The generator is 124.6 volts RMS.
A boy walks from point C to point D which is 50 m apart. Then, he walks back to point C. what is his displacement of his whole journey ?
A.25 m
B.75 m
C.50 m
D.0 m
Answer: D. 0 m
Explanation:
Concept:
Here, we need to know the concept of displacement.
Displacement is defined to be the change in position of an object.
The difference between displacement and distance is the total movement of an object without any regard to direction, while displacement is the pure change of position.
If you are still confused, please refer to the attachment below for a graphical explanation.
Solve:
STEP ONE: the boy walks from point C to point D (a distance of 50 m)
C ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ D
50 m
STEP TWO: the boy walks from point D to point C (a distance of 50 m)
D ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ C
50 m
STEP THREE: find the displacement
The boy started with point C
The boy ended with point C
He did not change his position throughout the journey.
Therefore, his displacement is 0 m.
Hope this helps!! :)
Please let me know if you have any questions
What star is known as the "cold planet"?
Explanation:
OGLE-2005-BLG-390Lb.
PSR B1620-26 b. Surface Temperature: 72 Kelvin. ...
Neptune. Surface Temperature: 72 Kelvin. ...
Uranus. Surface Temperature: 76 Kelvin. ...
Saturn. Surface Temperature: 134 Kelvin. ...
Jupiter. Image Courtesy: NASA. ...
OGLE-2016-BLG-1195Lb. Surface Temperature: Unknown
Is it true that as we gain mass the force of gravity on us decreases
Answer:
No. As we gain mass the force of gravity on us does not decrease
The 52-g arrow is launched so that it hits and embeds in a 1.50 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.47 m higher than the block's starting point. How fast was the arrow moving before it joined the block? What mechanical work must you do to lift a uniform log that is 3.1 m long and has a mass of 100 kg from the horizontal to a vertical position?
Answer:
[tex]v_1=87.40m/s[/tex]
Explanation:
From the question we are told that:
Mass of arrow [tex]m=52g[/tex]
Mass of rock [tex]m_r=1.50kg[/tex]
Height [tex]h=0.47m[/tex]
Generally the equation for Velocity is mathematically given by
[tex]v = \sqrt{(2gh)}[/tex]
[tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]
[tex]v= 3.035m/s[/tex]
Generally the equation for conservation of momentum is mathematically given by
[tex]m_1v_1=m_2v_2[/tex]
[tex]0.052kg * v = 1.5 * 3.03m/s[/tex]
[tex]v_1=87.40m/s[/tex]
Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. What is the effective "spring constant" of this simple harmonic motion?
Express your answer to three significant digits and include the appropriate units.
We have that the spring constant is mathematically given as
[tex]k=2.37*10^{11}N/m[/tex]
Generally, the equation for angular velocity is mathematically given by
[tex]\omega=\sqrt{k}{m}[/tex]
Where
k=spring constant
And
[tex]\omega =\frac{2\pi}{T}[/tex]
Therefore
[tex]\frac{2\pi}{T}=\sqrt{k}{n}[/tex]
Hence giving spring constant k
[tex]k=m((\frac{2 \pi}{T})^2[/tex]
Generally
Mass of earth [tex]m=5.97*10^{24}[/tex]
Period for on complete resolution of Earth around the Sun
[tex]T=365 days[/tex]
[tex]T=365*24*3600[/tex]
Therefore
[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]
[tex]k=2.37*10^{11}N/m[/tex]
In conclusion
The effective spring constant of this simple harmonic motion is
[tex]k=2.37*10^{11}N/m[/tex]
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a beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a focal length of 40 cm
Answer:
beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.
To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm
Solution:
As per the given criteria,
the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)
(a) lens is a convex lens with
focal length, f=20cm
object distance, u=12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
20
1
+
12
1
⟹
v
1
=
60
3+5
⟹v=7.5cm
Hence the image formed is real, at 7.5cm from the lens on its right side.
(b) lens is a concave lens with
focal length, f=−16cm
object distance, 12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
−16
1
+
12
1
⟹
v
1
=
48
−3+4
⟹v=48m
Hence the image formed is real, at 48 cm from the lens on the right side.
The cells lie odjacent to the sieve tubes
Answer:
Almost always adjacent to nucleus containing companion cells, which have been produced as sister cells with the sieve elements from the same mother cell.After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving
This question is incomplete, the complete question is;
A bat flies towards a moth at 7.1 m/s while the moth is flying towards the bat at 4.4 m/s. The bat emits a sound wave of 51.7 kHz.
After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving
Answer:
the second moth is moving at 5.062 m/s
Explanation:
Given the data in the question;
Using doppler's effect
[tex]f_{moth[/tex] = f₀( [tex]v_{s[/tex] ± [tex]v_{observer[/tex] / [tex]v_{s[/tex] ± [tex]v_{source[/tex] )
f₁ = f₀( ([tex]v_{s[/tex] + v₂) / ( [tex]v_{s[/tex] - v₁ ) )
frequency reflected from the moth,
Now, moth is the source and the bat is the receiver
f₂ = f₁( ([tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) )
hence, f = f₀[ ( ( [tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) ) ( ( [tex]v_{s[/tex] + u₂ ) / ( [tex]v_{s[/tex] - u₁ ) )
we know that, the velocity of sound [tex]v_{s[/tex] = 343 m/s.
given that v₁ and v₂ { velocity of bat } = 7.1 m/s, f₀ = 51.7 kHz and f = 55.5 kHz.
we substitute
55.5 = 51.7[ ( ( 343 + 7.1 ) / ( 343 - 7.1 ) ) ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 51.7[ ( 350.1 / 335.9 ) ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 51.7[ 1.04227 ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 53.885359 ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 / 53.885359 = ( 343 + u ) / ( 343 - u )
1.02996 = ( 343 + u₂ ) / ( 343 - u )
( 343 + u₂ ) = 1.02996( 343 - u )
343 + u = 353.27628 - 1.02996u
u + 1.02996u = 353.27628 - 343
2.02996u = 10.27628
u = 10.27628 / 2.02996
u = 5.062 m/s
Therefore, the second moth is moving at 5.062 m/s
An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that the equation is dimensionally correct.
Answer:
[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].
The unit of both sides of this equation are [tex]\rm s[/tex].
Explanation:
The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].
The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].
On the right-hand side of this equation:
[tex]\pi[/tex] has no unit.The unit of [tex]r[/tex] is [tex]\rm m[/tex].The unit of [tex]G[/tex] is [tex]\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}[/tex], which is equivalent to [tex]\rm m^{3} \cdot kg^{-1} \cdot s^{-2}[/tex].The unit of [tex]m[/tex] is [tex]\rm kg[/tex].[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].
Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].
A force of 1000N is used to kick a football of mass 0.8kg find the velocity with which the ball moves if it takes 0.8 sec to be kicked.
The velocity of the ball is 100m/s
The first step is to write out the parameters;
The force used to kick the ball is 1000N
The mass of the ball is 0.8 kg
Time is 0.8 seconds
Therefore the velocity can be calculated as follows
F= Mv-mu/t
1000= 0.8(v) - 0.8(0)/0.8
1000= 0.8v- 0.8/0.8
Cross multiply both sides
1000(0.8) = 0.8v
800= 0.8v
divide both sides by the coefficient of v which is 8
800/0.8= 0.8v/0.8
v= 1000m/s
Hence the velocity is 1000m/s
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a stone is thrown vertically upwards with a velocity of 20 m per second what will be its velocity when it reaches a height of 10.2 m
Answer:
Explanation:
Here's the info we have:
initial velocity is 20 m/s;
final velocity is our unknown;
displacement is -10.2 m; and
acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation
v² = v₀² + 2aΔx and filling in accordingly to solve for v:
[tex]v=\sqrt{(20)^2+2(-9.8)(-10.2)}[/tex] Rounding to the correct number of sig fig's to simplify:
[tex]v=\sqrt{400+2.0*10^2}[/tex] to get
v = [tex]\sqrt{600}=20\frac{m}{s}[/tex] If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.
So either 20 m/s or 24 m/s
Which item will be shipped third?
—-
Answer:
I know it's groceries
Explanation:
electronics ship before clothing
electronics ship after groceries
urgent items are first so
order:
1.) A/Electronics
2.) Clothing/B
3.) Groceries(since groceries aren't urgent)
thing is it's C or D I'm leaning to D since it says it ships last but i dont know so if I'm wrong sorry.
Which of the units of the following physical quantities are derived
Answer:
where is the attachment
Explanation:
A 20 N south magnetic force pushes a charged particle traveling with a velocity of 4 m/s west through a 5 T magnetic field pointing downwards . What is the charge of the particle ?
Answer:
Charge of the particle is 1 coulomb.
Explanation:
Force, F:
[tex]{ \bf{F=BeV}}[/tex]
F is magnetic force.
B is the magnetic flux density.
e is the charge of the particle.
V is the velocity
[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]
1. A 2.7-kg copper block is given an initial speed of 4.0 m/s on a rough horizontal surface. Because of friction, the block finally comes to rest. (a) If the block absorbs 85% of its initial kinetic energy as internal energy, calculate its increase in temperature.
Answer:
ΔT = 0.017 °C
Explanation:
According to the given condition, the change in internal energy of the block must be equal to 85% of its kinetic energy:
Change in Internal Energy = (0.85)(Kinetic Energy)
[tex]mC\Delta T = (0.85)\frac{1}{2}mv^2\\\\C\Delta T = (0.425)v^2\\\\\Delta T = \frac{0.425v^2}{C}[/tex]
where,
ΔT = increase in temperature = ?
v = speed of block = 4 m/s
C = specific heat capacity of copper = 389 J/kg.°C
Therefore,
[tex]\Delta T = \frac{(0.425)(4\ m/s)^2}{389}\\\\[/tex]
ΔT = 0.017 °C
When the drag force on an object falling through the air equals the force of gravity, the object has reached
terminal force.
terminal acceleration,
terminal illness.
terminal velocity
Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B
Explanation:
Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]
The sum of the two vectors is
[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]
[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]
The difference between the two vectors can be written as
[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]
[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]
A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from the center of the wire. (in Nm2/C, keep 3 significant figures)
Answer:
[tex]E=35921.96N/C[/tex]
Explanation:
From the question we are told that:
Radius [tex]r=0.321mm[/tex]
Charge Density [tex]\mu=0.100[/tex]
Distance [tex]d= 5.00 cm[/tex]
Generally the equation for electric field is mathematically given by
[tex]E=\frac{mu}{2\pi E_0r}[/tex]
[tex]E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}[/tex]
[tex]E=35921.96N/C[/tex]
An ideal spring is hung vertically from the ceiling. When a 2.0-kg mass hangs at rest from it the spring is extended 6.0 cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10 cm. While the spring is being extended by the force, the work done by the spring is:
a. -3.6 J
b. -3.3 F
c. -3.4 times 10^-5 J
d. 3.3 J
e. 3.6 J
Answer:
b) - 3.3 J
Explanation:
Given;
mass, m = 2 kg
initial extension of the spring, x = 6 cm = 0.06 m
The weight of the mass on the spring;
W = mg
where;
g is acceleration due to gravity = 9.81 m/s²
W = 2 x 9.81
W = 19.62 N
The spring constant is calculated as;
W = kx
k = W/x
k = 19.62 / 0.06
k = 327 N/m
The work done by the spring when it is extended to an additional 10 cm;
work done = force x distance
distance = extension, x = 10 cm = 0.1 m
The work done by the spring opposes the applied force by acting in opposite direction to the force.
W = - Fx
W = - (kx) x
W = - kx²
W = - (327) x (0.1)²
W = - 3.27 J
W ≅ - 3.3 J
Therefore, the work done by the spring by opposing the applied force is -3.3 J
If a car drives 10 mph South, this is an example of a:
A. Displacement
B. Velocity
C. Speed
D. Distance
Answer:
杰杰伊杜杜杜伊格富尔杰迪耶赫分离福音
Explanation:
莱德利 · 赫耶尔伊 3uritievrirjrirhruebwkwieheoo2hfjcbvi3hd
Answer:
B velocity
Explanation:
The masses of two heavenly bodies are 2×10‘16’ and 4×10 ‘22’ kg respectively and the distance between than is 30000km. find the gravitational force between them ? ans. 2.668× 10-9N
[tex]F = 5.93×10^{13}\:\text{N}[/tex]
Explanation:
Given:
[tex]m_1= 2×10^{16}\:\text{kg}[/tex]
[tex]m_2= 4×10^{22}\:\text{kg}[/tex]
[tex]r = 30000\:\text{km} = 3×10^7\:\text{m}[/tex]
Using Newton's universal law of gravitation, we can write
[tex]F = G\dfrac{m_1m_2}{r^2}[/tex]
[tex]\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}[/tex]
[tex]\:\:\:\:= 5.93×10^{13}\:\text{N}[/tex]
Which sequence shows the chain of energy transfers that create surface currents on the ocean?
Answer:
The correct answer is A. The sun is the energy source of the surface currents in the ocean
The energy transfer starts from solar energy , then wind energy and finally wind energy is the cause of surface current .
What is surface current ?Surface currents are currents that are located in the upper feet of the ocean , they are simply how water moves from one place to another . Pattern of surface current are determined by wind direction .
Surface currents are formed by global wind system that are fueled by energy from the sun . Because of heating effect of sun , the earth's atmosphere gets warmed up . As we know , warm air is lighter then cool air , it rises up and create low pressure near the equator because of this wind causes surface currents the ocean .
hence , The energy transfer starts from solar energy , then wind energy and finally wind energy is the cause of surface current .
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15 . A scientist who studies the whole environment as a working unit .
Botanist
Chemist
Ecologist
Entomologist
Answer:
Ecologist.
Your answer is Ecologist.
(Ecologist) is a scientist who studies the whole environment as a working unit.
Imagine you’re driving along a road and you approach a bridge. You notice a sign that reads, “Bridge freezes before road.” Why do bridges become covered with ice before roads do? Research this question and respond in depth, writing a full paragraph. Be sure to include examples. At the end of your response, provide at least two authoritative sources that you used in your research.
Answer:
During wet and freezing temperatures, ice is able to form at a faster pace on bridges because freezing winds blow from above and below and both sides of the bridge, causing heat to quickly escape. The road freezes slower because it is merely losing heat through its surface.Sources:
-- https://intblog.onspot.com/en-us/why-do-bridges-become-icy-before-roads
and
-- https://www.accuweather.com/en/accuweather-ready/why-bridges-freeze-before-roads/687262
I hope this helps you! ^^
Determine the density in kg \cm of solid whose Made is 1080 and whose dimension in cm are length=3 ,width=4,and height=3
Answer:
d = 30kg/cm³
Explanation:
d = m/v
d = 1080kg/(3cm*4cm*3cm)
d = 30kg/cm³
A Ball A and a Ball B collide elastically. The initial momentum of Ball A is -2.00kgm/s and the initial momentum of Ball B is -5.00kgm/s. Ball A has a mass of 4.00kg and is traveling at 2.50 m/s after the collision. What is the velocity of ball B if it has a mass of 6.50kg?
The velocity of B after the collision is obtained as -2.6 m/s.
What is the principle of conservation of momentum?Now we now that the principle of conservation of momentum states that the momentum before collision is equal to the momentum after collision.
Thus;
(-2.00kgm/s) + ( -5.00kgm/s) = ( 4.00kg * 2.50 m/s) + ( 6.50kg * v)
-7 = 10 + 6.5v
-7 - 10 = 6.5v
v = -7 - 10 /6.5
v = -2.6 m/s
Hence, the velocity of B after the collision is obtained as -2.6 m/s.
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The US currently produces about 27 GW of electrical power from solar installations. Natural gas, coal, and oil powered installations produce about 740 GW of electrical power. The average intensity of electromagnetic radiation from the sun on the surface of the earth is 1000 W/m2 . If solar panels are 30% efficient at converting this incident radiation into electrical power, what is the total surface area of solar panels responsible for the 27 GW of power currently produced
Answer:
The total surface area is "90 km²".
Explanation:
Given:
Power from solar installations,
= 27 GW
Other natural installations,
= 740 GW
Intensity,
[tex]\frac{F}{At}=\frac{P}{A}=1000 \ W/m^2[/tex]
%n,
= 30%
Now,
⇒ %n = [tex]\frac{out.}{Inp.}\times 100[/tex]
then,
⇒ [tex]Inp.=\frac{27}{30}\times 100[/tex]
[tex]=90 \ GW[/tex]
As we know,
⇒ [tex]I=\frac{P}{A}[/tex]
by substituting the values, we get
[tex]1000=\frac{90\times 10^9}{A}[/tex]
[tex]A = \frac{90\times 10^9}{10^3}[/tex]
[tex]=90\times 10^6[/tex]
[tex]=90 \ km^2[/tex]
How can I solve this?
You have three capacitors of values 40 F, 10 F and 50 F. What would their equivalent capacitance (in F) be if they were connected in parallel with each other? Enter your answer as a number only, to one decimal place.
Explanation:
The equivalent capacitance of capacitors in parallel can be determined as
[tex]C_{eq} = C_1 + C_2 + C_3[/tex]
[tex]\:\:\:\:\:= 40\:\text{F} + 10\:\text{F} + 50\:\text{F} = 100\:\text{F}[/tex]
A nearsighted person has a near point of 50 cmcm and a far point of 100 cmcm. Part A What power lens is necessary to correct this person's vision to allow her to see distant objects
Answer:
P = -1 D
Explanation:
For this exercise we must use the equation of the constructor
/ f = 1 / p + 1 / q
where f is the focal length, p and q is the distance to the object and the image, respectively
The far view point is at p =∞ and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image
[tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]
f = 1 m
P = 1/f
P = -1 D
Give examples of motion in which the directions of the velocity and acceleration vectors are (a) opposite, (b) the same, and (c) mutually perpendicular
Answer:
a) When moving body applies brake then velocity and acceleration would be in opposite direction
b) When body starts to increase velocity then velocity and acceleration would be in same direction
c) When body is circulating then velocity and acceleration would be perpendicular to each other
Explanation:
a) When body applies brake then its velocity starts decreasing, in this case its acceleration would try to stop the moving body. So direction of velocity would be same as direction of motion of body but direction of acceleration would be in opposite direction
b) When body starts to increase velocity, its acceleration would make the body to move faster. So direction of velocity would be the direction of motion of body and acceleration would also be in same direction
c) When body moves in circular path then its acceleration would be towards centre of circle and velocity would try to snap the body out of circle to straight line which in tangent to circle.
