A) 1 Year, if the Earth is larger in mass, but still the same distance from the sun, it will orbit at the same rate.
The Earth will continue to orbit at the same rate even if its mass increases but its distance from the sun stays the same, or 1 year. Hence, option A is correct.
What are planets?A huge, spherical celestial object that is not a star nor a remnant is called a planet. The emission nebula hypothesis, which holds that a cosmic cloud collapses out of a supernova to produce a young primitive orbited by a planetary system, is the best theory currently available for explaining planet formation. The slow accumulation of matter propelled by gravity—a process known as accretion—leads to the formation of planets in this disk.
The planetary systems Mercury, Venus, Earth, and Mars, as well as the giant planets Jupiter, Saturn, Uranus, and Neptune, make up the Solar System's minimum number of eight planets. Each of these planets revolves around an axis that is inclined with regard to its orbit pole.
Hence, option A is correct.
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A stretched string is fixed at both ends, 77.7 cm apart. If the density of the string is 0.014 g/cm and its tension is 600 N, what is the wavelength of the first harmonic?
Answer:
The wavelength of the first harmonic is 155.4 cm.
Explanation:
Given that,
The separation between a stretched string that is fixed at both ends, l = 77.7 cm
The density of the string, [tex]d=0.014\ g/cm^3[/tex]
Tension in the string, T = 600 N
We need to find the wavelength of the first harmonic. The wavelength of the nth harmonic on the string that is fixed at both ends is given by :
[tex]\lambda=\dfrac{2l}{n}[/tex]
Here, n = 1
[tex]\lambda=2l[/tex]
[tex]\lambda=2\times 77.7[/tex]
[tex]\lambda=155.4\ cm[/tex]
So, the wavelength of the first harmonic is 155.4 cm. Hence, this is the required solution.
The wavelength of the first harmonic will be "155.4 cm".
Wavelength and DensityAccording to the question,
Separation between stretched string, I = 77.7 cm
String's density, d = 0.014 g/cm³
String's tension, T = 600 N
We know the relation,
Wavelength, λ = [tex]\frac{2l}{n}[/tex]
here, n = 1
then, λ = 2l
By substituting the values,
= 2 × 77.7
= 155.4 cm
Thus the above answer is appropriate.
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The volume of a fixed mass of an ideal gas is doubled while the temperature is increased from 100 K to 400 K.
What is the final pressure in terms of its initial pressure?
a. 2 Pi
b. 3 Pi
c. 4 Pi
d. 1/2 Pi
Answer:
c. 4 Pi
Explanation:
The pressure law states that the pressure of a gas is directly proportional to its temperature provided volume and other physical quantities remain constant.
If t is the temperature and p is the pressure of a gas,
p ∝ t
p = kt where k s the constant of proportionality
k = p/t
If t1 and p1 are the initial temperature and pressure respectively and
t2, p2 are the final temperature and pressure
p1/t1 = p2/t2
t1 = 100k, t2 = 400k
p2 = p1 × 400/100
p2 = 4t1
The final pressure of the gas is 4 times the initial pressure.
A silver dollar is dropped from the top of a building that is 1344 feet tall. Use the position function below for free-falling objects. s(t) = -16t^(2) + v0t + s0
(a) Determine the position and velocity functions for the coin. s(t)=_____________ v(t)=_____________
(b) Determine the average velocity on the interval [3,4]. ______________= ft/s
(c) Find the instantaneous velocities when t = 3 s and t = 4 s. v(3)=________ v(4)=________
(d) Find the time required for the coin to reach the ground level. t=___________s
(e) Find the velocity of the coin at impact. vf =___________ft/s
Answer:
a. position function of the coin:
[tex]s=-16t^2+1344[/tex]
Now the velocity function:
[tex]v=-32t[/tex]
b. [tex]v_{avg}=-112\ m.s^{-1}[/tex]
c. [tex]v_3=-96\ m.s^{-1}[/tex] & [tex]v_4=-128\ m.s^{-1}[/tex]
d. [tex]t=9.165\ s[/tex]
e. [tex]v_f=293.285\ m.s^{-1}[/tex]
Explanation:
Given:
height of dropping the silver dollar, [tex]h=1344\ ft[/tex]Given position function associated with free falling objects:
[tex]s=-16t^2+v_0t+s_0[/tex]
here:
[tex]s_0=[/tex] initial height
[tex]v_0=[/tex]initial velocity
[tex]t=[/tex] time of observation
a)
position function of the coin:
[tex]s=-16t^2+1344[/tex]
∵the object is dropped it was initially at rest
Now the velocity function:
[tex]v=\frac{d}{dt} s[/tex]
[tex]v=-32t[/tex]
b)
we know average velocity is given as:
[tex]\rm v_{avg}=\frac{total\ displacement}{total\ time}[/tex]
Displacement in the given interval:
[tex]s_{_{3-4}}=s_4-s_3[/tex]
[tex]s_{_{3-4}}=(-16\times 4^2+1344)-(-16\times 3^2+1344)[/tex]
[tex]s_{_{3-4}}=-112\ ft[/tex]
Now,
[tex]v_{avg}=\frac{-112}{4-3}[/tex]
[tex]v_{avg}=-112\ m.s^{-1}[/tex]
c)
Instantaneous velocity at t = 3 s:
[tex]v_3=-32\times 3[/tex]
[tex]v_3=-96\ m.s^{-1}[/tex]
Instantaneous velocity at t = 4 s:
[tex]v_4=-32\times 4[/tex]
[tex]v_4=-128\ m.s^{-1}[/tex]
d)
At ground we have s=0:
Put this in position function:
[tex]0=-16t^2+1344[/tex]
[tex]t=9.165\ s[/tex]
e)
Velocity of the coin at impact:
[tex]v_f=-32\times 9.165[/tex]
[tex]v_f=293.285\ m.s^{-1}[/tex]
The position and velocity functions of the silver dollar are s(t) = -16t^2 + 1344 and v(t) = -32t, respectively. After a time of 9.6 seconds, the coin hits the ground with a velocity of -307.2 ft/s. The average speed between 3 and 4 seconds is -224 ft/s.
Explanation:
In this particular problem, we know that the silver dollar is dropped from rest from a height (s0) of 1344 feet. Thus, the initial velocity (v0) is 0.
(a) The position function with these initial conditions becomes s(t) = -16t^2 + 1344 and the velocity function is derived from the position function as v(t) = -32t.
(b) The average velocity on the interval [3,4] can be found by taking the change in position divided by the change in time in that interval and is calculated by [s(4)-s(3)]/[4-3] = -224 ft/s.
(c) The instantaneous velocities at t = 3 s and t = 4 s are found by substituting these times into the velocity function, yielding v(3) = -96 ft/s and v(4) = -128 ft/s.
(d) The time required for the coin to reach the ground level can be found by solving the position function s(t) = 0 for t: t = sqrt(1344/16) = 9.6 s.
(e) The velocity of the coin at impact is found by evaluating the velocity function at t = 9.6 s: vf = -32*9.6 = -307.2 ft/s.
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An electron is moving horizontally in a laboratory when a uniform electric field is suddenly turned on. This field points vertically upward. Which of the paths shown will the electron follow, assuming that gravity can be neglected?
Explanation:
We know that electron always tends to travel in the direction opposite the direction of electric field ( force in directed opposite to the field on electron).
So, since the field points vertically upward , it force the electron move downward. So, the electron will be accelerated in downward in the right direction.
Which of the following changes would increase the seperation between the bright fringes in the diffraction pattern formed by diffraction grating?
A. INCREASE THE WAVELENGTH OF THE LIGHT
B. IMMERSE THE APPARATUSE IN WATER
C. INCREASE THE SEPERATION BETWEEN THE SLITS
D. NONE OF THESE
E. MORE THAN ONE OF THESE
The increase in the separation between the bright fringes in the diffraction pattern formed by diffraction grating occurs when there is increase in the wavelength of the light.
The equation for diffraction grating that relates the wavelength and separation of bright fringes is given as;
[tex]d sin\theta = m \lambda \\\\sin \ \theta = \frac{m \lambda}{d}[/tex]
where;
λ is the wavelength d is the spacingThus, we can conclude that the increase in the separation between the bright fringes in the diffraction pattern formed by diffraction grating occurs when there is increase in the wavelength of the light.
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Increasing the wavelength of the light will increase the separation between the bright fringes in the diffraction pattern. The correct option is Option A. Immersion in water or changing slit separation do not contribute similarly.
To increase the separation between the bright fringes in a diffraction pattern, one effective approach is to increase the wavelength of the light. When the wavelength increases, the angle of diffraction becomes larger for each order of maximum, resulting in greater fringe separation. Therefore, the correct answer is Option A: INCREASE THE WAVELENGTH OF THE LIGHT.
During a hard sneeze, your eyes might shut for 0.50 s. If you are driving a car at 90 km/h during such a sneeze, how far does the car move during that time?
Answer:
12.5 m.
Explanation:
Speed: This can be defined as the rate of change of distance. The S.I unit of speed is m/s. Mathematically it can be expressed as,
Speed = distance/time
S = d/t......................... Equation 1
d = S×t ...................... Equation 2.
Where S = speed of the car, d = distance, t = time taken to shut the eye during sneezing
Given: S = 90 km/h, t = 0.50 s.
Conversion: 90 km/h ⇒ m/s = 90(1000/3600) = 25 m/s.
S = 25 m/s.
Substitute into equation 2.
d = 25×0.50
d = 12.5 m.
Hence the car will move 12.5 m during that time
Final answer:
To find out the distance a car covers during a sneeze while driving at 90 km/h, we first convert the speed to meters per second (25 m/s) and then multiply by the sneeze duration (0.50 s), resulting in a distance of 12.5 meters.
Explanation:
To calculate the distance a car moves during a sneeze, we can use the formula for distance traveled at a constant speed, which is distance = speed imes time. Given that the car's speed is 90 km/h, we first need to convert this speed into meters per second. Since 1 km = 1000 meters and 1 hour = 3600 seconds, the conversion yields:
90 km/h imes (1000 meters/km) imes (1 hour/3600 seconds) = 25 m/s.
Now, using the time duration of the sneeze, which is 0.50 seconds, the distance traveled during the sneeze can be calculated as:
Distance = 25 m/s imes 0.50 s = 12.5 meters.
Therefore, the car travels 12.5 meters during a sneeze lasting 0.50 seconds.
Joules are Nm, which is kg middot m^2/s^2 Use this relationship and 1 Btu = 1.0551 kJ to determine what Btu using primary English units (ft, lb_m, degree F, sec, etc.)
Answer:
[tex]23851.35840\ lbft^2/s^2[/tex]
Explanation:
[tex]1\ Btu=1.0051\ kJ\\\Rightarrow 1\ Btu=1.0051\times 1000 kgm^2/s^2[/tex]
The English units that are to be used here are pounds (lb) and feet (ft)
[tex]1\ kg=2.20462\ lb[/tex]
[tex]1\ m=3.28084\ ft[/tex]
[tex]1005.1\times 2.20462\times 3.28084^2=23851.35840\ lbft^2/s^2[/tex]
The answer is [tex]1\ Btu=23851.35840\ lbft^2/s^2[/tex]
consider a perfectly inelastic head-on collision between a small car and a large truck traveling at the same speed. Which vehicle has a greater change in kinetic energy as a result of the collision
Answer: The truck has a greater kinetic energy
Explanation:
The initial velocities of both the truck and car are the same. Likewise, the final velocity is the same. The change in kinetic energy is dependent only on mass. The truck has a greater mass, therefore, the change in its kinetic energy is greater.
The practical limit to an electric field in air is about 3.00 × 10^6 N/C . Above this strength, sparking takes place because air begins to ionize.
(a) At this electric field strength, how far would a proton travel before hitting the speed of light (100% speed of light) (ignore relativistic effects)?
(b) Is it practical to leave air in particle accelerators?
Answer:
(a) x=157 m
(b) No
Explanation:
Given Data
Mass of proton m=1.67×10⁻²⁷kg
Charge of proton e=1.6×10⁻¹⁹C
Electric field E=3.00×10⁶ N/C
Speed of light c=3×10⁸ m/s
For part (a) distance would proton travel
Apply the third equation of motion
[tex](v_{f})^{2} =(v_{i})^{2}+2ax[/tex]
In this case vi=0 m/s and vf=c
so
[tex]c^{2}=(0)^{2}+2ax\\ c^{2}=2ax\\x=\frac{c^{2} }{2a}[/tex]
[tex]x=\frac{c^{2}}{2a}--------Equation (i)[/tex]
From the electric force on proton
[tex]F=qE\\where\\ F=ma\\so\\ma=qE\\a=\frac{qE}{m}\\[/tex]
put this a(acceleration) in Equation (i)
So
[tex]x=\frac{c^{2} }{2(qE/m)}\\ x=\frac{mc^{2}}{2qE} \\x=\frac{(1.67*10^{-27})*(3*10^{8})^{2} }{2*(1.6*10^{-19})*(3*10^{6})}\\ x=157m[/tex]
For part (b)
No the proton would collide with air molecule
A 75 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 620 m/s . Part A What is the recoil speed of the hunter?
Express your answer to two significant figures and include the appropriate units.
To develop this problem we will apply the concepts related to the conservation of momentum. For this purpose we will define that the initial moment must be preserved and be equal to the final moment. Since the system was at a stationary moment, the moment before the bullet is fired will be 0, while the moment after the bullet is fired will be equivalent to the sum of the products of the mass and the velocity of each object. , this is,
[tex]m_1u_1+m_2u_2 = m_1v_1+m_2v_2[/tex]
Here,
m = mass of each object
u = Initial velocity of each object
v = Final velocity of each object
Our values are,
[tex]\text{Mass of hunter} ) m_1 = 75 kg = 75000 g[/tex]
[tex]\text{Mass of bullet} = m_2 = 42 g[/tex]
[tex]\text{Speed of bullet} = v_2 = 6200 m/s[/tex]
[tex]\text{Recoil speed of hunter} =v_1 = ?[/tex]
Applying our considerations we have that the above formula would become,
[tex]m_1v_1 + m_2v_2 = 0[/tex]
Solving for speed 1,
[tex]v_1 = - \frac{m_2v_2}{m_1 }[/tex]
[tex]v_1 = - \frac{42 * 620}{75000 }[/tex]
[tex]v_1 = - 0.3472 m/s[/tex]
Therefore the recoil speed of the hunter is -0.3472m/s, said speed being opposite to the movement of the bullet.
The recoil speed of the hunter standing on the frictionless ice is 0.35m/s².
Given the data in the question;
Mass of hunter; [tex]m_1 = 75kg[/tex]Mass of bullet; [tex]m_2 = 42g = 0.042kg[/tex]Velocity of bullet; [tex]v_2 = 620m/s[/tex]Recoil speed of the hunter; [tex]v_1 =\ ?[/tex]
From the law of conservation of momentum:
[tex]MV = mv[/tex]
Where m is mass and v is velocity
So
[tex]m_1v_1 = m_2v_2\\\\v_1 = \frac{m_2v_2}{m_1}[/tex]
We substitute our given values into the equation
[tex]v_1 = \frac{ 0.042kg \ *\ 620m/s^2}{75kg} \\\\v_1 = \frac{26.04kg.m/s^2}{75kg} \\\\v_1 = 0.35m/s^2[/tex]
Therefore, the recoil speed of the hunter standing on the frictionless ice is 0.35m/s².
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Two identical 2.0-kg objects are involved in a collision. The objects are on a frictionless track. The initial velocity of object A is 12 m/s and the initial velocity of object B is -6.0 m/s. What is the final velocity of the two objects
Answer:
final velocity of the objects will be 3 m/sec
Explanation:
We have given there are two identical objects of mass 2 kg
So [tex]m_1=m_2=2kg[/tex]
Initial velocity of object A [tex]v_1=12m/sec[/tex]
Initial velocity of object B [tex]v_2=-6m/sec[/tex]
According to conservation of momentum
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
[tex]2\times 12+2\times -6=(2+2)v[/tex]
[tex]4v=24-12[/tex]
[tex]4v=12[/tex]
v = 3 m/sec
So final velocity of the objects will be 3 m/sec
To find the final velocity of mass B after a collision, you can use the principle of conservation of momentum. The formula for conservation of momentum is (mass of A * initial velocity of A) + (mass of B * initial velocity of B) = (mass of A * final velocity of A) + (mass of B * final velocity of B). By substituting the given values and solving for the final velocity of mass B, we can determine its value.
Explanation:To find the final velocity of mass B after the collision, we can use the principle of conservation of momentum. Since the collision is elastic, the total momentum before the collision is equal to the total momentum after the collision.
Mass A has an initial velocity of 5.0 m/s in the +x-direction and after the collision, moves with a velocity of 3.0 m/s in the -x-direction. Mass B has an initial velocity of 3.0 m/s in the -x-direction.
Let's assume the final velocity of mass B is vB. Using the principle of conservation of momentum, we can set up the equation:
Initial momentum = Final momentum
(mass of A * initial velocity of A) + (mass of B * initial velocity of B) = (mass of A * final velocity of A) + (mass of B * final velocity of B)
Substituting the given values:
(5.0 kg * 5.0 m/s) + (5.0 kg * 3.0 m/s) = (5.0 kg * 3.0 m/s) + (mass of B * vB)
Simplifying the equation:
25.0 m/s + 15.0 m/s = 15.0 m/s + (mass of B * vB)
40.0 m/s = 15.0 m/s + (mass of B * vB)
Subtracting 15.0 m/s from both sides:
25.0 m/s = (mass of B * vB)
Dividing both sides by the mass of B:
vB = 25.0 m/s / mass of B
Therefore, the final velocity of mass B after the collision is 25.0 m/s divided by the mass of B.
An insulated beaker with negligible mass contains liquid water with a mass of 0.270 kg and a temperature of 82.5 ∘C . How much ice at a temperature of -22.3 C must be dropped into the water so that the final temperature of the system will be 34.0 C?
Answer:
Explanation:
Given
mass of water [tex]m_1=0.27\ kg[/tex]
Temperature of water [tex]T_{wi}=82.5^{\circ}C[/tex]
Initial Temperature of ice[tex]=-22.3^{\circ}C[/tex]
Final temperature of system [tex]T=34^{\circ}C[/tex]
specific heat of water [tex]c=4.18\ kJ/kg-K[/tex]
specific heat of ice [tex]c_i=2.108\ kJ/kg-K[/tex]
Latent heat of ice [tex]L=336\ kJ/kg[/tex]
Heat loss by Water is equal to heat gained by ice
Heat loss by water [tex]Q_1=m_w\times c\times \Delta T[/tex]
[tex]Q_1=0.27\times 4.18\times (82.5-34)=54.7371\ kJ[/tex]
Heat gained by ice [tex]Q_1=x\times c_i(0-(-22.3))+x\times L+x\times c\times (T-0)[/tex]
[tex]Q_2=x\times 2.108\times (22.3)+x\times 336+x\times 4.18\times 34[/tex]
[tex]Q_2=525.1284x\ kJ[/tex]
[tex]Q_1=Q_2[/tex]
[tex]x=\frac{54.73}{525.1284}[/tex]
[tex]x=0.104\ kg[/tex]
A certain first-order reaction is 58% complete in 95 s. What are the values of the rate constant and the half-life for this process?
Answer:
[tex]0.005734 s^{-1}[/tex] and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]a_o=a\times e^{-kt}[/tex]
where,
k = rate constant
t = age of sample
[tex]a_o[/tex] = let initial amount of the reactant
a = amount left after decay process
We have :
[tex]a_o=x[/tex]
[tex]a=58\%\times x=0.58x[/tex]
t = 95 s
[tex]0.58x=x\times e^{-k\times 95 s}[/tex]
[tex]\k= 0.005734 s^{-1}[/tex]
Half life is given by for first order kinetics::
[tex]t_{1/2}=\frac{0.693}{k}[/tex]
[tex]=\frac{0.693}{0.005734 s^{-1}}=120.86 s[/tex]
[tex]0.005734 s^{-1}[/tex] and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..
At time t in seconds, a particle’s distance s(t), in centimeters, from a point is given by s(t)=13sin t+40. What is the average velocity of the particle from t= π3 to t= 13π3?
Answer:
Explanation:
Given
distance [tex]s(t)=13\sin (t)+40[/tex]
at [tex]t=\frac{\pi }{3}[/tex]
[tex]s(\frac{\pi }{3})=13\sin (\frac{\pi }{3})+40[/tex]
[tex]s(\frac{\pi }{3})=13\times \frac{\sqrt{3}}{2}+40[/tex]
at [tex]t=\frac{13\pi}{3}=4\pi+\frac{\pi}{3}[/tex]
[tex]s(4\pi+\frac{\pi}{3})=13\sin (4\pi+\frac{\pi}{3})+40[/tex]
Average velocity is change in position w.r.t time
[tex]v_{avg}=\frac{s(4\pi+\frac{\pi}{3})-s(\frac{\pi }{3})}{4\pi+\frac{\pi}{3}-\frac{\pi }{3}}[/tex]
[tex]v_{avg}=0[/tex]
Suppose a thin conducting wire connects two conducting spheres. A negatively charged rod is brought near one of the spheres, the wire between them is cut, and the charged rod is taken away. Which one of the following is true? a. The spheres will attract each other. b. The spheres will repel each other. c. There will be no electrostatic force between the spheres
Answer:
a. The spheres will attract each other.
Explanation:
When two conducting spheres are connected by a conducting wire and a negatively charged rod is brought near it then this will induce opposite (positive) charge at the nearest point on the sphere and by the conservation of charges there will also be equal amount of negative charge on the farthest end of this conducting system this is called induced polarization.
When the conducting wire which joins them is cut while the charged rod is still in proximity to of one of the metallic sphere then there will be physical separation of the two equal and unlike charges on the spheres which will not get any path to flow back and neutralize.Hence the two spheres will experience some amount of electrostatic force between them.A 7500 kg open train car is rolling on frictionless rails at 25.0 m/s when it starts pouring rain. A few minutes later, the car's speed is 20.0 m/s. What mass of water has collected in the car?
Answer:
Mass of rain collected will be 1875 kg
Explanation:
We have given mass of the train [tex]m_i[/tex] = 7500 kg
Velocity of the car [tex]v_i[/tex] = 25 m/sec
After few minutes velocity of the car = 20 m/sec
From conservation of momentum we know that [tex]m_iv_i=m_fv_f[/tex]
So [tex]7500\times 25=m_f\times 20[/tex]
[tex]m_f=9375kg[/tex]
[tex]m_f[/tex] will be sum of mass of car and mass of rain collected
As mass of car is 7500 kg
So mass of rain collected = 9375 - 7500 = 1875 kg
Calculate the hoop stress at rupture if the burst (maximum internal) pressure is 136.4 MPa. The PV has a nominal thickness of 4 mm.
Answer:
[tex]\sigma_x = 1705 MPa[/tex]
Explanation:
Given data:
Internal pressure = 136.4 MPa
thickness is given as 4 mm
diameter is not given in the question hence we are assuming the diameter of given cylinder to be 100 mm
circumferential stress is given as
[tex]\sigma_x = \frac{P\times r}{t} [/tex]
[tex]\sigma_x = \frac{136.4 \times \frac{100}{2}}{4}[/tex]
[tex]\sigma_x = 1705 MPa[/tex]
the number of electrons in a copper penny is approximately 10 10^23. How large would the force be on an object if it carried this charge and were repelled by an equal charge one meter away?
Answer:
0.000230144 N
Explanation:
n = Number of electrons = [tex]10\times 10^{23}[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
r = Distance between obects = 1 m
The force would be
[tex]F=\dfrac{nkq^2}{r^2}\\\Rightarrow F=\dfrac{10\times 10^{23}\times 8.99\times 10^9\times (1.6\times 10^{-19})^2}{1^2}\\\Rightarrow F=0.000230144[/tex]
The force would be 0.000230144 N
7. A neutral aluminum rod is at rest on a foam insulating base. A negatively charged balloon is brought near one end of the rod but not in direct contact with it. In what way, if any, will the charges in the rod be affected?
Answer:
If a negatively charged balloon is brought near one end of the rod but not in direct contact, then the negative charges on the balloon repel the same amount of negative charges on the end of the rod that is close to the balloon, and the positive charges stay at the balloon-side of the rod. The total charge of the rod is still zero, but the distribution of the charges are now non-uniform.
The winning time for the 2005 annual race up 86 floors of the Empire State Building was 10 min and 49 s . The winner's mass was 60 kg . If each floor was 3.7 m high, what was the winner's change in gravitational potential energy? If the efficiency in climbing stairs is 25 %, what total energy did the winner expend during the race? How many food Calories did the winner "burn" during the race? What was the winner's metabolic power in watts during the race up the stairs? Express your answer using two significant figures.
Answer:
[tex]1.9\times 10^5\ J[/tex]
[tex]1.8\times 10^{2}\ kcal[/tex]
[tex]1.3\times 10^{2}\ kcal[/tex]
[tex]1.2\times 10^{3}\ W[/tex]
Explanation:
m = Mass of person = 60 kg
g = Acceleration due to gravity = 9.81 m/s²
t = Time taken = 10 min and 49 s
Total height
[tex]h=3.7\times 86\\\Rightarrow h=318.2\ m[/tex]
Potential energy is given by
[tex]U=mgh\\\Rightarrow U=60\times 9.81\times 318.2\\\Rightarrow U=187292.52\ J[/tex]
The gravitational potential energy is [tex]1.9\times 10^5\ J[/tex]
The energy in the climb
[tex]\dfrac{187292.52}{0.25}=749170.08\ J[/tex]
Converting to kcal or Cal
[tex]\dfrac{749170.08}{4184}=179.05594\ kcal[/tex]
The amount of energy used to climb [tex]1.8\times 10^{2}\ kcal[/tex]
Amount gone to heat
[tex]179.05594\times 0.75=134.291955\ kcal[/tex]
The amount burned [tex]1.3\times 10^{2}\ kcal[/tex]
Power is given by
[tex]P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{749170.08}{10\times 60+49}\\\Rightarrow P=1154.34526\ W[/tex]
The power is [tex]1.2\times 10^{3}\ W[/tex]
It has been suggested, and not facetiously, that life might have originated on Mars and been carried to Earth when a meteor hit Mars and blasted pieces of rock (perhaps containing primitive life) free of the surface. Astronomers know that many Martian rocks have come to Earth this way. One objection to this idea is that microbes would have to undergo an enormous, lethal acceleration during the impact. Let us investigate how large such an acceleration might be. To escape Mars, rock fragments would have to reach its escape velocity of 5.0 km/s, and this would most likely happen over a distance of about 4.0m during the impact.
What would be the acceleration, in m/s, of such a rock fragment?
Answer:
[tex]a=3125000 m/s^2\\a=3.125*10^6 m/s^2[/tex]
Acceleration, in m/s, of such a rock fragment = [tex]3.125*10^6m/s^2[/tex]
Explanation:
According to Newton's Third Equation of motion
[tex]V_f^2-V_i^2=2as[/tex]
Where:
[tex]V_f[/tex] is the final velocity
[tex]V_i[/tex] is the initial velocity
a is the acceleration
s is the distance
In our case:
[tex]V_f=V_{escape}, V_i=0,s=4 m[/tex]
So Equation will become:
[tex]V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2[/tex]
Acceleration, in m/s, of such a rock fragment = [tex]3.125*10^6m/s^2[/tex]
Final answer:
To find the acceleration of a rock fragment blasted from Mars, use the equation for acceleration, considering final velocity, initial velocity, and distance. Given the escape velocity of Mars, the acceleration would be approximately 6250 m/s².
Explanation:
To calculate the acceleration of a rock fragment blasted from Mars, we use the equation for acceleration: a = (v² - u²) / (2s), where v is the final velocity, u is the initial velocity, and s is the distance. Given that the final velocity is the escape velocity of Mars, 5.0 km/s, the initial velocity is 0 km/s (as it starts from rest), and the distance s is 4.0 m, we can compute the acceleration.
Substitute the values into the equation: a = ((5.0 km/s)² - (0 km/s)²) / (2 × 4.0 m) = 6.25 km/s². Converting the units to meters per second squared, we get an acceleration of 6250 m/s² for the rock fragment.
The parking brake on a 1000 kg Cadillac has failed, and it is rolling slowly, at 5 mph , toward a group of small children. Seeing the situation, you realize you have just enough time to drive your 2000 kg Volkswagen head-on into the Cadillac and save the children. Part A With what speed should you impact the Cadillac to bring it to a halt?
To solve this problem we will apply the concepts related to the conservation of the Momentum. The initial momentum must be equal to the final momentum. Momentum is described as the product between body mass and its respective velocity. Since there is no movement at the end of the collision, the final momentum will be zero.
Our values are given as
Mass of Cadillac
[tex]m_1 = 1000kg[/tex]
mass of VW
[tex]m_2 = 2000kg[/tex]
Let VW impact with speed v, then for conservation of momentum
[tex]m_1v_1+m_2v_2 = 0 \rightarrow[/tex] the Final speed is zero
Replacing,
[tex]1000kg*5mph+2000kg(-v) = 0[/tex]
[tex]v = \frac{1000kg*5mph}{2000kg}[/tex]
[tex]v = 2.5mph[/tex]
Therefore the speed to bring the cadillac to a halt is 2.5mph
The speed that can impact the Cadillac to bring it to a halt is 2.5 mph.
Velocity of the car can be calculated by the conservation of momentum formula,
[tex]\bold {m_1 v_1 =m_2v_2 = 0}[/tex] Since the final speed is zero,
where,
[tex]\bold {m_1 - mass\ of\ the\ cadillac = 1000 kg}\\\bold {m_2 - mass\ of\ the\ Volks\ Vagon = 2000 kg}[/tex]
[tex]\bold {v_1}[/tex] - velocity of the Cadillac - 5 mph.
Put the values in the formula
[tex]\bold {1000\ kg \times 5\ mph + 2000 kg (-v) = 0 }\\\\\bold {v = \dfrac {1000\ kg \times 5\ mph}{2000\ kg}}\\\\\bold {v = 2.5\ mph}[/tex]
Therefore, the speed that can impact the Cadillac to bring it to a halt is 2.5 mph.
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An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 0.40 m. What is the average angular acceleration of each wheel?
To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.
Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.
The average angular acceleration
[tex]\alpha = \frac{\omega_f - \omega_0}{t}[/tex]
Here
[tex]\alpha[/tex] = Angular acceleration
[tex]\omega_{f,i} =[/tex] Initial and final angular velocity
There is not initial angular velocity,then
[tex]\alpha = \frac{\omega_f}{t}[/tex]
We know that the relation between the tangential velocity with the angular velocity is given by,
[tex]v = r\omega[/tex]
Here,
r = Radius
[tex]\omega[/tex] = Angular velocity,
Rearranging to find the angular velocity
[tex]\omega = \frac{v}{r}}[/tex]
[tex]\omega = \frac{30}{0.20} \rightarrow[/tex] Remember that the radius is half te diameter.
Now replacing this expression at the first equation we have,
[tex]\alpha = \frac{30}{0.20*6}[/tex]
[tex]\alpha = 25 rad /s^2[/tex]
Therefore teh average angular acceleration of each wheel is [tex]25rad/s^2[/tex]
Final answer:
To find the average angular acceleration of each wheel, calculate it using the provided linear acceleration data and the relationship between linear and angular quantities.
Explanation:
Angular acceleration can be calculated using the formula:
α = Δω / Δt
Given the acceleration of the car from 0 to 30 m/s in 6.0 s, we can find the angular acceleration by first calculating the final and initial angular velocities using the relationship between linear and angular quantities.
Substitute the values into the formula and solve for the average angular acceleration of each wheel.
Suppose you designed a spacecraft to work by photon pressure. The sail was a completely absorbing fabric of area 1.0 km2 and you directed a laser beam of wavelength 650 nm onto it at a rate of 1 mol of photons per second from a base on the moon. The spacecraft has a mass of 1.0 kg. Given that, after a period of acceleration from standstill, speed = (force/mass) x time, how many minutes would it take for the craft to accelerate to a speed of 1.0 m/s (about 2.2 mph)?
Answer:
(a) F = 6.14 *10⁻⁴ N
(b) P = 6.14* 10⁻¹⁰ Pa
(c) t = 27.2 min
Explanation:
Area of sail A = 1.0 km² = 1.0 * 10⁶m²
Wavelength of light λ = 650 nm = 650 * 10⁻⁹ m
Rate of impact of photons R = 1 mol/s = 6.022 * 10²³ photons/s
(a)
Momentum of each photon is Ρ = h/λ = 6.625 * 10⁻³⁴ / 650 * 10⁻⁹
= 1.0192 * 10⁻²⁷ kg.m/s
Since the photons are absorbed completely, in every collision the above momentum is transferred to the sail.
Momentum transferred to the sail per second is product of rate of impact of photons and momentum transferred by each photon.
dp/dt = R * h/ λ
This is the force acting on the sail.
F = R * h/λ = 6.022 * 10²³ * 1.0192 * 10⁻²⁷ = 6.14 *10⁻⁴ N
F = 6.14 *10⁻⁴ N
b)
Pressure exerted by the radiation on the sail = Force acting on the sail / Area of the sail
P = F/A = 6.14 * 10⁻⁴ / 1.0 * 10⁶ = 6.14* 10⁻¹⁰ Pa
P = 6.14* 10⁻¹⁰ Pa
c)
Acceleration of spacecraft a = F/m = 6.14 * 10⁻⁴ /1.0 = 6.14 * 10⁻⁴m/s²
As the spacecraft starts from rest, initial speed u=0,m/s ,
final speed is u = 1.0 m/s after time t
v = u+at
t = 1.0 - 0/ 6.14 * 10⁻⁴ = 1629s = 27.2 min
t = 27.2 min
How many photons does a green (532nm) 1mW laser pointer emit? Can we still treat the light as a continuous electromagnetic wave in this case? (2 pts)
Answer:
[tex]n=2.68\times 10^{15}[/tex]photons/s
Yes,we can treat the light as a continuous electromagnetic wave in this case.
Explanation:
We are given that
Wavelength of photon=[tex]\lambda=532 nm=532\times 10^{-9}m[/tex]
1nm=[tex]10^{-9} m[/tex]
Power of photon=1mW=[tex]0.0010J/s[/tex]
We have to find the number of photons emit from laser.
In 1 s, photon emit energy=0.0010 J
We know that
[tex]E=nh\nu=nh\times \frac{c}{\lambda}[/tex]
Where c[tex]\c=3\times 10^8 m/s[/tex]
[tex]E=[/tex]Energy of photon
[tex]\lambda[/tex]=Wavelength of light
[tex]h=6.626\times 10^{-34}Js[/tex]
Substitute the values then, we get
[tex]0.0010=n\times 6.626\times 10^{-34}\times \frac{3\times 10^8}{532\times 10^{-9}}[/tex]
[tex]n=\frac{0.0010\times 532\times 10^{-9}}{6.626\times 10^{-34}\times 3\times 10^8}[/tex]
[tex]n=2.68\times 10^{15}[/tex]photons/s
Visible light is EM wave.
Range of wavelength of visible light =400nm- 700 nm
Green color lies in visible region because its wavelength lies in 400nm-700 nm.
Therefore, we can treat the light as a continuous electromagnetic wave in this case.
A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general there are an infinite number of such modes, each one with a distinctive frequency fi and associated pattern of oscillation.
Consider an example of a system with normal modes: a string of length L held fixed at both ends, located at x=0 and x=L. Assume that waves on this string propagate with speed v. The string extends in the x direction, and the waves are transverse with displacement along the y direction.
In this problem, you will investigate the shape of the normal modes and then their frequency.
The normal modes of this system are products of trigonometric functions. (For linear systems, the time dependance of a normal mode is always sinusoidal, but the spatial dependence need not be.) Specifically, for this system a normal mode is described by
yi(x,t)=Ai sin(2π*x/λi)sin(2πfi*t)
A)The string described in the problem introduction is oscillating in one of its normal modes. Which of the following statements about the wave in the string is correct?
The wave is traveling in the +x direction.
a) The wave is traveling in the -x direction.
b) The wave will satisfy the given boundary conditions for any arbitrary wavelength lambda_i.
c) The wavelength lambda_i can have only certain specific values if the boundary conditions are to be satisfied.
d) The wave does not satisfy the boundary condition y_i(0;t)=0.
B)Which of the following statements are true?
a)The system can resonate at only certain resonance frequencies f_i and the wavelength lambda_i must be such that y_i(0;t) = y_i(L;t) = 0.
b) A_i must be chosen so that the wave fits exactly on the string.
c) Any one of A_i or lambda_i or f_i can be chosen to make the solution a normal mode.
C) Find the three longest wavelengths (call them lambda_1, lambda_2, and lambda_3) that "fit" on the string, that is, those that satisfy the boundary conditions at x=0 and x=L. These longest wavelengths have the lowest frequencies.
D) The frequency of each normal mode depends on the spatial part of the wave function, which is characterized by its wavelength lambda_i.
Find the frequency f_i of the ith normal mode.
Answer:
Explanation:
(A)
The string has set of normal modes and the string is oscillating in one of its modes.
The resonant frequencies of a physical object depend on its material, structure and boundary conditions.
The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.
Given below are the incorrect options about the wave in the string.
• The wave is travelling in the +x direction
• The wave is travelling in the -x direction
• The wave will satisfy the given boundary conditions for any arbitrary wavelength [tex]\lambda_i[/tex]
• The wave does not satisfy the boundary conditions [tex]y_i(0;t)=0 [/tex]
Here, the string of length L held fixed at both ends, located at x=0 and x=L
The key constraint with normal modes is that there are two spatial boundary conditions,[tex]y(0,1)=0 [/tex]
and [tex]y(L,t)=0[/tex]
.The spring is fixed at its two ends.
The correct options about the wave in the string is
• The wavelength [tex]\lambda_i[/tex] can have only certain specific values if the boundary conditions are to be satisfied.
(B)
The key factors producing the normal mode is that there are two spatial boundary conditions, [tex]y_i(0;t)=0[/tex] and [tex]y_i(L;t)=0[/tex], that are satisfied only for particular value of [tex]\lambda_i[/tex] .
Given below are the incorrect options about the wave in the string.
• [tex]A_i[/tex] must be chosen so that the wave fits exactly o the string.
• Any one of [tex]A_i[/tex] or [tex]\lambda_i[/tex] or [tex]f_i[/tex] can be chosen to make the solution a normal mode.
Hence, the correct option is that the system can resonate at only certain resonance frequencies [tex]f_i[/tex] and the wavelength [tex]\lambda_i[/tex] must be such that [tex]y_i(0;t) = y_i(L;t)=0 [/tex]
(C)
Expression for the wavelength of the various normal modes for a string is,
[tex]\lambda_n=\frac{2L}{n}[/tex] (1)
When [tex]n=1[/tex] , this is the longest wavelength mode.
Substitute 1 for n in equation (1).
[tex]\lambda_n=\frac{2L}{1}\\\\2L[/tex]
When [tex]n=2[/tex] , this is the second longest wavelength mode.
Substitute 2 for n in equation (1).
[tex]\lambda_n=\frac{2L}{2}\\\\L[/tex]
When [tex]n=3[/tex], this is the third longest wavelength mode.
Substitute 3 for n in equation (1).
[tex]\lambda_n=\frac{2L}{3}[/tex]
Therefore, the three longest wavelengths are [tex]2L[/tex],[tex]L[/tex] and [tex]\frac{2L}{3}[/tex].
(D)
Expression for the frequency of the various normal modes for a string is,
[tex]f_n=\frac{v}{\lambda_n}[/tex]
For the case of frequency of the [tex]i^{th}[/tex] normal mode the above equation becomes.
[tex]f_i=\frac{v}{\lambda_i}[/tex]
Here, [tex]f_i[/tex] is the frequency of the [tex]i^{th}[/tex] normal mode, v is wave speed, and [tex]\lambda_i[/tex] is the wavelength of [tex]i^{th}[/tex] normal mode.
Therefore, the frequency of [tex]i^{th}[/tex] normal mode is [tex]f_i=\frac{v}{\lambda_i}[/tex]
.
The wave in the string travels in the +x direction and the wavelength lambda_i can have only certain specific values to satisfy the boundary conditions. The system can resonate at certain resonance frequencies and the wavelength must satisfy the boundary conditions. The frequency of each normal mode depends on the wavelength.
Explanation:For the given system of a string held fixed at both ends, the wave in the string travels in the +x direction. The boundary conditions for the wave to satisfy are that yi(0, t) = 0 and yi(L, t) = 0. Therefore, the wavelength lambda_i can have only certain specific values if the boundary conditions are to be satisfied.
It is not necessary for Ai to fit exactly on the string, so statement (b) is incorrect. However, statement (a) is true - the system can resonate at only certain resonance frequencies fi, and the wavelength lambda_i must be such that yi(0, t) = 0 and yi(L, t) = 0. Any one of Ai or lambda_i or fi can be chosen to make the solution a normal mode, so statement (c) is also true.
The three longest wavelengths that fit on the string and satisfy the boundary conditions are lambda_1 = 2L, lambda_2 = L, and lambda_3 = L/2. These wavelengths have the lowest frequencies.
The frequency fi of the ith normal mode depends on the wavelength lambda_i. The frequency can be calculated using the equation fi = (v/lambda_i) * (1/2), where v is the speed of propagation of waves on the string.
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A 2 kg rock is suspended by a massless
string from one end of a 7 m measuring stick.
What is the weight of the measuring stick
if it is balanced by a support force at the
1 m mark? The acceleration of gravity is
9.81 m/s^2
Answer in units of N.
Answer:
Weight = 7.848 N
Explanation:
Given:
Mass of the rock (m) = 2 kg
Length of the stick (L) = 7 m
Distance of support from the rock (x) = 1 m
The weight of the stick acts at the center, which is at a distance of 3.5 m from one end.
So, let 'M' be mass of the measuring stick.
Distance of 'W' from the supporting force (d) = 3.5 - 1 = 2.5 m
Now, for equilibrium, the sum of clockwise moments must be equal to the sum of anticlockwise moments.
Sum of clockwise moments = Sum of anticlockwise moments
[tex]m\times g\times x=M\times g\timesd\\\\M=\frac{mx}{d}[/tex]
Plug in the given values and solve for 'M'. This gives,
[tex]M=\frac{2\times 1}{2.5}=0.8\ kg[/tex]
Now, weight of the stick is given as:
[tex]W=Mg=0.8\times 9.81=7.848\ N[/tex]
Answer:
The weight at [tex]1[/tex]m mark is 7.848 N.
Explanation:
We have been given,
Mass of the rock= [tex]2[/tex] kg
Acceleration (gravity) = [tex]9.8[/tex] m/s^2
Length of the string = [tex]7[/tex] m
To find the weight at [tex]1[/tex] m mark let it be W.
So,
We know that torque will be balanced here.
Moment arm for the force (W) is [tex]3.5 - 1=2.5[/tex] m
As it is of [tex]7[/tex] m length so [tex]3.5[/tex] m is the COM (center of mass) of the string.
Let the moment be [tex]L_1\ and\ L_2[/tex] on clockwise and anti-clockwise direction.
[tex]L_1=L_2[/tex]
[tex]2.5*W = 2* 9.81*1[/tex]
Dividing both sides with [tex]2.5[/tex]
[tex]W=\frac{2*9.81*1}{2.5} =7.848\N[/tex]
So the weight is 7.848 N.
The coefficient of performance of a refrigerator is 5.0.
A. If the compressor uses 10 J of energy, how much heat is exhausted to the hot reservoir?
B. if the hot reservoir temperature is 27 degrees Celsius, what is the lowest possible temperature in degrees Celsius of the cold reservoir?
Answer:
[tex]Q_x=60\ J[/tex]
[tex]T_L=-23\ ^{\circ}C[/tex] is the lowest possible temperature of the cold reservoir.
Explanation:
Given:
Coefficient of performance of refrigerator, [tex]COP=5[/tex]
A)
We know that for a refrigerator:
[tex]\rm COP=\frac{ desired\ effect }{\: energy\ supplied }[/tex]
Given that 10 J of energy is consumed by the compressor:
[tex]5=\frac{desired\ effect}{10}[/tex]
[tex]\rm Desired\ effect=50\ J[/tex]
Now the by the conservation of energy the heat exhausted :
[tex]Q_x=50+10[/tex]
[tex]Q_x=60\ J[/tex]
B)
Also
[tex]COP=\frac{T_L}{T_H-T_L}[/tex]
where:
[tex]T_H\ \&\ T_L[/tex] are the absolute temperatures of high and low temperature reservoirs respectively.
[tex]5=\frac{T_L}{(273+27)-T_L}[/tex]
[tex]1500-5\times T_L=T_L[/tex]
[tex]T_L=250\ K[/tex]
[tex]T_L=-23\ ^{\circ}C[/tex] is the lowest possible temperature of the cold reservoir.
Final answer:
The coefficient of performance (COP) of a refrigerator is the ratio of heat removed from the cold reservoir to the work done by the compressor. Using this definition, we can solve for the amount of heat exhausted to the hot reservoir and the temperature of the cold reservoir.
Explanation:
The coefficient of performance (COP) of a refrigerator is a measure of its efficiency. It is defined as the ratio of heat removed from the cold reservoir to the work done by the compressor:
COP = heat removed / work done
(A) To find the amount of heat exhausted to the hot reservoir, we can use the formula:
heat removed = COP x work done
Substituting the given values, we get:
heat removed = 5.0 x 10 J = 50 J
(B) To find the lowest possible temperature of the cold reservoir, we can use the Carnot efficiency formula:
COP = hot temperature / (hot temperature - cold temperature)
Substituting the given values and rearranging the formula, we get:
cold temperature = hot temperature - (hot temperature / COP)
cold temperature = 27°C - (27°C / 5.0) = 21.6°C
If a rope loaded with a force of 100 lbf had a spring constant of 500 lbf/inch, how far would the rope stretch or elongate?
To solve this problem we will apply the concepts related to Hooke's law for which the force exerted on a spring is described as the product between the spring constant and its displacement, that is
[tex]F = kx[/tex]
Where k is the spring's constant and x is the elongation,
Rearranging to find the elongation we have
[tex]x = \frac{F}{k}[/tex]
Replacing,
[tex]x = \frac{100}{500}[/tex]
[tex]x = 0.2in[/tex]
[tex]x = 5.08mm[/tex]
Therefore the elongation produced in the rope from its original length is 0.2in or 5.08mm
Two point charges of equal magnitude are 7.3 cm apart. At the midpoint of the line connecting them, their combined electric field has a magnitude of 50 N/C .
Answer:
q₁ = q₂ = Q = 14.8 pC
Explanation:
Given that
q₁ = q₂ = Q = ?
Distance between charges = r =7.3 cm = 0.073 m
Combined electric field = E₁ + E₂ = E = 50 N/C
Using formula
[tex]E=2\frac{kQ}{r^2}[/tex]
Rearranging for Q
[tex]Q= \frac{Er^2}{2k}\\\\Q=\frac{(50)(.005329)}{2\times 9\times 10^9}[/tex]
[tex]Q=14.8\times 10^{-12}\, C\\\\Q=14.8\, pC[/tex]