Explanation:
Distance d=1.5×108 km=1.5×1011 m
Mass of the sun, m=2×1030 kg
Mass of the earth, M=6×1024 kg
Force of gravitation, F=G×d2m×M
F=6.7×10−11×(1.5×1011)22×1030×6×1024=3.57×1022 N
A fan is turned off, and its angular speed decreases from 10.0 rad/s to 6.3 rad/s in 5.0 s. What is the magnitude of the angular acceleration of the fan?
A) 0.37 rad/s2
B) 11.6 rad/s2
C) 0.74 rad/s2
D) 0.86 rad/s2
E) 1.16 rad/s2
Answer:
chk photo
Explanation:
The lines in the emission spectrum of hydrogen result from __________.
a. energy given off in the form of visible light when an electron moves from a higher energy state to a lower energy state
b. protons given off when hydrogen burns
c. electrons given off by hydrogen as it cools
d. electrons given off by hydrogen when it burns
e. decomposing hydrogen atoms.
Answer:
Option (a) is correct.
Explanation:
The lines in the emission spectrum of hydrogen is due to the transfer of electrons form higher energy levels to the lower energy levels.
When the electrons transfer from one level of energy that is higher level of energy to the other means to the lower level of energy then they emit some photons which having the frequency or the wavelength in the visible region.
One charge is fixed q1 = 5 µC at the origin in a coordinate system, a second charge q2 = -3.2 µC the other is at a distance of x = 90 m from the origin.
What is the potential energy of this pair of charges?
Answer:
5.4uC
Explanation:
A battery is two or more individual cells connected together. Some large trucks utilize large 24 volt lead acid batteries. How many lead acid cells would be required to construct a battery with this voltage
Answer:
#_pile = 12 celdas
Explanation:
Lead acid sulfur batteries generate each cell a potential of 2 volts. By colonato to reach the voltage of 24 volts
#_pile = 24/2
#_pile = 12 cledas
serially connected
A 25g rock is rolling at a speed of 5 m/s. What is the kinetic energy of the rock?
Answer:
The answer is 312.5j
Explanation:
The kinetic energy (KE):
KE=1/2*m*v^2
M= mass of the object
v= velocity of the object
We have;
m=25g
v=5m/s
KE=1/2*25g*5^2m/s
KE =312.5j
A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force of 9.14 × 10-17 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.
Answer:
a) The proton's speed is 5.75x10⁵ m/s.
b) The kinetic energy of the proton is 1723 eV.
Explanation:
a) The proton's speed can be calculated with the Lorentz force equation:
[tex] F = qv \times B = qvBsin(\theta) [/tex] (1)
Where:
F: is the force = 9.14x10⁻¹⁷ N
q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C
v: is the proton's speed =?
B: is the magnetic field = 3.28 mT
θ: is the angle between the proton's speed and the magnetic field = 17.6°
By solving equation (1) for v we have:
[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]
Hence, the proton's speed is 5.75x10⁵ m/s.
b) Its kinetic energy (K) is given by:
[tex] K = \frac{1}{2}mv^{2} [/tex]
Where:
m: is the mass of the proton = 1.67x10⁻²⁷ kg
[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]
Therefore, the kinetic energy of the proton is 1723 eV.
I hope it helps you!
a volcano that may erupt again at some time in the distant future is
When you exert 75 N on a jack to lift a 6000 N car, what is the jack’s actual mechanical advantage? Show your work.
Answer:
80
Explanation:
the mechanical advantage is the ratio of the load to the effort so it doesn't have units.to calculate it you use the formula
mechanical advantage=load/effort
in this case the load is 6000N and the effort is 75N
Ma=6000/75
=80
I hope this helps
A surveyor measures the distance across a straight river by the following method: Starting directly across from a tree on the opposite bank, he walks x = 106 m along the riverbank to establish a baseline. Then he sights across to the tree. The angle from his baseline to the tree is = 32.8°. How wide is the river?
Answer:
x = 68.3 m
Explanation:
tan 32.8 = x / 106
Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are infinitely far away. What is the closest distance that she can see an object clearly when she wears her contacts?
Answer:
the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm
Explanation:
Given the data in the question,
near point = 20 cm
far point = 2 m = 200 cm
Now, for an object that is infinitely far away, the image is at is its far point.
so using the following expression, we can determine the focal length
1/f = 1/i + 1/o
where f is the focal length, i is the image distance and o is the object distance.
here, far point i = 2 m = 200 cm and v is ∞
so we substitute
1/f = 1/(-200 cm) + 1/∞
f = -200 cm
Also, for object at its closest point, the image appear at near point,
so
1/f = 1/i + 1/o
we make o the subject of formula
o = ( i × f ) / ( i - f )
given that near point i = 20 cm
we substitute
o = ( -20 × -200 ) / ( -20 - (-200) )
o = 4000 / 180
o = 22.2 cm
Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm
Two streams merge to form a river. One stream has a width of 8.3 m, depth of 3.2 m, and current speed of 2.2 m/s. The other stream is 6.8 m wide and 3.2 m deep, and flows at 2.4 m/s. If the river has width 10.4 m and speed 2.8 m/s, what is its depth?
Answer:
The depth of the resulting stream is 3.8 meters.
Explanation:
Under the assumption that streams are formed by incompressible fluids, so that volume flow can observed conservation:
[tex]\dot V_{1} + \dot V_{2} = \dot V_{3}[/tex] (1)
All volume flows are measured in cubic meters per second.
Dimensionally speaking, we can determine the depth of the resulting stream ([tex]h_{3}[/tex]), in meters, by expanding (1) in this manner:
[tex]w_{1}\cdot h_{1}\cdot v_{1} + w_{2}\cdot h_{2}\cdot v_{2} = w_{3}\cdot h_{3}\cdot v_{3}[/tex]
[tex]h_{3} = \frac{w_{1}\cdot h_{1}\cdot v_{1}+w_{2}\cdot h_{2}\cdot v_{2}}{w_{3}\cdot v_{3}}[/tex] (2)
[tex]v_{1}, v_{2}[/tex] - Speed of the merging streams, in meters per second.
[tex]h_{1}, h_{2}[/tex] - Depth of the merging streams, in meters.
[tex]w_{1}, w_{2}[/tex] - Width of the merging streams, in meters.
[tex]w_{3}[/tex] - Width of the resulting stream, in meters.
[tex]v_{3}[/tex] - Speed of the resulting stream, in meters per second.
If we know that [tex]w_{1} = 8.3\,m[/tex], [tex]h_{1} = 3.2\,m[/tex], [tex]v_{1} = 2.2\,\frac{m}{s}[/tex], [tex]w_{2} = 6.8\,m[/tex], [tex]h_{2} = 3.2\,m[/tex], [tex]v_{2} = 2.4\,\frac{m}{s}[/tex], [tex]w_{3} = 10.4\,m[/tex] and [tex]v_{3} = 2.8\,\frac{m}{s}[/tex], then the depth of the resulting stream is:
[tex]h_{3} = \frac{(8.3\,m)\cdot (3.2\,m)\cdot \left(2.2\,\frac{m}{s} \right) + (6.8\,m)\cdot (3.2\,m)\cdot \left(2.4\,\frac{m}{s} \right)}{(10.4\,m)\cdot \left(2.8\,\frac{m}{s} \right)}[/tex]
[tex]h_{3} = 3.8\,m[/tex]
The depth of the resulting stream is 3.8 meters.
The Michelson-Morley experiment was designed to measure Group of answer choices the velocity of the Earth relative to the ether. the relativistic momentum of the electron. the relativistic mass of the electron. the acceleration of gravity on the Earth's surface. the relativistic energy of the electron.
Answer:
The Michelson-Morley was designed to detect the motion of the earth through the ether.
No such relation was found and the speed of light is assumed to be the same in all reference frames.
The Michelson-Morley experiment was designed to measure: A. the velocity of the Earth relative to the ether.
Michelson-Morley experiment is an experiment which was first performed in Germany by the American physicist named, Albert Abraham Michelson between 1880 to 1881.
However, the experiment was later modified and refined by Michelson and Edward W. in 1887.
The main purpose of the Michelson-Morley experiment was to measure the velocity of planet Earth relative to the luminiferous ether, which is a medium in space that is hypothetically said to carry light waves.
In conclusion, the Michelson-Morley experiment was designed to measure the velocity of the Earth relative to the hypothetical luminiferous ether.
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What is science?Give two examples of living beings?
Answer:
the study of the past
Explanation:
dogs and cats
Assume that a friend hands you a 10-newton box to hold for her. If you hold the box without moving it at a height of 10 meters above the ground, how much work do you do
Answer:
100 Joules
Explanation:
Applying,
W = mgh................... Equation 1
Where W = workdone to hold the box above the ground, mg = weight of the box, h = height of the box.
From the question,
Given: mg = 10 newtons, h = 10 meters.
Substitute these values into equation 1
W = 10×10
W = 100 Joules.
Hence the amount of workdone is 100 Joules
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 11.0 ft/s at point A and 18.0 ft/s at point C. The cart takes 5.00 s to go from point A to point C, and the cart takes 1.30 s to go from point B to point C. What is the cart's speed at point B
Answer:
The speed at B is 16.18 ft/s .
Explanation:
Speed at A, u = 11 ft/s
Speed at C, v' = 18 ft/s
Time from A to C = 5 s
Time from B to C = 1.3 s
Let the speed of car at B is v.
Let the acceleration is a.
From A to B
Use first equation of motion
v = u + a t
18 = 11 + a x 5
a = 1.4 ft/s^2
Let the time from A to B is t' .
t' = 5 - 1.3 = 3.7 s
Use first equation of motion from A to B
v = 11 + 1.4 x 3.7 = 16.18 ft/s
If a proton and electron both move through the same displacement in an electric field, is the change in potential energy associated with the proton equal in magnitude and opposite in sign to the change in potential energy associated with the electron?
a. The magnitude of the change is smaller for the proton.
b. The magnitude of the change is larger for the proton.
c. The signs Of the two changes in potential energy are opposite.
d. They are equal in magnitude.
e. The signs of the two changes in potential energy are the same.
Answer: They are equal in magnitude.
- The signs of the two changes in potential energy are opposite
Explanation:
When the proton and electron both move through the same displacement in an electric field, the change in potential energy that is associated with the proton is equal in magnitude.
Also, it should be noted that the signs of the two changes in potential energy are opposite.
Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes, subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is Q=ΔV/Δt what is the flow rate in pipe A? Viscosity: Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes, subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is what is the flow rate in pipe A?
a) Q√2
b) 16Q
c) 2Q
d) 4Q
e) 8Q
Answer:
c) 2Q
Explanation:
From the given information:
The pressure inside a pipe can be expressed by using the formula:
[tex]\Delta P = \dfrac{128 \mu L Q}{\pi D^4}[/tex]
Since the diameter in both pipes is the same, we can say:
[tex]D = D_A = D_B[/tex]
where;
length of the first pipe A [tex]L_A = L[/tex] and the length of the second pipe B [tex]L_B = 2L[/tex]
Since the difference in pressure is equivalent in both pipes:
Then:
[tex]\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}[/tex]
[tex]\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}[/tex]
[tex]\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}[/tex]
[tex]\mathbf{Q_1 = 2Q}[/tex]
The flow rate in pipe B is 2Q of the flow rate of the pipe A
What is flow rate?
The flow rate is defined as the flow of the fluid across the cross section in per unit time.
From the given information:
The pressure inside a pipe can be expressed by using the formula:
[tex]\Delta p=\dfrac{128\mu LQ}{\pi D^4}[/tex]
Since the diameter in both pipes is the same, we can say:
[tex]D=D_A=D_B[/tex]
where;
length of the first pipe A [tex]L_A=L[/tex] and the length of the second pipe B
[tex]L_B=2L[/tex]
Since the difference in pressure is equivalent in both pipes:
Then:
[tex]\dfrac{128\mu L_1Q_1}{\pi D_1^4}=\dfrac{128\mu L_2Q_2}{\pi D_2^4}[/tex]
[tex]\dfrac{L_1Q_1}{D_1^4}=\dfrac{L_2Q_2}{D_2^4}[/tex]
[tex]\dfrac{LQ_1}{D_1^4}=\dfrac{2LQ}{D_2^4}[/tex]
[tex]Q_1=2Q[/tex]
Hence the flow rate in pipe B is 2Q of the flow rate of the pipe A
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which one is more powerful hydrogen bomb or atom bomb and why?
Hydrogen bomb is more powerful than atom bomb
Hydrogen has a calorie value of 150000KJ .It is very much than nuclear bomb or atom bombScientists also told that Hydrogen bomb is more powerful.But both bombs are destructive.The valves in the heart open and close to move blood in between heart chambers. O a) three directions O b) two directions O c) four directions d) one direction
Hi,
The valves in the heart open and close to move blood in between heart chambers.
Answer:
d) one direction
In the diagram, the crest of the wave is show by:
A
B
C
D
Answer:
D.
Explanation:
The crest of a wave refers to the highest point of a wave. This is illustrated by D.
Why is oiling done time and again in a sewing machine?
Answer:
to prevent friction on the surfaces
Answer:
Explanation:
Oiling reduces friction between parts with relative motion between them.
Repeated oiling is needed as the film of oil reducing the friction becomes thinner with time as some of the oil gets pushed off of the areas of motion where it can no longer be useful.
Oil also becomes oxidized which reduces the oil's ability to decrease friction.
Oil can also be fouled by dirt, lint or other material. This added material becomes coated in oil and typically gets sequestered away from the moving parts reducing the oil available for lubrication purposes.
The Earth’s orbit around the Sun is slightly elliptical. At Earth's closest approach to the Sun (perihelion) the orbital radius is 1.471×10^11m, and at its farthest distance (aphelion) the orbital radius is 1.521×10^11m.
a. Find the difference in gravitational potential energy between when the Earth is at its aphelion and perihelion radii.
b. If the orbital speed of the Earth is 29,290 m/s at aphelion, what is its orbital speed at perihelion?
Answer:
1.25
Explanation:
True or false : conservation of energy gives a relationship between the speed of a falling object and the height from which it was dropped
Answer:
truee
Explanation:
A singly charged 7Li ion has a mass of 1.16 10-26 kg. It is accelerated through a potential difference of 523 V and subsequently enters a uniform magnetic field of magnitude 0.370 T perpendicular to the ion's velocity. Find the radius of its path.
Answer:
[tex]R=0.023m[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=1.16*10^{-26}[/tex]
Potential difference [tex]V=523V[/tex]
Magnitude [tex]m=0.370 T[/tex]
Generally the equation for Velocity is mathematically given by
[tex]\frac{1}{2}mv^2=ev[/tex]
[tex]v=\frac{2ev}{m}[/tex]
[tex]v=\frac{2*1.6*10^{-19}*542}{1.16*10^{-26}}[/tex]
[tex]v=12.22*10^4m/s[/tex]
Generally the equation for Force is mathematically given by
[tex]F=qvBsin \theta[/tex]
Where
[tex]qVB=m\frac{v^2}{R}[/tex]
[tex]F=m\frac{v^2}{R}sin\theta[/tex]
Therefore
[tex]R=\frac{mv}{qB sin \theta}[/tex]
[tex]R=\frac{1.6*10^{-26}*12.2*10^{4}}{1.60*10^{-19}*0.394 sin 90}[/tex]
[tex]R=0.023m[/tex]
Consider a 200-ft-high, 1200-ft-wide dam filled to capacity. Determine (a) the hydrostatic force on the dam and (b) the force per unit area of the dam near the top and near the bottom. Note: we will see that the resultant hydrostatic force will be
Answer:
a) [tex]F_g=1.5*10^9Ibf[/tex]
b) [tex]F_t=12490Ibf/ft^2[/tex]
[tex]F_b=0[/tex]
Explanation:
From the question we are told that:
Height [tex]h=200ft[/tex]
Width [tex]w=1200ft[/tex]
a)
Generally the equation for Dam's Hydro static force is mathematically given by
[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]
Where
[tex]\rho=Density\ of\ water[/tex]
[tex]\rho=62.4Ibm/ft^3[/tex]
Therefore
[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]
[tex]F_g=1.5*10^9Ibf[/tex]
b)
Generally the equation for Dam's Force per unit area is mathematically given by
[tex]F=\rho*g*h[/tex]
For Top
[tex]F_t=\rho*g*h[/tex]
[tex]F_t=62.4*32.2*200[/tex]
[tex]F_t=12490Ibf/ft^2[/tex]
For bottom
[tex]Here \\H=0 zero[/tex]
Therefore
[tex]F_b=0[/tex]
The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].
The force per unit area near the top is 86.74 psi.
The force per unit area near the bottom is zero.
Hydrostatic force
The hydrostatic force on the dam is the force exerted on the dam by the column of the water.
[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]
Force per unit area near the topThe force per unit area is the pressure exerted near the top of the dam.
[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]
where;
P is pressure in PSI
ρ is density of water in lb/gal
h is the vertical height in ft
[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]
The pressure near the bottom is zero, become the vertical height is zero.
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Assume that the car on the left makes a quick turn to the left. According to inertia, your body will resist a change and still want to go in the original direction. In which direction with the passenger slide?
Answer:
to the right
Explanation:
if the car turns to the lift, the body forces energy to the left side, so according to the first law of Newton, the body will move to the right side to resist the sudden motion.
Derive the dimension of coefficient of linear expansivity
Answer:
The SI unit of coefficient of linear expansion can be expressed as °C-1 or °K-1. ... The dimension of coefficient of linear expansion will be M0L0T0K−1.
2. A parallel-plate capacitor has a capacitance of C. If the area of the plates is doubled and
the distance between the plates is doubled, what is the new capacitance?
A) C/4
B) C/2
C)C
D) 4C
(C)
Explanation:
The capacitance C of a parallel plate capacitor is given by
[tex]C = \epsilon_0 \dfrac{A}{d}[/tex]
Let C' be the new capacitance where the area and the plate separation distance are doubled. This gives us
[tex]C' = \epsilon_0\dfrac{A'}{d'} = \epsilon_0\left(\dfrac{2A}{2d}\right) = \epsilon_0 \dfrac{A}{d} = C[/tex]
A long string is moved up and down with simple harmonic motion with a frequency of 46 Hz. The string is 579 m long and has a total mass of 46.3 kg. The string is under a tension of 3423 and is fixed at both ends. Determine the velocity of the wave on the string. What length of the string, fixed at both ends, would create a third harmonic standing wave
Answer:
a) [tex]v=206.896m/s[/tex]
b) [tex]L=6.749m[/tex]
Explanation:
From the question we are told that:
Frequency [tex]F=46Hz[/tex]
Length [tex]l=579m[/tex]
Total Mass [tex]T=4.3kg[/tex]
Tension [tex]T=3423[/tex]
a)
Generally the equation for velocity is mathematically given by
[tex]v=\sqrt{\frac{T}{\rho}}[/tex]
Where
[tex]\pho=m*l\\\\\pho=46*579\\\\\pho=0.0799kg/m[/tex]
Therefore
[tex]v=\sqrt{\frac{3423}{0.0799}}[/tex]
[tex]v=206.896m/s[/tex]
b)
Generally the equation for length of string is mathematically given by
[tex]L=\frac{3\lambda}{2}[/tex]
Where
[tex]\lambda=\frac{v}{f}[/tex]
[tex]\lambda=\frac{206.89}{46}[/tex]
[tex]\lambda=4.498[/tex]
Therefore
[tex]L=\frac{3*4.498}{2}[/tex]
[tex]L=6.749m[/tex]
A Geiger counter registers a count rate of 8,000 counts per minute from a sample of a radioisotope. The count rate 24 minutes later is 1,000 counts per minute. What is the half-life of the radioisotope?
11.54 minutes
Explanation:
The decay rate equation is given by
[tex]N = N_0e^{-\frac{t}{\lambda}}[/tex]
where [tex]\lambda[/tex] is the half-life. We can rewrite this as
[tex]\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}[/tex]
Taking the natural logarithm of both sides, we get
[tex]\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)[/tex]
Solving for [tex]\lambda[/tex],
[tex]\lambda = -\dfrac{t}{\ln \left(\frac{N}{N_0}\right)}[/tex]
[tex]\:\:\:\:= -\dfrac{(24\:\text{minutes})}{\ln \left(\frac{1000\:\text{counts/min}}{8000\:\text{counts/min}}\right)}[/tex]
[tex]\:\:\:\:=11.54\:\text{minutes}[/tex]
