There is an omission of some sentences in the question which affects the answering of question B and C, so we will based the omission of the sentences on assumption in order to solve the question that falls under it.
NOTE: The omitted sentences are written in bold format
A tennis ball has a mass of 0.059 kg. A professional tennis player hits the ball hard enough to give it a speed of 41 m/s (about 92 miles per hour.) The ball moves toward the left, hits a wall and bounces straight back to the right with almost the same speed (41 m/s).
As indicated in the diagram below, high-speed photography shows that the ball is crushed about d = 2.0 cm at the instant when its speed is momentarily zero, before rebounding.
A) What is the average speed of the ball during the period from first contact with the wall to the moment the ball's speed is momentarily zero?
B) How much time elapses between first contact with the wall, and coming to a stop?
C) What is the magnitude of the average force exerted by the wall on the bal dring contact?
D) In contrast, what is the magnitude of the gravitational force of the Earth on the ball?
Answer:
a) [tex]V_{avg} = 20.5m/s[/tex]
b) 9.76 × 10⁻⁴s
c) 247.9 N
d) 5.8 N
Explanation:
Given that;
Initial speed [tex](V_i)[/tex] = 0
Final speed [tex](V_f)[/tex] = 41 m/s
Distance (d) = 0.002
mass (m) = 0.059 kg
g = 9.8 m/s²
a)
The average speed of the ball can be calculated as;
[tex]V_{avg} = \frac{V_i+V_f}{2}[/tex]
[tex]V_{avg} = \frac{0+41}{2}[/tex]
[tex]V_{avg} = 20.5m/s[/tex]
b)
The time elapsed can be calculated by using the second equation of motion which is given as:
[tex]S=(\frac{V_i+V_f}{2})t[/tex]
If we make time (t) the subject of the formula; we have:
[tex](V_i+V_f)t=2S[/tex]
[tex]t= (\frac{2S}{V_I+V_f})[/tex]
[tex]=\frac{2(0.02)}{41+0}[/tex]
[tex]=\frac{0.04}{41}[/tex]
= 0.000976
=9.76 × 10⁻⁴s
c)
the magnitude of the average force (F) exerted by the wall on the bal dring contact can be determined using;
Force (F) = mass × acceleration
where acceleration [tex](a)= \frac{Vo}{t}[/tex]
[tex]\frac{41}{0.00976}[/tex]
acceleration (a) = 4200.82 m/s²
F = m × a
= 0.059 × 4200.82
= 247.85
≅ 247.9 N
d)
the magnitude of the gravitational force of the Earth on the ball
Force (F) = mass (m) × gravity (g)
= 0.059kg × 9.8 m/s²
= 5.782 N
≅ 5.8 N
The average speed of the tennis ball during contact with the wall is zero, and without the time of contact, we cannot determine the time elapsed or the average force exerted by the wall. However, the gravitational force on the ball is 0.5782 N.
Explanation:The question relates to the change in momentum and the forces involved when a tennis ball bounces off a wall. Specifically, a tennis ball with a mass of 0.059 kg is hit at a speed of 41 m/s, bounces off a wall, and comes back at the same speed. To tackle the posed questions, it is essential to apply concepts from Newton's laws of motion and the conservation of momentum.
Part AThe average speed of the ball during contact is zero since the speed decreases uniformly from 41 m/s to zero.
Part BWithout the time of contact with the wall, this cannot be determined. Previous examples of collisions show time of contact can vary, so it must be provided to answer this part of the question.
Part CTo calculate the magnitude of the average force exerted by the wall on the ball, we would need the time of contact with the wall. Since it is not given, this cannot be calculated accurately.
Part DThe magnitude of the gravitational force of the Earth on the ball is calculated as the product of the mass of the ball and the acceleration due to gravity (9.8 m/s²), which is 0.059 kg * 9.8 m/s² = 0.5782 N.
The atmosphere of the sun consists mostly of hydrogen atoms (not molecules) at a temperature of 6000 K. What are (a) the average translational kinetic energy per atom and (b) the rms speed of the atoms
Explanation:
(a) Formula for average translational kinetic energy of a particle is as follows.
U = [tex]\frac{3}{2}(\frac{RT}{N})[/tex]
where, R = Reydberg's constant
T = absolute temperature
N = Avogadro's number
Therefore, we will calculate value of average translational kinetic energy as follows.
U = [tex]\frac{3}{2}(\frac{RT}{N})[/tex]
= [tex]\frac{3}{2}(\frac{8.314 J/mol K \times 6000 K}{6.023 \times 10^{23} mol^{-1}})[/tex]
= [tex]1.24 \times 10^{-19}[/tex] J
Therefore, value of average translational kinetic energy is [tex]1.24 \times 10^{-19}[/tex] J.
(b) Formula for average kinetic energy is as follows.
K.E = [tex]\frac{1}{2}(\frac{M}{N})v^{2}_{rms}[/tex]
Here, M = molar mass = 1 kg/K mol
And, the average kinetic energy is equal to the average translational kinetic energy.
Hence, K.E = U
[tex]\frac{1}{2}(\frac{M}{N})v^{2}_{rms}[/tex] = [tex]\frac{3}{2}(\frac{RT}{N})[/tex]
[tex]v^{2}_{rms} = \frac{3RT}{M}[/tex]
[tex]v = \sqrt{\frac{3RT}{M}}[/tex]
therefore, we will calculate r.m.s speed of the given atom as follows.
[tex]v = \sqrt{\frac{3RT}{M}}[/tex]
[tex]v = \sqrt{\frac{3 \times 8.314 J/mol K \times 6000 K}{1 kg/K mol}}[/tex]
= 386.84 m/s
Hence, value of r.m.s speed of the given atom is 386.84 m/s.
Following are the solution to the given points:
Given:
Temperature [tex]= 6000\ K[/tex]
To find:
kinetic energy =?
rms speed of the atoms=?
Solution:
For point a)
[tex]\to Avg \ KE =\frac{3}{2}\ KT \\\\[/tex]
[tex]\to KE_{avg} =1.5 \times 1.38 \times 10^{-23} \times 6000 \\\\[/tex]
[tex]=1.5 \times 1.38 \times 10^{-23} \times 6000 \\\\=2.07 \times 10^{-23} \times 6000 \\\\=2.07 \times 10^{-23} \times 6000 \\\\=12.420 \times 10^{-20}\ J[/tex]
For point b)
[tex]\to[/tex] rms speed [tex]V_{rms}[/tex]:
[tex]\to v_{rms}=\sqrt{\frac{3RT}{M}}[/tex]
[tex]=\sqrt{\frac{3 \times 8314 \times 6000}{1}} \\\\=\sqrt{\frac{24942\times 6000}{1}} \\\\=\sqrt{149652000}\\\\=12233.233\\\\ = 12233\ \frac{m}{s}[/tex]
Therefore, the final answer is "[tex]12.420 \times 10^{-20} \ J\ \ and \ \ 12233 \ \frac{m}{s}[/tex]".
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A library of e-books contains metadata for each book. The metadata is intended to help a search feature find
books that users are interested in. Which of the following is LEAST likely to be contained in the metadata of each e-book?
a. An archive containing previous versions of the e-book
b. The author and title of the e-book
c. The date the e-book was first published
d. The genre of the e-book (e.g., comedy, fantasy, romance, etc.)
The metadata is intended to help a search feature the least likely feature contained in the metadata is An archive containing previous versions of the e-book so, option A is correct.
What are e-books?The term "e-book" refers to a digital file with a body of text and graphics that is suited for electronic distribution and displays on-screen similarly to a printed book.
E-books can be made by converting a printer's source files into forms that are simple to download and read on a screen, or they can be taken from a database or a collection of text files that weren't made just for printing.
When businesses like Peanut Press started offering book content for reading on personal digital assistants (PDAs), handheld devices that were the forerunners of today's smartphones and tablet computers, the market for buying and selling e-books first became a mainstream industry in the late 1990s.
Thus, metadata does not contain a feature to find an archive containing previous versions of the e-book.
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A bicyclist is moving toward a sheer wall while holding a tuning fork rated at 488 Hz. 1) If the bicyclist detects a beat frequency of 8 Hz (between the waves coming directly from the tuning fork and the echo waves coming from the sheer wall), calculate the speed of the bicycle. Assume the speed of sound is 343 m/s.
Answer:
speed of the bicycle = 2.81 m/s
Explanation:
given data
tuning fork rated f = 488 Hz
beat frequency fb = 8 Hz
speed of sound vs = 343 m/s
solution
we get here frequency that rider is listening that is
rider hearing frequency = 488 Hz + 8 Hz
rider hearing frequency = 496 Hz
and when it bouncing from sheer wall at frequency of related to stationary object
frequency = 488 Hz + 4 Hz
frequency = 492 Hz
so now we get speed of the bicycle that is
speed of the bicycle = ( [tex]\frac{492}{488}[/tex] × 343 ) - 343 Hz
speed of the bicycle = 2.81 m/s
The speed of the bicycle is 2.81 m/s.
Based on the given information,
• The tuning fork rated at frequency (f) 488 Hz.
• The beat frequency (fb) given is 8Hz.
• The speed of the sound is 343 m/s.
Now with the help of the given information, the rider is hearing a frequency of,
[tex]=(f+fb)\\=(488+8)\\=496 Hz[/tex]
However, it is bounding from sheer wall at the frequency of,
[tex]=(488+\frac{8}{2})\\= 492 Hz[/tex]
Now the speed of the bicycle will be,
[tex]=(\frac{492}{488} * 343)-343\\=2.811 m/s\\= 2.81 m/s[/tex]
Thus, the speed of the bicycle is 2.81 m/s.
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As a science project, you drop a watermelon off the top of the Empire State Building, 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a speed of 39.0 m/s.
How fast is the watermelon going when it passes Superman?
Answer:
The velocity of watermelon when it passes Superman is 78 m/s.
Explanation:
Height of the building, d = 320 m
Speed of the superman, v = 39 m/s
We need to find the speed of watermelon when it passes Superman. Let t is the time taken by the watermelon. So,
[tex]d=ut+\dfrac{1}{2}gt^2[/tex]
Here u = 0
[tex]d=\dfrac{1}{2}gt^2[/tex]
[tex]39t=\dfrac{1}{2}\times 9.8\times t^2[/tex]
t = 7.95 seconds
Let v is the speed of the watermelon. It is given by :
[tex]v=gt[/tex]
[tex]v=9.8\times 7.95[/tex]
v = 77.91 m/s
or
v = 78 m/s
So, the velocity of watermelon when it passes Superman is 78 m/s.
A stack of bricks weighs 170 KN (Kilo-newtons). The stack exerts 180 KPa (Kilo-pascals) of pressure on the ground. What is the area upon which this pressure is exerted (in square ft)?
Answer:
10.12square feet
Explanation:
Pressure exerted on the object is defined as the ratio of the force exerted on it to its unit area. Mathematically, Pressure = Force/Area
Given the force = 170kN
Pressure = 180KPa
Area = Force /Pressure
Area = 170kN/180KPa
Area = 0.94N/Pa
Note that 1Newton/Pascal = 10.764square feet
Therefore 0.94N/Pa = x
x = 0.94× 10.764
x = 10.12square feet
Therefore the area upon which this pressure is exerted is 10.12sqft.
We divide the electromagnetic spectrum into six major categories of light, listed below. Rank these forms of light from left to right in order of increasing wavelength. To rank items as equivalent, overlap them.a.gamma raysb.X raysc.ultravioletd.visible lighte.infraredf.radio waves
Answer:
(1.) gamma rays. (2.) X-rays. (3.) ultraviolet. (4.) visible light. (5.) infrared.
(6.) radio waves
Explanation:
Notice that these wavelengths span an enormous range. The wavelengths of gamma rays can be smaller than the size of an atomic nucleus, while the wavelengths of radio waves can be many meters (or even milometers) long. Visible light spans only a very narrow range of wavelengths, from about 400 nano meters at the blue (violet) end to about 700 nano meters at the red end.
The types of light in the electromagnetic spectrum, ranked from shortest to longest wavelength, are gamma rays, X-rays, ultraviolet, visible light, infrared, and radio waves.
Explanation:The electromagnetic spectrum is made up of different types of waves, each characterized by specific wavelengths. The categories of light in the electromagnetic spectrum from shortest to longest wavelength are:
Gamma raysX-raysUltravioletVisible lightInfraredRadio wavesAs we move from gamma rays to radio waves, the wavelength increases. Gamma rays have the shortest wavelength, while radio waves have the longest.
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The partial pressure of ethanol in the air above a sample of liquid ethanol in a closed container is 19.5 torr, at a given experimental Temperature. Assuming the pressure inside the container is 758.6 torr at the experimental T, what is the molar percent of ethanol in the air?
Answer:
The molar percent of ethanol in the air is 2.6%
Explanation:
Molar percent of ethanol in the air = partial pressure of ethanol/total pressure × 100 = 19.5 torr/758.6 torr × 100 = 2.6%
A student does 60. joules of work pushing a 3.0-kilogram box up the full length of a ramp that is 5.0 meters long. What is the magnitude of the force applied to the box to do this work?
Answer:
F= 12 N
Explanation:
Given that
Work done by student ,W= 60 J
The mass of the box ,m = 3 kg
Length ,x = 5 m
We know that ,The work done by a force a force F is given as
W= F .x
x=Displacement
F=Force
W=Work
Now by putting the values
60 = F x 5
[tex]F=\dfrac{60}{5}\ N[/tex]
F= 12 N
That is why the magnitude of the force will be 12 N.
The magnitude of the force applied to the box to do the work is 12 N
Definition of workWorkdone is defined as the product of force and distance moved in the direction of the force. Mathematically, it can be expressed as:
Workdone (Wd) = force (F) × distance (d)
Wd = Fd
With the above formula, we can obtain the force used to do the work.
Determination of the force•Workdone (Wd) = 60 J
•Distance (d) = 5 m
•Force (F) =?
Wd = Fd
60 = F × 5
Divide both side by 5
F = 60 / 5
F = 12 N
Thus, 12 N is required to do the work.
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Ask Your Teacher It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an altitude of 340 m the field has magnitude 60.0 N/C. At an altitude of 220 m, the magnitude is 100 N/C. Find the net amount of charge contained in a cube 120 m on edge, with horizontal faces at altitudes of 220 and 340 m. Neglect the curvature of Earth.
Answer:q=3.536+10^-6C
Explanation:
wooden floor at a constant speed of 1.0 m/s. The coefficient of kinetic friction is 0.15. Now you double the force on the box. How long would it take for the velocity of the crate to double to 2.0 m/s?
Answer:
Time = 1.36s
Explanation:
coefficient of kinetic friction = μk
μk = F/N = F / mg
where m = mass of the body
g = acceleration due to gravity.
Doubling the force on the box we have,
μk = 2ma/mg
hence μk = 0.15/2 = 0.075
To determine the time it takes to reach a velocity of 2m/s from 1m/s .
From Newton law of motion,
v = u + a*t
for a deceleration of a = μk x g
2m/s = 1m/s + 0.075 x 9.8m/s² x t
t = (2m/s - 1m/s)/0.735m/s²
t = 1.36 seconds..
it will take the 1.36s for the crate to double to 2.0m/s
An Olympic-class sprinter starts a race with an acceleration of 5.10 m/s2. What is her speed 2.40 s later?
Answer:
12.24 m/s
Explanation:
Speed: This can be defined as the rate of change of distance with time. The S.I unit of speed is m/s.
Using the formula,
a = v/t................ Equation 1
Where a = acceleration of the sprinter, v = speed of the sprinter, t = time.
making v the subject of the equation,
v = at ................. Equation 2
Given: a = 5.1 m/s², t = 2.4 s.
Substitute into equation 2
v = 5.1(2.4)
v = 12.24 m/s.
Hence, the speed of the sprinter = 12.24 m/s
If a word problem gives the separation distance between two charges in centimeters (cm), you should ____. Question 4 options:
1) enter that value directly since cm is the standard unit
2) convert that distance to inches, the standard unit
3) convert that distance to meters, the standard unit
4) convert that distance to whatever units are given in the second distance
5) look to see whether the charges are like or opposite to know what to do
Answer:3
Explanation:
Convert the distance to meters,the standard units.
Final answer:
When a word problem provides the separation distance between two charges in centimeters, you should (3) convert that distance to meters, the standard SI unit for measurements in physics. Hence, (3) is the correct option.
Explanation:
If a word problem gives the separation distance between two charges in centimeters (cm), you should convert that distance to meters, the standard unit. Physicists use the International System of Units (SI units) for calculations to ensure consistency and comparison of data.
Since the SI unit for displacement is the meter (m), converting from centimeters to meters is necessary for performing electrostatic calculations involving Coulomb's law or any other physics formula.
For instance, when using Coulomb's law to find the force between two charges, the formula is F = k * (q₁ * q₂ / r²), where k is Coulomb's constant (8.987 x 10⁹ N m²/C²), q₁ and q₂ are the charges, and r is the separation distance in meters. If the problem states the distance in centimeters, you would convert the distance to meters by dividing by 100, since there are 100 centimeters in a meter.
Newton's second law A. describes how an object accelerates when a force is applied B. says that objects eventually stop unless a force is applied C. objects with mass attract each other D. forces come in action/reaction pairs E. an object will remain in uniform motion unless acted upon by a force F. like charges repel, opposite charges attract
Answer:
A. describes how an object accelerates when a force is applied
Explanation:
Newton's second law of motion concerns the behavior of objects that do not have a stability between all established forces. The second law states that an object's acceleration depends on two factors: the net force on the entity and the entity's mass.
The doppler effect is the change in observed frequency due to
Answer: relative motion between observer and the sound source.
Explanation: The Doppler effect states that when there is a relative motion between an observer and a sound source the frequency of sound perceived by the observer is different in frequency from the original from the source.
The mathematical back up for this claim is given below.
f' = (v+v') /(v-vs) × f
Where f' = observed frequency
v = speed of sound in air
v' = velocity of observer
vs = velocity of source
f = frequency of sound source.
From the formulae, it can be seen that a change in the value of the velocity of observer (v') and source (vs) produces different value of observed frequency (f').
Note, frequency of sound (f) is a constant.
The current through two identical light bulbs connected in a series is 0.25 amps. The voltage across both bulbs is 100 volts. Find the resistance of a single light bulb.
Answer: 200ohms
Explanation:
According to Ohm's law, the current (I) passing through a metallic conductor is directly proportional to the potential difference (v) between its ends at constant temperature and pressure. Mathematically, E=IRt
Where E is the voltage across both bulbs, I is the current, Rt is the total equivalent resistance
E = 100V, I = 0.25amps Rt = R+R(since they are 2 identical bulbs in series, we will add them together)
Rt = 2R
Substituting this values in the formula to get R, we have;
100 = 0.25(2R)
100 = 0.5R
R = 100/0.5
R = 200ohms
The resistance of a single light bulb will be 200ohms.
Note that the unit of resistance is ohms
A nonlinear spring is used to launch a toy car. The car is pushed against the spring, compressing the spring 2.5 cm. The force the spring exerts on the car is given by the equation F=−Kx2, where K=5000 Nm2. The potential energy stored in the spring when the car is pushed against it is most nearly:________________
The potential energy stored in the spring when the car is pushed against it is most nearly is 0.026 J.
The calculation is as follows:Non linear spring is
[tex]F = -kx^2 \rightarrow force[/tex]
The potential energy
[tex]\frac{dU}{dX} = -F\\\\U = \int\limits {-Fdx}\\\\U = + \int\limits^x_0 {kx^2dx}\\\\U = \frac{kx^3}{3} \\\\U = 5000 \div 3 (\frac{2.5}{100}^)2\\\\U = \frac{5000 \times 2.5\times 2.5\times 2.5}{3\times 100\times 100\times 100}[/tex]
= 0.026 J
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A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.50 s to reach the water.
A. How high above the water was the preserver released?
B. List the knowns in this problem.
Answer:
13.125 m
Explanation:
b) The quantity which are known as
Initial velocity of life preserver, u = 1.40 m/s
time to reach, t = 1.50 s
final velocity of the life preserver, v = 0 m/s
acceleration due to gravity = 9.8 m/s²
a) height of the preserver above water
using equation of motion
[tex]h = u t + \dfrac{1}{2}gt^2[/tex]
[tex]h = 1.4\times 1.50 + \dfrac{1}{2}\times 9.8\times 1.5^2[/tex]
h = 13.125 m
Hence, the height from where preserver is thrown is equal to 13.125 m
Final answer:
The life preserver was released from approximately 8.925 meters above the water. This calculation was based on the initial velocity, time it took to reach the water, and the acceleration due to gravity, making use of kinematic equations.
Explanation:
To find out how high above the water the life preserver was released, we can use the kinematic equations for uniformly accelerated motion. Since there is an initial velocity and the acceleration due to gravity is applicable, the following kinematic equation can be used:
h = vot + ½at²
Where:
h is the height from which the preserver is released (unknown)vo is the initial velocity = 1.40 m/s (given)t is the time the preserver takes to hit the water = 1.50 s (given)a is the acceleration due to gravity = -9.8 m/s²Plugging in the known values into the kinematic equation, we get:
h = (1.40 m/s)(1.50 s) + ½(-9.8 m/s2)(1.50 s)²
Calculating this gives us:
h = 2.10 m - 11.025 m = -8.925 m
Since height can't be negative, we take the absolute value which is 8.925 m. Hence, the life preserver was released approximately 8.925 m above the water.
Knowns
Initial velocity (vo) = 1.40 m/sTime (t) = 1.50 sAcceleration due to gravity (a) = -9.8 m/s²Oxygen initially is at 40 F and 16.8 Ibf/in?. It fills a closed, rigid 5 ft³ tank filled with a paddle wheel. During the process, paddle wheel provides 4 Btu of energy transfer by work to the gas. During the process, Gas temperature increases up to 90 F. Assuming ideal gas behavior and ignoring K.E and P.E effects, determine the Heat transfer in Btu and the mass of oxygen.
Explanation:
The given data is as follows.
Initial temperature ([tex]T_{1}[/tex]) = [tex]40^{o}F[/tex] = 499.67R
Initial pressure ([tex]P_{1}[/tex]) = 16.8 Psi
Volume (V) = 5 [tex]ft^{3}[/tex]
Work done = 4 Btu
Final temperature ([tex]T_{2}[/tex]) = 90 F
So, R of [tex]O_{2}[/tex] = 0.3353 Psi ft^{3}/lbm R
Therefore, we will calculate the mass of oxygen as follows.
M = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{16.8 \times 5}{0.3353 \times 499.67}[/tex]
= 0.50137 lbm
Therefore, mass of oxygen is 0.50137 lbm.
Now, we will calculate the change in internal energy as follows.
[tex]\Delta U = mC_{v} [T_{2} - T_{1}][/tex]
= [tex]0.50137 \times 0.157 \times (90 - 40)[/tex]
= 3.936 BTU
Relation between heat energy and internal energy is as follows.
[tex]\Delta Q = \Delta u + W.d[/tex]
= [tex]3.936 + (-4)[/tex]
= -0.0642 BTU
Therefore, amount of heat transfer in BTU is 0.0642 BTU.
You are operating a powerboat at night. You see green and white lights on another boat. What should you do?
Answer:
Just maintain the course you are on and your speed.
Explanation:
When you see a green and a white light, then from the port (left) side you will start coming across another craft. You are the stand-alone craft in this scenario and therefore should stay consistent with your pace and direction. The other craft must take important and early measures to keep the craft away from you.
Which one of the following statements is true?
1. Burning coal contributes to sulfur aerosols in the stratosphere; these sulfur aerosols have moderated global warming.
2. Sulfur compounds released into the atmosphere have relatively short lifetimes of only an hour or two.
3. Volcanoes are an anthropogenic source of sulfur compounds.
4. The widespread use of nuclear energy has released significant amounts of sulfur dioxides that have made their way into the stratosphere.
Answer:
1. Burning coal contributes to sulfur aerosols in the stratosphere; these sulfur aerosols have moderated global warming.
Explanation:
Sulfur aerosols both absorb as well as scatter solar radiation coming from the sun , thus saving the earth from getting wormed up . In this way it has moderating effect on the global warming . This is also called global dimming effect. Less of the solar radiation is able to reach earth surface due to these aerosol.
If the L-shaped rod has a moment of inertia I=9kgm2, F1=12N, F2=27N, and again F3=0, how long a time t would it take for the object to move through 45∘ ( π/4 radians)? Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. Express the time in seconds to two significant figures.
Answer:
Time, t = 2.80seconds
Explanation:
Calculating net Torque, T
Torque(net)= F2(0.08m) - F1(0.03)
Torque(net) = (27×0.08)-(12×0.03)
Torque(net)= 2.16 - 0.36 =1.8N/m
Pi/4= Torque(net)/ Inertia
3.142/4= 1.8/9 = 0.2rad/s^2=a
Pi/4= 1/2 at^2
3.142/4=1/2×0.2 ×t^2
0.7855= 0.1 t^2
t = sqrt(0.7856/0.1)
t= sqrt(7.855)
t=2.80 seconds
By Assuming that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. then the time in seconds is 2.80seconds
What are the types of force ?Force can be denoted as the parameter obtained by pushing or pulling of any object which result in object’s interaction or movement, without force the objects can not move, it can be stopped or change the direction.
Force is a quantitative property and it is an interaction between two physical bodies, means an object and its environment, there are different types of forces in nature.
If an object is in its moving state then it will be either static or motion, if it is pushed or pulled the object will change its position and The external push or pull upon the object called as Force.
There are two types of forces such as contact force which are the force occurs when we apply some effort on the object such as Spring Force, Applied Force, Air Resistance Force, Normal Force, Tension Force, Frictional Force
Non-Contact forces can be defined as the force occur from a distance such as Electromagnetic Force, Gravitational Force, Nuclear Force
Calculating net Torque, T, Torque(net)= F2(0.08m) - F1(0.03)
Torque(net) = (27×0.08)-(12×0.03)
Torque(net)= 2.16 - 0.36 =1.8N/m
Pi/4= Torque(net)/ Inertia
3.142/4= 1.8/9 = 0.2rad/s^2=a
Pi/4= 1/2 at^2
3.142/4=1/2×0.2 ×t^2
0.7855= 0.1 t^2
t = sqrt(0.7856/0.1)
t= sqrt(7.855)
t=2.80 seconds
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On the Moon, gravitational field is around 1.6 N/kg. How does this affect gravitational potential and kinetic energy of objects dropped from a height on the Moon as compared with Earth?
Answer:An object with mass in a gravitational field experiences a force called weight. This weight on the moon is about 1/6 but the mass of that object remains the same due to the moon's gravitational field which is less than the earth.
Explanation:There are factors that affect the gravitational potential and kinetic energy . They are:
1. The height of the object in the moon
2.The mass of the object
3. The strength of the gravitational field
4. The change in position of the object.
If you put two 60 W bulbs in series across a 120 V outlet, how much power would each consume if its resistance were constant?
Answer:15 watts
Explanation:
To get the current
Power P=iV
(Where P is power of bulb 60watts, i is current, and V is the voltage 120V
i=P/V
i = 60/120
i = 1/2A
But
V=iR where R is the resiance
Therefore R=V/i
=120÷1/2
=120×2
=240ohms
For series connection, each bulb draws the same current since,
Since the two bulbs are 60-watt bulbs, they will have the same resistance, therefore the voltage across each bulb is the same and equals halve of the applied voltage, 120/2 = 60 volts.
Taken the resistance 240 ohm is constant across the series, we can roughly estimate the current flow and calculate power dissipation in the series connection.
We have 120 V/(240 + 240 ) ohms = 1/4 A.
The power dissipated in each bulb is (1/4) × 60 = 15 watts.
When two 60 W bulbs are connected in series across a 120 V outlet, each bulb consumes 30 W of power because they share the same current, and the voltage is divided equally among them due to their series connection.
When two 60 W bulbs are connected in series across a 120 V outlet, they share the same current because they are in a series circuit. In a series circuit, the current is constant throughout, and the voltage is divided among the components. Since power is related to both voltage and current, we can use the formula P = VI to find out how much power each bulb consumes.
Total Resistance (R_total): In a series circuit, the total resistance is the sum of the individual resistances. Let's assume the resistance of each bulb is R.
Since P = VI, we can rearrange the formula to find current (I):
I = P / V
The power of each bulb is 60 W, and the voltage is 120 V, so the current passing through the circuit is:
I = 60 W / 120 V = 0.5 A
Voltage Across Each Bulb: In a series circuit, the voltage is divided among the components. Each bulb receives a portion of the total voltage. So, the voltage across each bulb is 60 V.
Power Consumed by Each Bulb: Now, we can use the formula P = VI to find the power consumed by each bulb. We know the voltage (60 V) and the current (0.5 A) passing through each bulb:
P = (0.5 A) * (60 V) = 30 W
Each bulb consumes 30 W of power when connected in series across the 120 V outlet.
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An automobile moves at a constant speed over the crest of a hill traveling at a speed of 88.5 km/h. At the top of the hill, a package on a seat in the rear of the car barely remains in contact with the seat. What is the radius of curvature (m) of the hill?
Answer:
[tex]r=61.65m[/tex]
Explanation:
Since the package remains in contact with the car's seat, the package's speed is equal to the car's speed. At the top on the mountain the package's centripetal force must be equal to its weight:
[tex]mg=F_c[/tex]
The centripetal force is defined as:
[tex]F_c=ma_c=\frac{mv^2}{r}[/tex]
Here v is the linear speed of the object and r is the radius of curvature. We need to convert the linear speed to [tex]\frac{m}{s}[/tex]:
[tex]88.5\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=24.58\frac{m}{s}[/tex]
Now, we calculate r:
[tex]mg=\frac{mv^2}{r}\\r=\frac{v^2}{g}\\r=\frac{(24.58\frac{m}{s})^2}{9.8\frac{m}{s^2}}\\\\r=61.65m[/tex]
A beam of light is traveling through a medium at 200,000 km/s. It enters a different medium and speeds up to almost 250,000 km/s. Finally, it enters a third medium and halts. What were the media the light traveled through, in order? Hurry!!!!!!!!!!!
Answer
From the question we can see that the in medium 1 speed of light is 200,000 Km/s where as in Medium 2 speed of light is 250,000 km/s.
We can conclude that medium 1 is denser than Medium 2.
In third medium Light speed halts so, the third medium is opaque.
We can say that
Medium 1 can be either water, glass.
Medium 2 will be gas
Medium 3 will be an opaque material.
Answer:
D
Explanation:
An elevator cab is pulled upward by a cable. The cab and its single occupant have a combined mass of 1510 kg. When that occupant drops a coin, its acceleration relative to the cab is 7.40 m/s2 downward. What is the tension in the cable?
Answer:
Explanation:
Given
Combined mass of cab and elevator is [tex]M=1510\ kg[/tex]
acceleration of coin with respect to elevator [tex]a_{ce}=-7.4\ m/s^2[/tex]
[tex]a_{ce}[/tex] can be written as
[tex]a_{ce}=a_{cg}-a_{eg}[/tex]
where
[tex]a_{cg}[/tex]=acceleration of coin with respect to ground
[tex]a_{eg}[/tex]=acceleration of elevator with respect to ground
[tex]a_{ce}=a_{cg}-a_{eg}[/tex]
[tex]-7.4=-9.8-a_{eg}[/tex]
[tex]a_{eg}=-9.8+7.4=-2.4\ m/s^2[/tex]
Forces on elevator
[tex]T-Mg=Ma_{eg}[/tex]
[tex]T=M(g+a_{eg})[/tex]
[tex]T=1510\times (9.8-2.4)[/tex]
[tex]T=11,174\ N[/tex]
Here we considered downward direction as negative and upward as positive
Charge A is sitting in an electric field you know the following information:________
the magnitude of charge A and the magnitude of both the field and the potential at charge A position.
What would you do to get the potential energy of charge A?
Answer:
The equation that will relate all the given parameters, in other to calculate the potential energy of charge A is:
∆V = ∆U/q, ∆V is potential at charge A position, q is magnitude of charge A, ∆U will be made the subject of the relation, which is the Potential Energy of charge A. The notation "∆" show, the quantities have both in values and final values, in the electric field.(Change in Electric potential and potential energy, due to the effect of the field)
Explanation:
The potential energy of a charged particle (Charge A) in an electric field depends on the magnitude of the charge(Known as stated in the question). However, the potential energy per unit charge has a unique value at any point in the electric field.
normal hearing elsewhere. How much more intense is a 6000 Hz tone than a 80 Hz tone if they are both barely audible to the child?
Explanation:
There is no any direct relationship between the frequency of a sound relative the intensity of the sound.
Frequency is a perception of pitch of the sound whereas intensity determines the loudness of the sound. Intensity is proportional to amplitude of the sound.
[tex]\beta= log_{10}\frac{I}{I_o}[/tex]
So, it would be hard to determine how much more intense is a 6000 Hz tone than a 80 Hz tone.
A hanging titanium wire with diameter 2.0 mm (2.0 × 10-3 m) is initially 2.5 m long. When a 9 kg mass is hung from it, the wire stretches an amount 0.605 mm. A mole of titanium has a mass of 48 grams, and its density is 4.51 g/cm3. Find the approximate value of the effective spring stiffness of the interatomic bond.
To find the approximate value of the effective spring stiffness of the interatomic bond, calculate the stiffness constant of the titanium wire using Young's modulus.
Explanation:To find the approximate value of the effective spring stiffness of the interatomic bond, we need to calculate the stiffness constant of the titanium wire. The stiffness constant, or Young's modulus (Y), is given by the formula Y = F/A/L, where F is the force applied, A is the cross-sectional area of the wire, and L is the original length of the wire.
First, we need to find the force applied. The force can be calculated using the equation F = mg, where m is the mass and g is the acceleration due to gravity.
Next, we need to calculate the cross-sectional area of the wire. The cross-sectional area can be calculated using the formula A = π(r^2), where r is the radius of the wire.
Finally, we can substitute the values into the formula Y = F/A/L to find the Young's modulus of the titanium wire.
A solid conducting sphere with radius 0.75 m carries a net charge of 0.13 nC. What is the magnitude of the electric field at a point located 0.50 from the sphere's center 0.25 beneath the sphere's surface)?
Answer:
Explanation:
given that
Radius =0.75m
Cnet=0.13nC
a. Electric field inside the sphere located 0.5m from the center of the sphere.
The electric field located inside the sphere is zero.
b. The electric field located 0.25m beneath the sphere.
Since the radius is 0.75m
Then, the total distance of the electric field from the centre of the circle is 0.75+0.25=1m
Then
E=kq/r2
K=9e9Nm2/C2
q=0.13e-9C
r=1m
Then,
E= 9e9×0.13e-9/1^2
E=1.17N/C. Q.E.D
Answer:
Magnitude of Electric field E at at point 0.50m which is within the sphere is Zero( i.e E = 0)
Explanation:
It is understand from Guass' law, the electric field in a region enclosed by a conducting sphere is Zero.
It is given that the radius of the sphere is 0.75m, therefore, a point located at 0.50m from the sphere centre 0.25m before the sphere surface still falls inside the sphere, therefore making the electric field at that point zero in magnitude.