Answer:
The answer to the question is
A 100 lb person would weigh 300.33 lbf at the upper atmosphere of Jupiter
Explanation:
To solve the question we note that
Mass of object = 100 lb = 45.35924 kg
Mass of Jupiter = 300×Mass of Earth = 300×5.972 × 10²⁴ kg =1.7916×10²⁷ kg
Radius of Jupiter = 10× Radius of Earth = 10×6,371 km = 63710 km
Gravitational constant, G = 6.67408 × 10⁻¹¹ m³ kg-1 s-2
Gravitational force is given by [tex]F_G= \frac{Gm_1m_2}{r^2}[/tex]
Plugging in the values we get
[tex]F_G[/tex] = [tex]\frac{6.67408*10^{-11}*45.35924*1.7916*10^{27}}{63710^2}[/tex] = 1335.93 N
Converting into lbf gives 1335.93 N *0.2248 lbf/N = 300.33 lbf
On Jupiter, you would weigh 2.5 times more than on Earth due to its stronger gravity. Therefore, if you weighed 100 lb on Earth, you would weigh approximately 250 lb in the upper atmosphere of Jupiter though your weight may slightly vary with altitude. It is important to note that while weight changes with location, mass remains constant.
Explanation:If you weighed 100 lb on Earth, it is interesting to calculate what you would weigh in the upper atmosphere of Jupiter. Since gravitational force is what we perceive as weight, and gravity varies with the mass of the planet and the distance from its center, we can determine the comparative weight. Jupiter has a mass approximately 318 times that of Earth and a radius about 11 times greater. However, for simplicity, the student question references Jupiter's mass as 300 times that of Earth and a radius 10 times that of Earth. This would normally impact the surface gravity calculation significantly.
Now, since it's stated that on Jupiter you would weigh 2.5 times more than on Earth, if you weigh 100 lb on Earth, you would weigh 250 lb on Jupiter. But note that in the upper atmosphere, your weight might be slightly less than this because you would not be at Jupiter's surface, and gravity decreases with altitude.
Lastly, remember the important conceptual distinction: While your weight changes depending on the gravity of the celestial body you are on, your mass remains constant. Your mass represents the amount of matter within you and does not change with location.
slader A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s and it bounces back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball?
Answer: 800N
Explanation:
Given :
Mass of ball =0.8kg
Contact time = 0.05 sec
Final velocity = initial velocity = 25m/s
Magnitude of the average force exerted on the wall by the ball is can be calculated using the relation;
Force(F) = mass(m) * average acceleration(a)
a= (initial velocity(u) + final velocity(v))/t
m = 0.8kg
u = v = 25m/s
t = contact time of the ball = 0.05s
Therefore,
a = (25 + 25) ÷ 0.05 = 1000m/s^2
Therefore,
Magnitude of average force (F)
F=ma
m = mass of ball = 0.8
a = 1000m/s^2
F = 0.8 * 1000
F = 800N
How much work did the movers do (horizontally) pushing a 46.0-kg crate 10.4 m across a rough floor without acceleration, if the effective coefficient of friction was 0.60? Express your answer using two significant figures.
Answer:
The total work done by the mover is 2.81 kJ.
Explanation:
Given the 46 kg crate is displaced by 10.4 meters.
And the acceleration is zero. Also, [tex]\mu_k=0.60[/tex]
let [tex]P[/tex] is applied force, [tex]F_N[/tex] is the net force, [tex]m[/tex] is the mass and [tex]g=9.81\ m/s^2[/tex]
and [tex]\mu_k=0.60[/tex]
[tex]F_N=P- \mu_k\times mg[/tex]
As the acceleration is zero, the net force will also be zero.
[tex]0=P- \mu_k\times mg\\P=\mu_k\times mg[/tex].
[tex]P=0.6\times 46\times 9.81=270.76\ N[/tex]
Now, we know the work done is force times displacement.
So,
[tex]W=P\times d\\W=270.76\times 10.4=2815.90\ J\\W=2.81\ kJ[/tex]
So, the total work done by the mover to displace 46 kg crate by 10.4 meters 2.81 kJ.
Awning windows can be 100% openable and are best used where extreme weather conditions require a tight seal when the window is closed, although these windows are in common use everywhere.1. True2. False
Answer:
The correct option is false
Explanation:
Awning windows are easy to open windows (made of glass) that are hinged at the top and are opened from the bottom. These windows form a slant (of about 45 degree) after opening hence do not open 100%; this slant opening is advantageous in preventing rain from entering the building.
The windows can also be tightly sealed (from the inside) during extreme/cold weather conditions but are not commonly used everywhere in the world because of there limitations (slight openings) which can prevent proper ventilation in hot regions.
The coefficient of performance of a residential heat pump is 1.6. Calculate the heating effect in kJ/s this heat pump will produce when it consumes 4 kW of electrical power.
Answer:
[tex]Q_{H}=6.4kJ/s[/tex]
Explanation:
Given data
Coefficient of performance of a residential heat pump=1.6
Electrical power P=4kW
Required
Heat Q
Solution
The rate of heat produced is given as
[tex]Q_{H}=COP_{HP}Win\\[/tex]
Substitute the given values
So
[tex]Q_{H}=4kW*1.6\\Q_{H}=6.4kJ/s[/tex]
Find the magnitude of the torque produced by a 4.5 N force applied to a door at a perpendicular distance of 0.26 m from the hinge. Answer in units of N · m.
Answer:
The required torque is 1.17 N-m.
Explanation:
The given data :-
The magnitude of force ( F ) = 4.5 N.
The length of arm ( r ) = 0.26 m.
Here given that force is applied at perpendicular means ( ∅ ) = 90°.
The torque ( T ) is given by
T = F * r * sin∅
T = 4.5 * 0.26 * sin 90°
T = 4.5 * 0.26 * 1
T = 1.17 N-m.
The magnitude of the torque produced by the given force perpendicular to the lever is 1.17N.m.
Given the data in the question;
Force; [tex]F = 4.5N[/tex]Perpendicular distance or radius; [tex]r = 0.26m[/tex]Since the force is perpendicular to the lever, Angle; [tex]\theta = 90^o[/tex]Torque; [tex]T = \ ?[/tex]
Torque simply the measure of the force that can cause an object to rotate about an axis. It is expressed as:
[tex]T = rFsin\theta[/tex]
Where r is radius, F is force applied and θ is the angle between the force and the lever arm.
We substitute our given values into the equation;
[tex]T = 0.26m * 4.5N * sin90^o\\\\T = 0.26m * 4.5N * 1\\\\T = 1.17N.m[/tex]
Therefore, the magnitude of the torque produced by the given force perpendicular to the lever is 1.17N.m.
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The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is constant. At 15 degrees Celsius the pressure is 101.3 kPa at sea level and 87.14 kPa at h = 1000m. What is the pressure at an altitude of 3000 m?
Answer:
64.59kpa
Explanation:
See attachment
A pentagon is a polygon with 5 sides of equal length. Imagine a pentagon in the xy-plane with its base (one of the 5 identical sides) extending from the origin along the x-axis (for which y=0) to x = 5.55 cm. The rest of the pentagon has y>0. Now imagine that a current, i = 3.33 Amps, runs along the 4 sides of the pentagon that have y>0 (ie there is no current along the x-axis side). If a uniform magnetic field of magnitude 0.222 Tesla pointing in the positive y-direction pervades the space, then what is the force on the 4-sided wire that carries the current. Answer = _______________ Newtons.
Answer:
Explanation:
Imagine that current runs in all the 5 sides . In that case the pentagon will have a magnetic moment directed in z- direction . So in the magnetic field it will experience a torque equal to MB where M is magnetic moment and B is magnetic field. But net force on it will be zero.
So , force on four sides due to magnetic field will be equal and opposite to force on fifth side if current runs through it.
force on fifth side if current runs through it
= BIL
= .222 x 3.33 x .0555
= 41.03 x 10⁻³ N .
Be sure to answer all parts. Isooctane (C8H18; d = 0.692 g/mL) is used as the fuel during a test of a new automobile engine. How much energy (in kJ) is released by complete combustion of 21.8 gal of isooctane to gases (ΔH o rxn = −5.44 × 103 kJ/mol)? Enter your answer in scientific notation.
Answer:
[tex]2.72\times 10^{6} J[/tex]
Explanation:
Molar mass of isooctane will be 12*8+8=114 g/mole
Combustion of 21.8 gal yields energy of
21.8*3780*0.692*-5440/114=-2721124.6484210 J
In scientific notation, this is [tex]2.72\times 10^{6} J[/tex]
You are driving home from school steadily at 95 km/h for 180km. It then begins to rain and you slow to 65 km/h. You arrive home after driving 4.5 h. (a) How far is your hometown from school? (b) What was your average speed?
Answer:
Explanation:
Given
Person travel 180 km with [tex]v_1=95\ km/h[/tex]
and slow down to [tex]v_2=65\ km/h[/tex] due to rain
time taken to cover 180 km is
[tex]t_1=\dfrac{180}{95}[/tex]
[tex]t_1=1.89\ hr[/tex]
total time [tex]T=4.5\ hr[/tex]
suppose [tex]t_2[/tex] is the time for which it travels with [tex]v_2=65\ km/h[/tex]
[tex]t_2=T-t_1[/tex]
[tex]t_2=4.5-1.89[/tex]
[tex]t_2=2.61\ hr[/tex]
so distance travel during this time [tex]d_2=v_2\times t_2[/tex]
[tex]d_2=65\times 2.61[/tex]
[tex]d_2=169.65\ km[/tex]
total distance [tex]d=180+d_2[/tex]
[tex]d=180+169.65[/tex]
[tex]d=349.65\ km[/tex]
(b)Average speed
[tex]v_{avg}=\dfrac{Distance}{time}[/tex]
[tex]v_{avg}=\dfrac{349.65}{4.5}[/tex]
[tex]v_{avg}=77.7\ km/h[/tex]
What is its maximum altitude above the ground? The answer is the maximum height above the ground
Answer:
Maximum altitude above the ground = 1,540,224 m = 1540.2 km
Explanation:
Using the equations of motion
u = initial velocity of the projectile = 5.5 km/s = 5500 m/s
v = final velocity of the projectile at maximum height reached = 0 m/s
g = acceleration due to gravity = (GM/R²) (from the gravitational law)
g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)
g = -9.82 m/s² (minus because of the direction in which it is directed)
y = vertical distance covered by the projectile = ?
v² = u² + 2gy
0² = 5500² + 2(-9.82)(y)
19.64y = 5500²
y = 1,540,224 m = 1540.2 km
Hope this Helps!!!
A typical flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Take the mass of the insect to be 7.0g , and assume the wings move an average downward distance of 1.5cm during each stroke. Assuming 117 downward strokes per second, estimate the average power output of the insect.
Answer:
Average power = 0.240Watts
Explanation:
The insect requires double the force on its wings to counteract the gravitational force pulling it down.it generates upward force as below
Upward Force=2Gravitational force
force =2x(mass x gravitational pull)
F=(0.007x9,8)x 2
F=0.137 Nm
Power generated is =work/time
power =force *displacement/time
convert 1.5cm to SI units meters=0.015meters
displacement in one second is=0.015 x number of strokes
displacement =0.015*117
displacement =1.755 meters
power=(0.137*1.755)/1second
power=0.240 Watts
A proton moves 10.0 cm on a path parallel to the direction of a uniform electric field of strength 3.0 N/C. What is the change in electrical potential energy?
Answer:
ΔPE= -4.8×10⁻²⁰J
Explanation:
Given data
Electric field of strength E=3.0 N/C
Charge of proton q=1.60×10⁻¹⁹C
Proton moves distance d=10 cm=0.10 m
To find
Change in electrical potential energy ΔPE
Solution
As we know that:
ΔPE= -qEd
[tex]=-(1.60*10^{-19}C )(3.0N/C)(0.10m)\\=-4.8*10^{-20}J[/tex]
ΔPE= -4.8×10⁻²⁰J
Answer:
Explanation:
Given:
D = 10 cm
= 0.1 m
E = 3.0 N/C
Qp = 1.602 × 10^-19 C
U = Q × V
But,
V = E × D
= 3 × 0.1
= 0.3 V
U = 1.602 × 10-19 × 0.3
= 4.806 × 10^-20 J.
Who performed classic experiments that supported the semiconservative model of dna replication?
Meselson and Stahl
Explanation:
The classic experiment that supported the semiconservative model of dna replication was performed by Matthew Meselson and Franklin W. Stahl. In this model, the two strands of DNA unwind from each other, and each acts as a template for synthesis of a new, complementary strand. This results in two DNA molecules with one original strand and one new strand. They used E. coli bacteria as a model system.
Two identical loudspeakers separated by distance d emit 200Hz sound waves along the x-axis. As you walk along the axis, away from the speakers, you don't hear anything even though both speakers are on.What are the three lowest possible values for d? Assume a sound speed of 340m/s.
Answer:
The three lowest possible values for d are 0.85 m, 2.55 m and 4.25 m.
Explanation:
Given that,
Distance = d
Frequency = 200 Hz
Speed of sound = 340 m/s
We need to calculate the wave length
Using formula of frequency
[tex]f= \dfrac{v}{\lambda}[/tex]
[tex]\lambda=\dfrac{v}{f}[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{340}{200}[/tex]
[tex]\lambda=1.7\ m[/tex]
We need to calculate the three lowest possible values for d
Using formula of destructive interference
[tex]\Delta\phi=2\pi\dfrac{\Delta x}{\lambda}[/tex]
[tex]2\pi\dfrac{\Delta x}{\lambda}=(m+\dfrac{1}{2})2\pi[/tex]
Where, [tex]\Delta x[/tex] = distance
Put the value into the formula
For m = 0,
[tex]\dfrac{\Delta x}{1.7}=(0+\dfrac{1}{2})[/tex]
[tex]\Delta x=\dfrac{1.7}{2}[/tex]
[tex]\Delta x=0.85\ m[/tex]
For m =1 ,
[tex]\dfrac{\Delta x}{1.7}=(1+\dfrac{1}{2})[/tex]
[tex]\Delta x=\dfrac{1.7\times3}{2}[/tex]
[tex]\Delta x=2.55\ m[/tex]
For m=2,
[tex]\dfrac{\Delta x}{1.7}=(2+\dfrac{1}{2})[/tex]
[tex]\Delta x=\dfrac{1.7\times5}{2}[/tex]
[tex]\Delta x=4.25\ m[/tex]
Hence, The three lowest possible values for d are 0.85 m, 2.55 m and 4.25 m.
The three lowest possible values for d are 0.85 m, 2.55 m and 4.25 m.
Destructive interference
Since we are looking for the points of no sound, these are points of destructive interference. So, the path difference, ΔL which is the distance between the two speakers, d is
ΔL = d = (n + 1/2)λ where
n = integer and λ = wavelength = v/f where v = speed of sound = 340 m/s and f = frequency of sound waves = 200 HzSo, d = (n + 1/2)v/f
d = (n + 1/2)340m/s ÷ 200 Hz
d = (n + 1/2)1.7 m
The lowest possible values of d
The lowest possible values of d are when n = 0, 1 and 2.
So,
When n = 0d = (n + 1/2)1.7 m
d = (0 + 1/2)1.7 m
d = (1/2)1.7 m
d = 0.85 m
When n = 1d = (n + 1/2)1.7 m
d = (1 + 1/2)1.7 m
d = (3/2)1.7 m
d = (1.5)1.7 m
d = 2.55 m
When n = 2d = (n + 1/2)1.7 m
d = (2 + 1/2)1.7 m
d = (5/2)1.7 m
d = 2.5 × 1.7 m
d = 4.25 m
So, the three lowest possible values for d are 0.85 m, 2.55 m and 4.25 m.
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Strategic planning is long-range, formulated by top management, and made as if the company operated in a vacuum Group of answer choices True False
Answer:
THE ANSWER IS: TRUE
Explanation:
Strategic planning is indeed long-range and formulated by top management but not under the assumption that the company operates in a vacuum. Effective strategic planning acknowledges and incorporates external factors.
Explanation:The statement is partially true and partially false. It is accurate that strategic planning is long-range and formulated by top management. These types of plans typically look several years into the future to set comprehensive goals for the company. However, the statement is false in suggesting that strategic planning is done as if a company operates in a vacuum. In truth, effective strategic planning must incorporate external influences like market trends, competitive activities, and regulatory changes.
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A uniform sphere of radius R rotates about a diameter with an angular momentum of magnitude L. Under the action of internal forces the sphere collapses to a uniform sphere of radius R/2. The magnitude of its new angular momentum is:_______.A. L/4.B. L/2.C. L.D. 2L.E. 4L.
Answer: C. L
Explanation:
Internal forces are forces that are produced from external forces which are acting on structure members. Example of such members include beams, and columns. There are three types of internal forces which are, axial, shear and moment.
Angular momentum is the rotational or angular equivalent of linear momentum. It is a conserved quantity.
As angular momentum is a conserved quantity, internal forces are unable to change it.
The magnitude of the new angular momentum for the uniform sphere of radius ([tex]\frac{R}{2}[/tex]) is: C. L.
An internal force refer to a force that is typically generated from an external force which acts on structure members such as columns, poles and beams.
Also, internal forces are exchanged between the objects in a system.
Generally, there are three (3) main types of internal forces and these include:
Shear force.Axial (normal) force.Moment.
Angular momentum is specific to rotational motion and it is the product of an object's moment of inertia and its angular velocity.
According to law of conservation of momentum, the initial angular momentum of an object is always equal to the final angular momentum.
This ultimately implies that, angular momentum is a conserved quantity.
In this context, the magnitude of the new angular momentum for the uniform sphere of radius ([tex]\frac{R}{2}[/tex]) is equal to L because it is conserved.
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The circuit to the right consists of a battery ( V 0 = 64.5 V) (V0=64.5 V) and five resistors ( R 1 = 711 (R1=711 Ω, R 2 = 182 R2=182 Ω, R 3 = 663 R3=663 Ω, R 4 = 534 R4=534 Ω, and R 5 = 265 R5=265 Ω). Find the current passing through each of the specified points. -g
Answer:
The current in R₁ is 0.0816 A.
The current at H point is 0.0243 A.
Explanation:
Given that,
Voltage = 64.5
Resistance is
[tex]R_{1}=711\ \Omega[/tex]
[tex]R_{2}=182\ \Omega[/tex]
[tex]R_{3}=663\ \Omega[/tex]
[tex]R_{4}=534\ \Omega[/tex]
[tex]R_{5}=265\ \Omega[/tex]
Suppose, The specified points are R₁ and H.
According to figure,
R₂,R₃,R₄ and R₅ are connected in parallel
We need to calculate the resistance
Using parallel formula
[tex]\dfrac{1}{R}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dfrac{1}{R_{4}}+\dfrac{1}{R_{5}}[/tex]
Put the value into the formula
[tex]\dfrac{1}{R}=\dfrac{1}{182}+\dfrac{1}{663}+\dfrac{1}{534}+\dfrac{1}{265}[/tex]
[tex]\dfrac{1}{R}=\dfrac{35501}{2806615}[/tex]
[tex]R=79.05\ \Omega[/tex]
R and R₁ are connected in series
We need to calculate the equilibrium resistance
Using series formula
[tex]R_{eq}=R_{1}+R[/tex]
[tex]R_{eq}=711+79.05[/tex]
[tex]R_{eq}=790.05\ \Omega[/tex]
We need to calculate the equivalent current
Using ohm's law
[tex]i_{eq}=\dfrac{V}{R_{eq}}[/tex]
Put the value into the formula
[tex]i_{eq}=\dfrac{64.5}{790.05}[/tex]
[tex]i_{eq}=0.0816\ A[/tex]
We know that,
In series combination current distribution in each resistor will be same.
So, Current in R and R₁ will be equal to [tex]i_{eq}[/tex].
The current at h point will be equal to current in R₅
We need to calculate the voltage in R
Using ohm's law
[tex]V=I_{eq}\timesR[/tex]
Put the value into the formula
[tex]V=0.0816\times79.05[/tex]
[tex]V=6.45\ Volt[/tex]
In resistors parallel combination voltage distribution in each part will be same.
So, [tex]V_{2}=V_{3}=V_{4}=V_{5}=6.45 V[/tex]
We need to calculate the current at H point
Using ohm's law
[tex]i_{h}=\dfrac{V_{5}}{R_{5}}[/tex]
Put the value into the formula
[tex]i_{h}=\dfrac{6.45}{265}[/tex]
[tex]i_{h}=0.0243\ A[/tex]
Hence, The current in R₁ is 0.0816 A.
The current at H point is 0.0243 A.
________ is the characteristic color bands that represent the range of radiations emitted by an element when it is heated
Answer:
Emission spectrum
Explanation:
The emmison spectrum is unique to every elements
Light consists of electromagnetic radiation of different wavelengths. Therefore, when the elements or their compounds are heated either on a flame or by an electric arc they emit energy in the form of light.
The color band indicate the amount of energy emitted in form of electromagnetic radiation.
When electrons in an atom gain energy, the become excited, on falling and losing their energy, they return to their ground state releasing the measure of energy they absorbed in transitioning.
The characteristic color bands emitted by an element when heated are called line spectra, which act as a unique fingerprint for the element and are important for identifying various physical properties.
The characteristic color bands that represent the range of radiations emitted by an element when it is heated are known as the line spectra. When elements are heated to incandescence, like in Bunsen and Kirchhoff's experiments, they emit light with a series of sharp wavelengths characteristic of that element's atomic composition.
This light, when analyzed using a spectrometer or a simple prism, results in visible narrow bands of colors, each corresponding to a specific wavelength. For instance, hydrogen emits a red light with a strong line at 656 nm, and sodium is known to emit in the yellow portion of the spectrum at about 589 nm.
The line spectra serve as a unique fingerprint for each element and are crucial for understanding not only the chemical composition but also various physical properties such as temperature and density of the radiant gas. In terms of non-visible bands, color palettes are often chosen to represent different levels of brightness, energies of photons, and inferred physical properties from data modeling, which are essential in the study of astronomy and spectroscopy.
A solid cylinder and a hollow cylinder of equal mass and radius are at rest at the top of an inclined plane. They are released simultaneously and roll down the plane without slipping. Which object reaches the bottom of the incline first?
Answer:
Solid cylinder
Explanation:
Solid cylinder will reach first
If radius of each cylinder is r
mass is m
then,
Moment of inertia I= dm[tex]r^{2}[/tex]
Here d = measure as how close the mass is to the edge
Velocity of rolling cylinder is given by
v= [tex]\sqrt} \frac{2gh}{1+d}[/tex]
where,
g= 9.8 m/s2
h= height from ground
So from formula of velocity we can say that velocity will be maximum if denominator is minimum i-e if k is of small value -- or in other word if mass is away from edge i-e if mass is closer to the center !
As all the mass in hollow cylinder is near the edge so k value will be higher for it and hence it will have low velocity value. So it will reach later as compared to the solid cylinder in which mass is closer to the center and hence k is greater for solid cylinder.
A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 15-m-long cable. You are (unwisely!) standing directly beneath the point from which the wrecking ball is hung when you notice that the ball has just been released and is swinging directly toward you.
Answer:
1.9 s
Explanation:
We are given that
Length of cable=l=15 m
We have to find the time you have to move out of the way.
We know that
Time period,T=[tex]2\pi\sqrt{\frac{l}{g}}[/tex]
Where g=[tex]9.8m/s^2[/tex]
By using the formula
[tex]T=2\pi\sqrt{\frac{15}{9.8}}[/tex]
[tex]T=2\times 3.14\times \sqrt{\frac{15}{9.8}}=7.77 s[/tex]
Time you have to move out
[tex]t=\frac{T}{4}=\frac{7.77}{4}=1.9 s[/tex]
Hence,time you have to move out of the way=7.77 s
Matter is __________. A. the amount of energy contained within an object B. a measure of the force gravity exerts on objects C. something that has mass and occupies space D. a measure of an object's inertial force Please
Answer: C
Explanation:
Matter is anything that has mass and takes up space. People, paper, goats, chairs and even water is Matter!
Matter can be defined as something that has mass and occupies space. They are present in the atmosphere in different phases. Thus, the correct option is C.
What is Matter?
Matter is any substance which is made up of various types of particles that occupies some physical space and has mass. According to the modern physics, there are various types of particles with each having a specific mass and size. The most common examples of these material particles are electrons, protons and neutrons.
Matter is present in different phases in the atmosphere. The three phases are solids, liquids, and gases. All of these phases show mass and occupies some space.
Therefore, the correct option is C.
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A force of 250 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 20 centimeters to 50 centimeters?
Answer:
Work done = 87.5 J
Explanation:
Given:
Force required to stretch the spring (F) = 250 N
Extension of the spring (x) = 30 cm = 0.30 m [1 cm = 0.01 m]
So, spring constant (k) of the spring is given as:
[tex]k=\frac{F}{x}\\\\k=\frac{250\ N}{0.30\ m}=833.33\ N/m[/tex]
Also given:
Initial length of the spring (x₁) = 20 cm = 0.20 m
Final length of the spring (x₂) = 50 cm = 0.50 m
Now, work done in stretching the spring from an initial length (x₁) to final length (x₂) is given as:
[tex]W=\frac{1}{2}k(x_2^2-x_1^2)[/tex]
Plug in the given values and solve for 'W. This gives,
[tex]W=\frac{1}{2}\times \frac{250}{0.3}\times (0.50^2-0.20^2)\\\\W=\frac{1250}{3}\times (0.25-0.04)\\\\W=\frac{1250\times 0.21}{3}=87.5\ J[/tex]
Therefore, work is done in stretching the spring from 20 centimeters to 50 centimeters is 87.5 J
The work done in stretching a spring can be calculated using Hooke's law. Here, to find the work done in stretching a spring from 20 centimeters to 50 centimeters, we first find the spring constant using the given force and distance, then compute the work using the difference in the squares of the final and initial stretching distances. The work done is 104.1 Joules.
Explanation:The subject of this question pertains to the physics concept of work done on an object, in this case, a spring. When a force is applied to a spring, it stretches, and in doing so, work is done. The work done can be calculated using Hooke's Law, which states that the force (F) needed to extend or compress a spring by some distance X is proportional to that distance. This is represented by the equation F = kX, where k is the spring constant.
First, we need to find the spring constant k using the provided force (250 newtons) and the initial stretching distance (30 centimeters or 0.3 meters), so k = F/X = 250/0.3 = 833.33 N/m.
Next, we compute the work done in stretching the spring from 20 centimeters (0.20 meters) to 50 centimeters (0.50 meters). The work done (W) on a spring is given by the equation W =1/2*k*(X2^2-X1^2), where X2 and X1 are the final and initial stretching distances respectively. So, W = 1/2 * 833.33 * ((0.5)^2 - (0.2)^2) = 104.1 Joules.
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The playing time of 16 popular songs is found to have a standard deviation of 54.5 seconds. Use a 0.05 significance level to test the claim that the songs are from a population with a standard deviation less than one minute (60 seconds). State the initial and final conclusion.
Answer:
fail to reject the null hypothesis; there is not sufficient evidence to support the claim that the songs are from a population with a standard deviation less than one minute
Explanation:
A bob of mass m is suspended from a fixed point with a massless string of length L (i.e., it is a pendulum). You are to investigate the motion in which the string moves in a cone with half-angle θ.
What tangential speed, v, must the bob have so that it moves in a horizontal circle with the string always making an angle from the vertical?
How long does it take the bob to make one full revolution (one complete trip around the circle)?
Express your answer in terms of m, L, θ and acceleration due to gravity g.
The tangential speed v that the bob must have is: v = [tex]\sqrt{gLsin(\theta)tan(\theta)[/tex]
And the time for one full revolution is: T = [tex]\frac{2\pi \sqrt{Lsin(\theta)}}{\sqrt{g tan(\theta)}}[/tex]
To solve this problem, we need to find the tangential speed, v, that the bob must have to move in a horizontal circle and the time, T, it takes to make one complete revolution. Here’s a step-by-step approach to the solution:
Determine the radius of the circular motion (R):
Given that the string length is L and the angle from the vertical is θ, the radius of the horizontal circle can be expressed as:
R = L sin(θ)
Analyze the forces acting on the bob:
The forces acting on the bob are:
The tension in the string (T), which has components both in the radial (horizontal) and vertical directionsThe gravitational force (mg), acting vertically downwardEquations of motion:
In the vertical direction, since there is no vertical acceleration, the vertical component of the tension must balance the weight of the bob:
Tcos(θ) = mg
In the horizontal direction, the horizontal component of the tension provides the centripetal force necessary for the circular motion:
Tsin(θ) = mv²/R
Solve for the tension (T):
From the vertical force balance equation:
T = mg /cos(θ)
Substitute T into the horizontal force equation and solve for v:
mgsin(θ) /cos(θ) = mv²/R
Simplifying, we get:
gtan(θ) = v² /R
Substitute R = L sin(θ):
gtan(θ) = Lsin(θ)v²
Solving for v, we get:
v = [tex]\sqrt{gLsin(\theta)tan(\theta)}[/tex]
Determine the period of the motion (T):
The time for one complete revolution (T) is the circumference of the circle divided by the tangential speed v:
T = 2πR /v
Substitute R and v:
T = [tex]\frac{2\pi Lsin(\theta)}{\sqrt{gLsin(\theta)tan(\theta)}}[/tex]
Simplifying further:
T = [tex]\frac{2\pi \sqrt{Lsin(\theta)}}{\sqrt{g tan(\theta)}}[/tex]
Therefore, the tangential speed v: v = [tex]\sqrt{gLsin(\theta)tan(\theta)[/tex] ,and the period of revolution is: T = [tex]\frac{2\pi \sqrt{Lsin(\theta)}}{\sqrt{g tan(\theta)}}[/tex].
Which one of the following crystal structures has the fewest slip directions and therefore the metals with this structure are generally more difficult to deform at room temperature?
a. BCC
b. FCC
c. HCP
d. BCT
The HCP structure has the fewest slip directions, making HCP metals like Mg and Zn more difficult to deform at room temperature compared to those with BCC or FCC structures.
Explanation:The crystal structure that has the fewest slip directions and is thus generally more difficult to deform at room temperature is the hexagonal close-packed (HCP) structure. Metals with the HCP structure include Cd, Co, Li, Mg, Na, and Zn. The nature of HCP's stacking arrangement, with alternating type A and type B close-packed layers (ABABAB...), results in fewer slip systems compared to body-centered cubic (BCC) and face-centered cubic (FCC) structures. BCC structures, found in metals like K, Ba, Cr, Mo, W, and Fe, have a coordination number of 8, whereas FCC structures have a coordination number of 12 and include metals like Ag, Al, Ca, Cu, Ni, Pb, and Pt. The closer atomic packing found in FCC, identified as CCP or cubic close-packed, is reflected in the ABCABCABC... stacking sequence, which affords a higher number of slip directions.
Three bulbs are connected by tubing, and the tubing is evacuated. The volume of the tubing is 45.0 mL. The first bulb has a volume of 77.0 mL and contains 8.89 atm of argon, the second bulb has a volume of 250 mL and contains 2.82 atm of neon, and the third bulb has a volume of 21.0 mL and contains 8.42 atm of hydrogen. If the stopcocks (valves) that isolate all three bulbs are opened, what is the final pressure of the whole system in atm
Answer:
The final pressure of the whole system is 34.80 atm.
Explanation:
Given that,
Volume = 45.0 ml
Volume of first bulb = 77.0 mL
Pressure = 8.89 atm
Volume of second bulb = 250 mL
Pressure = 2.82 atm
Volume of third bulb = 21.0 mL
Pressure = 8.42 atm
We need to calculate the final pressure of the whole system
Using formula of pressure
[tex]P_{1}V_{1}+P_{2}V_{2}+P_{3}V_{3}+P_{t}V_{t}=P_{f}V_{f}[/tex]
Where, [tex]P_{1}[/tex]= pressure of first bulb
[tex]P_{2}[/tex]= pressure of second bulb
[tex]P_{3}[/tex]= pressure of third bulb
[tex]P_{4}[/tex]= initial pressure of tube
[tex]V_{1}[/tex]= Volume of first bulb
[tex]V_{2}[/tex]=Volume of second bulb
[tex]V_{3}[/tex]= Volume of third bulb
[tex]V_{4}[/tex]= Initial volume of tube
Put the value into the formula
[tex]8.89\times77.0+250\times2.82+21.0\times8.42+0=P_{f}\times45[/tex]
[tex]P_{f}=\dfrac{1566.35}{45}[/tex]
[tex]P_{f}=34.80\ atm[/tex]
Hence, The final pressure of the whole system is 34.80 atm.
A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 6.5 m/s relative to the water. However, the local ocean current is 1.50 m/s in a direction 40.0º north of east and changes the ship's intended motion. What is the velocity of the ship relative to the Earth?
Answer:
Explanation:
velocity of ship with respect to water = 6.5 m/s due north
[tex]\overrightarrow{v}_{s,w}=6.5 \widehat{j}[/tex]
velocity of water with respect to earth = 1.5 m/s at 40° north of east
[tex]\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)[/tex]
velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth
[tex]\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}[/tex]
[tex]\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j} \right )[/tex]
[tex]\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}[/tex]
The magnitude of the velocity of ship relative to earth is [tex]\sqrt{1.15^{2}+5.54^{2}}[/tex] = 5.66 m/s
Two hypothetical discoveries in Part A deal with moons that, like Earth's moon, are relatively large compared to their planets. Which of the following best explains why finding 1 planet with such a moon is consistent with the nebular theory, while finding 6 planets with such moons is not consistent?
Answer:
Unusually large moons form in giant impacts, which are relatively rare events
Explanation:
Solution:
- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.
- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible. The transition from an undifferentiated cloud to a star system complete with planets and moons takes about 100 million years.
Before we understood that objects have a tendency to maintain their velocity in a straight line unless acted upon by a net force, people thought that objects have a tendency to stop on their own. This happened because a specific force was not yet understood. What was that force?
Answer:Friction force
Explanation:
The frictional force is responsible for stopping the moving object.
The friction force is provided by nature in the form of air resistance, fluid resistance, Surface resistance.
For surface resistance, the friction force is of two types namely kinetic friction force and static friction force.
Static friction acts when the object is stationary and a force is applied to cause the motion while kinetic friction acts when the body starts moving.
kinetic friction is lesser is in magnitude as compared to static friction.
The misunderstood force that causes objects to seem to stop on their own is known as friction. This force opposes the motion of objects. The misconception was fixed with Newton's first law of motion.
Explanation:The force that people did not understand before they realized that objects have a tendency to maintain their velocity in a straight line unless acted upon by a net force is friction. Friction is a force that opposes movement. It's the reason why objects seem to stop on their own when in contact with other objects or surfaces. The perception that an object moves and then eventually stops due to its own 'nature' occurs because friction was at play, causing it to slow down and finally stop. It was Sir Isaac Newton who, in his first law of motion, clarified this misperception by stating that an object in motion stays in motion unless acted upon by an external force.
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Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 9 m into 32 cm. of water. Assuming that he stops just as he reaches the bottom of the water and estimating his mass to be 62 kg, find the magnitudes of the impulse on him from the water.
Answer:
823.46 kgm/s
Explanation:
At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.
So, mgh = 1/2mv²
From here, his velocity just as he reaches the surface of the water is
v = √2gh
h = 9 m and g = 9.8 m/s²
v = √(2 × 9 × 9.8) m/s
v = √176.4 m/s
v₁ = 13.28 m/s
So his velocity just as he reaches the surface of the water is 13.28 m/s.
Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.
So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.
His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,
J = (62 × 0 + 62 × 13.28) kgm/s = 0 + 823.46 kgm/s = 823.46 kgm/s
So the magnitude of the impulse J of the water on him is 823.46 kgm/s