The area of a square is given by:
A = s²
A is the square's area
s is the length of one of the square's sides
Let us take the derivative of both sides of the equation with respect to time t in order to determine a formula for finding the rate of change of the square's area over time:
d[A]/dt = d[s²]/dt
The chain rule says to take the derivative of s² with respect to s then multiply the result by ds/dt
dA/dt = 2s(ds/dt)
A) Given values:
s = 14m
ds/dt = 3m/s
Plug in these values and solve for dA/dt:
dA/dt = 2(14)(3)
dA/dt = 84m²/s
B) Given values:
s = 25m
ds/dt = 3m/s
Plug in these values and solve for dA/dt:
dA/dt = 2(25)(3)
dA/dt = 150m²/s
When the side of the square is 14 m, the rate at which the area is changing is 84 m²/s.
When the side of the square is 25 m, the rate at which the area is changing is 150 m²/s.
The given parameters;
rate at which the side of the square is increasing, [tex]\frac{dl}{dt}[/tex] = 3 m/sThe area of the square is calculated as;
A = L²
The change in the area is calculated as;
[tex]\frac{dA}{dt} = 2l\frac{dl}{dt}[/tex]
When the side of the square is 14 m, the rate at which the area is changing is calculated as;
[tex]\frac{dA}{dt} = 2l \frac{dl}{dt} \\\\\frac{dA}{dt} = 2 \times l \times \frac{dl}{dt}\\\\\frac{dA}{dt} = 2 \times 14 \times 3\\\\\frac{dA}{dt} = 84 \ m^2/s[/tex]
When the side of the square is 25 m, the rate at which the area is changing is calculated as;
[tex]\frac{dA}{dt} = 2l \frac{dl}{dt} \\\\\frac{dA}{dt} = 2 \times l \times \frac{dl}{dt}\\\\\frac{dA}{dt} = 2 \times 25 \times 3\\\\\frac{dA}{dt} = 150 \ m^2/s[/tex]
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Calculate the approximate radius of the 31P nucleus (assume r0 = 1.3x10-15 m).
Answer:
4.08 x 10⁻¹⁵ m
Explanation:
r₀ = constant of proportionality = 1.3 x 10⁻¹⁵ m
r = radius of the nucleus 31P = ?
A = mass number of the nucleus 31P = 31
Radius of the nucleus is given as
[tex]r=r_{o}A^{\frac{1}{3}}[/tex]
Inserting the values
[tex]r=(1.3\times 10^{-15})(31)^{\frac{1}{3}}[/tex]
[tex]r=(1.3\times 10^{-15})(3.14)[/tex]
r = 4.08 x 10⁻¹⁵ m
A 64.9 kg sprinter starts a race with an acceleration of 3.89 m/s2. She keeps this acceleration for 17 m and then maintains the velocity for the remainder of the 100-m dash, what will be her time for the race?
Answer:
Time of race = 10.18 s
Explanation:
She keeps this acceleration for 17 m and then maintains the velocity for the remainder of the 100-m dash
Time to travel 17 m can be calculated
s = ut + 0.5at²
17 = 0 x t + 0.5 x 3.89 x t²
t = 2.96 s
Velocity after 2.96 seconds
v = 3.89 x 2.96 = 11.50 m/s
Remaining distance = 100 - 17 = 83 m
Time required to cover 83 m with a speed of 11.50 m/s
[tex]t=\frac{83}{11.50}=7.22s[/tex]
Time of race = 2.96 + 7.22 = 10.18 s
A laboratory uses a laser with a light wavelength of 910nm. If the lab turns the laser on for 20 picoseconds, how many wavelengths can be found in the resulting wave pulse?
Answer:
6593.4
Explanation:
wavelength, λ = 910 nm = 910 x 10^-9 m
Speed of laser = 3 x 10^8 m/s
t = 20 pico seconds
Distance traveled by laser in this time, d = c x t
d = 3 x 10^8 x 20 x 10^-12 = 6 x 10^-3 m
number of wavelengths, n = d / λ
n = (6 x 10^-3) / (910 x 10^-9)
n = 6593.4
A solid cylinder of mass 7 kg and radius 0.9 m starts from rest at the top of a 20º incline. It is released and rolls without slipping to the bottom of the incline. Assume g = 9.81 m/s2. If the difference in height between the top of the incline and the bottom is 2.3 m, the total energy of the cylinder at the bottom is:
Answer:
157.8 J
Explanation:
m = mass of the cylinder = 7 kg
h = height difference in top and bottom of the incline = 2.3 m
g = acceleration due to gravity = 9.8 m/s²
TE = Total Energy at the bottom
PE = Gravitational potential energy at the top
Using conservation of energy
Total Energy at the bottom = Gravitational potential energy at the top
TE = PE
TE = m g h
TE = (7) (9.8) (2.3)
TE = 157.8 J
A vector in the xy plane has a magnitude of 25 m and x component of 12 m. What is its angle with the positive x axis?
Final answer:
The angle a vector makes with the positive x-axis can be found using the cosine ratio; the angle is approximately 58.2 degrees for the vector with a magnitude of 25 m and an x-component of 12 m.
Explanation:
To determine the angle a vector makes with the positive x-axis, we can use trigonometry since we have the magnitude of the vector and the length of the x-component. Specifically, the cosine of the angle (θ) that the vector makes with the x-axis is equal to the x-component divided by the magnitude of the vector.
Here is the formula:
cos(θ) = (x-component) / (magnitude)
By substituting the given values:
cos(θ) = 12 m / 25 m
Now calculate the angle:
θ = cos⁻¹(12/25)
θ = 58.2° (approx.)
Therefore, the vector makes an angle of approximately 58.2 degrees with the positive x-axis.
The center of the Milky Way galaxy lies in the direction of the _constellation, aboutlight-years away.
Answer:
Sagittarius
Explanation:
The center of the Milky Way galaxy lies in the direction of the Sagittarius constellation, about 26,000 light-years away.
The center of the Milky Way galaxy is in the direction of the Sagittarius constellation and is about 26,000 light-years away from us.
Explanation:The center of the Milky Way galaxy lies in the direction of the Sagittarius constellation and is approximately 26,000 light-years away. When we look towards Sagittarius, we are looking towards the densest part of the Milky Way, which is why it appears brighter in the night sky.
A light-year, by definition, is the distance that light can travel in one year. Hence, saying that the center of the Milky Way is 26,000 light-years away means it would take 26,000 years for the light from the center of the galaxy to reach us.
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An airplane is attempting to land on a runway when the wind is blowing at a velocity of 10 m/s perpendicular to the runway. Given that the airplane is flying at an airspeed of 47 m/s, at what angle relative to the runway direction must the pilot keep the nose pointed into the wind to maintain a flight path aligned with the runway?
Answer:
The pilot must keep the tip pointed at 12.1 degrees to the right with respect to the direction of the runway to align the flight path with the runway.
Explanation:
x= -10m/s
y= 47m/s
r= √(x²)+(y²)
r=48.05 m/s
β= tan⁻¹(y/x)
β=102.01°
the runway is at 90 degrees. Considering the wind, the airplane is flying at 102.01 º direction. Must fly at 12.1 degrees to the right with respect to the direction of the runway to contrarest the wind effect.
A 1m length of wire carrying a current of 7A lies on a horizontal table with a rectangular top of dimensions 0.6m x 0.8m. The ends of the wire are attached to opposite ends of a diagonal of the rectangle. A vertical magnetic field of 0.IT is present. What magnetic force acts on this segment of wire? a) 0.7N b) 0.98N c) 7N d) zero
Answer:
F = 0.7 N
Explanation:
As we know that the wire touch the ends of the diagonal of the rectangle
so here the length of the wire is given as
[tex]L = \sqrt{x^2 + y^2}[/tex]
[tex]L = \sqrt{0.6^2 + 0.8^2}[/tex]
[tex]L = 1 m[/tex]
now force due to magnetic field on current carrying wire is given as
[tex]F = iLB[/tex]
now we have
[tex]F = (7)(1)(0.1)[/tex]
[tex]F = 0.7 N[/tex]
A 1200 kg frictionless roller coaster starts from rest at a height of 19 m. What is its kinetic energy when it goes over hill that is 13 m high?
To solve this question, we need to use the concept of the conversion between potential and kinetic energy.
The roller coaster's initial potential energy before it goes down the hill can be calculated using the formula for gravitational potential energy: `PE = m * g * h`,
where `m` is the mass,
`g` is the acceleration due to gravity, and
`h` is the height.
Substituting the given values, where `m` is 1200 kg, `g` is 9.81 m/s² and the initial height `h` is 19 m,
`PE_initial = 1200 kg * 9.81 m/s² * 19 m = 223668 J`
This is the energy the roller coaster has due to its position at the top of the 19m high hill before it starts to move.
When the roller coaster reaches the hill that is 13 m high, we can calculate its potential energy at this point the same way we calculated the initial potential energy, with `h` being now the final height of 13 m,
`PE_final = 1200 kg * 9.81 m/s² * 13 m = 153036 J`
This is the energy the coaster has due to its position at the top of the 13m high hill.
The kinetic energy (KE) of the roller coaster at this point is gained by the conversion of some of the initial potential energy into kinetic energy. This conversion is equal to the difference between the initial potential energy and the final potential energy:
`KE = PE_initial - PE_final = 223668 J - 153036 J = 70632 J`
So, when the roller coaster goes over the hill that is 13 m high, its kinetic energy is 70632 J. This is the energy the roller coaster has due to its speed at this point.
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Two objects attract each other with a gravitational force of magnitude 0.98 x 10-8 N when separated by 19.5 cm. If the total mass of the two objects is 5.09 kg, what is the mass of each? kg (heavier mass) kg (lighter mass)
Final answer:
The question involves calculating the individual masses of two objects that collectively weigh 5.09 kg and attract each other with a gravitational force of 0.98 x 10^-8 N, separated by 19.5 cm. The gravitational formula F = G×(m1×m2)/r^2 and a system of equations are used to find the solution.
Explanation:
The student's question is about determining the mass of two objects that attract each other with a known gravitational force when separated by a certain distance. The given force is 0.98 x 10-8 N, and the total mass of the two objects is 5.09 kg. The distance between them is 19.5 cm (or 0.195 m in SI units).
To solve this problem, we'll use the formula for the gravitational force between two masses, which is:
F = G×(m1×m2)/r2,
where:
The intensity of solar radiation near the earth is 1.4 × 103 W/m2 . What force is exerted by solar radiation impinging normally on a 5.0 m2 perfectly reflecting panel of an artificial satellite orbiting the earth?
Answer:
4.665×10⁻¹¹ N
Explanation:
Intensity of solar radiation=I=1.4×10³ W/m²
Area of panel=A=5.0 m²
speed of sound=c=3×10⁸ m/s
[tex]P_t=\text {Total\ Solar\ Radiation\ Pressure}[/tex]
[tex]P_t=2\frac{I}{c}\\\Rightarrow P_t=2\frac{1.4\times 10^{-3}}{3\times 10^{8}}\\\Rightarrow P_t=0.933\times 10^{-11}[/tex]
∴Total Solar Radiation Pressure=0.933×10⁻¹¹ Ws/m³
Force= Pressure×Area
Force=0.933×10⁻¹¹×5
Force=4.665×10⁻¹¹ Ws/m=4.665×10⁻¹¹ N
∴Force is exerted by solar radiation impinging normally on a perfectly reflecting panel of an artificial satellite orbiting the earth is 4.665×10⁻¹¹ N
The force exerted by solar radiation on the panel is 7.0 × 10³ N.
Explanation:
The force exerted by solar radiation can be calculated using the formula: force = power/area. In this case, the power is 1.4 × 10³ W/m² and the area is 5.0 m². Substituting these values in, the force exerted by solar radiation on the panel is: force = (1.4 × 10³ W/m²) x (5.0 m²) = 7.0 × 10³ N.
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An astronaut leaves Earth in a spaceship at a speed of 0.96 c relative to an observer on Earth. The astronaut's destination is a star system 14.4 light-years away (one light-year is the distance light travels in one year.) According to the astronaut, how long does the trip take?
Answer:
t=4.2 years
Explanation:
velocity v= 0.96c
destination star distance = 14.4 light year
According to the theory of relativity length contraction
[tex]l= \frac{l_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
[tex]l_0= l{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
putting values we get
[tex]l_0= 14.4{\sqrt{1-\frac{(0.96c)^2}{c^2}}}[/tex]
[tex]l_0= 4.032 light years[/tex]
now distance the trip covers
D= vt
[tex]4.032\times c= 0.96c\times t[/tex]
[tex]t= \frac{4.032c}{0.96c}[/tex]
t= 4.2 years
so the trip will take 4.2 years
Use the worked example above to help you solve this problem. Find the energy transferred in 1 h by conduction through a concrete wall 1.6 m high, 4.00 m long, and 0.20 m thick if one side of the wall is held at 20°C and the other side is at 5°C.
Answer:
The energy transferred is 2.93 MJ.
Explanation:
Given that,
Time = 1 h = 3600 s
Area[tex]A = 1.6\times4.00[/tex]
Width [tex]\Delta x=0.20 m[/tex]
Temperature [tex] T_{1}=20^{\circ}\ C[/tex]
[tex]T_{2}=5^{\circ}\ C[/tex]
For concrete wall,
The thermal conductivity
k = 1.7 W/mK
Using Fourier's law
[tex]Q=kA\dfrac{dt}{\Delta x}[/tex]
Where, A = area
[tex]\Delta x[/tex] = width
[tex]\Delta t[/tex] = change in temperature
k = thermal conductivity
Put the value into the formula
[tex]Q=1.7\times 1.6\times 4.00\times\dfrac{15}{0.20}[/tex]
[tex]Q=816\ W[/tex]
The energy transferred in 1 hour
[tex]E = Q\times t[/tex]
[tex]E=816\times3600[/tex]
[tex]E=2937600[/tex]
[tex]E = 2.93\times10^{6}\ J[/tex]
[tex]E = 2.93\ MJ[/tex]
Hence, The energy transferred is 2.93 MJ.
n an oscillating LC circuit, L = 3.76 mH and C = 3.13 μF. At t = 0 the charge on the capacitor is zero and the current is 2.95 A. (a) What is the maximum charge that will appear on the capacitor? (b) At what earliest time t > 0 is the rate at which energy is stored in the capacitor greatest, and (c) what is that greatest rate?
Answer:
Part a)
[tex]Q = 320 \mu C[/tex]
Part b)
[tex]t = 8.52 \times 10^{-5} s[/tex]
Part c)
Rate of energy = 301.5 J/s
Explanation:
Part a)
Since energy is always conserved in LC oscillating system
So here for maximum charge stored in the capacitor is equal to the magnetic field energy stored in inductor
[tex]\frac{1}{2}Li^2 = \frac{Q^2}{2C}[/tex]
now we have
[tex]Q = \sqrt{LC} i[/tex]
[tex]Q = \sqrt{(3.76 \times 10^{-3})(3.13 \times 10^{-6})} (2.95)[/tex]
[tex]Q = 320 \mu C[/tex]
Part b)
Energy stored in the capacitor is given as
[tex]U = \frac{q^2}{2C}[/tex]
now rate of energy stored is given as
[tex]\frac{dU}{dt} = \frac{q}{C}\frac{dq}{dt}[/tex]
so here we also know that
[tex]q = Q sin(\omega t)[/tex]
[tex]\frac{dq}{dt} = Q\omega cos(\omega t)[/tex]
now from above equation
[tex]\frac{dU}{dt} = \frac{Qsin(\omega t)}{C} (Q\omega cos\omega t)[/tex]
so maximum rate of energy will be given when
[tex]sin\omega t = cos\omega t[/tex]
[tex]\omega t = \frac{\pi}{4}[/tex]
[tex]t = \frac{\pi}{4}\sqrt{LC}[/tex]
[tex]t = 8.52 \times 10^{-5} s[/tex]
Part c)
Greatest rate of energy is given as
[tex]\frac{dU}{dt} = \frac{Q^2\omega}{C}[/tex]
[tex]\frac{dU}{dt} = \frac{(320 \mu C)^2 \sqrt{\frac{1}{(3.76 mH)(3.13 \mu C)}}}{3.13 \mu C}[/tex]
[tex]\frac{dU}{dt} = 301.5 J/s[/tex]
A circular loop of wire is rotated at constant angular speed about an axis whose direction can be varied. In a region where a uniform magnetic field points straight down, what must be the orientation of the loop's axis of rotation if the induced emf is to be zero?
Answer:
here the coil must be oriented in such a way that its plane is perpendicular to the magnetic field
Explanation:
As we know by Faraday's law of electromagnetic induction
Rate of change in magnetic flux will induce EMF in the coil
so here we will have
[tex]EMF = \frac{d\phi}{dt}[/tex]
here we know that
[tex]\phi = NB.A[/tex]
now if the magnetic flux will change with time then it will induce EMF in the coil
[tex]EMF = N\frac{d}{dt}(B.A)[/tex]
so here induced EMF will be zero in the coil if the flux linked with the coil will remain constant
so here the coil must be oriented in such a way that its plane is perpendicular to the magnetic field
In such a way when coil will rotate then the flux linked with the coil will remains constant and there will be no induced EMF in it
To have zero induced emf in a circular loop rotating in a magnetic field, the loop's axis of rotation should be parallel to the magnetic field lines.
The orientation of the loop's axis of rotation must be parallel to the magnetic field lines in a region where a uniform magnetic field points straight down for the induced emf to be zero. When the loop's axis is perpendicular to the magnetic field, the maximum emf is induced. Thus, to minimize the induced emf, the axis should be parallel to the magnetic field.
(a) A woman climbing the Washington Monument metabolizes 6.00×102kJ of food energy. If her efficiency is 18.0%, how much heat transfer occurs to the environment to keep her temperature constant? (b) Discuss the amount of heat transfer found in (a). Is it consistent with the fact that you quickly warm up when exercising?
Answer:
a)
492 kJ
b)
Consistent
Explanation:
Q = Heat stored by woman from food = 600 k J
η = Efficiency of woman = 18% = 0.18
Q' = heat transferred to the environment
heat transferred to the environment is given as
Q' = (1 - η) Q
Inserting the values
Q' = (1 - 0.18) (600)
Q' = 492 kJ
b)
Yes the amount of heat transfer is consistent. The process of sweating produces the heat and keeps the body warm
A woman climbing the Washington Monument metabolizes food energy with 18% efficiency, meaning 82% of the energy is lost as heat. When we calculate this value, we find that 492 kJ of energy is released as heat, which is consistent with the fact that people quickly warm up when exercising.
Explanation:The woman climbing the Washington Monument metabolizes 6.00×10² kJ of food energy with an efficiency of 18%. This implies that only 18% of the energy consumed is used for performing work, while the remaining (82%) is lost as heat to the environment.
To calculate the energy lost as heat:
Determine the total energy metabolized, which is 6.00 × 10² kJ.Multiply this total energy by the percentage of energy lost as heat (100% - efficiency), which gives: (6.00 × 10² kJ) * (100% - 18%) = 492 kJ.The released heat of 492 kJ is consistent with the fact that a person quickly warms up when exercising, because a significant portion of the body's metabolic energy is lost as heat due to inefficiencies in converting energy from food into work.
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A long solenoid that has 1 170 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00 10-4 T at its center. What current is required in the windings for that to occur?
I = 28.6mA.
The magnetic field in the center of a solenoid is given by:
B = μ₀NI/L
Clear I from the equation above, we obtain:
I = BL/μ₀N
With B = 1.00 x 10⁻⁴T, L = 0.42m, μ₀ = 4π x 10⁻⁷T.m/A and N = 1170turns
I = [(1.00 x 10⁻⁴T)(0.42m)]/[(4π x 10⁻⁷T.m/A)(1170turns)]
I = 0.0286A
I = 28.6mA
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3.5 × 10-4 mm (1.378 × 10-5 in.) and a crack length of 4.5 × 10-2 mm (1.772 × 10-3 in.) when a tensile stress of 170 MPa (24660 psi) is applied?
Given:
applied tensile stress, [tex]\sigma[/tex] = 170 MPa
radius of curvature of crack tip, [tex]r_{t}[/tex] = [tex]3.5\times 10^{-4}[/tex] mm
crack length = [tex]4.5\times 10^{-2}[/tex] mm
half of internal crack length, a = [tex]\frac{crack length}{2} = \frac{4.5\times 10^{-2}}{2}[/tex]
a = [tex]2.25\times 10^{-2}[/tex]
Formula Used:
[tex]\sigma _{max} = 2\times\sigma \sqrt{\frac{a}{r_t}}[/tex]
Solution:
Using the given formula:
[tex]\sigma _{max} = 2\times170 \sqrt{\frac{2.25\times 10 ^{-2}}{3.5\times 10^{-4}}}[/tex]
[tex]\sigma _{max}[/tex] = 2726 MPa (395372.9 psi)
The magnitude of the maximum stress at the tip of an internal crack can be determined using the stress concentration factor formula.
Explanation:The magnitude of the maximum stress at the tip of an internal crack can be determined using the formula for the stress concentration factor, which is the ratio of the maximum stress to the applied stress. The stress concentration factor (Kt) for an internal crack can be calculated using the following equation:
Where Kt is the stress concentration factor, a is the crack length, and r is the radius of curvature of the crack.
Plugging in the given values:
Kt = 1 + 2 * (4.5 x 10-2 mm) / (3.5 x 10-4 mm) = 127
The magnitude of the maximum stress can be calculated by multiplying the stress concentration factor by the applied stress:
Maximum Stress = 127 * 170 MPa = 21,590 MPa
A particle is located on the x axis 4.9 m from the origin. A force of 38 N, directed 30° above the x axis in the x-y plane, acts on the particle. What is the torque about the origin on the particle? Round your answer to the nearest whole number. Use a minus sign to indicate a negative direction and no sign to indicate a positive direction.
Answer:
Torque is 93 Nm anticlockwise.
Explanation:
We have value of torque is cross product of position vector and force vector.
A force of 38 N, directed 30° above the x axis in the x-y plane.
Force, F = 38 cos 30 i + 38 sin 30 j = 32.91 i + 19 j
A particle is located on the x axis 4.9 m and we have to find torque about the origin on the particle.
Position vector, r = 4.9 i
Torque, T = r x F = 4.9 i x (32.91 i + 19 j) = 4.9 x 19 k = 93.1 k Nm
So Torque is 93 Nm anticlockwise.
Marcus can drive his boat 24 miles down the river in 2 hours but takes 3 hours to return upstream. Find the rate of the boat in still water and the rate of the current.
Answer:
speed of boat as
[tex]v_b = 10 mph[/tex]
river speed is given as
[tex]v_r = 2 mph[/tex]
Explanation:
When boat is moving down stream then in that case net resultant speed of the boat is given as
since the boat and river is in same direction so we will have
[tex]v_1 = v_r + v_b[/tex]
Now when boat moves upstream then in that case the net speed of the boat is opposite to the speed of the river
so here we have
[tex]v_2 = v_b - v_r[/tex]
as we know when boat is in downstream then in that case it covers 24 miles in 2 hours
[tex]v_1 = \frac{24}{2} = 12 mph[/tex]
also when it moves in upstream then it covers same distance in 3 hours of time
[tex]v_2 = \frac{24}{3} = 8 mph[/tex]
[tex]v_b + v_r = 12 mph[/tex]
[tex]v_b - v_r = 8 mph[/tex]
so we have speed of boat as
[tex]v_b = 10 mph[/tex]
river speed is given as
[tex]v_r = 2 mph[/tex]
The rate of the boat in still water is 10 mph
The rate of the current is 2 mph
Further explanationAcceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
a = acceleration (m / s²)v = final velocity (m / s)
u = initial velocity (m / s)
t = time taken (s)
d = distance (m)
Let us now tackle the problem!
Given:
distance covered = d = 24 miles
time for driving down the river = td = 2 hours
time for driving up the river = tu = 3 hours
Unknown:
velocity of the boat in still water = vs = ?
velocity of the current = vc = ?
Solution:
When Marcus drive his boat down the river , the velocity of the boat is in the same direction to the velocity of the current.
[tex]v_s + v_c = \frac{d}{t_d}[/tex]
[tex]v_s + v_c = \frac{24}{2}[/tex]
[tex]v_s + v_c = 12[/tex]
[tex]v_s = 12 - v_c[/tex] → Equation 1
When Marcus drive his boat up the river , the velocity of the boat is in the opposite direction to the velocity of the current.
[tex]v_s - v_c = \frac{d}{t_d}[/tex]
[tex]v_s - v_c = \frac{24}{3}[/tex]
[tex]v_s - v_c = 8[/tex]
[tex]12 - v_c - v_c = 8[/tex] ← Equation 1
[tex]12 - 2v_c = 8[/tex]
[tex]2v_c = 12 - 8[/tex]
[tex]2v_c = 4[/tex]
[tex]v_c = 4 \div 2[/tex]
[tex]\large {\boxed {v_c = 2 ~ mph} }[/tex]
[tex]v_s = 12 - v_c[/tex]
[tex]v_s = 12 - 2[/tex]
[tex]\large {\boxed {v_s = 10 ~ mph} }[/tex]
Learn moreVelocity of Runner : https://brainly.com/question/3813437Kinetic Energy : https://brainly.com/question/692781Acceleration : https://brainly.com/question/2283922The Speed of Car : https://brainly.com/question/568302 Answer detailsGrade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate
What is direct current? In which direction does current go according to the electron flow convention?
Final answer:
Direct current (DC) is the flow of electric current in only one direction. According to the electron flow convention, the current flows from the positive terminal to the negative terminal.
Explanation:
Direct current (DC) is the flow of electric current in only one direction. It refers to systems where the source voltage is constant. The direction of current flow in a direct current circuit is from the positive terminal to the negative terminal. This is according to the electron flow convention, where the movement of negative charges (electrons) is considered as the flow of current.
During strengthening heat treatment, the _______ step traps the material in an unstable crystalline structure. a)-Quenching, b)-Tempering c)-Solution treatment
Answer: A) Quenching
Hope this helps
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. Part A What is the magnitude of the current in wire 3?
Answer:
The current in wire 3 is 0.25 A.
Explanation:
It is given that, three wires meet at a junction.
Current in wire 1, I₁ = 0.4 A
Current in wire 2, I₂ = -0.65 A (out of the junction)
We need to find the magnitude of the current in wire 3 (I₃). Applying Kirchhoff's current law which states that the sum of current in the circuit at a junction is equal to zero.
[tex]I_1+I_2+I_3=0[/tex]
[tex]I_3=-(I_2+I_1)[/tex]
[tex]I_3=-(-0.65+0.4)\ A[/tex]
I₃ = 0.25 A
So, the current in wire 3 is 0.25 A. Hence, this is the required solution.
Using the principle of Kirchhoff's junction rule, we calculate the current in Wire 3 as being 0.25 amperes.
Explanation:The understanding of this question relies on the principle of Kirchhoff's junction rule in Physics, which states the sum of currents entering a junction equals the sum of currents leaving the junction. In this case scenario, Wire 1 has a current of 0.40 A into the junction and wire 2 has a current of 0.65 A out of the junction. The current of Wire 3 can be solved via the equation: I1 + I3 = I2, where I1, I2, and I3 are the currents on Wires 1, 2, and 3 respectively. Plugging in the given values, we get: 0.40 A + I3 = 0.65 A. Solving the equation for I3, we find that I3 = 0.25 A. Hence, the magnitude of the current in Wire 3 is 0.25 A.
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A plastic rod that has been charged to â14 nC touches a metal sphere. Afterward, the rod's charge is â1.0 nC . How many charged particles were transferred?
Express your answer using two significant figures.
Answer:
[tex]N = 8.1 \times 10^{10}[/tex]
Explanation:
Initial charge on the rod is
[tex]Q_i = 14 nC[/tex]
final charge on the rod is
[tex]Q_f = 1 nC[/tex]
now the charge transferred from to the sphere is given as
[tex]\Delta Q = Q_i - Q_f[/tex]
[tex]\Delta Q = 14 - 1 = 13 nC[/tex]
now we also know that
Q = Ne
so number of particles transferred is
[tex]N = \frac{\Delta Q}{e}[/tex]
[tex]N = \frac{13 \times 10^{-9}}{1.6 \times 10^{-19}}[/tex]
[tex]N = 8.1 \times 10^{10}[/tex]
When the charged plastic rod touches the metal sphere, it transfers around 13 nC of charge to the sphere. This amounts to about 8.125 x 10^10 electrons, as each electron carries a charge of approximately -1.6 x 10^-19 C.
Explanation:When the charged plastic rod touches the metal sphere, charge transfer occurs until both objects have the same charge. Initially, the plastic rod had a charge of -14 nC and after touching it had a charge -1.0 nC, which means that 13 nC was transferred to the metal sphere.
The elementary charge (charge of an electron) is approximately -1.6 x 10^-19 C. Therefore, the number of electrons transferred would be the total transferred charge divided by the charge of one electron, which results in approximately 8.125 x 10^10 electrons.
The charge is negative indicating electrons, which have negative charge, were transferred.
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What is the wavelength of the radio waves from an FM station operating at a frequency of 99.5 MHz
Answer:
Wavelength, [tex]\lambda=3.01\ m[/tex]
Explanation:
It is given that,
Frequency, f = 99.5 MHz = 99.5 × 10⁶ Hz
We need to find the wavelength of the radio waves from an FM station operating at above frequency. The relationship between the frequency and the wavelength is given by :
[tex]c=f\lambda[/tex]
[tex]\lambda=\dfrac{c}{f}{[/tex]
c = speed of light
[tex]\lambda=\dfrac{3\times 10^8\ m/s}{99.5\times 10^6\ Hz}[/tex]
[tex]\lambda=3.01\ m[/tex]
So, the wavelength of the radio waves from an FM station is 3.01 m. Hence, this is the required solution.
A rocket sled accelerates at a rate of 49.0 m/s2 . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body.
Explanation:
It is given that,
Mass of the passenger, m = 75 kg
Acceleration of the rocket, [tex]a=49\ m/s^2[/tex]
(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :
F = m a
[tex]F=75\ kg\times 49\ m/s^2[/tex]
F = 3675 N
Ratio, [tex]R=\dfrac{F}{W}[/tex]
[tex]R=\dfrac{3675}{75\times 9.8}=5[/tex]
So, the ratio between the horizontal force and the weight is 5 : 1.
(b) The magnitude of total force the seat exerts against his body is F' i.e.
[tex]F'=\sqrt{F^2+W^2}[/tex]
[tex]F'=\sqrt{(3675)^2+(75\times 9.8)^2}[/tex]
F' = 3747.7 N
The direction of force is calculated as :
[tex]\theta=tan^{-1}(\dfrac{W}{F})[/tex]
[tex]\theta=tan^{-1}(\dfrac{1}{5})[/tex]
[tex]\theta=11.3^{\circ}[/tex]
Hence, this is the required solution.
The horizontal component of the force the seat exerts against the passenger's body is 3675 N. The ratio of this force to the passenger's weight is 5. The total force the seat exerts has a magnitude of 3793 N.
Explanation:(a) To calculate the horizontal component of the force the seat exerts against the passenger's body, we can use Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the passenger is 75.0 kg and the acceleration of the rocket sled is 49.0 m/s2. So the force exerted by the seat is:
Force = mass * acceleration
Force = 75.0 kg * 49.0 m/s2
Force = 3675 N
Now let's compare this force to the passenger's weight. The weight of an object is given by the formula:
Weight = mass * gravitational acceleration
Weight = 75.0 kg * 9.8 m/s2
Weight = 735 N
To find the ratio, we divide the force exerted by the seat by the weight of the passenger:
Ratio = Force / Weight
Ratio = 3675 N / 735 N
Ratio = 5
(b) The total force the seat exerts against the passenger's body has both a horizontal and vertical component. The direction of the total force is the same as the direction of the acceleration of the rocket sled. The magnitude of the total force can be found using the Pythagorean theorem:
Total Force = √(horizontal component2 + vertical component2)
Total Force = √(36752 + 7352)
Total Force = 3793 N
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F16–5. A wheel has an angular acceleration of a = (0.5 u) rad>s 2 , where u is in radians. Determine the magnitude of the velocity and acceleration of a point P located on its rim after the wheel has rotated 2 revolutions. The wheel has a radius of 0.2 m and starts at v0 = 2 rad>s.
Using a formula linking initial and final angular velocities, angular acceleration, and total angle, the final angular velocity can be obtained. Multiplying the final angular velocity and radius gives the final linear velocity. Angular acceleration and linear acceleration on the rim of a wheel are related by their radius.
Explanation:Given the angular acceleration, a, the number of revolutions, and the radius of the wheel, r, you can determine the angular velocity at the end of 2 revolutions using the formula w_f = sqrt(w_i^2 + 2*a*θ), where w_i is the initial angular velocity, w_f is the final angular velocity, a is angular acceleration, and θ is the total angle swept out in radians. The angular velocity can then be used to find the final linear velocity, v = r*w_f.
Angular acceleration is directly given by a = α*r, where α is angular acceleration and r is the radius. Angular acceleration is also related to the linear (tangential) acceleration, a_t, by a = r*α, where α is angular acceleration and r is the radius. Tangential acceleration involves not only changes in speed (which causes tangential acceleration) but also changes in direction (which causes radial or centripetal acceleration).
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Electric Field of Ring of Charge A ring shaped charge has a radius of 0.250 meters and bears a total charge of +5.00 C What is the magnitude of the electric field along the axis of symmetry of the ring a distance of 0.500 meters from the center of the ring? ? A. 1.29 × 105 N/C 0 B. 1.67 × 105 N/C O C. 2.59 x 105 N/C ? D. 3.95 × 105 N/C
Answer:
6.44 × 10^10 N/C
Explanation:
Electric field due to the ring on its axis is given by
E = K q r / (r^2 + x^2)^3/2
Where r be the radius of ring and x be the distance of point from the centre of ring and q be the charge on ring.
r = 0.25 m, x = 0.5 m, q = 5 C
K = 9 × 10^9 Nm^2/C^2
E = 9 × 10^9 × 5 × 0.25 / (0.0625 + 0.25)^3/2
E = 6.44 × 10^10 N/C
Two train whistles, and , each have a frequency of 392 Hz. is stationary and is moving toward the right (away from ) at a speed of 35.0 m/s. A listener is between the two whistles and is moving toward the right with a speed of 15.0 m/s (). No wind is blowing. (a) What is the frequency from as heard by the listener? (b) What is the frequency from as heard by the listener? (c) What is the beat frequency detected by the listener?
(a) The frequency from the stationary whistle as heard by the listener is approximately [tex]\(409.62 \, \text{Hz}\)[/tex].
(b) The frequency from the moving whistle as heard by the listener is approximately [tex]\(454.55 \, \text{Hz}\).[/tex]
(c) The beat frequency detected by the listener is approximately [tex]\(44.93 \, \text{Hz}\).[/tex]
To solve this problem, we can use the Doppler effect formula for sound frequencies.
The formula for the apparent frequency [tex](\(f'\))[/tex] heard by an observer when the source and/or observer is in motion relative to the medium is given by:
[tex]\[ f' = \frac{f \cdot (v + v_o)}{(v + v_s)} \][/tex]
where:
- [tex]\( f \)[/tex] is the frequency emitted by the source,
- [tex]\( v \)[/tex] is the speed of sound in the medium (assumed to be constant),
- [tex]\( v_o \)[/tex] is the speed of the observer relative to the medium,
- [tex]\( v_s \)[/tex] is the speed of the source relative to the medium.
Given:
- [tex]\( f = 392 \, \text{Hz} \)[/tex] (frequency emitted by both whistles),
-[tex]\( v = 343 \, \text{m/s} \)[/tex] (speed of sound in air),
- [tex]\( v_o = 15.0 \, \text{m/s} \)[/tex] (speed of the listener),
- [tex]\( v_s = 35.0 \, \text{m/s} \)[/tex] (speed of the moving whistle).
Let's calculate each part of the problem:
(a) Frequency from the stationary whistle [tex](\(f_1'\))[/tex] as heard by the listener:
[tex]\[ f_1' = \frac{f \cdot (v + v_o)}{(v + v_s)} \]\[ f_1' = \frac{392 \, \text{Hz} \cdot (343 \, \text{m/s} + 15.0 \, \text{m/s})}{(343 \, \text{m/s} + 0 \, \text{m/s})} \]\[ f_1' = \frac{392 \, \text{Hz} \cdot (358.0 \, \text{m/s})}{343 \, \text{m/s}} \]\[ f_1' \approx 409.62 \, \text{Hz} \][/tex]
(b) Frequency from the moving whistle [tex](\(f_2'\))[/tex] as heard by the listener:
[tex]\[ f_2' = \frac{f \cdot (v + v_o)}{(v - v_s)} \]\[ f_2' = \frac{392 \, \text{Hz} \cdot (343 \, \text{m/s} + 15.0 \, \text{m/s})}{(343 \, \text{m/s} - 35.0 \, \text{m/s})} \]\[ f_2' = \frac{392 \, \text{Hz} \cdot (358.0 \, \text{m/s})}{308 \, \text{m/s}} \]\[ f_2' \approx 454.55 \, \text{Hz} \][/tex]
(c) Beat frequency detected by the listener:
The beat frequency is the difference in frequencies between the two whistles as heard by the listener:
[tex]\[ \text{Beat frequency} = |f_1' - f_2'| \]\[ \text{Beat frequency} = |409.62 \, \text{Hz} - 454.55 \, \text{Hz}| \]\[ \text{Beat frequency} \approx 44.93 \, \text{Hz} \][/tex]
So, the answers are:
(a) The frequency from the stationary whistle as heard by the listener is approximately [tex]\(409.62 \, \text{Hz}\)[/tex].
(b) The frequency from the moving whistle as heard by the listener is approximately [tex]\(454.55 \, \text{Hz}\).[/tex]
(c) The beat frequency detected by the listener is approximately [tex]\(44.93 \, \text{Hz}\).[/tex]
Our galaxy, the Milky Way, has a diameter of about 100,000 light years. How many years would it take a spacecraft to cross the galaxy if it could travel at 99% the speed of light? A. 1.4 × 103 yrs B. 1.0 × 105 yrs C. 1.4 × 104 yrs D. 7.2 × 103 yrs E. None of these is correct.
Answer :
The time is [tex]1\times10^{5}\ yrs[/tex]
(B) is correct option.
Explanation :
Given that,
Diameter = 100000 light year
Velocity = 0.99 c
We need to calculate the time
Using formula of time
[tex]t = \dfrac{d}{v}[/tex]
Put the value into the formula
[tex]t =\dfrac{100000 c}{0.99c}[/tex]
[tex]t =1\times10^{5}\ yrs[/tex]
Hence, The time is [tex]1\times10^{5}\ yrs[/tex]