15
Explanation:
a = 12t − 3t^½
Integrate to find velocity.
v = ∫ a dt
v = ∫ (12t − 3t^½) dt
v = 6t² − 2t^³/₂ + C
Use initial condition to find C.
0 = 6(0)² − 2(0)^³/₂ + C
C = 0
v = 6t² − 2t^³/₂
Integrate to find position.
s = ∫ v dt
s = ∫ (6t² − 2t^³/₂) dt
s = 2t³ − ⅘t^⁵/₂ + C
Use initial condition to find C.
15 = 2(0)³ − ⅘(0)^⁵/₂ + C
15 = C
s = 2t³ − ⅘t^⁵/₂ + 15
A box with an initial speed of 15 m/s slides along a surface where the coefficient of sliding friction is 0.45. How long does it take for the block to come to rest
Answer:
t = 3.4 s
The box will come to rest in 3.4 s
Explanation:
For the block to come to rest, the friction force must become equal to the unbalanced force. Therefore:
Unbalanced Force = Frictional Force
but,
Unbalanced Force = ma
Frictional Force = μR = μW = μmg
Therefore,
ma = μmg
a = μg
where,
a = acceleration of box = ?
μ = coefficient of sliding friction = 0.45
g = 9.8 m/s²
Therefore,
a = (0.45)(9.8 m/s²)
a = -4.41 m/s² (negative sign due to deceleration)
Now, for the time to stop, we use first equation of motion:
Vf = Vi + at
where,
Vf = Final Speed = 0 m/s (since box stops at last)
Vi = Initial Speed = 15 m/s
t = time to stop = ?
Therefore,
0 m/s = 15 m/s + (-4.41 m/s²)t
(-15 m/s)/(-4.41 m/s²) = t
t = 3.4 s
The box will come to rest in 3.4 s
A 100 kg lead block is submerged in 2 meters of salt water, the density of which is 1096 kg / m3. Estimate the value of the hydrostatic pressure.
Answer:
21,920 Pascals
Explanation:
P = ρgh
P = (1096 kg/m³) (10 m/s²) (2 m)
P = 21,920 Pa
If 62.9 cm of copper wire (diameter = 1.15 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at the constant rate of 8.43 mT/s, at what rate is thermal energy generated in the loop?
Answer:
The answer is "[tex]\bold{7.30 \times 10^{-6}}[/tex]"
Explanation:
length of the copper wire:
L= 62.9 cm
r is the radius of the loop then:
[tex]r=\frac{L}{2 \pi}\\[/tex]
[tex]=\frac{62.9}{2\times 3.14}\\\\=\frac{62.9}{6.28}\\\\=10.01\\[/tex]
area of the loop Is:
[tex]A_L= \pi r^2[/tex]
[tex]=100.2001\times 3.14\\\\=314.628[/tex]
change in magnetic field is:
[tex]=\frac{dB}{dt} \\\\ = 0.01\ \frac{T}{s}[/tex]
then the induced emf is: [tex]e = A_L \times \frac{dB}{dt}[/tex]
[tex]=314.628 \times 0.01\\\\=3.14\times 10^{-5}V[/tex]
resistivity of the copper wire is: [tex]\rho =[/tex] 1.69 × 10-8Ω·m
diameter d = 1.15mm
radius (r) = 0.5mm
[tex]= 0.5 \times 10^{-3} \ m[/tex]
hence the resistance of the wire is:
[tex]R=\frac{\rho L}{\pi r^2}\\[/tex]
[tex]=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times (0.5 \times 10^{-3})^2}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.5 \times 0.5 \times 10^{-6}}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.25 \times 10^{-6}}\\\\=135.41 \times 10^{-2}\\=1.35\times 10^{-4}\\[/tex]
Power:
[tex]P=\frac{e^2}{R}[/tex]
[tex]=\frac{3.14\times 10^{-5}\times 3.14\times 10^{-5}}{1.35 \times 10^{-4}}\\\\=7.30 \times 10^{-6}[/tex]
The final answer is: [tex]\boxed{7.30 \times 10^{-6} \ W}[/tex]
Julie is playing with a toy car and is pushing it around on the floor. The little car has a mass of 6.3 g. The car has a velocity of 2.5 m/s. What is the car's momentum?
Answer:
Momentum of the car = [tex]1.575\times 10^{-2}[/tex] kg meter per second
Explanation:
Julie is playing with a car which has mass = 6.3 g = [tex]6.3\times 10^{-3}[/tex] kg
Velocity of the car is 2.5 meter per second
Since formula to calculate the momentum of an object is,
p = mv
Where, p = momentum of the object
m = mass of the object
v = velocity of the object
By substituting these values in the formula,
p = [tex](6.3\times 10^{-3})\times 2.5[/tex]
= [tex]1.575\times 10^{-2}[/tex] Kg meter per second
Therefore, momentum of the car will be [tex]1.575\times 10^{-2}[/tex] Kg meter per second.
Which of the following statements is accurate? A) Compressions and rarefactions occur throughout a transverse wave. B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave. C) Sound waves passing through the air will do so as transverse waves, which vibrate vertically and still retain their horizontal positions. D) Amplitude of longitudinal waves is measured at right angles to the direction of the travel of the wave and represents the maximum distance the molecule has moved from its normal position.
Answer:
B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave
Answer:
B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave.
Explanation: hope this helps ;)
Which of the following changes will increase the frequency of the lowest frequency standing sound wave on a stretching string?Choose all that apply.A. Replacing the string with a thicker stringB. Plucking the string harderC. Doubling the length of the string
Answer:
A, C
Explanation:
Since the frequency is inversely proportional to the length of a string, then I want to increase the frequency of the lowest
A. Replacing the string with a thicker string.
Thicker strings have more density. The more density the string has, the lower the sound.
Mathematically, we can see the proportionality (direct and inverse) by looking at those formulas for Frequency and Speed, when combined:
For:
[tex]f=\frac{v}{\lambda}[/tex]
[tex]f=\frac{v}{\lambda}*\sqrt{\frac{T}{D} }[/tex]
See above, how density (D) and [tex](\lambda)[/tex] wave length are inversely proportional.
C. Doubling the length of the string.
Because the length of the string is inversely proportional to the frequency.
The longer the string, the lower the frequency.
So, if we double string, we'll hear lower sounds in any string instrument
--
In short, for A, and C We can justify both since length and density are inversely proportional to the Frequency, we need longer or thicker string.
A 1,470-N force pushes a 500-kg piano up along a ramp. What is the work done by the 1,470-N pushing force on the piano as it moves 10 m up the ramp
Answer:
W = 14700 J
Explanation:
This is an exercise on Newton's second law.
To solve it we must fix a coordinate system, the most common is an axis parallel to the ramp and the other perpendicular axis, we write Newton's second law
Y Axis . Perpendicular to the ramp
N - Wy = 0
X axis. Parallel to the ramp, we assume it is positive when the ramp is going up
F - Wx = m a
in this case F = 1470 N and it is parallel to the plane.
Work is defined by
W = F .d
boldface indicates vectors
W = F d cos θ
let's calculate
W = 1470 10 cos 0
W = 14700 J
A solenoid with 500 turns, 0.10 m long, carrying a current of 4.0 A and with a radius of 10-2 m will have what strength magnetic field at its center
Answer:
B = 0.025T
Explanation:
In order to calculate the strength of the magnetic field at the center of the solenoid, you use the following formula:
[tex]B=\frac{\mu N i}{L}[/tex] (1)
μ: magnetic permeability of vacuum = 4π*10^-7 T/A
N: turns of the solenoid = 500
i: current = 4.0A
L: length of the solenoid = 0.10m
You replace the values of the parameters in the equation (1):
[tex]B=\frac{(4\pi*10^{-7}T/A)(500)(4.0A)}{0.10m}=0.025T[/tex]
The strength of the magnetic field at the center of the solenoid = 0.025T
Answer:
Magnetic field strength at the center is 2.51x10^-2T
Explanation:
Pls see attached file for step by step calculation
3. Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.
Answer:
Object 2 has the larger drag coefficient
Explanation:
The drag force, D, is given by the equation:
[tex]D = 0.5 c \rho A v^2[/tex]
Object 1 has twice the diameter of object 2.
If [tex]d_2 = d[/tex]
[tex]d_1 = 2d[/tex]
Area of object 2, [tex]A_2 = \frac{\pi d^2 }{4}[/tex]
Area of object 1:
[tex]A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2[/tex]
Since all other parameters are still the same except the drag coefficient:
For object 1:
[tex]D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2[/tex]
For object 2:
[tex]D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2[/tex]
Since the drag force for the two objects are the same:
[tex]0.5 c_1 \rho (\pi d^2) v^2 = 0.5 c_2 \rho (\pi d^2/4) v^2\\4c_1 = c_2[/tex]
Obviously from the equation above, c₂ is larger than c₁, this means that object 2 has the larger drag coefficient
What is the length (in m) of a tube that has a fundamental frequency of 108 Hz and a first overtone of 216 Hz if the speed of sound is 340 m/s?
Answer:
Length of a tube = 1.574 m
Explanation:
Given:
Fundamental frequency (f1) = 108 Hz
First overtone (f2) = 216 Hz
Speed of sound (v) = 340 m/s
Find:
Length of a tube
Computation:
We know that,
f = v / λ
f = nv / 2L [n = number 1,2,3]
So,
f1 = 1(340) / 2L
f1 = 170 / L
L = 170 / 108 = 1.574 m
f2 = 2(340) / 2L
L = 340 / 216
L = 1.574 m
A 750 kg car is moving at 20.0 m/s at a height of 5.0 m above the bottom of a hill when it runs out of gas. From there, the car coasts. a. Ignoring frictional forces and air resistance, what is the car’s kinetic energy and velocity at the bottom of the hill
Answer:
Explanation:
Kinetic energy at the height = 1/2 m v²
= 1/2 x 750 x 20²
= 150000 J
Its potential energy = mgh
= 750 x 9.8 x 5
=36750 J
Total energy = 186750 J
Its total kinetic energy will be equal to 186750 J , according to conservation of mechanical energy
If v be its velocity at the bottom
1/2 m v² = 186750
v = √498
= 22.31 m /s
When a nerve cell fires, charge is transferred across the cell membrane to change the cell's potential from negative to positive. For a typical nerve cell, 9.2pC of charge flows in a time of 0.52ms .What is the average current through the cell membrane?
Answer:
The average current will be "17.69 nA".
Explanation:
The given values are:
Charge,
q = 9.2 pC
Time,
t = 0.52ms
The equivalent circuit of the cell surface is provided by:
⇒ [tex]i_{avg}=\frac{charge}{t}[/tex]
Or,
⇒ [tex]i_{avg}=\frac{q}{t}[/tex]
On substituting the given values, we get
⇒ [tex]=\frac{9.2\times 10^{-12}}{0.52\times 10^{-3}}[/tex]
⇒ [tex]=17.69^{-9}[/tex]
⇒ [tex]=17.69 \ nA[/tex]
A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m=0.249 kg , are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is ????=0.929 kg·m2 ?
Answer:
The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]
Explanation:
The Diagram for this question is gotten from the first uploaded image
From the question we are told that
The mass of the rod is [tex]M = 3.41 \ kg[/tex]
The mass of each small bodies is [tex]m = 0.249 \ kg[/tex]
The moment of inertia of the three-body system with respect to the described axis is [tex]I = 0.929 \ kg \cdot m^2[/tex]
The length of the rod is L
Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as
[tex]I = I_r + 2 I_m[/tex]
Where [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as
[tex]I_r = \frac{ML^2 }{12}[/tex]
And [tex]I_m[/tex] the moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented as
[tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]
Thus [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]
Hence
[tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]
=> [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]
substituting vales we have
[tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]
[tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]
[tex]L = 1.5077 \ m[/tex]
Copper wire of diameter 0.289 cm is used to connect a set of appliances at 120 V, which draw 1850 W of power total. The resistivity of copper is 1.68×10−8Ω⋅m.
A. What power is wasted in 26.0 m of this wire?
B. What is your answer if wire of diameter 0.417 cm is used?
Answer:
(a) The power wasted for 0.289 cm wire diameter is 15.93 W
(b) The power wasted for 0.417 cm wire diameter is 7.61 W
Explanation:
Given;
diameter of the wire, d = 0.289 cm = 0.00289 m
voltage of the wire, V = 120 V
Power drawn, P = 1850 W
The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m
Area of the wire;
A = πd²/4
A = (π x 0.00289²) / 4
A = 6.561 x 10⁻⁶ m²
(a) At 26 m of this wire, the resistance of the is
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 6.561 x 10⁻⁶
R = 0.067 Ω
Current in the wire is calculated as;
P = IV
I = P / V
I = 1850 / 120
I = 15.417 A
Power wasted = I²R
Power wasted = (15.417²)(0.067)
Power wasted = 15.93 W
(b) when a diameter of 0.417 cm is used instead;
d = 0.417 cm = 0.00417 m
A = πd²/4
A = (π x 0.00417²) / 4
A = 1.366 x 10⁻⁵ m²
Resistance of the wire at 26 m length of wire and 1.366 x 10⁻⁵ m² area;
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 1.366 x 10⁻⁵
R = 0.032 Ω
Power wasted = I²R
Power wasted = (15.417²)(0.032)
Power wasted = 7.61 W
A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region is 5.40 mlong and reduces the toboggan's speed by 1.20 m/s .
a) What average friction force did the rough region exert on the toboggan?
b) By what percent did the rough region reduce the toboggan's kinetic energy?
c) By what percent did the rough region reduce the toboggan's speed?
Answer:
a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.
Explanation:
a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:
[tex]K_{1} = K_{2} + W_{f}[/tex]
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] are the initial and final translational kinetic energies of the tobbogan, measured in joules.
[tex]W_{f}[/tex] - Dissipated work due to friction, measured in joules.
By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:
[tex]f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})[/tex]
Where:
[tex]f[/tex] - Friction force, measured in newtons.
[tex]\Delta s[/tex] - Distance travelled by the toboggan in the rough region, measured in meters.
[tex]m[/tex] - Mass of the toboggan, measured in kilograms.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the toboggan, measured in meters per second.
The friction force is cleared:
[tex]f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}[/tex]
If [tex]m = 375\,kg[/tex], [tex]v_{1} = 4.50\,\frac{m}{s}[/tex], [tex]v_{2} = 1.20\,\frac{m}{s}[/tex] and [tex]\Delta s = 5.40 \,m[/tex], then:
[tex]f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}[/tex]
[tex]f = 653.125\,N[/tex]
The average friction force exerted on the toboggan is 653.125 newtons.
b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:
[tex]\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%[/tex]
[tex]\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%[/tex]
[tex]\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%[/tex]
[tex]\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%[/tex]
[tex]\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%[/tex]
If [tex]v_{1} = 4.50\,\frac{m}{s}[/tex] and [tex]v_{2} = 1.20\,\frac{m}{s}[/tex], then:
[tex]\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%[/tex]
[tex]\%K_{loss} = 92.889\,\%[/tex]
The rough region reduced the kinetic energy of the toboggan in 92.889 %.
c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:
[tex]\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%[/tex]
[tex]\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%[/tex]
If [tex]v_{1} = 4.50\,\frac{m}{s}[/tex] and [tex]v_{2} = 1.20\,\frac{m}{s}[/tex], then:
[tex]\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%[/tex]
[tex]\%v_{loss} = 73.333\,\%[/tex]
The speed of the toboggan is reduced in 73.333 %.
The average frictional force exerted on the toboggan by the rough surface is 661.5 N.
The percentage of the toboggan kinetic energy reduction is 7.11%.
The percentage of the toboggan speed reduction is 26.67%.
The given parameters;
mass of the toboggan, m = 375 kginitial speed of the toboggan, u = 4.5 m/slength of the rough region, d = 5.4 mfinal speed of the toboggan, v = 1.2 m/sThe normal force on the toboggan is calculated as follows;
Fₙ = mg
Fₙ = 375 x 9.8 = 3675 N
The acceleration of the toboggan is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\a = \frac{v^2 - u^2 }{2s} \\\\a = \frac{(1.2)^2 - (4.5)^2 }{2(5.4)}\\\\a = -1.74 \ m/s^2[/tex]
The coefficient of friction is calculated as follows;
[tex]\mu_k = \frac{a}{g} \\\\\mu_k = \frac{1.74}{9.8} \\\\\mu_k = 0.18[/tex]
The average frictional force exerted on the toboggan by the rough surface;
[tex]F_k = \mu_k F_n\\\\F_k = 0.18 \times 3675\\\\F_k = 661.5 \ N[/tex]
The initial kinetic energy of the toboggan is calculated as follows;
[tex]K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 375\times 4.5^2\\\\K.E_i = 3,796.88 \ J[/tex]
The final kinetic energy of the toboggan is calculated as follows;
[tex]K.E_f = \frac{1}{2} mv^2\\\\K.E_f = \frac{1}{2} \times 375\times 1.2^2\\\\K.E_f = 270 \ J[/tex]
The percentage of the toboggan kinetic energy reduction is calculated as follows;
[tex]\frac{K.E_f}{K.E_i} \times 100\% = \frac{270}{3796.88} \times 100\% = 7.11 \%[/tex]
The percentage of the toboggan speed reduction is calculated as follows;
[tex]\frac{1.2}{4.5} \times 100\% = 26.67 \%[/tex]
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Two Earth satellites, A and B, each of mass m = 980 kg , are launched into circular orbits around the Earth's center. Satellite A orbits at an altitude of 4100 km , and satellite B orbits at an altitude of 12100 km The radius of Earth RE is 6370 km.
(a) What is the ratio of the potential energy of satellite B to that of satellite A, in orbit?
(b) What is the ratio of the kinetic energy of satellite B to that of satellite A, in orbit?
(c) Which satellite has the greater total energy if each has a mass of 14.6 kg?
(d) By how much?
Answer:
Do u have a picture of the graph?
Explanation:
I can solve it with refraction
a block of wood is pulled by a horizontal string across a rough surface at a constant velocity with a force of 20N. the coefficient of kinetic friction between the surfaces is 0.3 the force of the friction is
Answer:
6 N
Explanation:
From the laws of friction
F = ¶R = 0.3 × 20 = 6 N
The force of friction opposing the block's motion is 6 N.
The given parameters;
force applied on the block, F = 20 Ncoefficient of kinetic friction = 0.3The force of friction which opposes the motion of the block is obtained by applying Newton's second law of motion.
F = ma
Fₓ = μF
Substitute the given parameters to calculate the frictional force on the object.
Fₓ = 0.3 x 20
Fₓ = 6 N
Thus, the force of friction opposing the block's motion is 6 N.
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In a two-slit experiment, monochromatic coherent light of wavelength 500 nm passes through a pair of slits separated by 1.30 x 10-5 m. At what angle away from the centerline does the first bright fringe occur
Answer:
2.20°
Explanation:
For the central bright spot, we will use the constructive pattern for a double slit interference,
[tex]m\times w = d \times Sin\beta[/tex]
where w indicates the wavelength
and [tex]\beta[/tex] indicates the angle between the bright spot and center line.
now we will use the given values,
1 × 500 × 10^-9 = 1.3 × 10^-5 × Sin [tex]\beta[/tex]
Solving for [tex]\beta[/tex],
[tex]\beta[/tex] = 2.204° ~ 2.20°
Therefore the correct answer is 2.20°