Answer:
Energy production requires the setting up of a complete interconnected chain from generation of energy from the root source of the energy to the storage of the generated energy and the eventual utilization of the energy when required
Solar energy, indirectly, continues to be the main source of energy, however, the direct use of solar energy to power the systems we use in our everyday life, require the development of technologies, such as high efficiency solar cells, means of energy storage, and compatible efficient energy usage which are industrial areas that are seeing good progress but in which the current developed equipment are expensive to produce, and due to their efficiency, are undergoing further research and development
Therefore, due to the continuous increasing improvement in solar technology which can observed, the use of the produced energy through solar is evolving, and therefore, will continue to play a continuously increasing but lower role compared to other sources of energy which have been developed to satisfactory level that can drive an industry, considering the financial investment involved
Explanation:
a small object is placed between two plane mirrors inclined at an angle of 60° to each other in a dark room how many images are seen explain
Answer:
nothing
Explanation:
bocouse of darkness
please answer these diagrammatic questions ASAP and please no spam answers
Answer:
i. The pressure of due to the water, P, is given according to the following equation;
P = ρ·g·h
Where;
ρ = The density of the water (a constant) = 997 kg/m³
g = The acceleration due to gravity = 9.81 m/s²
h = The height of the water (minimum h = h₁, maximum h = h₂)
The pressure is directly proportional to the water height, and we have;
The pressure, P, will be maximum when the water height, h, is maximum or h = h₂, which is the level DC
ii. The thrust = The force acting on the body = Pressure × Area
The maximum areas exposed to the water are on side AB and DC
However, the pressure at level DC, which is the location of the maximum pressure, is larger than the pressure at level AB, therefore, the maximum thrust will be at the level DC
Explanation:
We reduce friction in machines? why
Answer:
friction reduces the efficiency of machines, thus we must reduce the friction force that is acting upon it.
Answer:
Because it causes a lot of wear and tear in machine parts that move against each other. It erodes the surfaces and destroys their symmetries
Explanation:
The atomic bomb dropped on Hiroshima converted about 7.00x10-4kg of mass to energy. How much energy did that bomb produce?
A)2.10x10^5J
B)7.78x10^-21J
C)6.30x10^13J
D)2.10x10^61J
Answer:
[tex] \sf \: given \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \: mass \: \: \: m \: = 7.00 \times {10}^{ - 4} \: kg \\ \\ \bf \: E=mc^2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > E=7.00 \times {10}^{ - 4} \times ({3 \times {10}^{8} })^{2} \\ \\ = > \green{ \boxed{ E = 6.3 \times {10}^{13} \: J}}[/tex]
Which of the following best describes reverberation?
A.The wave fronts become mixed and broken up due to contact with a rough or
irregular surface.
B. A change in the sound wave velocity causes the wave to bend in a different
direction.
C. A fraction of the sound waves are absorbed by an object and converted to heat
energy.
D. A single sound undergoes several reflections due to multiple reflecting surfaces.
Reverberation, in psychoacoustics and acoustics, is a persistence of sound after the sound is produced
Explanation:
I think it is right hope its helps
Answer:
D. A single sound undergoes several reflections due to multiple reflecting surfaces.
Explanation:
Sometimes, the source of a sound is surrounded by multiple reflecting surfaces. The waves traveling from the source strike these different surfaces, causing multiple reflections. For example, a single clap of thunder reflects on several clouds and the earth's surface, causing you to hear a rolling rumble instead of a single sound.
A tightrope walker is walking between two buildings holding a pole with length L=14.0 m, and mass mp=17.5 kg. The daredevil grips the pole with each hand a distance d=0.595 m from the center of the pole. A bird of mass mb=560 g lands on the very end of the left‑hand side of the pole. Assuming the daredevil applies upward forces with the left and right hands in a direction perpendicular to the pole, what magnitude of force Fleft and Fright must the left and right hand exert to counteract the torque of the bird?
Answer:
F = 32.28 N
Explanation:
For this exercise we must use the rotational equilibrium relation
Σ τ = 0
In the initial configuration it is in equilibrium, for which all the torque and forces are compensated. By the time the payment lands on the bar, we assume that the counter-clockwise turns are positive.
W_bird L / 2 - F_left 0.595 - F_right 0.595 = 0
we assume that the magnitude of the forces applied by the hands is the same
F_left = F_right = F
W_bird L / 2 - 2 F 0.595 = 0
F = [tex]\frac{m_{bird} \ g L} { 4 \ 0.595}[/tex]
we calculate
F = 0.560 9.8 14.0 /2.38
F = 32.28 N
Suppose the charged sphere is made from a conductor, rather than an insulator. Do you expect the magnitude of the force between the point charge and the conducting sphere to be greater than, less than, or equal to the force between the point charge and an insulating sphere
Answer:
* Point charge outside the radius of the sphere r> R, the force in the two systems is the same
* Point charge inside the sphere r <R, therefore the force in the system with the insulating sphere is greater
Explanation:
To answer this question let's use the relation
F = q E
with q being the point charge and E the electric field created by the sphere.
If we use Gauss's law
The electric field flux is proportional to the wax charge within the surface.
Let's analyze our situation.
* Point charge outside the radius of the sphere
r> R
where R is the radius of the sphere and r the distance from the center of the sphere to the point charge
in this case the waxed charge for the insulating and conducting sphere is the same, therefore the force in the two systems is the same
* Point charge inside the sphere
r <R
conductive sphere.
As the charges are mobile, they are located on the surface of the sphere and there is no waxed charge within a Gaussian surface that passes through the point charge, therefore the electric field is zero and consequently the force
F = 0
insulating sphere
Charges cannot move therefore there is a fraction of charge within a surface that passes through the point charge, consequently the electric field is different from zero
Fe> 0
for this second position the force on the conducting sphere is zero
therefore the force in the system with the insulating sphere is greater
A plane mirror produces a _____.
virtual image
refracted image
real image
Answer:
Explanation:
A plane mirror is the kind you look into when you look into a "regular" mirror. The image you see is right-side-up. These images are virtual. Real images are always upside down and are made by mirrors that are "parabolic" in shape. Virtual images are always right-side-up.
A 2.0kg object is dropped from a height of 30m.
After it drops for 2.0 seconds, what is its kinetic
energy and what is its potential energy?
(Assume no air resistance.)
Answer:
1) The kinetic energy of the object after it drops for 2.0 seconds is approximately 384.9 Joules
2) The potential energy of the object after it drops for 2.0 seconds is approximately 204 J
Explanation:
1) The given mass of the object, m = 2.0 kg
The height from which the object is dropped, h = 30 m
The kinetic energy of the object after it drops for 2.0 seconds = Required
Kinetic energy, K.E. = (1/2)·m·v²
Where;
v = The velocity of the object
The kinematic equation for finding the velocity of the object is presented as follows;
v = u + g·t
Where;
u = The initial velocity of the object = 0
g = The acceleration due to gravity of the object ≈ 9.81 m/s²
t = The time of motion of the object = 2.0 seconds
∴ The velocity after 2 seconds, v ≈ 0 + 9.81 m/s² × 2 s = 19.62 m/s
The kinetic energy, K.E. after 2 seconds as the object drops is given as follows;
[tex]K.E._{(after \ two \ seconds)}[/tex] = (1/2) × 2.0 kg × (19.62 m/s)² = 384.9444 J ≈ 384.9 J
2) The total energy, M.E. of the object at the top, h = 30 m, u = 0, is given as follows;
The total mechanical energy, M.E. = P.E. + K.E.
M.E. = m·g·h + (1/2)·m·u²
∴ M.E. = 2.0 kg × 9.81 m/s² × 30 m + 0 = 588.6 J
M.E. = 588.6 J
Given that the total mechanical energy, M.E., is constant, we have;
At 2.0 seconds, M.E. = 588.6 J , K.E. ≈ 384.9 J, P.E. = M.E. - K.E.
∴ P.E. = 588.9 J - 384.9 J ≈ 204 J
The potential energy after it drops 2.0 seconds, P.E. ≈ 204 J
A red car has a head-on collision with an approaching blue car with the same magnitude of momentum. A green car driving with the same momentum as the other cars collides with an enormously massive wall. Which of the three cars will experience the greatest impulse
All three cars experience the same impulse.
Impulse is equal to change in momentum.
Each car starts with the same amount of momentum and ends up with zero, so the magnitudes of all three changes are equal.
Indoor pollution experts conducted an analysis of the paint used in many office buildings which revealed that the paint contains traces of lead. The lead seems to rise to the surface of the paint and escape into the building's air supply. Tests show that because of the chemical drying process, the lead is not discernible on the surface until the paint has been on the walls for at least six months. To meet safety standards, owners should repaint walls at least every six months, or cover the walls with a different material.Which of the following, if true, would most weaken the conclusion above?A) The indoor pollution experts had no clear understanding of why it took six months for the lead to become discernible on the paint's surface.B) The indoor pollution experts neglected to examine the paint for traces of other toxic substances such as cadmium and mercury.C) The amount of lead found on the surface of the paint after six months remained constant for the next two years.D) The indoor pollution experts found that even in those offices painted with a different brand of paint, traces of lead were still found in the air workers breathed.E) The indoor pollution experts' research shows that the amounts of lead that come into contact with the air people breathe, even in the office buildings that used the greatest amount of paint, are too low to affect workers.
Answer:
E) The indoor pollution experts' research shows that the amounts of lead that come into contact with the air people breathe, even in the office buildings that used the greatest amount of paint, are too low to affect workers
Explanation:
The conclusion that would weaken the above statement if true was that the amount of lead found in air was not hazardous for the people living in building and the effects of such paint was quite low o affect workers.If the velocity of a motorcycle increases from 30 mis too 50m/s in 10 seconds what will be the acceleration of motorcycle?
Answer:
2 m/s^2
Explanation:
a = ∆v / ∆t
a = (50 - 30) / 10
a = 20 / 10
a = 2 m/s^2
acceleration = u-v/t
50-30/10
=2m/s
You observe that you see more mockingbirds in small trees and more hawks in large trees. Which of the following is an appropriate scientific question based on this observation?
How does the size of a tree affect the bird species that prefer to live in it?
How do birds fly?
What type of food do birds eat?
What time of year to birds mate?
Answer:
A) How the size of a tree affect the bird species that prefer to live in it
Explanation:
I took the quiz
What is the decay constant for Oxygen-19 if it has a half-life of 26.5s?
A)0.0262/s
B)18.4
C)0.0377/s
C)38.2/s
Answer:
Option A.
Explanation:
We define the half time T as the time such that an initial quantity A reduces to its half.
So we can model the quantity as a function of time like:
P(t) = A*e^(-k*t)
Then for the half time, T, we will have:
P(T) = A/2 = A*e^(-k*T)
solving for k, we get:
A/2 = A*e^(-k*T)
1/2 = e^(-k*T)
ln(1/2) = ln( e^(-k*T)) = -k*T
-ln(1/2)/T = k
Here we know that the half time is T = 26.5s
if we input that in the above equation, we get:
-ln(1/2)/26.5s = k = 0.0262 s^-1
Then the correct option is A
If the power of a working system(machine) is 2088.8 watts then its power in horsepower will be: A. 1hp B. 2.7hp C. 2.9hp D. 2.8hp
Answer:
D. 2.8 hp
Explanation:
1 hp=746 watts
2088.8÷746= 2.8
the atomic number of phosphorus is
Answer:
The atomic number of phosphorus is 15.
Explanation:
It’s found after Si(Silicon) and before S(sulphur)
84. Three resitors each of value 30 respectively are connected in a parallel
combination across a 10 V battery the current through each resitor is
Answer:
each resistor draws 1/3 of an amp or 0.33333 amps
Explanation:
V = I * R
V = 10 volts
R = 30 ohms
10 = I * 30 Divide by 30
10/30 = I
I = 0.33333
10. Explain the principle of electric motor. Write its uses.
Explanation:
The principle of an electric motor is based on the current carrying conductor which produces magnetic field around it. A current carrying conductor is placed perpendicular to the magnetic field so that it experiences a force.
The largest electric motors are used for ship propulsion, pipeline compression and pumped-storage applications with ratings reaching 100 megawatts. Electric motors are found in industrial fans, blowers and pumps, machine tools, household appliances, power tools and disk drives.
A railroad car (with a mass of 3250 kg) moves at 8.1 m/s . It collides with and couples to another car that was initially at rest. After the collision, the two cars move together at a speed of 4.50 m/s. What is the mass of the second car?
Answer:
m_2 = 2600kg
Explanation:
P_1 = P_2
P = (m_1)*(v_1)+(m_2)(v_2)
P_1 = (3250kg)(8.1m/s)+(m_2)(0m/s)
P_1 = 26,325 kg*m/s
P_2 = (3250kg)(4.5m/s)+(m_2)(4.5m/s)
P_2 = 14,625kg*m/s+(4.5m/s)m_2
26,325 kg*m/s = 14,625kg*m/s+(4.5m/s)m_2
11,700kg*m/s = (4.5m/s)m_2
m_2 = 2600kg
1. Find the temperature when the degrees of the Celsius scale will be one fifth of the corresponding degrees of the Fahrenheit scale
2. How much heat is necessary to warm 500g of water from 20°C to 65°C?
Answer:
F = 9/5 C + 32 conversion from C to F
F = 9/5 * F/5 + 32
25 F = 9 F + 800
16 F = 800
F = 50
Check:
C = 5/9 ( F - 32) = 5/9 (50 - 32) = 10 as requested
Q = c m change in temp
Q = 1 cal/gm-deg C * 500 gm * 45 deg C = 22,500 calories
50 Fahrenheit heat required.
How much heat required?Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy between physical systems.
F = 9/5 C + 32 conversion from C to F
F = 9/5 * F/5 + 32
25 F = 9 F + 800
16 F = 800
F = 50
The answer is 50 Fahrenheit.
Learn more about heat transfer
https://brainly.com/question15086449
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how is one standard kilogram defined in SI system?
Answer:
One standard kilogram is define as the mass of platinium iridium cylinder having equal diameter and height kept at the particular condition of international bureo of weigh and measure sevre near paris.
find the expression for the displacement covered in nth or in last one second
Answer:
Snth = u + a/2 ( 2n - 1)
Explanation:
Do you need explanation based on graph, integration or other method?
Calculate the relative atomic mass of MgO
Answer:
MgO relative atomic mass is 40
Explanation:
Mg=24
O=16
7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ball h, in metres can be represented, after t seconds, is modelled by the equation h = –4.8t² + 8.2t + 1.8. (a) Determine the height of the ball after 1.7 seconds.
Answer:
8392
Explanation:
d=s/t
Two stones are dropped from the edge of a 60m cliff , the second stone 1.6secon after the first . How far below the top of the cliff is the second stone when the separation between the two stone is 36m?
Answer:
The separation between the two stones is 36 m, when the second stone is approximately 10.9 m below the top of the cliff
Explanation:
The given parameters are;
The height of the cliff from which the stones are dropped, h = 60 m
The time at which the second stone is dropped = 1.6 seconds after the first
The distance below the top of the cliff when the distance between the two stones is 36 m = Required
We have;
The kinematic equation of motion that can be used is s = u·t - (1/2)·g·t²
For the first stone, we have, s₁ = u·t₁ - (1/2)·g·t₁²
For the second stone, we get; s₂ = u·t₂ - (1/2)·g·t₂²
t₁ = t₂ + 1.6
g = The acceleration due to gravity ≈ 9.81 m/s²
s = The distance below the cliff top
The initial velocity of the stones, u = 0
Let t represent the time from which the second stone is dropped at which the distance between the two stones is 36 m, we have;
s₁ = u·(t + 1.6) + (1/2)·g·(t + 1.6)²
s₂ = u·t + (1/2)·g·t²
u = 0
∴ s₁ - s₂ = 36 = (1/2)·g·(t + 1.6)² - (1/2)·g·t²
2 × 36/(g) = (t + 1.6)² - t² = t² + 3.2·t + 2.56 - t² = 3.2·t + 2.56
2 × 36/(9.81) = 3.2·t + 2.56
t = (2 × 36/(9.81) - 2.56)/3.2 = ≈ 1.49 s
t ≈ 1.49 s
s₂ = (1/2)·g·t²
∴ s₂ = (1/2) × 9.81 × 1.49² ≈ 10.9
The distance below the top of the cliff of the second stone when the the separation between the two stones is 36 m, s₂ ≈ 10.9 m.
In a lunar experiment, a 950-g aluminum (920 J/(°Ckg)) sphere is dropped from the space probe while is 75 m above the Lunar ground. If the sphere’s temperature increased by 0.11°C when it hits the ground, what percentage of the initial mechanical energy was absorbed as thermal energy by the aluminum sphere?
Answer:
13.759 % of the initial mechanical energy is lost as thermal energy.
Explanation:
By the First Law of Thermodynamics we know that increase in internal energy of the object ([tex]U[/tex]), in joules, is equal to the lost amount of the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:
[tex]\frac{x}{100} \cdot \Delta U_{g} = \Delta U[/tex] (1)
Where [tex]x[/tex] is the percentage of the energy loss, no unit.
By definition of the gravitational potential energy and internal energy, we expand this equation:
[tex]\frac{x\cdot m \cdot g \cdot h}{100} = m\cdot c\cdot \Delta T[/tex] (1b)
Where:
[tex]m[/tex] - Mass of the object, in kilograms.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]h[/tex] - Initial height of the object above the lunar ground, in meters.
[tex]c[/tex] - Specific heat of aluminium, in joules per degree Celsius-kilogram.
[tex]\Delta T[/tex] - Temperature increase due to collision, in degree Celsius.
If we know that [tex]m = 0.95\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]h = 75\,m[/tex], [tex]c = 920\,\frac{J}{kg\cdot ^{\circ}C}[/tex] and [tex]\Delta T = 0.11\,^{\circ}C[/tex], then the percentage of energy loss due to collision is:
[tex]x = \frac{100\cdot c\cdot \Delta T}{g\cdot h}[/tex]
[tex]x = \frac{100\cdot \left(920\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (0.11\,^{\circ}C)}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (75\,m)}[/tex]
[tex]x = 13.759\,\%[/tex]
13.759 % of the initial mechanical energy is lost as thermal energy.
2- A copper wire of 3mm diameter with conductivity of 6.7 10' (0.M), and electron mobility of 0.0064 m2 /V sec. Is subjected to an electric field of 30 mV/m. Find (a) the charge density of free electrons, (b) the current density, (c) current flowing in the wire, (d) the electron draft velocity.
Answer:
a) [tex]n=5.98*10^{26}/m^3[/tex]
b) [tex]i=2010000A/m^2[/tex]
c) [tex]I_w=14.207A[/tex]
d) [tex]V_e=1.92*10^{-4}m/s[/tex]
Explanation:
From the question we are told that:
Diameter [tex]d=3mm=>3*10^{-3}[/tex]
Conductivity [tex]\sigma= 6.7 10^7 (0.M),[/tex]
Electron mobility [tex]\phi= 0.0064 m2 /V sec[/tex]
Electric field [tex]E= 30 mV/m[/tex]
a)
Generally the equation for Charge Density is mathematically given by
[tex]\phi=\frac{\sigma}{n e}[/tex]
Therefore
[tex]n=\frac{\sigma}{\phi e}[/tex]
[tex]n=\frac{6.7 10^7}{1.6*10^{-19} *0.0064}[/tex]
[tex]n=5.98*10^{26}/m^3[/tex]
b)
Generally the equation for current density is mathematically given by
[tex]i=\sigma*E[/tex]
[tex]i= 30*10^{-3] *6.7 10^7[/tex]
[tex]i=2010000A/m^2[/tex]
c)
Generally the equation for current in wire is mathematically given by
[tex]I_w=iA[/tex]
[tex]I_w=i*\pi r^2[/tex]
[tex]I_w=(2010000)*\pi( 1.5*10^{-3})^2[/tex]
[tex]I_w=14.207A[/tex]
d)
Generally the equation for electron draft velocity. is mathematically given by
[tex]V_e=\phi E[/tex]
[tex]V_e=(0.0064)*(30*10^{-3})[/tex]
[tex]V_e=1.92*10^{-4}m/s[/tex]
(Q022) A negative magnetic anomaly a. occurs when the Earth's magnetic field measured in ancient rocks is the same as it is today. b. is created when weak magnetic forces in basalt grains add to the force produced by the Earth's dipole. c. describes the sawtooth pattern of magnetic signal strength measured along the Atlantic Ocean seafloor. d. is indicated when a magnetometer measures intervals of magnetism that are weaker than expecte
Answer:
d. Is indicated when a magnetometer measures intervals of magnetism that are weaker than expected
Explanation:
Magnetic field intensity is measured with a magnetometer on the surface of the Earth. Areas in which the magnetic field strength is lower or more than average are known as areas with magnetic anomalies, which may be due to the presence of rocks that have a different magnetic characteristics
Where the magnetic anomaly is negative, it is indicative of a magnetic field strength reading that is lower than average magnetic field which is generally obtainable
Therefore, the correct option is indicated when a magnetometer measures intervals of magnetism that are weaker than expected
Discuss the role of globalization in the development of sI unit
Answer:
Sharing of informationExplanation:
The development of SI unit has helped in the sharing of scientific as well as techical information internationally.Answer:
It was created during the French Revolution in 1799 and has enabled for the international exchange of scientific and technical information. Calculating with SI units is also a lot easier than using the English system.
why a person feel weightlessness in a spacecraft orbiting around a heavenly body
Answer:
The orbital velocity an aircraft orbiting around a heavenly body is found as follows;
At the orbital velocity, [tex]F_G[/tex] = [tex]F_C[/tex]
Where;
[tex]F_G[/tex] = The gravitational force = [tex]\dfrac{G \cdot M \cdot m}{R_E^2}[/tex]
[tex]F_C[/tex] = The centripetal force = [tex]\dfrac{m \cdot v_0^2}{R_E}[/tex]
Therefore
[tex]v_0 = \sqrt{\dfrac{G \cdot M}{R_E} }[/tex]
Therefore, at the orbital velocity of the spacecraft, the centripetal force attracting the person away from the central region heavenly body is equal to the gravitational force pulling the person towards the center of the heavenly body (which was felt as her or his weight), and the person feels weightless while inside the orbiting spacecraft
Explanation:
