Answer:
Approximately [tex]2.03 \times 10^{7}\; {\rm m}[/tex].
Explanation:
Assume that the radius of this orbit is [tex]r[/tex].
Let [tex]m[/tex] denote the mass of this satellite and let [tex]M[/tex] denote the mass of the Earth. At a distance of [tex]r[/tex] from the center of the earth, the magnitude of the gravitational attraction on this satellite would be [tex]G\, m\, M / (r^{2})[/tex].
The question implies that the gravitational pull from the earth is the only significant force on this satellite. Hence, the net force on this satellite would be also [tex]G\, m\, M / (r^{2})[/tex].
The acceleration of this satellite would thus be [tex]a = (\text{net force}) / (\text{mass}) = G\, M / (r^{2})[/tex].
Let [tex]\omega[/tex] denote the angular velocity of this satellite. Since this satellite in in a circular motion, the acceleration on this satellite would need to satisfy [tex]a = \omega^{2} \, r[/tex].
In other words:
[tex]\begin{aligned} \frac{G\, M}{r^{2}} = a = \omega^{2} \, r \end{aligned}[/tex].
[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3}\end{aligned}[/tex].
The question asks for a rotation of [tex]3\times (2\, \pi) = 6\, \pi\; {\text{rad}}[/tex] within a day, which is [tex]24 \times 3600\; {\rm s}[/tex]. The angular velocity of this satellite should be:
[tex]\begin{aligned}\omega = \frac{6\, \pi}{24 \times 3600\; {\rm s}} \end{aligned}[/tex].
Substitute this value into the expression for [tex]r[/tex] and evaluate:
[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3} \\ &= \left(\frac{(6.67 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}}) \times (5.97 \times 10^{24}\; {\rm kg})}{((6\, \pi) / (24 \times 3600\; {\rm s}))^{2}}\right)^{1/3} \\ &\approx 2.03 \times 10^{7}\; {\rm m}\end{aligned}[/tex].
(Note that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].)
A solid cylinder of uniform density of 0.85 g/cm3 floats in a glass of water tinted light blue by food coloring.
Its circular surfaces are horizontal. What effect will the following changes, each made to the initial system, have on X, the height of the upper surface above the water? The liquids added do not mix with the water, and the cylinder never hits the bottom.
1. The cylinder is replaced with one that has the same density and diameter, but with half the height.
2. Some of the water is removed from the glass.
3. A liquid with a density of 1.06 g/cm3 is poured into the glass.
4. The cylinder is replaced with one that has the same height and diameter, but with density of 0.83 g/cm3.
5. A liquid with a density of 0.76 g/cm3 is poured into the glass.
6. The cylinder is replaced with one that has the same density and height, but 1.5× the diameter.
Options are: Increase, Decrease, No change
Each side has to have at least 44 horses
F61160 N. This is further explained below.
What is the force?Generally, We are only interested in the component that operates horizontally since the vertical components all cancel each other out. The pressure difference works on the hemisphere to generate a normal force all over the surface, but we are only concerned with that force's horizontal component. This may be determined by supposing the hemispheres to be two flat circular plates of the same radius as the hemispheres that have been forced together.
Therefore, force is equal to pressure multiplied by area, which is
F= (970 -15 )( * (0.45 m)2)
F=60754 N for each side.
Therefore, each side has to have at least 44 horses
44* 1390 = 61160 N
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Give two examples of situations or application where electronic circuits are used
PLEASE read this carefully.
A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of ρ0 is placed in a container of water. Initially the sphere floats and the water level is marked on the side of the container. What happens to the water level, when the original sphere is replaced with a new sphere which has different physical parameters? Notation: r means the water level rises in the container, f means falls, s means stays the same. Combination answers like 'r or f or s' are possible answers in some of the cases.
The new sphere has a radius of r < r0 and a mass of m > m0.
The new sphere has a mass of m > m0 and a density of ρ < ρ0.
The new sphere has a density of ρ < ρ0 and a radius of r > r0.
For each of the given scenarios and based on the volume, mass, and density of the spheres, the water level is given as follows:
Scenario 1: The water level will rise; r
Scenario 2: The water level may rise or fall; r or f
Scenario 3: The water level will fall; f
What is the relationship between volume, density, and mass of the objects?The density, volume and mass of an object are related by the formula below:
Density = mass/volumeThe rise in the level of fluid when an object is placed in that fluid depends on the weight, volume and density of that object.
This is given by Archimedes' principle which states that the upthrust acting on a body immersed fully or partially in a fluid, is equal to the weight of the fluid displaced.
Based on the above, the water level in each of the scenarios is described thus:
Scenario 1: The new sphere has a radius of r < r0 and a mass of m > m0.
The new has a smaller volume and a greater mass and density. The water level will rise.
Scenario 2: The new sphere has a mass of m > m0 and a density of ρ < ρ0.
The new sphere has a greater volume and will float. Therefore, the water level may rise or fall.
Scenario 3: The new sphere has a density of ρ < ρ0 and a radius of r > r0.
The new sphere has a greater volume but less mass. Therefore, the water level will fall.
In conclusion, the rise or fall of the water level in the container depends on the volume, mass, and density of the spheres.
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The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5 m. If the box has a mass of 1 kg, what is the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s?
The average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.
Acceleration of the box
The acceleration of the box is calculated as follows;
vf² = vi² + 2as
a = (vf² - vi²)/2s
a = (11.5² - 13²) / (2 x 8.5)
a = -2.16 m/s²
Time of motion of the boxThe time taken for the box to travel is calculated as follows;
a = (vf - vi)/t
t = (vf - vi) / a
t = (11.5 - 13) / (-2.16)
t = 0.69 s
Average power supplied by the frictionP = Fv
P = (ma)(vf - vi)
P = (1 x -2.16) x (11.5 - 13)
P = 3.24 W
Thus, the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.
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The density of molybdenum is 10.28 g/cm^3 and it crystallizes in the face centered cubic unit cell. Calculate the edge length of the unit cell. (The atomic mass of Mo is 95.96 g/mole)
The edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.
Volume of molybdenumV = (zm/ρN)
where;
z is 2 for cubic unit cellm is mass of the molybdenumρ is density of the molybdenumV = (2 x 95.96) / (10.28 x 6.02 x 10²³)
V = 3.10 x 10⁻²³ cm³
Edge length of the unit cella³ = V
a = (V)^¹/₃
a = ( 3.10 x 10⁻²³)^¹/₃
a = 3.142 x 10⁻⁸ cm
a = 3.142 x 10⁻¹⁰ m
a = 314.2 x 10⁻¹² m
a = 314.2 pm
Thus, the edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.
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If a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the
direction 20° south of east, and finally walks 28 m in the direction 30° west of north.(2pt)
a) How far and at what angle is the Aster's final position from her initial position?
b) In what direction would she has to head to return to her initial position
The Aster's final position from her initial position is determined as 45.3 m.
The direction that she has to head to return to her initial position is 30° west of north.
Distance of the person from her initial position
The distance traveled by the person can be determined by making a sketch of the person's journey.
From the initial distance traveled and the second distance distance the angle between the two position is calculated as;
θ = 37⁰ + 20⁰ = 57⁰
Distance opposite to the angleThe distance opposite to the angle is resultant displacement and it is calculated as follows;
r² = (70²) + (82²) - (2 x 70 x 82) x cos(57)
r² = 5,371.54
r = √5,371.54
r = 73.3 m
Haven walked 28 m in the same direction to her initial position, the remaining distance is calculated as follows;
d = 73. 3 m - 28 m
d = 45.3 m
Thus, the Aster's final position from her initial position is determined as 45.3 m.
The direction that she has to head to return to her initial position is 30° west of north.
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Answer:
See below
Explanation:
I will break the legs into vertical and horizontal components then add them
Horizontal = 70 cos 37 + 82 cos 340 + 28 cos 120 = 118.96 m
Vertical = 70 sin 37 + 82 sin 340 + 28 sin 120 = 38.33 m
Resultant distance from origin is found via pythag theorem
d^2 = 118.96^2 + 38.33^2 d = 124.98 m
direction FROM origin is arctan ( 38.33/118.96) = 17.86°
to get BACK to the origin , will have to walk 180 degrees from this = 197.86°
A sine bar is used to determine the angle of a part feature. The length of the sine
bar=8.000in. The rolls have a diameter of 1.000 in. All inspection is performed on a surface plate. In
order for the sine bar to match the angle of the part, the following gage blocks must be stacked: 2.0000,
0.5000, 0.2500, and 0.0050. Determine the angle of the part feature.
A sine bar is used to get the angular measurement of a part feature. The angle of the part feature is 18.24°
From the question
Given that,
Length of the sine bar, L = 8.000 in
Diameter of the rolls = 1.000 in
Height under the roll, H = 2.0000 + 0.5000 + 0.0050
= 2.505 in
From sine bar formula,
we know that,
H = sin A x L
sin A = H ÷ L
where,
A ⇒ angle of part feature
H ⇒ height under the roll
L ⇒ length of the sine bar
Substituting values in the above equation,
sin A = H / L
A = sin⁻¹ ( 2.505 ÷ 8 )
A = sin⁻¹ (0.3131)
A = 18.24⁰
Hence the angle of the part feature = 18.24°
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A ball is released from the top of a tower of height h meters. it takes T seconds to reach the ground . what is the position of the ball in T/3 seconds ?
Answer:the
8/9 h
Explanation:
Height = 1/2 a T^2 now change to T/3
now height = 1/2 a (T/3)^2 = 1/9 1/2 a T^2 <===== it is 1/9 of the way down or 8/9 h
HELP PLS
The minimum frequency for the photoelectric effect of a zinc plate is 9.7 × 10^14 Hz. If a green photon with a wavelength of 532nm strikes the plate, will electrons be emitted? Show your work to support your answer.
yes electrons will be emitted with the kinetic energy of
[tex]30.87 \times 10 { }^{ - 19} j[/tex]
what is photoelectric effect?
In photoelectric effect, the electrically charged particles are released from or within a material when it absorbs electromagnetic radiation.
what is kinetic energy?
It is a form of energy that an object or a particle has by reason of its motion. If work, which transfers energy, is done on an object by applying a net force, the object speeds up and thereby gains kinetic energy.
Given:
wavelength,λ
[tex] = 5.32 \times 10 {}^{ - 8} m[/tex]
Frequency
[tex] = 9.7 \times {10}^{14} [/tex]
Einstein's photoelectric equation
€= w + KE
€ = hc/ λ
[tex] = \frac{6.63 \times {10}^{ - 34} \times 3 \times {10}^{8} } {5.32 \times 10 {}^{ - 8} } [/tex]
[tex]€= 3.73 \times {10}^{ - 18} [/tex]
w=h.f
[tex] = 6.63 \times {10}^{ - 34} \times 9.7 \times {10}^{14} [/tex]
[tex] = 6.43 {10}^{ - 19} j[/tex]
KE = € - W
[tex]37.3 \times {10}^{ - 19} - 6.43 \times {10}^{ - 19} [/tex]
[tex]kE = 30.87 \times {10}^{ - 19} [/tex]
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16) Find the orbital speed of an ice cube in the rings of Saturn. The mass of Saturn is 5.68 x 10^26 kg, and use an orbital radius of 3.00 x 105 km. (G= 6.67 × 10-11 N·m2/kg2) A) 19.5 km/s B) 27.5 km/s *C) 11.2 km/s D) 20.5 km/. orbital speed?
The orbital speed of an ice cube in the rings of Saturn is 11.2 Km/s. The correct answer is option C
What does Orbital speed depend on ?The speed of an object travelling around a circle depends on two quantities namely;
Its angular velocity wIts distance from the center of the circle.Given that an ice cube in the rings of Saturn. The mass of Saturn is 5.68 x 10^26 kg, and use an orbital radius of 3.00 x 105 km. (G= 6.67 × 10-11 N·m2/kg2)
The given parameters are:
The mass of Saturn = 5.68 x 10^26 kgThe orbital radius = 3.00 x 105 kmG = 6.67 × 10-11 N·m2/kg2Let us first calculate the gravitational field strength on the Saturn.
g = GM/r²
Substitute all the necessary parameters and convert km to m
g = (6.67 × [tex]10^{-11}[/tex] × 5.68 × [tex]10^{26}[/tex]) ÷ (300000 × 1000)²
g = 3.79 × [tex]10^{16}[/tex] ÷ 9 × [tex]10^{16}[/tex]
g = 0.421 m/s²
The orbital speed will be
V² = gr
V² = 0.4211 × 300000 × 1000
V² = 126333333.3
V = √126333333.3
V = 11239.8 m/s
Convert it to Km/s by dividing the answer by 1000
V = 11239.8/1000
V = 11.2 Km/s
Therefore, the orbital speed of an ice cube in the rings of Saturn is 11.2 Km/s
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What does the Q stand for in the enthalpy of fusion equation?
The Q in the enthalpy of fusion equation stands for heat energy absorbed or released.
What is enthalpy?In thermodynamics, enthalpy is a measure of the heat content of a chemical or physical system.
The enthalpy of fusion is the amount of heat energy required to convert a unit mass of a solid at its melting point into a liquid without an increase in temperature.
Its units are usually Joules per gram (J/g) or calories per gram (cal/g).
The enthalpy of fusion equation is given as follows:
Q = m·ΔHf
Where;
Q = heat energym = massΔHf = heat of fusionTherefore, the Q in the enthalpy of fusion equation stands for heat energy absorbed or released.
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A 22-g bullet traveling 265 m/s penetrates a 1.9 kg block of wood and emerges going 125 m/s .
If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?
Express your answer to two significant figures and include the appropriate units.
The body moves at a velocity of 1.62m/s after the bullet emerges.
Given:Mass of bullet, [tex]m_1[/tex] = 22g
= 0.022 kg
Mass of the block, [tex]m_2[/tex] = 1.9 kg
Velocity of bullet , [tex]v_1[/tex] = 265 m/s
[tex]v_2 = 0[/tex]
According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.
After penetration;
[tex]v^{'}_1 =125 m/s[/tex]
[tex]v^{'}_2=?[/tex]
The formula for calculating the collision of a body is expressed as:
p = mv
m is the mass of the body
v is the velocity of the body
∴ Momentum before = Momentum after
Substitute the given parameters into the formula as shown:
[tex]m_1v_1+ m_2v_2 = m_1v^{'}_1+ m_2v^{'}_2\\0.022* 265 + 0 = 0.022*125+1.9*v^{'}_2\\5.83 = 1.9 v^{'}_2\\v^{'}_2 = 1.62 m/s[/tex]
Therefore, It moves with a velocity of 1.62 m/s.
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4. A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface? how to draw the question?
It is to be noted that the magnitude of the electric flux through the hemisphere surface is πR²E. (Option D)
What is an Electric Flux?
The characteristic of an electric field known as electric flux may be thought of as the quantity of electric lines of force (or electric field lines) that cross a certain region.
According to this theory, electric field lines begin with positive electric charges and end with negative charges.
What is the calculation for the above?
Recall that the A hemispherical surface (half of a spherical surface) of radius R.
In order to derive the Magnitude of the electric flux we use the following formula:
Ф = ∈ x Area of the Curve Surface
⇒ Ф = ∈ x πR²
⇒ Ф = πR²∈
Thus, option D is the correct answer.
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Full Question:
A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface?
A) 2πR²∈
B) 0
C) 4 πR²∈/3
D) πR²∈
A body of mass 80kg moving with a velocity of 6m/s hits a stationary body
of mass of 40kg. If the two bodies stick after the impact, calculate their
common velocity.
Answer:
4 m/s
Explanation:
Conservation of momentum , mv
Before collision 80 kg * 6 m/s = 480 kg-m/s
after the collision,
the momentum is the same but the mass is 80+40 kg =120 kg
mv = 480
120 * v = 480
v = 4 m/s
The Sun subtends an angle of about 0.5∘ to us on Earth, 150 million km away.
Estimate the radius of the Sun.
Express your answer to two significant figures and include the appropriate units.
The radius of the Sun is approximately 654497.9 kilometers.
A subtended angle in geometry is an angle created by a common point (in this case, the Earth) that crosses two points of a nearby circular arc (the Sun in this case). Below is a visual illustration of the subtended angle.
The law of cosine can be used to calculate the sun's radius, which is 654497.950 kilometers in kilometers.
The Sun's radius is nearly 109 times bigger than the Earth's. The Earth's radius is 6378 kilometers. Or to put it another way, both dimensions are 1: 109 ratios.
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A -4.00 nC point charge is at the origin, and a second -6.00 nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x -axis at x = 0.200 m'
The net electric force that the two charges would exert on an electron placed at point on the x -axis is 1.68 x 10⁻¹⁶ N.
Force on electron due to charge 1
The force exerted on the electron due to the charge q1 placed at the origin is calculated as follows;
F = kq₀q₁/r²
where;
k is coulomb's constantq0 is charge at the originq1 is the charge at 0.2 m (electron)r is the distance between the chargesF(01) = (9 x 10⁹ x 4 x 10⁻⁹ x 1.6 x 10⁻¹⁹)/(0.2²)
F(01) = 1.44 x 10⁻¹⁶ N
Force on electron due to charge 2The force exerted on the electron due to the charge q1 placed at the origin is calculated as follows;
F = kq1q2/r²
where;
k is coulomb's constantq2 is charge at the 0.8 mq1 is the charge at 0.2 m (electron)r is the distance between the charges = 0.8 m - 0.2 m = 0.6 mF(12) = (9 x 10⁹ x 6 x 10⁻⁹ x 1.6 x 10⁻¹⁹)/(0.6²)
F(12) = 2.4 x 10⁻¹⁷ N
Net force on the electronF(net) = 2.4 x 10⁻¹⁷ N + 1.44 x 10⁻¹⁶ N
F(net) = 1.68 x 10⁻¹⁶ N
Thus, the net electric force that the two charges would exert on an electron placed at point on the x -axis is 1.68 x 10⁻¹⁶ N.
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High-power laser in factories are used to cut through cloth and metal. One such laser has a beam diameter of 1.00mm and generates an electric field having an amplitude 0.800MV/m at the target. Find(a) the amplitude of the magnetic field produced,(b) the intensity of the laser and (c) the power delivered by the laser.
(a) The amplitude of the magnetic field produced is 2.667 mV/m.
(b) The intensity of the laser is 2.832 W/m².
(c) The power delivered by the laser is 2.22 x 10⁻⁶ W.
Amplitude of the magnetic field producedIn electromagnetic waves, the amplitude of magnetic field is the maximum field strength of the magnetic fields.
B₀ = E₀/c
where;
E₀ is the amplitude electric fieldB₀ is the amplitude magnetic fieldc is speed of lightB₀ = (0.8 MV/m) / (3 x 10⁸)
B₀ = (0.8 x 10⁶ V/m) / (3 x 10⁸)
B₀ = 2.667 x 10⁻³ V/m
B₀ = 2.667 mV/m
Intensity of laserThe intensity of the laser is calculated as follows;
I = ¹/₂ε₀E₀²
I = (0.5)(8.85 x 10⁻¹²)(0.8 x 10⁶)²
I = 2.832 W/m²
power delivered by the laserP = IA
where;
A is the area of the beamA = πd²/4
where;
d is diameterA = π(1 x 10⁻³)²/4
A = 7.854 x 10⁻⁷ m²
Power = (2.832 W/m²) x (7.854 x 10⁻⁷ m²)
Power = 2.22 x 10⁻⁶ W
Thus, the amplitude of the magnetic field produced is 2.667 mV/m.
The intensity of the laser is 2.832 W/m².
The power delivered by the laser is 2.22 x 10⁻⁶ W.
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A weight lifter lifts a mass of 250 kg with a force of 5000N to a height of 5m. a. What is the workdone by the weight lifter? b. What is the workdone by the gravity? c. What is the net workdone on the object?
hii friends
Answers are:
a. The work done by the weight lifter W = 25000 Joule
b. The work done by the gravity W = 12250 Joule
c. The net work done on the object W = 12750 Joule
What is Work ?
Work is the product of force and distance in the direction of the force applied. That is, W = F x h
The S.I unit is Joule.
Given that a weight lifter lifts a mass of 250 kg with a force of 5000N to a height of 5m.
The given parameters are;
Mass m = 250 kgForce F = 5000 NHeight h = 5 ma. The work done by the weight lifter will be
W = F x h
W = 5000 x 5
W = 25000 Joule
b. The work done by the gravity will be
W = mg x h
W = 250 x 9.8 x 5
W = 12250 Joule
c. The net work done on the object will be
W = 25000 - 12250
W = 12750 Joule
Since the work done by the gravity is in opposite direction to the work done by the weight lifter, the net work done on the object will be the difference between the two.
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Lam Lesson Name: Uncovering Your Personality
m number: 700047RR
Exam Guidelines
Exam Instructions
Question 10 of 20:
Select the best answer for the question.
10. Which characteristic of a turbulent person causes them to always strive for self-improvement, and to never see accomplishing a goal as good enough?
O A. Confident
B. Self-critical
O C. Ignorant
O D. Calm
Mark for review (Will be highlighted on the review page)
ex Previous Question
Next Questin
Review My F
The correct answer is Self-critical.
Why is self-improvement?
Enhancing strengths, mental health, and even mending relationships benefit self-improvement. Simple actions like reading a book, trying something new, meditating, or even getting up early are some ways to improve oneself. There are so many easy, efficient methods to begin the process of improving oneself.A self-improvement strategy enables you to build the life you want for yourself. It enables you to maintain perspective on your priorities and the things most important to you in life to experience greater meaning and fulfillment.Self-development is taking steps to better yourself, such as by learning new skills or overcoming bad habits. An example of self-development is taking courses at the university to learn new skills and interesting things.Self-critical:
Self-critical causes them to always strive for self-improvement and never to see accomplishing a goal as good enough.
The characteristic of a turbulent person causes them to always strive for self-improvement and to never see accomplishing a goal as good enough is Self-critical.
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Two charged particles are a distance of 1.72 m from each other. One of the particles has a charge of 7.03 nC, and the other has a charge of 4.02 nC.
(A) What is the magnitude (in N) of the electric force that one particle exerts on the other?
______N
(B) Is the force attractive or repulsive?
The magnitude of the electric force that one particle exerts on the other is 8.59 x [tex]10^{-8}[/tex] N and the force is attractive.
How is Coulomb's Law Stated ?Coulomb law states that the attractive or repulsive force between two point charges is proportional to the product of the charges and inversely proportional to square of the distance between them. That is,
F = KQq / r²
Given that two charged particles are a distance of 1.72 m from each other. One of the particles has a charge of 7.03 nC, and the other has a charge of 4.02 nC.
Where
K = constant of proportionality = 9 x [tex]10^{9}[/tex] Nm²/c²Q = 7.03 x [tex]10^{-9}[/tex] Cq = 4.02 x [tex]10^{-9}[/tex] Cr = 1.72 m(A) The magnitude of the electric force that one particle exerts on the other will be
F = (9 x [tex]10^{9}[/tex] x 7.03 x [tex]10^{-9}[/tex] x 4.02 x [tex]10^{-9}[/tex]) / 1.72²
F = 2.54 x [tex]10^{-7}[/tex] / 2.96
F = 8.59 x [tex]10^{-8}[/tex] N
(B) Since our answer is positive, the force is therefore attractive.
Therefore, the magnitude of the electric force that one particle exerts on the other is 8.59 x [tex]10^{-8}[/tex] N and the force is attractive.
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A. The magnitude (in N) of the electric force that one particle exerts on the other is 8.60×10⁻⁸ N
B. The force is repulsive
A. How to determine the magnitude of the electric forceFrom the question given above, the following data were obtained:
Charge 1 (q₁) = 7.03 nC = 7.03×10¯⁹ CCharge 2 (q₂) = 4.02 nC = 4.02×10¯⁹ CElectric constant (K) = 9×10⁹ Nm²/C²Distance apart (r) = 1.72 mForce (F) =?The magnitude of the electric force can be obtained by using the Coulomb's law equation as shown below:
F = Kq₁q₂ / r²
F = (9×10⁹ × 7.03×10¯⁹ × 4.02×10¯⁹) / (1.72)²
F = 8.60×10⁻⁸ N
B. How to determine whether the force is attractive or repulsiveFrom the question given, we were told that:
Charge 1 (q₁) = 7.03 nC Charge 2 (q₂) = 4.02 nCSince both charge are positive, then the force attraction between them is repulsive as like charges repels and unlike charges attracts
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2. What energy source produces most of the electrical energy in the United States?
natural gas
O oil
O coal
Answer:
Nat gas
Explanation:
Approx:
Natural gas 38% coal 21% oil 19%
A thin-walled hollow sphere has a radius 4cm from the center
of the sphere, the eletric field points radially inward and
has a magnitude of 1.5 × 10^4 NC". How much charge is on the surface
The electric charge on surface is 2.66×10-⁹NC
What is an electric charge?The electric charge is the property of subatomic particle that causes it to experience a force when placed in an electric and magnetic field.
We have given ,
Electric field= 1.5×10⁴N/C
Radius = 4cm = 0.04m
E = k× q/ r²
1.5×10⁴×16×10-⁴/9×10⁹=q
q= 2.66×10-⁹C
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the apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ? use negative is the acceleration vector is pointing downwards
Answer:
-.457 m/s^2
Explanation:
Actual weight = 60 .3 (9.81) = 591.54 N
Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift
60.3 ( 9.81 - L ) = 564
solve for L = .457 m/s^2 DOWNWARD
so L = - .457 m/s^2
Calculate the frequency if the number of revolutions is 300 and the paired poles are 50.
15 kHz
150 kHz
1500 kHz
150 Hz
Answer: A
Explanation: We know that f=p*n
f=50*300=15000 Hz = 15kHz.
Have a great day! <3
If the number of revolutions is 300 and the paired poles are 50 , then the frequency would be 15 kHz, therefore the correct answer is option A.
What is the frequency ?It can be defined as the number of cycles completed per second. It is represented in hertz and inversely proportional to the wavelength.
The frequency of a pendulum is the reciprocal of the time period can be given by the following relation,
F = 1 / T
As given in the problem, we have to calculate frequency if the number of revolutions is 300 and the paired poles are 50.
F = 300 × 50
= 1500 kHz
Thus, If the number of revolutions is 300 and the paired poles are 50, then the frequency would be 15 kHz, therefore the correct answer is option A.
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Warm-Up
Type your response in the box.
When the Moon comes directly between the Sun and Earth, a solar eclipse occurs. The eclipse may be a total eclipse,
where the Sun is completely blocked by the Moon, or an annular eclipse, where you can see a ring of sunlight
surrounding the Moon. What do you think makes a total eclipse occur instead of an annular eclipse?
Answer: Both total and annular eclipses depend on the distances between Earth, the Moon, and the Sun. A total solar eclipse occurs when the Moon is closer to Earth, thus taking up a larger portion of the sky. The Moon is then able to cover the entire Sun.
Explanation: Edmentum answer
A thin-walled hollow sphere has a radius 4cm from the center
of the sphere, the eletric field points radially inward and
has a magnitude of 1.5 × 10^4NC^-1 How much charge is on the surface
A thin-walled hollow sphere has a radius of 4 cm from the center
of the sphere, the electric field points radially inward and
has a magnitude of 1.5 × 10⁴ N/C then the charge on the surface would be 2.6690×10⁻⁹ C
What is an electric charge?Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. You can have a positive or negative electric charge.
The electric field inside a spherical shell is given by the formula
E = q/4πεr²
where q is the charge on the surface
r is the radius of the sphere
ε is the electrical permeability
By substituting the respective values in the given formula
1.5 × 10⁴ = q/4π(8.85✕ 10⁻⁻¹²)(.04²)
q = 2.6690×10⁻⁹ C
Thus, the charge on the surface would be2.6690×10⁻⁹ C
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Lifting a 81-kg barbell and weights from rest up to a speed of 1.0 m/s in 0.50 s, resisted by the combined weight of the barbell and weights, requires what applied force in N to two significant digits?
Answer:
960 N
Explanation:
answer should have two signif digits
a = change in velocity divided by change in time
1 m/s / .5 s = 2 m/s^2
F = ma a = g + 2 m/s^2 = 9.81 + 2 = 11.81 m/s^2
81 kg * 11.81 m/s^2 = 956.61 N = ~ 960 N (two sig digits)
(Please let me know if this is incorrect....I am not SURE )
On space missions with international teams, every country involved wants the flight
director to be chosen from their country because their national space organization
receives a significant amount of funding if they lead the mission.
True
False
False, their national space organization receives a significant amount of funding if they lead the mission.
The team typically sends about 30,000 remote control commands from the ground to the space station to manage the platform. Highly interactive, hands-on, tactile, and highly rewarding.
A mission control center (MCC, sometimes called a flight control center or operations center) is a facility that typically manages spaceflight from the launch point to landing or the end of the mission. This is part of the ground segment of spacecraft operations.
The station serves as a microgravity and space laboratory where scientific research in fields such as astrobiology, astronomy, meteorology, and physics is conducted.
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The figure illustrates flow through a pipe with diameters of 1.0 mm and 2.0 mm and with different elevations. Px is the pressure in the pipe, and Vx is the speed of a non-viscous incompressible fluid at locations x = Q,R,S,T, or U. Options are: Greater than, Less than, Equal to
PU is ... PQ
VU is ... 2VT
PR is ... PU.
VR is ... VS
VQ is ... VU
PR is ... PS
By Bernoulli's equation as well as the Equation of continuity the relation between pressure and velocities at different points are,
[tex]P_u > P_q[/tex] [tex]V_u > 2V_t[/tex] [tex]P_u > P_r[/tex] [tex]V_r=V_s[/tex] [tex]V_q=V_u[/tex][tex]P_r > P_s[/tex]To find the answer, we need to know about the Bernoulli's equation as well as the Equation of continuity.
How to find the solution?1) We have to find the relation between pressure at U and Q.
We have the Bernoulli's equation,[tex]P+\frac{1}{2}dV^2+dgh=constant.[/tex]
where, P is the pressure, V is the velocity, d is density, g is acceleration due to gravity and h is the height of the flow.
By using the equation, we can find the pressure at U and Q.[tex]P_u+\frac{1}{2}dV_u^2+dg*0= P_q+\frac{1}{2}dV_q^2+dgh\\where,\\V_u=V_q, \\since, D_p=D_q\\Thus,\\P_u=P_q+dgh[/tex]
[tex]P_u > P_q[/tex]
2) We have to find the relation between velocity at U and T.
For this, we have the equation of continuity as,[tex]AV=constant\\A_1V_1=A_2V_2[/tex]
From the diagram, we have,[tex]A_u=\pi r^2=\pi *(0.5)^2=0.25\pi *mm^2\\A_t=\pi *1=\pi mm^2\\V_u=V\\V_t=?[/tex]
Thus, the relation between velocity at U and T is,[tex]V_t=\frac{A_uV_u}{A_t}=\frac{V}{4}[/tex]
[tex]2V_t=2*\frac{V}{4} =\frac{V}{2}\\V_u=V\\[/tex]
[tex]V_u > 2V_t[/tex]
3) We have to find the relation between pressure at R and U
[tex]P_u+\frac{1}{2}dV_u^2+dg*0= P_r+\frac{1}{2}dV_r^2+dg*0\\\\V_u=V , then\\V_r=V_t=\frac{V}{4}=\frac{V_u}{4} \\\\P_u=P_r+\frac{1}{16}[/tex]
[tex]P_u > P_r[/tex]
4) We have to find the relation between velocity at R and S
Both points R and S, have same area, thus same velocity.[tex]V_r=V_s[/tex]
5) We have to find the relation between velocity at Q and U
Both points Q and U, have same area, thus same velocity.[tex]V_q=V_u[/tex]
6) We have to find the relation between pressure at R and S
Both points R and S, have same area and thus, same velocities.[tex]P_r=P_s+dgh[/tex]
[tex]P_r > P_s[/tex]
Thus, we can conclude that, By Bernoulli's equation as well as the Equation of continuity the relation between pressure and velocities at different points are,
[tex]P_u > P_q[/tex] [tex]V_u > 2V_t[/tex] [tex]P_u > P_r[/tex] [tex]V_r=V_s[/tex] [tex]V_q=V_u[/tex][tex]P_r > P_s[/tex]Learn more about the Bernoulli's equation here:
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A child rolls a ball on a level floor 5.2 m to another child.
If the ball makes 15.0 revolutions, what is its diameter?
Express your answer to two significant figures and include the appropriate units.
Diameter of the ball = 11.03cm
The ball rolled by the child must cover a distance in a linear motion which will be then equal to the circumference of the ball after one complete revolution.
Linear distance = 5.2m
d = diameter of the ball =?
Total revolutions = 15
S = 2πr
S = πd
For 15 revolutions
S = 15πd
The Diameter of a ball is double the radius of the ball thus we use D instead of 2r in the formula
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