Answer:
1,860 g
Explanation:
In a system, the coefficient of static friction is usually higher than the coefficient of kinetic friction. This means that the kinetic friction is usually less than the static friction. From the question, since the book is already sliding, it means that kinetic friction is the friction in play. This means that before the reading on the scale that could have been possible at the moment the student overcame the static friction must be greater than the reading on the scale during sliding. The only option above 1580 g is 1860 g
An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. However, an observer moving at high speed parallel to one of the object's edges and knowing that the object's mass is 2.0 kg determines its density to be 7300 kg/m3, which is much greater than the density of glass. What is the moving observer's speed (in units of c) relative to the cube
Answer:
The velocity is [tex]v = 2.6*10^{8} \ m/s[/tex]
Explanation:
From the question we are told that
The side of the cube is [tex]l = 0.13 \ m[/tex]
The mass of the object is [tex]m = 2.0 \ kg[/tex]
The density of the object is [tex]\rho = 7300 \ kg / m^3[/tex]
Generally the volume of the object according to the moving observer is mathematically represented as
[tex]V =\frac{m}{\rho}[/tex]
[tex]V =\frac{2}{7300}[/tex]
[tex]V = 2.74*10^{-4} \ m^3[/tex]
Therefore the length of the side as observed by the observer on high speed is mathematically represented as
[tex]L = \sqrt[3]{V}[/tex]
[tex]L = \sqrt[3]{2.74 *10^{-4}}[/tex]
[tex]L =0.065[/tex]
Now the original length of side is mathematically represented as
[tex]L= l * \sqrt{ (1 - ( \frac{ v}{c})^2 )}[/tex]
Where c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
So
[tex]v = \sqrt{1 - [\frac{L}{l}]^2} * c[/tex]
=> [tex]v = \sqrt{1 - [\frac{0.065}{0.13}]^2} * c[/tex]
=> [tex]v = 2.6*10^{8} \ m/s[/tex]
A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)sin[(500rad/s)t]. What is the voltage across the resistor at 2.09 x 10-3 s? Group of answer choices 205 V 515 V 636 V 542 V
Answer:
205 V
V[tex]_{R}[/tex] = 2.05 V
Explanation:
L = Inductance in Henries, (H) = 0.500 H
resistor is of 93 Ω so R = 93 Ω
The voltage across the inductor is
[tex]V_{L} = - IwLsin(wt)[/tex]
w = 500 rad/s
IwL = 11.0 V
Current:
I = 11.0 V / wL
= 11.0 V / 500 rad/s (0.500 H)
= 11.0 / 250
I = 0.044 A
Now
V[tex]_{R}[/tex] = IR
= (0.044 A) (93 Ω)
V[tex]_{R}[/tex] = 4.092 V
Deriving formula for voltage across the resistor
The derivative of sin is cos
V[tex]_{R}[/tex] = V[tex]_{R}[/tex] cos (wt)
Putting V[tex]_{R}[/tex] = 4.092 V and w = 500 rad/s
V[tex]_{R}[/tex] = V[tex]_{R}[/tex] cos (wt)
= (4.092 V) (cos(500 rad/s )t)
So the voltage across the resistor at 2.09 x 10-3 s is which means
t = 2.09 x 10⁻³
V[tex]_{R}[/tex] = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))
= (4.092 V) (cos (500 rads/s)(0.00209))
= (4.092 V) (cos(1.045))
= (4.092 V)(0.501902)
= 2.053783
V[tex]_{R}[/tex] = 2.05 V
Can a person break a wall with a punch?(I mean technically,maybe with inhuman speed)
Answer:
yes, shaolin monks can do. you can find tutorials on Youtb
A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.5 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation?
Answer:
0.99Hz
Explanation:
Using F= -mx ( spring force)
At equilibrium the gravitational force will be balanced by the spring force so mg= kx
K= mg/ 0.25 N/m
But
Frequency f= 1/2pi √g/0.25
Frequency is 0.99Hz
The block is pulled down slightly and released so, Frequency of oscillation is 3.15 Hz
Frequency of oscillation based problem:What information do we have?
Length starched = 2.5 cm
F = Kx
We know that
F = mg
So,
mg = Kx
K/m = g/x
[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{x} }\\f=\frac{1}{2\pi}\sqrt{\frac{9.8}{0.025} }[/tex]
Frequency of oscillation = 3.15 Hz
Find out more information about 'Oscillation'.
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Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen if the spacing between the first and second dark fringes is to be 4.2 mm?
Answer:
The distance is [tex]D = 2.6 \ m[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 476.1 \ nm = 476.1 *10^{-9} \ m[/tex]
The distance between the slit is [tex]d = 0.29 \ mm = 0.29 *10^{-3} \ m[/tex]
The between the first and second dark fringes is [tex]y = 4.2 \ mm = 4.2 *10^{-3} \ m[/tex]
Generally fringe width is mathematically represented as
[tex]y = \frac{\lambda * D }{d}[/tex]
Where D is the distance of the slit to the screen
Hence
[tex]D = \frac{y * d}{\lambda }[/tex]
substituting values
[tex]D = \frac{ 4.2 *10^{-3} * 0.29 *10^{-3}}{ 476.1 *10^{-9} }[/tex]
[tex]D = 2.6 \ m[/tex]
15. Food chain always start with
a. Photosynthesis
Decay
b. Respiration
d. N2 Fixation
C.Photosynthesis
A 5.0-µC point charge is placed at the 0.00 cm mark of a meter stick and a -4.0-µC point charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field due to these charges equal to zero?
Answer:
Electric field is zero at point 4.73 m
Explanation:
Given:
Charge place = 50 cm = 0.50 m
change q1 = 5 µC
change q2 = 4 µC
Computation:
electric field zero calculated by:
[tex]E1 =k\frac{q1}{r^2} \\\\E2 =k\frac{q2}{R^2} \\\\[/tex]
Where electric field is zero,
First distance = x
Second distance = (x-0.50)
So,
E1 = E2
[tex]k\frac{q1}{r^2}=k\frac{q2}{R^2} \\\\[/tex]
[tex]\frac{5}{x^2}=\frac{4}{(x-50)^2} \\\\[/tex]
x = 0.263 or x = 4.73
So,
Electric field is zero at point 4.73 m
A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0 m/s (about 22 mi/h). Just as the motorist passes, a police officer on a motorcycle stopped at the comer starts off in pursuit with constant acceleration of 3.0 m/S2. (a) How much time elapses before the officer catches up with the motorist? (b) What is the officer's speed at that point? (c) What is the total distance each vehicle has traveled at that point? Please help me
Answer:
(a) 10 s
(b) 30 m/s
(c) 150 m
Explanation:
The motorist's position at time t is:
x = 15t
The officer's position at time t is:
x = ½ (3) t² = 1.5 t²
(a) When they have the same position, the time is:
15t = 1.5 t²
t = 0 or 10 s
(b) The officer's speed is:
v = 3t
v = 30 m/s
(c) The position is:
x = 15t = 150 m
I MIND TRICK PLZ HELP LOL
Troy and Abed are running in a race. Troy finishes the race in 12 minutes. Abed finishes the race in 7 minutes and 30 seconds. If Troy is running at an average speed of 3 miles per hour and speed varies inversely with time, what is Abed’s average speed for the race?
Answer:
Explanation:
Let the race be of a fixed distance x
[tex]Average Speed = \frac{Total Distance}{Total Time}[/tex]
Troy's Average speed = 3 miles/hr = x / 0.2 hr
x = 0.6 miles
Abed's Average speed = 0.6 / 0.125 = 4.8 miles/hr
A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force
Complete question is;
A force stretches a wire by 0.60 mm. A second wire of the same material has the same cross section and twice the length.
a) How far will it be stretched by the same force?
b) A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force?
Answer:
0.15 mm
Explanation:
According to Hooke's Law,
E = Stress(σ)/Strain(ε)
Where E is youngs modulus
Formula for stress is;
Stress(σ) = Force(F)/Area(A)
Formula for strain is;
Strain(ε) = Change in length/original length = (Lf - Li)/Li
We are also told that a second wire of the same material has the same cross section and twice the length.
Thus;
Rearranging Hooke's Law to get the constants on one side, we have;
F/(AE) = ε
Thus from the conditions given;
ε1 = 0.6/Li
ε2 = (Change in length)/(2*Li)
And ε1 = ε2
Thus;
0.6/Li = Change in length/(2*Li)
Li will cancel out and we now have;
Change in length = 2 × 0.6 = 1.2 mm
Finally, we are told A third wire of the same material has the same length and twice the diameter as the first.
Area of a circle;A1 = πd²/4
Now, we are told d is doubled.
Thus, new area of the new circle is;
A2 = π(2d)²/4 = πd²
Rearranging Hooke's Law,we have;
F/A = εE
Since F and E are now constants, we have;
F/E = constant = Aε
Thus;
A1(ε1) = A2(ε2)
A1 = πd²/4
e1 = 0.60/Li
A2 = πd²
e2 = Change in length/Li
Thus;
((πd²/4) × 0.6)/Li = (πd² × Change in length)/ Li
Rearranging, Li and πd² will cancel out to give;
0.6/4 = Change in length
Change in length = 0.15 mm
A small helium-neon laser emits red visible light with a power of 5.40 mW in a beam of diameter 2.30 mm.
Required:
a. What is the amplitude of the electric field of the light? Express your answer with the appropriate units.
b. What is the amplitude of the magnetic field of the light?
c. What is the average energy density associated with the electric field? Express your answer with the appropriate units.
d. What is the average energy density associated with the magnetic field? Express your answer with the appropriate units.
E) What is the total energy contained in a 1.00-m length of the beam? Express your answer with the appropriate units.
Answer:
A. 990v/m
B.330x10^-8T
C.2.19x10^-6J/m³
D.1.45x10^-11J
Explanation:
See attached file
light of wavelength 550 nm is incident on a diffraction grating that is 1 cm wide and has 1000 slits. What is the dispersion of the m = 2 line?
Answer:
The dispersion is [tex]D = 2.01220 *10^{5} \ rad/m[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 550 \ = 550 *10^{-9} \ n[/tex]
The width of the grating is[tex]k = 1\ cm = 0.01 \ m[/tex]
The number of slit is N = 1000 slits
The order of the maxima is m = 2
Generally the spacing between the slit is mathematically represented as
[tex]d = \frac{k}{N}[/tex]
substituting values
[tex]d = \frac{ 0.01}{1000}[/tex]
[tex]d = 1.0 *10^{-5} \ m[/tex]
Generally the condition for constructive interference is
[tex]d\ sin(\theta ) = m * \lambda[/tex]
substituting values
[tex]1.0 *10^{-5} sin (\theta) = 2 * 550 *10^{-9}[/tex]
[tex]\theta = sin^{-1} [\frac{ 2 * 550 *10^{-9}}{ 1.0 *10^{-5}} ][/tex]
[tex]\theta = 6.315^o[/tex]
Generally the dispersion is mathematically represented as
[tex]D = \frac{ m }{d cos(\theta )}[/tex]
substituting values
[tex]D = \frac{ 2 }{ 1.0 *10^{-5} cos(6.315 )}[/tex]
[tex]D = 2.01220 *10^{5} \ rad/m[/tex]
what is the average flow rate in of gasoline to the engine of a plane flying at 700 km/h if it averages 100.0 km/l
Answer:
1.94cm³/s
Explanation:
1L = 1000cm³
Ihr = 3600s
So
Using
Average flow rate
Fr= 1L/100Km x 700Km/1hr x 1hr/3600s x 1000cm³/ 1L
= 1.94cm³/s
The sentence, "The popcorn kernels popped twice as fast as the last batch," is a(n) _____. experiment hypothesis observation control
The correct answer is C. Observation
Explanation:
An observation is a statement a describes a phenomenon, which is the result of measuring the phenomenon or using the senses to collect information about it. Additionally, observations are part of the Scientific method because through observations it is possible to understand phenomena.
The sentence presented is an observation because this statement is the result of the researcher observing or measuring how fast kernels pops, which means the statement derives from studying a phenomenon. Also, this cannot be classified as a hypothesis because a hypothesis is a probable explanation, and it cannot be classified as an experiment because the experiment is the general method to prove or disprove a hypothesis.
An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase the magnitude of the induced emf? A. Straighten the wire out to be flat B. Reduce the resistance of the wire of which the loop is made C. Turning the plane of the loop to be parallel to the magnetic field D. Reducing the diameter of the loop
Answer:
changing the magnetic field more rapidly
Explanation:
According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.
Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.
Five wheels are connected as shown in the figure. Find the velocity of the block “Q”, if it is known that: RA= 5 [m], RB= 10 [m], RD= 6 [m], RE=12 [m].
Answer:
-5 m/s
Explanation:
The linear velocity of B is equal and opposite the linear velocity of E.
vB = -vE
vB = -ωE rE
10 m/s = -ωE (12 m)
ωE = -0.833 rad/s
The angular velocity of E is the same as the angular velocity of D.
ωE = ωD
ωD = -0.833 rad/s
The linear velocity of Q is the same as the linear velocity of D.
vQ = vD
vQ = ωD rD
vQ = (-0.833 rad/s) (6 m)
vQ = -5 m/s
Do an Internet search to determine what minerals are extracted from the ground in order to manufacture the following products:
a. Stainless steel utensils
b. Cat litter
c. Tums brand antacid tablets
d. Lithium batteries
e. Aluminum beverage cans
Answer:
Raw materials are most times gotten from the earth through various forms of extraction procedures.
A) Stainless steel utensils is made up of mainly Iron and other elements such as chromium , carbon etc.
B) Cat litter comprises of ceramic products which is made up of clay.
C) Tums brand antacid tablets comprises of calcium carbonate, magnesium hydroxide, aluminum hydroxide and sodium bicarbonate which could be extracted from the earth.
D)Lithium batteries are made up of elements in the earth such as lithium and carbon.
E)Aluminum beverage cans are made up of aluminum extracted from the ground.
The elastic portion of the downward-sloping straight-line demand curve lies:_______
a. at the intersection with the supply curve.
b. anywhere to the right of the current market price.
c. above the point of unit elasticity.
d. below the point where total revenue is maximized.
Answer:
c. above the point of unit elasticity.
Explanation:
The elastic portion of the downward-sloping straight-line demand curve lies above the point of unit elasticity. Supply and demand are fundamental concept in economics. The demand curve shows how much of a good people will want at a different prices. The demands curves illustrates the intuition why people purchase a good for a lower price. For the demand curve, the price is always shown on the vertical axis and the demand curve is shown on the horizontal axis. Thus , the quantity demanded increases as the price gets lower. However, the price elasticity of the demand curve varies along the demand curve. This is because there is a key distinction between the gradient and the elasticity. The gradient which is the slope of the line is always the same in the demand curve but elasticity of the demand changes in the percentage of the quantity demand. Therefore, elasticity will vary along the downward-sloping straight - line demand curve. So, in a downward-sloping straight-line demand curve, the elastic portion is usually above the point of unit elasticity
1) True or False:
Atomic mass number is the number of neutrons and protons.
2) True or False:
Fe (iron) has 26 protons. Hint: protons equal what number?
3) True or False:
A photon of infrared light has less energy than a photon of red light.
Answer:
1.true
Explanation:
Answer:
1. True
2. True
3. True
Explanation for Question 1.
A nucleus consists of a bunch of protons and neutrons; these are known as nucleons. The atomic mass number, which is the total number of nucleons;
So, this sentence says that the atomic number is the number of protons and neutrons.
Explanation for Question 2.
Iron has 26 protons.
The number of protons = the atomic number.
So, the atomic number should be 26 also,
When we see the periodic table, Iron's atomic number is 26, so the statement is true.
Explanation for Question 3.
Red photons of light carry about 1.8 electron volts of energy. Infrared radiation has longer waves than red light, and thus oscillates at a lower frequency and carries less energy.
So, the above statement proves that the photon of infrared light has less energy than the photon of red light.
An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance of 1 cm and there is vacuum in the region between the plates. The electron is initially found midway between the plates with a kinetic energy of 11.2 eV and with its velocity directed toward the negative plate. How close to the negative plate will the electron get if the potential difference between the plates is 100 V? (1 eV = 1.6 x 10-19 J)
Answer:
The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.
Explanation:
Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:
[tex]F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}[/tex]
and since the electric field E in between parallel plates separated a distance d and under a potential difference [tex]\Delta V[/tex], is given by:
[tex]E=\frac{\Delta\,V}{d}[/tex]
then :
[tex]a=\frac{q\,\Delta V}{m\,d}[/tex]
We want to find when the particle reaches velocity zero via kinematics:
[tex]v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a[/tex]
We replace this time (t) in the kinematic equation for the particle displacement:
[tex]\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}[/tex]
Replacing the values with the information given, converting the distance d into meters (0.01 m), using [tex]\Delta V=100\,V[/tex], and the electron's kinetic energy:
[tex]\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J[/tex]
we get:
[tex]\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters[/tex]Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:
0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Answer:
[tex]I=2.71\times 10^{-5}\ A[/tex]
Explanation:
A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Let given is,
The diameter of a parallel plate capacitor is 6 cm or 0.06 m
Separation between plates, d = 0.046 mm
The potential difference across the capacitor is increasing at 500,000 V/s
We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :
[tex]C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}[/tex], r is radius
Let I is the displacement current. It is given by :
[tex]I=C\dfrac{dV}{dt}[/tex]
Here, [tex]\dfrac{dV}{dt}[/tex] is rate of increasing potential difference
So
[tex]I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A[/tex]
So, the value of displacement current is [tex]2.71\times 10^{-5}\ A[/tex].
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.165×105 km and the planet has a radius of 4175 km and a mass of 6.70×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N·m2/kg2 .
Answer:
364days
Explanation:
Pls see attached file
Explanation:
The moon will take 112.7 days to make one revolution around the planet.
What is Kepler's third law?The period of the satellite around any planet only depends upon the distance between the planet's center and satellite and also depends upon the planet's mass.
Given, the distance from the moon's center to the planet's surface,
h = 2.165 × 10⁵ km,
The radius of the planet, r = 4175 km
The mass of the planet = 6.70 × 10²² kg
The total distance between the moon's center to the planet's center:
a = r +h = 2.165 × 10⁵ + 4175
a = 216500 + 4175
a = 220675
a = 2.26750 × 10⁸ m
The period of the planet can be calculated as:
[tex]T =2\pi \sqrt{\frac{a^3}{Gm} }[/tex]
[tex]T =2\3\times 3.14 \sqrt{\frac{(2.20675 \times 10^8)^3}{(6.67\times 10^{-11}).(6.70\times 10^{22})} }[/tex]
T = 9738253.26 s
T = 112.7 days
Learn more about Kepler's law, here:
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A foot is 12 inches and a mile is 5280I ft exactly. A centimeter is exactly 0.01m or mm. Sammy is 5 feet and 5.3tall. what is Sammy's height in inches?
Answer:
65.3 Inches tall
Explanation:
If Sammy is 5 feet and 5.3 inches tall, we simply need to convert the feet to inches, and sum the remaining inches from his height to determine his overall height in inches.
So, 5 feet = (12 inches/1foot) * (5 feet) = 60 inches
And 60 inches + 5.3 inches = 65.3 inches.
Hence, Sammy is 65.3 inches tall.
Cheers.
The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature in degrees Celsius when 1.50 kcal of heat enters 1.50 kg of the following, originally at 15.0°C.(a) water
(b) concrete
(c) steel
(d) mercury
Answer:
Final temperature Water = 20.99-degree celsius
Final temperature Concrete = 24.98 degree celsius
Final temperature Steel = 50.1 degree celsius
Final temperature Mercury = 29.26 degree celsius
Explanation:
Given the mass of each substance = 1.50 kg
Ti = 15
Q = 1.5 kcal = 6276 joule
We have to use the heat capacity of each object so find the heat capacity from the table.
Heat capacity of water = 4186 J/kg degree celsius.
Heat capacity of concrete = 840 J/kg degree celsius.
Heat capacity of steel = 452 J/kg degree celsius.
Heat capacity of mercury = 139 J/kg degree celsius.
Use the below formula to find the final temperature.
[tex]T_f = T_i + \frac{Q}{mc_w} \\[/tex]
[tex]\text{Temperature in the case of water.} \\= 20 + \frac{6276}{1.5 \times 4186 } \\= 20.99 \ degree \ celsius \\\text{Temperature in the case of concrete.} \\= 20 + \frac{6276}{1.5 \times 840 } \\= 24.98 \ degree \ celsius \\\text{Temperature in the case of steel.} \\= 20 + \frac{6276}{1.5 \times 452 } \\= 29.26 \ degree \ celsius \\\text{Temperature in the case of mercury.} \\= 20 + \frac{6276}{1.5 \times 139 } \\= 50.1 \ degree \ celsius \\[/tex]
Explain why the two plates of a capacitor are charged to the same magnitude when a battery is connected to the capacitor.
Answer:
This is because the same electron removed from the positively charged plate is what is taken to the negatively charged plate, maintaining the same amount of electron according to the conservation of charge in an electric circuit.
Explanation:
In any circuit, electrons are neither created nor destroyed according to the laws of conservation of charge, but are transferred from one point to another on the circuit. When the plates of a capacitor are connected to battery, the battery pushes the electron to move due to its potential difference. Electrons are then moved from the positive plate, at a steady rate, to the negative plate. The removal of electrons from the positive plate is what leaves it positively charged from deficiency of electrons, and the addition of electrons at the negatively charged plate is what leaves the plate negatively charge from excess of electrons. From this, we can see that the same electrons removed from the positively charged plate are taken to the negatively charged plate.
Which best identifies the requirements for work to be performed? an object that has a force acting on it an object that is moving and has no net force a force acting on a motionless object a force that moves an object
Answer:
a force that moves an object
Explanation:
the formula for work is force * distance
This question involves the concepts of work, force, and displacement.
The statement that best identifies the requirements for work to be performed is "a force that moves an object".
Work is defined as the product of force applied on an object and the distance moved by the object. Mathematically,
Work = (Force)(Displacement)
Hence, both the applied force and the displacement of the object as a result of the application of the force is necessary for the work to be done. If any one of these values becomes zero, the work automatically becomes zero, which means no work is performed.
Learn more about work here:
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What type of energy conversion occurs when you place your feet near the fireplace and they become warm?
O Radiant to thermal
o Thermal to mechanical
O Mechanical to chemical
O Nuclear to thermal
Answer:
It is the first one RADIENT TO THERMAL
Explanation:
The heat emitted from the campfires is an an example of radiant energy and thermal energy is refers to the energy contained within a system that is responsible for its tempreture with in this case is the campfires and heat energy being reflected upon your feet.
Answer:
A
Explanation:
Choose only one correct option. Explanation needed.
Answer:
[tex]\large \boxed{\mathrm{C. \ \ \frac{500}{7 \times 15 \times 8} \ g/cm^3 }}[/tex]
Explanation:
[tex]\displaystyle \sf Density = \frac{mass}{volume}[/tex]
[tex]\displaystyle \rho = \frac{m}{V}[/tex]
[tex]\sf mass=500 \ g[/tex]
[tex]\sf volume \ of \ a \ cuboid=width \times length \times height=( 7 \times 15 \times 8) \ cm^3[/tex]
[tex]\displaystyle \rho = \frac{500}{7 \times 15 \times 8}[/tex]
Take an electric field sensor and move it in a straight line, crossing the equipotential lines. Describe the relationship between the distance between the equipotential lines and the strength of the electric field.
Answer:
E = - dV / dx
Explanation:
The equipotential lines are lines or surfaces that have the same power, therefore we can move in them without carrying out work between equipotential lines, work must be carried out, therefore the electric field changes.
The electric field and the potential are related by
E = - dV / dx
therefore when the change is faster, that is, the equipotential lines are closer, the greater the electric field must be.
A thin film of soap with n = 1.37 hanging in the air reflects dominantly red light with λ = 696 nm. What is the minimum thickness of the film?
Answer:
The thickness is [tex]t = 1.273 *10^{-7} \ m[/tex]
Explanation:
From the question we are told that
The refractive index of the film is [tex]n = 1.37[/tex]
The wavelength is [tex]\lambda = 696 \ nm = 696 *10^{-9 } \ m[/tex]
Generally the condition for constructive interference in a film is mathematically represented as
[tex]2 * t = [m + \frac{1}{2} ] \lambda_k[/tex]
Here t is the thickness of the film , m is the order number (0, 1, 2, 3 ... )
[tex]\lambda _k[/tex] is the wavelength of light that is inside the film , this is mathematically evaluated as
[tex]\lambda _k = \frac{ \lambda }{ n}[/tex]
[tex]\lambda _k = \frac{ 696 *10^{-9}}{ 1.37}[/tex]
[tex]\lambda _k = 5.095 *10^{-7 } \ m[/tex]
So for m = 0
[tex]t = [ 0 + \frac{1}{2} ] \lambda _k * \frac{1}{2}[/tex]
substituting values
[tex]t = [ 0 + \frac{1}{2} ] (5.095 *10^{-7}) * \frac{1}{2}[/tex]
[tex]t = 1.273 *10^{-7} \ m[/tex]
