Completing the square gives the answer right away.
[tex]-16t^2+64t+23=-16(t^2-4t)+23=-16(t^2-4t+4-4)+23=-16((t-2)^2-4)+23[/tex]
[tex]\implies h(t)=-16(t-2)^2+87[/tex]
which indicates a maximum height of 87 when [tex]t=2[/tex].
Answer:
The maximum height of the arrow is 87 meters.
Explanation:
If we look at the height function of the arrow
[tex]h(t)=-16t^{2} +64t+23[/tex]
we see that its a parabola whose principal coefficient is negative, that means is inverted or upside down.
When the arrow reaches maximum height its velocity will be zero. The velocity of an object is the derivative of the position function, in this case the so called height function.
So we we derivate the height function to get that
[tex]h'(t)=-32t+64[/tex]
we must find the t that makes this equation equal to zero:
[tex]-32t+64=0[/tex]
[tex]32t=64[/tex]
[tex]t=2s[/tex]
we replace this value of t in the height function:
[tex]h(2 s)=-16.(2s)^{2} +64.(2s)+23[/tex]
we get that
[tex]h(2s)=87m[/tex]
The maximum height of the arrow is 87 meters.
We have used the MKS system which uses the meter, kilogram and second as base units.
If a ball is thrown vertically upward from the roof of a 32 foot tall building with a velocity of 48 ft/sec, its height in feet after T seconds is ______.
Answer:
The ball's maxium height after 1.49 seconds will be 103.52 feet.
Explanation:
Vi= 48 ft/sec
hi= 32 ft
hf= ?
time of maxium height :
t= Vi/g t= 48 ft/sec / 32.17 ft/sec²
t= 1.49 sec
maxium height:
hf= hi + Vi * t
hf= 32ft + 48 ft/sec * 1.49 sec
hf= 103.52 ft
A length change 0.08 m will occur for an object that is L= 56 m long. If the coefficient of thermal expansion is5.3 x 10 /C and if the original temperature is 253 C, find the increase in temperature.
Answer:
Increase in temperature = 269.54 °C
Explanation:
We have equation for thermal expansion
ΔL = LαΔT
Change in length, ΔL = 0.08 m
Length, L = 56 m
Coefficient of thermal expansion, α = 5.3 x 10⁻⁶ °C⁻1
Change in temperature, ΔT = T - 253
Substituting
0.08 = 56 x 5.3 x 10⁻⁶ x (T - 253)
(T - 253) = 269.54
T = 522.54 °C
Increase in temperature = 269.54 °C
Under metric prefixes, if a meter were divided into 100 equal-sized subunits, a single unit would be called a
Answer:
In metric prefixes, if a meter were divided into 100 equal-sized subunits, a single unit would be called a centimeter.
Explanation:
Here we asked to divide 1 meter in to 100 equal parts.
Let us find out what is the length of piece when 1 m is divided in to 100 equal parts.
Length
[tex]l = \frac{1}{100} = 0.01m[/tex]
That is length of 1 m divided into 100 equal parts is 0.01m.
We know that 0.01 m is 1 centimeter.
So, in metric prefixes, if a meter were divided into 100 equal-sized subunits, a single unit would be called a centimeter.
Final answer:
A meter divided into 100 equal-sized subunits is called a centimeter, with 'centi-' indicating one-hundredth of a meter in the metric system.
Explanation:
If a meter were divided into 100 equal-sized subunits, a single unit would be called a centimeter. The prefix centi- in the metric system indicates a division by 100.
Therefore, one centimeter is one-hundredth of a meter. This is consistent with metric prefixes being based on powers of ten, facilitating easy conversion and understanding of different scales of measurement.
The International System of Units, or SI, uses these prefixes to create a unified system for measuring length, and indeed all types of measurements.
A plane electromagnetic wave, with wavelength 6 m, travels in vacuum in the positive x direction with its electric vector E, of amplitude 214.6 V/M, directed along y axis. What is the frequency f of the wave? 250 MHz 100 MHz 50 MHz 105 MHz 25 MHz [2 points] (b) A plane electromagnetic wave, with wavelength 6 m, travels in vacuum in the positive x direction with its electric vector E, of amplitude 214.6 V/M, directed along y axis. What is the direction of the magnetic field associated with the wave? (i,j,k are the unit vectors corresponding to x, y and z directions) -j
a) For an EM wave traveling in a vacuum, this equation holds true:
c = fλ
c is the speed of light in a vacuum, f is the frequency, and λ is the wavelength.
Given values:
c = 3×10⁸m/s
λ = 6m
Plug in the values and solve for f:
3×10⁸ = f(6)
f = 50MHz
b) The direction of an EM wave's Poynting vector determines the direction of the wave's propagation.
S = 1/μ₀(E×B)
S is the Poynting vector, μ₀ is the magnetic constant, E is the electric field vector, and B is the magnetic field vector. Note that we are taking the cross product between E and B, not taking the product of two scalar quantities.
Since S depends on the cross product of E and B, you may use the right hand rule in the following way to determine the direction of B:
The EM wave travels in the +x direction, therefore S points in the +x direction. Face your palm in the +x direction.E points along the y axis. Let us arbitrarily choose +y. Point your thumb in the +y direction.This will constrain your other fingers to point in the +z direction. If you point your thumb in the -y direction then your fingers will point in the -z direction.You can conclude that B must point along the z axis, so you can represent B with the k unit vector.
a) The frequency f of the wave is 50 MHz.
b) The direction of the magnetic field associated with the wave is along Z-axis with the k unit vector.
What is Poynting vector?
The Poynting vector is a measurement that expresses the strength and direction of the energy flow in electromagnetic waves.
Mathematically;
S = 1/μ₀(E×B)
Where: S = Poynting vector, μ₀ = magnetic constant, E = electric field vector, and B = the magnetic field vector
a) For an EM wave traveling in a vacuum, it can be written that:
c = fλ
Where c is the speed of light in a vacuum, f is the frequency, and λ is the wavelength.
Speed of light in vacuum: c = 3×10⁸m/s
Wavelength of the given EM Wave: λ = 6m
Hence, the frequency of the EM wave: f = 3×10⁸m/s/ 6m = 50MHz.
b) The EM wave travels in the x direction, therefore S points in the x direction.
Electric field acts along the y axis.
Hence, The direction of the magnetic field associated with the wave is along Z-axis with the k unit vector.
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What is the fundamental frequency on a 8 m rope that is tied at both ends if the speed of the waves is 16 m/s?
Answer:
1 Hz
Explanation:
For rope fixed on both ends the length corresponds to λ/2 (λ is wavelength)\
Thus L = λ/2
=> λ = 16 m
We know that frequency and wavelength are related as
f x λ = v where f is frequency and v is speed of the wave
thus f = v/λ
f = 16/16 =1 Hz
The fundamental frequency of an 8 m rope tied at both ends with waves traveling at a speed of 16 m/s is 1 Hz, calculated using the formula for the fundamental frequency of a standing wave.
Explanation:The fundamental frequency of a string fixed at both ends (like an 8 m rope with waves traveling at 16 m/s) can be found using the formula for the fundamental frequency of a standing wave, which is given by f = v / (2L), where f is the fundamental frequency, v is the speed of the wave, and L is the length of the string. In this case, L = 8 m and v = 16 m/s. Plugging these values into the formula gives us:
f = 16 m/s / (2 × 8 m) = 1 Hz.
Therefore, the fundamental frequency of the 8 m rope tied at both ends with a wave speed of 16 m/s is 1 Hz.
A copper rod 0.570 � long and with a mass 0.05900 �� is suspended from two thin wire. At right angle to the rod is a uniform magnetic field of 0.670 � pointing into the page. Find (a) the direction and (b) magnitude of the electric current to levitate the copper rod’s gravitation force.
Answer:
a) Left to right
b) 1.51 A
Explanation:
a)
The gravitational force on the rod due to its mass is in downward direction. hence to levitate the rod, the magnetic force on the rod must be in upward direction.
The magnetic field is inward to page and magnetic force must be upward. Using right hand rule, the current must be flowing from left to right.
Left to right
b)
L = length of the copper rod = 0.570 m
m = mass of the rod = 0.059 kg
B = magnitude of magnetic field in the region = 0.670 T
θ = Angle between the magnetic field and rod = 90
i = current flowing throw the rod = ?
The magnetic force on the rod balances the gravitational force on the rod. hence
Magnetic force = gravitational force
mg = i B L Sinθ
(0.059) (9.8) = i (0.670) (0.570) Sin90
i = 1.51 A
In a physics laboratory experiment, a coil with 200 turns enclosing an area of 12 cm2 is rotated in 0.040 s from a position where its plane is perpendicular to the earth’s magnetic field to a position where its plane is parallel to the field. The earth’s magnetic field at the lab location is 6.0 x 10-5 T. What is the total magnetic flux through the coil before it is rotated?
Final answer:
The total magnetic flux through the coil before rotation is calculated using the formula Φ = B × A × cos(θ), with B being the Earth's magnetic field strength, A the area of the coil, and θ the angle between the field and the normal to the coil, resulting in a flux of 7.2 × 10⁻⁸ Tm².
Explanation:
The total magnetic flux Φ through the coil before it is rotated is given by the formula Φ = B × A × cos(θ), where B is the magnetic field strength, A is the area enclosed by the coil, and θ is the angle between the magnetic field and the normal to the plane of the coil. Before the rotation, the plane of the coil is perpendicular to Earth's magnetic field, making θ = 0° (or cos(θ) = 1), so the cos(θ) term does not affect the calculation.
To calculate the magnetic flux, convert the area from cm² to m² by multiplying by 10⁻⁴, and use the given values:
B = 6.0 × 10⁻⁵ T (Earth's magnetic field)A = 12 cm² × 10⁻⁴ = 1.2 × 10⁻³ m² (area of the coil)Φ = B × A × cos(0°) = 6.0 × 10⁻⁵ T × 1.2 × 10⁻³ m² × 1 = 7.2 × 10⁻⁸ Tm²Thus, the total magnetic flux through the coil before it is rotated is 7.2 × 10⁻⁸ Tm².
Two small plastic spheres each have a mass of 2 grams and a charge of -50 nC. They are placed 2 cm apart (center to center). A) What is the magnitude of the electric force on each sphere? B) If one sphere is held in place what is the resulting acceleration magnitude of the other?
Answer:
Part a)
F = 0.056 N
Part b)
a = 28.13 m/s/s
Explanation:
Part a)
Electrostatic force between two charged balls is given as
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
now we will have
[tex]q_1 = q_2 = 50 nC[/tex]
r = 2 cm
now we will have
[tex]F = \frac{(9\times 10^9)(50 \times 10^{-9})(50 \times 10^{-9})}{0.02^2}[/tex]
so here we have
[tex]F = 0.056 N[/tex]
Part b)
Now due to above force the acceleration of one ball which is released is given as
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{0.056}{2 \times 10^{-3}}[/tex]
[tex]a = 28.13 m/s^2[/tex]
A 1500-kg car accelerates from 0 to 30 m/s in 6.0 s. What is the minimum average power delivered by the engine?
Answer:
Power, P = 112500 watts
Explanation:
It is given that,
Mass of the car, m = 1500 kg
Initial velocity of the car, u = 0
Final velocity of the car, v = 30 m/s
Time taken, t = 6 s
The minimum average power delivered by the engine is given by :
P = W/t
Where
W = work done
t = time taken
Work done = change in kinetic energy
So, [tex]P=\dfrac{\Delta K}{t}=\dfrac{\dfrac{1}{2}m(v^2-u^2)}{t}[/tex]
[tex]P=\dfrac{\dfrac{1}{2}\times 1500\ kg\times (30\ m/s)^2}{6\ s}[/tex]
P = 112500 watts
So, the minimum average power delivered by the engine is 112500 watts. Hence, this is the required solution.
A train 400 m long is moving on a straight track with a speed of 81.4 km/h. The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of 17.6 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.
Answer:
86.5m
Explanation:
first convert km/h
then
81.4*1000/60*60=22.6
17.6*1000/60*60=4.89
then, x1/t1=x2/t2
we get
x2=400*4.89/22.6=86.5//
The current in a long solenoid of radius 6 cm and 17 turns/cm is varied with time at a rate of 5 A/s. A circular loop of wire of radius 8 cm and resistance 4 Ω surrounds the solenoid. Find the electrical current induced in the loop (in µA).
The current induced in the loop can be found using Faraday's law of electromagnetic induction. By calculating the rate of change of magnetic flux through the loop and using Ohm's law, we can determine the induced current in the loop. The given values for the solenoid's current, number of turns, radius, and the loop's resistance are used in the calculations.
Explanation:The current induced in the loop can be found using Faraday's law of electromagnetic induction.
Based on the given information, the rate of change in current (di/dt) in the solenoid is 5 A/s. The number of turns per unit length of the solenoid is 17 turns/cm, so the total number of turns in the solenoid is 17 * 2π * 6 cm. The radius of the loop is 8 cm and its resistance is 4Ω.
Using Faraday's law, the induced emf in the loop is given by the equation ε = -N * (dΦ/dt), where N is the total number of turns in the solenoid and (dΦ/dt) is the rate of change of magnetic flux through the loop. The magnetic flux through the loop depends on the magnetic field produced by the solenoid and the area of the loop.
By substituting the given values and using the formula for magnetic flux, we can find the induced emf. Finally, Ohm's law can be used to calculate the induced current in the loop by dividing the induced emf by the loop's resistance.
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A car accelerates uniformly from rest to 27.2 m/s in 8.30 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is 8.26 x 103 N, and (b) the weight of the car is 1.08 x 104 N.
Answer:
(a): The power required to accelerate the car is F= 2756.12 N.
(b): The power required to accelerate the car is F= 3603.67 N.
Explanation:
V= 27.2 m/s
Vi= 0 m/s
t= 8.3 s
g= 9.8 m/s²
W1= 8.26 * 10³ N
W2= 1.08 * 10⁴ N
a= (V-Vi)/t
a= 3.27 m/s²
m1= W1/g
m1= 842.85 kg
m2= W2/g
m2= 1102.04 kg
F= mi*a
F1= 2756.12 N (a)
F2= 3603.67 N (b)
(a) What net external force is exerted on a 1100.0-kg artillery shell fired from a battleship if the shell is accelerated at 2.40 × 10 4 m/s 2 ? (b) What is the magnitude of the force exerted on the ship by the artillery shell, and why?
Explanation:
It is given that,
Mass of the artillery shell, m = 1100 kg
Acceleration of he shell, [tex]a=2.4\times 10^4\ m/s^2[/tex]
(a) The magnitude of force exerted on the shell can be calculated using Newton's second law of motion as :
F = ma
[tex]F=1100\ kg\times 2.4\times 10^4\ m/s^2[/tex]
F = 26400000 N
or
[tex]F=2.6\times 10^7\ N[/tex]
(b) The magnitude of the force exerted on the ship by the artillery shell will be same but the direction is opposite as per Newton's third law of motion. i.e [tex]2.6\times 10^7\ N[/tex]
The magnitude of the force exerted on the ship by the artillery shell is 2.64 × 10^7 N, and it acts in the opposite direction of the shell's motion.
(a) Net External Force on the Artillery Shell:
We can use Newton's second law of motion (ΣF = ma) to solve this problem. Here, ΣF represents the net external force acting on the object (artillery shell), m is the mass of the shell (1100.0 kg), and a is the acceleration of the shell (2.40 × 10^4 m/s²).
Force (F) = mass (m) × acceleration (a)
F = 1100.0 kg × 2.40 × 10^4 m/s²
F = 2.64 × 10^7 N
Therefore, the net external force exerted on the artillery shell is 2.64 × 10^7 N.
(b) Force on the Ship and Explanation:
According to Newton's third law of motion, for every action, there is an equal and opposite reaction. In this case, the force exerted on the artillery shell to accelerate it (calculated in part (a)) is equal in magnitude but opposite in direction to the force exerted by the shell on the ship.
Therefore, the magnitude of the force exerted on the ship by the artillery shell is 2.64 × 10^7 N, and it acts in the opposite direction of the shell's motion.
point) A tank in the shape of an inverted right circular cone has height 5 meters and radius 4 meters. It is filled with 4 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. The density of hot chocolate is δ=1080 kg/m3. Your answer must include the correct units.
Answer:
W = 907963.50 J = 907.96 J
Explanation:
Note: Refer to the figure attached
Now, from the figure we have similar triangles ΔAOB and ΔCOD
we have
[tex]\frac{5}{4}=\frac{x}{r}[/tex]
or
[tex]r=\frac{4x}{5}[/tex]
Now, the work done to empty the tank can be given as:
[tex]W = \int\limits^4_0 {(5-x)\rho\times g A} \, dx[/tex]
or
[tex]W = \int\limits^4_0 {(5-x)1080\times 9.8 (\pi r^2)} \, dx[/tex]
or
[tex]W = \int\limits^4_0 {(5-x)\times10584\times (\pi (\frac{4x}{5})^2)} \, dx[/tex]
or
[tex]W = 6773.76\pi\int\limits^4_0 {(5-x)x^2)} \, dx[/tex]
or
[tex]W = 6773.76\pi[\frac{5}{3}x^3-\frac{1}{4}x^4]^4_0 [/tex]
or
[tex]W = 6773.76\pi[\frac{128}{3}] [/tex]
or
W = 907963.50 J = 907.96 J
The work required to empty the tank by pumping the hot chocolate over the top of the tank is 907.96 J.
What is work done?Work done is the force applied on a body to move it over a distance. The work required to lift a body through a height h is given as,
[tex]W=Fh[/tex]
Here, (F) is the magnitude of force and (f) is the height.
It can be rewritten as,
[tex]W=mgh[/tex]
Here, (m) is the mass of the body.
The tank is in the shape of an inverted right circular cone with height 5 meters and radius 4 meters. As, It is filled with 4 meters of hot chocolate.
Here the ratio of height to radius should be equal to the height at another point (say x), to the radius of that point. Thus,
[tex]\dfrac{5}{4}=\dfrac{x}{r}\\r=\dfrac{4x}{5}[/tex]
For the distance x the work done to empty the cone can be given as,
[tex]W=\int^4_0 (5-x)\rho g \pi r^2 dx\\W=\int^4_0 (5-x)1080\times9.81 \pi (\dfrac{4x}{5})^2 dx\\W=22280.4\int^4_0(5-x)x^2 dx\\W=22280.4[\dfrac{128}{3}]\\W=907.96\rm J[/tex]
Thus, the work required to empty the tank by pumping the hot chocolate over the top of the tank is 907.96 J.
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A far-sighted person has a near-point of 80 cm. To correct their vision so that they can see objects that are as close as 10 cm to their eye, what should be the focal length of the prescribed lens? Assume the lens will be 2 cm from the eye.
Answer:
[tex]f = 8.89 cm[/tex]
Explanation:
As we know that Far sighted person has near point shifted to 80 cm distance
so he is able to see the object 80 cm
now the distance of lens from eye is 2 cm
and the person want to see the objects at distance 10 cm
so here the image distance from lens is 80 cm and the object distance from lens is 8 cm
now from lens formula we have
[tex]\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}[/tex]
[tex]-\frac{1}{80} + \frac{1}{8} = \frac{1}{f}[/tex]
[tex]f = 8.89 cm[/tex]
An electric field of 8.30 x 10^5 V/m is desired between two parallel plates, each of area 31.5 cm^2 and separated by 2.45 mm. There's no dielectric. What charge must be on each plate?
Answer:
Charge on each plate = 2.31 x 10⁻⁸ C
Explanation:
We have the equations
[tex]E=\frac{V}{d}\texttt{ and }V=\frac{Q}{C}[/tex]
Combining both equations
[tex]E=\frac{\left (\frac{Q}{C}\right )}{d}=\frac{Q}{Cd}[/tex]
We also have the equation for capacitance
[tex]C=\frac{\epsilon A}{d}[/tex]
That is
[tex]E=\frac{Q}{\frac{\epsilon A}{d}\times d}=\frac{Q}{\epsilon A}\\\\Q=\epsilon AE[/tex]
Substituting
[tex]Q=8.85\times 10^{-12}\times 31.5\times 10^{-4}\times 8.30\times 10^5=2.31\times 10^{-8}C[/tex]
Charge on each plate = 2.31 x 10⁻⁸ C
A string is wrapped around a pulley of radius 3.60 cm , and a weight hangs from the other end. The weight falls with a constant acceleration 3.00 m/s^2 What's the angular acceleration of the pulley?
Answer:
83.3 rad/s^2
Explanation:
r = 3.6 cm = 0.036 m
a = 3 m/s^2
the relation between the linear acceleration and angular acceleration is given by
a = r α
where, α is angular acceleration, a be the linear acceleration.
α = a / r
α = 3 / 0.036 = 83.3 rad/s^2
A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the location of the center of mass xcom and the location of the center of gravity xcogof the two masses? Are the locations the same? Why or why not?
Answer:
center of mass of the two masses will lie at x = 2.52 cm
center of gravity of the two masses will lie at x = 2.52 cm
So center of mass is same as center of gravity because value of gravity is constant here
Explanation:
Position of centre of mass is given as
[tex]r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}[/tex]
here we have
[tex]m_1 = 10.2 kg[/tex]
[tex]m_2 = 4.6 kg[/tex]
[tex]r_1 = (0, 0)[/tex]
[tex]r_2 = (8.1cm, 0)[/tex]
now we have
[tex]r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}[/tex]
[tex]r_{cm} = {(37.26, 0)}{14.8}[/tex]
[tex]r_{cm} = (2.52 cm, 0)[/tex]
so center of mass of the two masses will lie at x = 2.52 cm
now for center of gravity we can use
[tex]r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}[/tex]
here we have
[tex]m_1 = 10.2 kg[/tex]
[tex]m_2 = 4.6 kg[/tex]
[tex]r_1 = (0, 0)[/tex]
[tex]r_2 = (8.1cm, 0)[/tex]
now we have
[tex]r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}[/tex]
[tex]r_g_{cm} = {(37.26, 0)}{14.8}[/tex]
[tex]r_g_{cm} = (2.52 cm, 0)[/tex]
So center of mass is same as center of gravity because value of gravity is constant here
The center of mass and center of gravity for the two given masses are both located at x = 2.49 cm from the origin. In typical situations where gravitational forces are uniform, the center of gravity will coincide with the center of mass.
Explanation:The location of the center of mass (xcom) and the center of gravity (xcog) for two bodies or particles can be calculated using the following formula: xcom = m1x1 + m2x2 / (m1 + m2) where m1, m2 are the masses and x1, x2 are their respective positions. In this case, m1 = 10.2 kg, x1 = 0 cm (since it's at the origin), m2 = 4.6 kg, and x2 = 8.1 cm.
Using these values, the formula becomes xcom = (10.2 kg * 0 cm + 4.6 kg * 8.1 cm) / (10.2 kg + 4.6 kg) = 36.86 cm / 14.8 kg = 2.49 cm. So, the center of mass is at x = 2.49 cm from the origin.
The concept of center of gravity is a specific application of the center of mass in the presence of a gravitational field. In everyday circumstances, where gravitational forces are uniform, the center of gravity coincides with the center of mass. Therefore, for this case, the center of gravity (xcog) will also be at x = 2.49 cm from the origin.
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What will an object weigh on the Moon's surface if it weighs 100 N on Earth's surface? (b) How many Earth radii must this same object be from the center of Earth if it is to weigh the same as it does on the Moon?
Answer:
Part a)
W = 16.7 N
Part b)
r = 2.45 R
Explanation:
Part a)
As we know that acceleration due to gravity on the surface of moon is 1/6 times the gravity on the surface of earth
So the force due to gravity will decrease by the factor of 6
so we will have
[tex]W_{moon} = \frac{1}{6}W_{earth}[/tex]
[tex]W_{moon} = \frac{1}{6}(100)[/tex]
[tex]W_{moon} = 16.7 N[/tex]
Part b)
For the same value of the weight as the surface of moon the acceleration due to gravity of earth must be 1/6 times
so we have
[tex]\frac{GM}{r^2} = \frac{GM}{6R^2}[/tex]
[tex]r^2 = 6R^2[/tex]
[tex]r = 2.45 R[/tex]
The object must be [tex]\( \sqrt{6} \)[/tex] Earth radii from the center of Earth to weigh the same as it does on the Moon.
The weight of an object on the Moon's surface is approximately 1/6 of its weight on Earth's surface due to the Moon's lower gravitational acceleration. Therefore, the object that weighs 100 N on Earth will weigh:
[tex]\[ W_{\text{Moon}} = \frac{1}{6} \times W_{\text{Earth}} = \frac{1}{6} \times 100 \text{ N} = 16.67 \text{ N} \][/tex]
For the second part of the question, we need to find the Earth radius multiple where the gravitational acceleration is the same as that of the Moon. The gravitational acceleration g on Earth is approximately [tex]\( 9.81 \, \text{m/s}^2 \)[/tex], and on the Moon, it is [tex]\( \frac{9.81}{6} \, \text{m/s}^2 \)[/tex]. The gravitational acceleration ( g' ) at a distance r from the center of Earth is given by:
[tex]\[ g' = g \left( \frac{R_{\text{Earth}}}{r} \right)^2 \][/tex]
where [tex]\( R_{\text{Earth}} \)[/tex] is the radius of Earth. We want to find r such that ( g' ) is equal to the gravitational acceleration on the Moon:
[tex]\[ g \left( \frac{R_{\text{Earth}}}{r} \right)^2 = \frac{g}{6} \][/tex]
Solving for r:
[tex]\[ \left( \frac{R_{\text{Earth}}}{r} \right)^2 = \frac{1}{6} \] \[ \frac{R_{\text{Earth}}}{r} = \sqrt{\frac{1}{6}} \] \[ r = R_{\text{Earth}} \sqrt{6} \][/tex]
What is the intensity of a sound with a sound intensity level (SIL) 67 dB, in units of W/m^2?
Answer:
The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²
Explanation:
We have expression for sound intensity level (SIL),
[tex]L=10log_{10}\left ( \frac{I}{I_0}\right )[/tex]
Here we need to find the intensity of sound (I).
[tex]L=10log_{10}\left ( \frac{I}{I_0}\right )\\\\log_{10}\left ( \frac{I}{I_0}\right )=0.1L\\\\\frac{I}{I_0}=10^{0.1L}\\\\I=I_010^{0.1L}[/tex]
Substituting
L = 67 dB and I₀ = 10⁻¹² W/m² in the equation
[tex]I=I_010^{0.1L}=10^{-12}\times 10^{0.1\times 65}\\\\I_0=10^{-12}\times 10^{6.5}=10^{-5.5}=3.16\times 10^{-6}W/m^2[/tex]
The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²
A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 35.0 m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?
Answer:
v = 14.35 m/s
Explanation:
As we know that crate is placed on rough bed
so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate
So here we have
[tex]\mu mg = \frac{mv^2}{R}[/tex]
here we have
[tex]\mu g = \frac{v^2}{R}[/tex]
now we know that
[tex]v = \sqrt{\mu Rg}[/tex]
here we have
[tex]\mu = 0.600[/tex]
R = 35 m
g = 9.81 m/s/s
now plug in all values in above equation
[tex]v = \sqrt{(0.600)(35)(9.81)}[/tex]
[tex]v = 14.35 m/s[/tex]
At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball's acceleration is 900 m/s^2 and the vertical or y component of its acceleration is 870 m/s^2. The ball's mass is 0.31 kg. What is the magnitude of the net force acting on the soccer ball at this instant?
Answer:
390 N
Explanation:
The net acceleration is:
a² = aₓ² + aᵧ²
a² = (900 m/s²)² + (870 m/s²)²
a = 1250 m/s²
So the net force is:
F = ma
F = (0.31 kg) (1250 m/s²)
F = 390 N
The Large Magellanic Cloud is a small galaxy that orbits the Milky Way. It is currently orbiting the Milky Way at a distance of roughly 160000 light-years from the galactic center at a velocity of about 300 km/s.
1) Use these values in the orbital velocity law to get an estimate of the Milky Way's mass within 160000 light-years from the center. (The value you obtain is a fairly rough estimate because the orbit of the Large Magellanic Cloud is not circular.)
Express your answer using one significant figure.
Answer: [tex]2(10)^{42}kg [/tex]
Explanation:
Approaching the orbit of the Large Magellanic Cloud around the Milky Way to a circular orbit, we can use the equation of velocity in the case of uniform circular motion:
[tex]V=\sqrt{G\frac{M}{r}}[/tex] (1)
Where:
[tex]V=300km/s=3(10)^{5}m/s[/tex] is the velocity of the Large Magellanic Cloud's orbit, which is assumed as constant.
[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
[tex]M[/tex] is the mass of the Milky Way
[tex]r=160000ly=1.51376(10)^{21}m[/tex] is the radius of the orbit, which is the distance from the center of the Milky Way to the Large Magellanic Cloud.
Now, if we want to know the estimated mass of the Milky Way, we have to find [tex]M[/tex] from (1):
[tex]M=\frac{V^{2} r}{G}[/tex] (2)
Substituting the known values:
[tex]M=\frac{(3(10)^{5}m/s)^{2}(1.51376(10)^{21}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex] (3)
[tex]M=\frac{1.362384(10)^{32}\frac{m^{3}}{s^{2}}}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex]
Finally:
[tex]M=2.0416(10)^{42}kg\approx 2(10)^{42}kg[/tex] >>>This is the estimated mass of the Milky Way
To estimate the Milky Way's mass within 160,000 light-years from the center, we can use the orbital velocity law. The estimated mass of the Milky Way within 160,000 light-years from the center is approximately 4.12 × 10^41 kg.
Explanation:To estimate the Milky Way's mass within 160,000 light-years from the center, we can use the orbital velocity law. The formula for this law is v = sqrt(GM/r), where v is the velocity, G is the gravitational constant, M is the Milky Way's mass, and r is the distance from the center. Rearranging the equation to solve for M, we get M = v^2 * r / G.
Using the given values of v = 300 km/s and r = 160,000 light-years, we need to convert them to appropriate units. 1 light-year is approximately 9.461 × 10^15 meters. After performing the necessary conversions and calculations, the estimated mass of the Milky Way within 160,000 light-years from the center is approximately 4.12 × 10^41 kg.
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he work function of a certain metal is 1.90 eV. What is the longest wavelength of light that can cause photoelectron emission from this metal? (1 eV = 1.60 × 10-19 J, c = 3.00 × 108 m/s, h = 6.626 × 10-34 J ∙ s)
Answer:
6538.8 Angstrom
Explanation:
work function, w = 1.9 eV = 1.9 x 1.6 x 10^-19 J = 3.04 x 10^-19 J
Let the longest wavelength is λ.
W = h c / λ
λ = h c / W
λ = (6.626 x 10^-34 x 3 x 10^8) / (3.04 x 10^-19)
λ = 6.5388 x 10^-7 m = 6538.8 Angstrom
Thus, the longest wavelength is 6538.8 Angstrom.
A 2.0-kg block is on a perfectly smooth (frictionless) ramp that makes an angle of 30^\circ30 ∘ with the horizontal. What is the force of the ramp on the block?
The force of the ramp on the block is 2.45 N.
Explanation:To find the force of the ramp on the block, we need to consider the forces acting on the block. The only force acting on the block along the ramp is the component of the weight of the block that is parallel to the ramp. This force can be calculated using the equation:
F = mg sin(theta)
where F is the force, m is the mass of the block, g is the acceleration due to gravity, and theta is the angle of the ramp. Plugging in the values, we get:
F = (2.0 kg)(9.8 m/s^2) sin(30 degrees) = 4.9 N sin(30 degrees) = 2.45 N
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A battery can be modeled as a source of emf in series with both a resistance of 10 Ω and an inductive reactance of 5 Ω. For a maximum power delivered to the load, the load should have a resistance of RL = 10 Ω, an inductive reactance of zero, and a capacitive reactance of 5 Ω. With this load, is the circuit in resonance?
Answer:
Yes circuit is in resonance
Explanation:
As we know that the circuit consist of a resistance of 10 ohm and an inductive reactance of 5 ohm
This circuit is connected to a load with having value of resistance of 10 ohm and capacitive reactance of 5 ohm
now they are all in series so as we know that capacitor and inductor is in opposite phase so here reactance of inductor is cancelled by the reactance of capacitor due to the opposite phase of them also they are same in magnitude
so we have
[tex]x_L = x_C[/tex]
so the circuit is purely resistive and hence we can say that it must be in resonance
An ion jet is accelerated by a potential difference of 10000 V, before entering a magnetic field of 1T. If the ions describe a circular path of 5 cm radius, determine the charge-mass relationship.
Answer:
Ratio of charge to mass is 8 x 10⁶
Explanation:
q = charge on the ion
V = potential difference = 10000 Volts
B = magnitude of magnetic field = 1 T
m = mass of the ion
[tex]v[/tex] = speed gained by the ion due to potential difference
using conservation of energy
kinetic energy gained by ion = electric potential energy lost
[tex](0.5)mv^{2}=qV[/tex]
[tex]mv^{2} = 2qV[/tex] eq-1
r = radius of the circular path described = 5 cm = 0.05 m
radius of circular path is given as
[tex]r = \frac{mv}{qB}[/tex]
taking square both side
[tex]r^{2} = \frac{m^{2}v^{2}}{q^{2}B^{2}}[/tex]
using eq-1
[tex]r^{2} = \frac{2mqV}{q^{2}B^{2}}[/tex]
[tex]r^{2} = \frac{2mV}{qB^{2}}[/tex]
[tex]\frac{q}{m} = \frac{2V}{r^{2}B^{2}}[/tex]
inserting the values
[tex]\frac{q}{m} = \frac{2(10000)}{(0.05)^{2}(1)^{2}}[/tex]
[tex]\frac{q}{m}[/tex] = 8 x 10⁶
Ratio of charge to mass is 8 x 10⁶
Two identical loudspeakers are some distance apart. A person stands 5.20 m from one speaker and 4.10 m from the other. What is the third lowest frequency at which destructive interference will occur at this point? The speed of sound in air is 344 m/s.
Answer:
125.09 Hz
Explanation:
Given : A person stands 5.20 from one speaker and 4.10 m away from the other speaker.
Distance between the speakers is 5.20 - 4.10 =1.10 m
We know that
For destructive interferences
n . λ / 2
where n =1,3,5,7....
Therefore difference between the speakers is
5.10 - 4.20 = 1 X 0.5 λ
λ = 2.2
Given velocity of sound is V = 344 m/s
Therefore frequency, f = [tex]\frac{v}{\lambda }[/tex]
= [tex]\frac{344}{2.2 }[/tex]
= 156.36 Hz
Now, the third lowest frequency is given by
λ = (5.20-4.10) X 5 X 0.5
= 2.75 m
Therefore frequency, f = [tex]\frac{v}{\lambda}[/tex]
= [tex]\frac{344}{2.75}[/tex]
= 125.09 Hz
Therefore third lowest frequency is 125.09 Hz
Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. What is the voltage across an 8.47-nm-thick membrane if the electric field strength across it is 8.76 MV/m? You may assume a uniform E-field.
Answer:
Voltage, V = 0.0741 volts
Explanation:
It is given that,
Width of membrane, [tex]d=8.47\ nm=8.47\times 10^{-9}\ m[/tex]
Electric field strength, [tex]E=8.76\ MV/m=8.76\times 10^6\ V/m[/tex]
We need to find the voltage across membrane. Electric field is given by :
[tex]E=\dfrac{V}{d}[/tex]
[tex]V=E\times d[/tex]
[tex]V=8.76\times 10^6\ V/m\times 8.47\times 10^{-9}\ m[/tex]
V = 0.0741 V
So, the voltage across the membrane is 0.0741 volts. Hence, this is the required solution.
. In an elastic collision, what happens to the change in kinetic energy? A) It is transformed into heat and also used to deform colliding objects. B) It is converted into potential energy. C) It is transformed into momentum such that momentum is conserved. D) All of the above. E) None of the above.
Explanation:
There are two types of collision. First one is elastic and other one is inelastic collision.
The linear momentum of two objects before and after the collision remains the same in case of elastic collision. Also, the kinetic energy in elastic collision is conserved. But in case of inelastic collision, the momentum of two objects before and after the collision remains constant but the kinetic energy is in conversed. It changes form one form of energy to another.
Hence, the correct option is (E) "None of the above".
C) It is transformed into momentum such that momentum is conserved. (This statement is incorrect. Conservation of momentum applies to all collisions, not just the change in kinetic energy)
In an ideal elastic collision, the following happens:
Total Kinetic Energy is Conserved: The total kinetic energy of the system (colliding objects) before the collision is equal to the total kinetic energy after the collision. No energy is lost or gained in terms of kinetic energy.
Here's why the other options are incorrect:
A) Heat and Deformation: While real collisions might involve some energy loss due to heat and deformation, an ideal elastic collision is considered perfectly elastic, meaning no energy is lost in these ways.
B) Potential Energy: The collision doesn't necessarily involve a change in potential energy of the objects.
C) Momentum: Momentum is always conserved in any collision, elastic or inelastic.
D) All of the Above: Only the conservation of total kinetic energy applies to an ideal elastic collision.
E) None of the Above: In an ideal elastic collision, kinetic energy is conserved.
Therefore, the correct answer is:
C) It is transformed into momentum such that momentum is conserved. (This statement is incorrect. Conservation of momentum applies to all collisions, not just the change in kinetic energy)