Option- E. the appearance of specific constellations is the correct answer.
What is astronomy ?Astronomy, a science that includes the study of all extraterrestrial objects and phenomena. Until the invention of the telescope in the 17th century and the discovery of the laws of motion and gravity, astronomy was primarily concerned with finding and predicting the positions of the sun, moon, and planets. Initially for calendar and astrological purposes, later for navigation and astronomy. Scientific use Interest. The catalog of celestial bodies currently being studied is much more extensive, including in order of increasing distance, the solar system, the stars that make up the Milky Way, and other more distant galaxies. With the advent of scientific space probes, the Earth has also been studied as one of the planets, but more detailed study remains the realm of Earth science.
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A man pushing a crate of mass
m = 92.0 kg
at a speed of
v = 0.880 m/s
encounters a rough horizontal surface of length
ℓ = 0.65 m
as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.359 and he exerts a constant horizontal force of 299 N on the crate.
(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
magnitude _______N
Direction?:
1. Same as the motion of the crate
2. opposite as the motion of the crate
(b) Find the net work done on the crate while it is on the rough surface.
___________J
(c) Find the speed of the crate when it reaches the end of the rough surface.
_________m/s
(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 24.67 N, opposite as the motion of the crate.
(b) The net work done on the crate while it is on the rough surface is -16.04 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.65 m/s.
Magnitude of net force on the crateF(net) = F - μFf
F(net) = 299 - 0.351(92 x 9.8)
F(net) = -24.67 N
Net work done on the crateW = F(net) x L
W = -24.67 x 0.65
W = - 16.04 J
Acceleration of the cratea = F(net)/m
a = -24.67/92
a = - 0.268 m/s²
Speed of the cratev² = u² + 2as
v² = 0.88² + 2(-0.268)(0.65)
v² = 0.426
v = √0.426
v = 0.65 m/s
Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 24.67 N, opposite as the motion of the crate.
The net work done on the crate while it is on the rough surface is -16.04 J.
The speed of the crate when it reaches the end of the rough surface is 0.65 m/s.
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What is the formula that relates current and voltage?
A. V = I/R
B. V = R/I
C. V = IR
D. V = I^2/R
"V = IR" is the equation that relates current to voltage.
Relationship between current and voltage:
Georg Simon Ohm, a German scientist, and mathematician carried out an experiment in 1827 using multiple circuits with variable wire lengths. He took measurements of both the voltage across the electrical component and the current flowing through the circuit.Ohm's law outlines the connection between voltage, current, and resistance. According to the equation,V = IR or,
I = V/R,
the amount of current (I) flowing through a circuit is inversely
related to the amount of resistance (r) and directly proportional to
the voltage (v).
Therefore the correct answer is option C i.e., "V = IR".
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Help me <3 please
Thank you :)
Answer:
11,890
Explanation:
First we need to know what is considered a significant figure.
A significant figure is a value that is not a zero at the start OR end of a value.
Which means, the 0 in the value of 90 or 0.363 are not considered a significant figure.
The 0 in the value of 3056 is considered a significant figure.
So from the table, we can deduce:
0.275 has 3 significant figures
750 has 2 significant figures
[tex]10.4 \times {10}^{5} = 1040000[/tex]
has 3 significant figures.
11,890 has 4 significant figures.
320,050 has 5 significant figures.
So from the above, we can already see the answer.
A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and 272.0 N (54.4 lbs) stand on the board so that the board is balanced.
What is the upward force exerted on the board by the support?
The upward force exerted on the board by the support is 530.8 N.
Upward force exerted on the board by the supportThe upward force exerted on the board by the support is calculated as follows;
F(up) = 52.8 N + 206.0 N + 272.0 N
F(up) = 530.8 N
Thus, the upward force exerted on the board by the support is 530.8 N.
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For a projectile launched horizontally at 650 m/s and travels for 5.75 seconds, What is the range?
The range of horizontal projectile motion is 3737.5 m.
A projectile's horizontal velocity is constant (a never changing value), Gravity causes a downward vertical acceleration with a magnitude of 9.8 m/s/s. A projectile's vertical velocity fluctuates by 9.8 m/s per second, A projectile's vertical motion is unrelated to its horizontal motion. we know that sinθ is maximum at 90°. Therefore horizontal range will maximum at 45°.
To calculate the range of horizontal projectile motion we use;
Δx = vₓ t
where , Δx = Range
vₓ = Velocity
t = Time
Initial horizontal velocity, vₓ = 650 m/s
Time = 5.75 sec
Δx = 650 × 5.75
= 3737.5 m
Therefore, the range of horizontal projectile motion is 3737.5 m.
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In Milikan’s experiment, a drop of radius of 1.64 μm and density 0.851 g/cm3 is suspended in the lower chamber when a downward-pointing electric field of 1.92 * 105 N/C is applied.
What is the weight of the drop?
Find the charge on the drop, in terms of e.
How many excess or deficit electrons does it have?
(a) The weight of the drop is 1.54 x 10⁻²⁵ N,
(b) The the charge on the drop, in terms of e is 5 x 10⁻¹²e and
(c) The excess electrons is 5 x 10⁻¹² electron.
Weight of the dropThe weight of the drop is calculated as follows;
Volume of the drop; V = ⁴/₃πr³
V = ⁴/₃π(1.64 x 10⁻⁶)³ = 1.845 x 10⁻¹⁷ m³
mass = density x volume
mass = 0.851 g/cm³ x 1.845 x 10⁻²³ cm³ = 1.572 x 10⁻²³ g = 1.57 x 10⁻²⁶ kg.
Weight = 1.57 x 10⁻²⁶ kg x 9.8 m/s² = 1.54 x 10⁻²⁵ N.
Charge on the dropq = F/E
q = (1.54 x 10⁻²⁵ N)/(1.92 x 10⁵)
q = 8.01 x 10⁻³¹ C
1.6 x 10⁻¹⁹ C = 1e
8.01 x 10⁻³¹ C = ?
= 5 x 10⁻¹²e
Excess electron on the drop1.6 x 10⁻¹⁹ C ------- 1 electron
8.01 x 10⁻³¹ C ------- ?
= 5 x 10⁻¹² electron
Thus, the weight of the drop is 1.54 x 10⁻²⁵ N, the the charge on the drop, in terms of e is 5 x 10⁻¹²e and the excess electrons is 5 x 10⁻¹² electron.
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I need to figure out how to extend a single paragraph's worth of answers that I can make into 3 paragraphs, requesting a LOT of help here.
Question:
Discuss one feature of a spacesuit that protects astronauts outside the spacecraft.
An astronaut's spacesuit is considerably more than just a pair of garments they put on during spacewalks. An entire spacesuit is actually a single-person spacecraft. The Extravehicular Mobility Unit, or EMU, is the official name for the spacesuit used on the International Space Station and Space Shuttle. "Extravehicular" refers to something that is outside of a vehicle or spaceship. To be "mobile" implies to be able to move around while wearing an astronaut suit. The astronaut is shielded by the spacesuit from the perils of being outdoors in space.
The human body cannot survive in space's hostile atmosphere without protection. A spacesuit must shield the astronaut from the vacuum of space, the drastic temperature changes in space, and, if at all possible, it must lessen the astronaut's exposure to radiation. Therefore, the suit's primary function is to act as a pressure vessel. It must keep an air environment close to the skin constant, deliver a constant supply of clean air to the lungs, and expel stale, carbon dioxide-rich air. Every 90 minutes or so, an astronaut in low Earth orbit (LEO) experiences day and night. They can fast chill to -250 F during the night and reach 250 F while the sun is shining on them (-156 C). The body's normal temperature of 98.6 F must be maintained by the suit (37 C).
To do all this, the current NASA suit has 14 layers. The liquid ventilation and cooling garment is the first three layers. It resembles a body-hugging spandex garment that has tubes inside that carry cool water across the body to dissipate extra heat. Air below the next layer cannot escape since it is a pressure vessel made of nylon coated with urethane. The following layer resembles a tent since it is constructed of Dacron. Its goal is to exert pressure on the pressure garment so that it keeps its form and doesn't expand. Neoprene coasted nylon ripstop is the following layer. It is quite resilient to tears. Seven layers of mylar film and foil blanket are used to cover it. These restrict warmth transfer into or out of the suit by acting like a thermos. Due to its several layers, it also provides defense against very small micrometeroids, which drain energy as they pierce and break each layer. The top layer is made of orthofabric, a goretex, nomex, and kevlar mixture. It is very impervious to tearing and thermally reflecting to help regulate temperature. The suits are made by ILC Dover. Incorporating materials that minimize radiation penetration through the suit is something they are always working on.
A 80-kg base runner begins his slide into second base when he is moving at a speed of 3.0 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
(a) How much mechanical energy is lost due to friction acting on the runner?
______J
(b) How far does he slide?
________m
The mechanical energy lost due to friction is 360 J.
The distance the runner slides is, s = 0.655 m
What is the mechanical energy of the runner?
The mechanical energy lost due to friction acting on the runner is equal to the change in kinetic energy.
Change in Kinetic energy = 1/2m(v -u)²
Change in Kinetic energy = 80 * (0 - 3.0)²/2
Change in Kinetic energy = 360 J
Mechanical energy lost = 360 J
Distance he slide is determined using the formula below as follows:
Acceleration of runner = coefficient of friction * acceleration due to gravity
acceleration, a = 0.7 * 9.8 = 6.86
v² = u² - 2as
s = v² + u²/2a
s = 0 + 3³/2 * 6.86
s = 0.655 m
In conclusion, the mechanical energy lost due to friction is the loss in kinetic energy.
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deriving projectile motion formulas
Answer:
Projectile motion formula or equations derived (In Tabular format)Motion Path equation:
y = (tanθ) x – (1/2) g . x2/(V0 cosθ)2
Explanation:
Compare and contrast visible light, infrared light, and ultraviolet light.
Answer:
NE BİLİM
Explanation:
PUAN İÇİN YAZIYORUM SEN BENİ GÖRMEZDEN GEL ANLDIN MI DOSTUM bu arada Suriye >>>>>
Derive the following equations of motion
1. v = at
2. s = ut + at²
3. v² = u² + 2as
according to definition of acceleration
a=v-u/t
t=v-u/a(equation 1)
according to the formula of average velocity
v+u/2*s/t
s=v+u/2*t(equation 2)
now putting the value of t in equation 2
s=v+u/2*v-u/a
s=v^2-u^2/2a
v^2=u^2+2as
please make me brainlist
Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N.
Calculate the magnitude of the tension in the string marked A.
The magnitude of the tension in the string marked A and B is mathematically given as
A = 52.5 N
What is the magnitude of the tension in the string marked A?Generally, the equation for is mathematically given as
tan=3/8
negative x
[tex]B= tan\phi \\tan \phi=5/4[/tex]
negative x
C= tan=1/6
Hence, considering the Horizontal components
74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)
A = 78.9 - 0.668B
Vertical components
74.9*sin(9.46) + Asin(20.6) = Bsin(51.3)
40.07 = 1.015B
B = 39.5 N
In conclusion, Sub the value of B is the equation of A
A = 78.9 - 0.668B
Sub
A = 78.9 - 0.668( 39.5 N)
A = 52.5 N
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By how much does the gravitational potential energy of a 55- kg pole vaulter change if her center of mass rises about 4.0 m during the jump?
Express your answer to two significant figures and include the appropriate units.
The change in the gravitational potential energy of the pole vaulter is 2,156 J.
Change in the gravitational potential energy
The change in the gravitational potential energy of the pole vaulter is calculated as follows;
ΔP.E = mg(Δh)
where;
m is massΔh is change in heightΔP.E = (55)(9.8)(4)
ΔP.E = 2,156 J
Thus, the change in the gravitational potential energy of the pole vaulter is 2,156 J.
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9.80 1. The density of mercury is 13600 kg/m³. What is this value in g/cm³? 2. Find the mass of water which will fit in a large tank measuring 2 m x 1 m x 20 cm. Density of water is 1000 kg/m³ or 1.0 g/cm³. 3. Find the volume of a lump of softwood whose mass is 120 g. Density of softwood is 0.6 gcm-3 or 600 kgm-³. -3
Answer:
1. 13..6 grams per centimeters cubed.
2. Mass = 400kg
3. Volume = 200cm = 2m
Explanation:
1. The conversion for kg/m^3 to g/cm^3 is divide by 1000.
2. [tex]density=\frac{mass}{volume}[/tex]
[tex]1000=\frac{mass}{2 * 1 * 0.2}[/tex]
[tex]1000*0.4=mass[/tex]
[tex]400kg = mass[/tex]
3. [tex]density=\frac{mass}{volume}[/tex]
[tex]0.6=\frac{120}{volume}[/tex]
[tex]volume=\frac{120}{0.6}[/tex]
[tex]volume= 200cm[/tex]
Aster 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 64 m
The angle is 300° and She has to head in West north direction to return to her initial position
What is Displacement ?The displacement is the distance travelled in a specific direction. Displacement is a vector quantity.
Given that Aster 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.
This can be solved by using bearing method. Cosine formula will be the best to solve for the distance D.
[tex]D^{2}[/tex] = [tex]70^{2}[/tex] + [tex]82^{2}[/tex] - 70 x 82 x Cos (37 + 20)
[tex]D^{2}[/tex] = 4900 + 6724 - 5740Cos57
[tex]D^{2}[/tex] = 11624 - 3126.23
[tex]D^{2}[/tex] = 8497.8
D = [tex]\sqrt{8497.8}[/tex]
D = 92.2 m
a) The Aster's final position from her initial position is 92.2 - 28 = 64 m
The angle = 270° + 30° = 300°
b) She has to head in West north direction to return to her initial position
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A rectangle measuring 30.0 cm by 40.0 cm is located inside a region of a spatially uniform magnetic field of 1.65 T, with the field perpendicular to the plane of the coil (Figure 1). The coil is pulled out at a steady rate of 2.00 cm/s traveling perpendicular to the field lines. The region of the field ends abruptly as shown.
a) Find the emf induced in this coil when it is all inside the field.
b) Find the emf induced in this coil when it is partly inside the field.
c) Find the emf induced in this coil when it is all outside the field.
The solution for the questions below is mathematically given as
induced emf=0induced emf=0.0132Vinduced emf=0What is the emf induced in this coil when it is all inside the field.?(A)
Generally, the equation for the flux is mathematically given as
[tex]\Phi _{B}=B.A[/tex]
[tex]\Phi _{B}=B.A\\\\\Phi _{B}=B(40*30*10^{-4}) ,[/tex]
So, the magnetic flux through the coil is constant.
From faradays law,
[tex]\varepsilon =-\frac{\mathrm{d} \Phi _{B}}{\mathrm{d} t}[/tex]
---(1) for the induced emf. Since magnetic flux is constant, LHS. of (1) =0
induced emf=0
(B)
Let x be the length of the coil's magnetic field area.
[tex]then, \Phi _{B}=B.A=B(40*x*10^{-4}) \\\\\frac{\mathrm{d}\Phi _{B} }{\mathrm{d} t}=40\\B*10^{-4}*\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
induced emf=0.0132V
(C)
In conclusion, Therefore, there is no variation in the magnetic flux across the coil when magnetic flux=0.
induced emf=0
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A roller-coaster car shown in the figure below is pulled up to point 1 where it is released from rest. Take y = 39 m .
(Figure 1)
Assuming no friction, calculate the speed at point 2.
Express your answer to two significant figures and include the appropriate units.
Calculate the speed at point 3.
Express your answer to two significant figures and include the appropriate units.
Calculate the speed at point 4.
Express your answer to two significant figures and include the appropriate units.
The speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.
Speed of the roller coasterThe speed of the roller coaster at any position is calculated by applying the principle of conservation of energy.
K.E = P.E
¹/₂mv² = mgh
v = √2gh
where;
h is vertical displacementSpeed at point 2v(2) = √[(2 x 9.8)(39 - 0)]
v(2) = 28 m/s
Speed at point 3v(3) = √[(2 x 9.8)(39 - 26)]
v(3) = 16 m/s
Speed at point 4v(4) = √[(2 x 9.8)(39 - 14)]
v(4) = 22 m/s
Thus, the speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.
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please need answers ASAP.
Answer:
overloading can be avoided if two many appliances are not connected to a single socket short circuiting is a name given to a situation in which they live and the natural voice accidentally coming contact
This technique allowed multiple peoples DNA to be compared to look for
similarities and differences in order to solve crime
DNA Examination
DNA Elimination
ODNA Sequencing
ODNA Fingerprinting
*2
DNA sequencing is the technique which allows multiple peoples DNA to be compared to look for similarities and differences in order to solve crime and is denoted as option C.
What is DNA?This is referred to as the deoxyribonucleic acid and contains the genetic components of organisms and is also located in the nucleus. On the other hand, DNA sequencing refers to the process of determining the nucleic acid sequence.
Each organisms has a unique DNA sequence which is why it is used to identify individuals through the use of bodily fluids such as saliva, blood etc.
It is used to compare the similarities and differences in order to solve crime thereby making it the most appropriate choice.
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Define a dipole. Hence, write the expression for calculating the electric moment of a dipole
Answer:
Explanation:
An electric dipole is formed by two point charges +q and −q connected by a vector a. The electric dipole moment is defined as p = qa
A man exerts a horizontal force of 123 N on a crate with a mass of 40.2 kg.
(A) If the crate doesn't move, what's the magnitude of the static friction force (in N)?
______N
(B) What is the minimum possible value of the coefficient of static friction between the crate and the floor? (Assume the crate remains stationary.)?
The magnitude of the static friction force = 123 N
The coefficient of static friction = 0.31
What is static friction?Static friction is the frictional force that must be overcome in other for a body to that moving over another.
Since the crate does not move, the magnitude of the static frictional force is equal to the applied force.
The magnitude of the static frictional force = 123 N
The coefficient of static friction = frictional force/normal reaction
The coefficient of static friction = 123/(40* 9.8)
The coefficient of static friction = 0.31
In conclusion, the static frictional force on the on the crate is equal to the applied force on the crate.
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If the magnetic flux through a loop increases according to the relation; where , is in milliweber (mWb) and t is in seconds. Calculate the magnitude of e.m.f induced in the loop when t = 2s
The magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.
emf induced in the loopThe magnitude of e.m.f induced in the loop is calculated as follows;
emf = dФ/dt
Ф = 6t² + 7t
dФ/dt = 12t + 7
at t = 2 seconds
emf = dФ/dt = 12(2) + 7 = 31 V
Thus, the magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.
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Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N.
Calculate the magnitude of the tension in the string marked A. (You'll need to get the various positions from the graph. The ends of the strings are exactly on one of the tic marks.)
The magnitude of the tension in the string marked A is 52.5N
Generally, the equation for is mathematically given as
Let's take θ be an angle at A
So, tanθ = 3/8
Let's take α be an angle at B (Below X)
tanα = 5/4
Let's take β be an angle at C (Below x)
tanβ = 1/6
First we take the Horizontal Components
74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)
By solving the equation, we get
A = 78.9 - 0.668B … (1)
Now, we take the vertical components
74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)
By solving the equation, we get
40.07 = 1.015B
B = 39.5N
By substituting the value of B in equation (1)
A = 78.9 - 0.6668× 39.5
A = 52.5N
Hence, the magnitude of the tension in the string marked A is 52.5N
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200 g of water is heated and its temperature goes from 280 K to 300 K. What was the change in temperature for this process?
A. 280 K
B. 20 F
C. 20 K
D. 300 K
The change in temperature is 20 kelvin
What is Temperature?
Temperature can simply be described as how hot or how cold an object is at a particular period in time. The unit of temperature is Kelvin.
The formula for calculating change in temperature is
Final temperature - Initial temperature
Final temperature = 300 kelvin
Initial temperature= 280 kelvin
Change in temperature= 300-280
= 20 kelvin
Thus the change in temperature for this process is 20 kelvin
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A block with a mass of 33.0 kg is pushed with a horizontal force of 150 N. The block moves at a constant speed across a level, rough surface a distance of 4.95 m.
(a) What is the work done (in J) by the 150 N force?
_________J
(b) What is the coefficient of kinetic friction between the block and the surface?
________
The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.
What is the work done?The work done is given by the use of the formula;
W = F * x
Where;
F = force applied
x = distance covered
W = 150 N * 4.95 m = 742.5 J
Now;
The coefficient of kinetic friction is given by;
μ = F/mg
μ = 150/ 33 * 9.8
μ = 0.46
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a. A light wave moves through glass (n = 1.5) at an angle of 15°. What angle will it have when it moves from the glass into water (n = 1.33)? (4 points)
b. Draw a ray diagram to locate the image of the arrow, as refracted through the lens shown. Write 2 - 3 sentences describing the type of image and its size relative to the object. What type of mirror could be used to form an image of the same type and size? (8 points)
c. An object is located 65 cm from a concave mirror with a focal length of 45 cm. What is the image distance? Is the image real or virtual? (6 points)
Answer in detail with the correct units and steps to solve. Will mark brainliest.
A wave is a phenomenon that does not cause a permanent displacement in the particles of the medium through which it passes. And it transfers energy from one end of the medium to the other. Examples of waves include light waves, sound waves, water waves, x-rays, radiowaves, etc. Thus the required answers for each part of the question are:
a. The angle that the light wave would have is [tex]7.5^{o}[/tex].
b. The type of mirror that can be used is a plane mirror.
c. The image distance is 146.3 cm.
ii. The image formed by the mirror is a real image.
a. When a ray of light passes from one medium to another, then refraction occurs. The refraction depends on the refractive index of the medium considered.
Thus from Snell's law, we have:
refractive index, n, = [tex]\frac{Sin i}{Sin r}[/tex]
where: i is the angle of incidence, and r is the refracted angle.
Now given that n = 1.5, and i = 15.
Then;
n = [tex]\frac{Sin i}{Sin r}[/tex]
1.5 = [tex]\frac{Sin 15}{Sin r}[/tex]
Sin r = [tex]\frac{0.2588}{1.5}[/tex]
= 0.17253
r = [tex]Sin^{-1}[/tex] 0.17253
= 9.936
r ≅ [tex]10^{o}[/tex]
Since the light wave now moves from the glass into water, the determined refracted angle now becomes its angle of incidence in water. So that;
n = [tex]\frac{Sin i}{Sin r}[/tex]
1.33 = [tex]\frac{Sin 10}{Sin r}[/tex]
Sin r = [tex]\frac{0.17365}{1.33}[/tex]
= 0.1306
r = [tex]Sin^{-1}[/tex] 0.1306
= 7.504
r = [tex]7.5^{o}[/tex]
Therefore, the angle that the light wave would have is [tex]7.5^{o}[/tex].
b. The image formed would be the same size as that of the object. And also the same distance as that of the object to the pole of the lens.
The type of mirror that can be used is a plane mirror.
The ray diagram is attached to this answer.
c. From the mirror formula;
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]
where; f is the focal length of the mirror, u is the object's distance to the mirror, and v is the image's distance to the mirror.
Given; u = 65 cm, and f = 45 cm, then:
[tex]\frac{1}{45}[/tex] = [tex]\frac{1}{65}[/tex] + [tex]\frac{1}{v}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{1}{45}[/tex] - [tex]\frac{1}{65}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{20}{2925}[/tex]
v = [tex]\frac{2925}{20}[/tex]
v = 146.25 cm
The image distance is 146.3 cm.
ii. The image formed by the mirror is a real image.
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Forces always act in ________. A. Solitude B. Pairs C. Unpredictable ways D. Interesting ways
The four tires of an automobile are inflated to a gauge pressure of 2.02×10^5 Pa. Each tire has an area of 217 cm^2 in contact with the ground. Determine the weight of the automobile.
The weight of the automobile is 17,533.6 N.
Weight of the automobile
The weight of the automobile is calculated as follows;
P = F / A
F = (W/4)
P = (W/4) / A
P = W/4A
W = 4AP
where;
P is pressure A is areaW = (4)(217 x 10⁻⁴ m²)(2.02 x 10⁵ Pa)
W = 17,533.6 N
Thus, the weight of the automobile is 17,533.6 N.
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8. The pulley is assumed massless and frictionless and rotates freely about its axle. The blocks have masses m = 40 g and m₂ = 20 g, and block mi is pulled to the right by a horizontal force of magnitude F = 0.03 N. Find the magnitude of the acceleration of block m2 and the tension in the cord if the surface is frictionless. (2pt) m₂ a₂ T₂ T₂ T₁ m₁
The magnitude of the acceleration of block m2 is 0.09 m/s² and the tension in the cord is 1.8 x 10⁻³ N.
Acceleration of the blocksThe acceleration of the blocks is calculated from the net force on the blocks.
∑F = ma
a = ∑F/m
a = (F) / (m₁ + m₂)
where;
F is the horizontal force appliedm₁ is mass of first block = 40 g = 0.04 kgm₂ is mass of the second block = 20 g = 0.02 kga = (0.03)/(0.04 + 0.02)
a = 0.09 m/s²
Tension due to block m₂T = m₂a
T = (0.02 x 0.09) = 1.8 x 10⁻³ N
Thus, the magnitude of the acceleration of block m2 is 0.09 m/s² and the tension in the cord is 1.8 x 10⁻³ N.
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b. Calculate the total resistance of the circuit below. (4 points)
c. In the circuit diagram above, the meters are labeled 1 and 2. Write 2 - 3 sentences identifying each type of meter and how it is connected with the 12 Ω resistor. (4 points)
d. In the circuit diagram above, predict which resistors (if any) will stop working when the switch is opened. Write 2 - 3 sentences explaining your reasoning. (4 points)
Please answer in complete sentences. Will mark brainliest.
The total resistance in the circuit, R is 4 Ω .
The meters connected to the 12 Ω resistance are:
Voltmeter - connected in parallelAmmeter - connected in seriesIf the switch is opened, only the 4 Ω and 2 Ω resistances will stop working.
What is the equivalent or total resistance in the circuit?The resistances in the circuit are connected both in series and in parallel
The resistances in series are the 4 Ω and the 2 Ω resistances.
Equivalent resistance = 4 + 2 = 6 Ω
The 4 Ω and the 2 Ω resistances are then connected in parallel with the 12 Ω resistance.
Total resistance, R is calculated as follows:
1/R = 1/12 + 1/6
1/R = 3/12
R = 12/3
R = 4Ω
The meters connected to the 12 Ω resistance are:
Voltmeter - connected in parallelAmmeter - connected in seriesIf the switch is opened, only the 4 Ω and 2 Ω resistances will stop working because the circuit connecting them to the cell is broken whereas the circuit to the 12 Ω resistance is continuous.
In conclusion, resistances can be connected in parallel or in series.
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