Answer:
Everything in the Universe is made up of matter and energy. Matter is anything that has mass and occupies space. ... Energy is the ability to cause change or do work. Some forms of energy include light, heat, chemical, nuclear, electrical energy and mechanical energy.
Explanation:
If the x-position of a particle is measured with an uncertainty of 1.00×10-10 m, then what is the uncertainty of the momentum in this same direction? (Useful constant: h-bar = 1.05×10-34 Js.)
Answer:
The uncertainty in momentum is 5.25x 10^25Jsm
Explanation:
We know that
h bar = h/2π
So
1.05x 10^34=h/2pπ
h=1.05x 10^ 34(2π)=6.597x 10^-34Js
dp=(6.597x10^-34/4pπ)/(1x10^-10)
=5.25x10^-25 Jsm
An electric device delivers a current of 5.0 A to a circuit. How many electrons flow through this circuit in 5 s?
Answer:
1.6×10²⁰
Explanation:
An ampere is a Coulomb per second.
1 A = 1 C / s
The amount of charge after 5 seconds is:
5.0 A × 5 s = 25 C
The number of electrons is:
25 C × (1 electron / 1.6×10⁻¹⁹ C) = 1.6×10²⁰ electrons
Convert 7,348 grams to kilograms
An electric train operates on 800 V. What is its power consumption when the current flowing through the train's motor is 2,130 A?
Answer:
1704 kWExplanation:
To solve for the power consumed by the trains motor we have to employ the formula for power which is
Power= current * voltage
Given that
voltage V= 800 V
current I= 2130 A
Substituting in the formula for power we have
Power= 2130*800= 1704000 watt
Power = 1704 kW
This is the amount of energy consumed, transferred or converted per unit of time
Hence the power consumed by the trains motor is 1704 kW
Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/2 its former radius. a. What are the pressure and velocity of the water after the contraction
Answer:
Explanation:
Using the Continuity equation
v X A = v' xA'
so if A is 1/2of A' then A velocity must be 2 times the A'
after-contraction v = 2 x 5.0m/s = 10m/s
Using the Bernoulli equation
p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂
, the "h" terms cancel
3.5 x 10^ 5Pa + ½ x 1000kg/m³x (5.0m/s)² = p₂ + ½ x 1000kg/m³ x (10m/s)²
p₂ = 342500pa
Consider a series RLC circuit where R=25.0 Ω, C=35.5 μF, and L=0.0940 H, that is driven at a frequency of 70.0 Hz. Determine the phase angle ϕ of the circuit in degrees.
Answer:
137.69°Explanation:
The phase angle of an RLC circuit ϕ is expressed as shoen below;
ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]
Xc is the capacitive reactance = 1/2πfC
Xl is the inductive reactance = 2πfL
R is the resistance = 25.0Ω
Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz
Xl = 2π * 70*0.0940
Xl = 41.32Ω
For the capacitive reactance;
Xc = 1/2π * 70*35.5*10⁻⁶
Xc = 1/0.0156058
Xc = 64.08Ω
Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]
ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]
[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]
Since tan is negative in the 2nd quadrant;
[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]
Hence the phase angle ϕ of the circuit in degrees is 137.69°
The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°
Phase angle:Given that:
capacitance C = 35.5 μF,
Inductance L = 0.0940 H,
The resistance R = 25.0Ω
and frequency f = 70.0Hz
The capacitive reactance is given by:
Xc = 1/2πfC
Xc = 1/2π × 70 × 35.5× 10⁻⁶
Xc = 1/0.0156058
Xc = 64.08Ω
The inductive reactance is given by:
Xl = 2πfL
Xl = 2π × 70 × 0.0940
Xl = 41.32Ω
The phase angle of an RLC circuit ϕ is given by:
[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]
Ф = -42.31°
Since tan is negative in the 2nd quadrant, thus:
ϕ = 180° - 42.31°
ϕ = 137.69°
Learn more about RLC circuit:
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PLEASE HELP ANSWER FAST As the vibration of molecules decreases, the _____ of the substance decreases. 1.temperature 2.internal energy 3.kinetic energy 4.all of the above
A speeding car has a velocity of 80 mph; suddenly it passes a cop car but does not stop. When the speeding car passes the cop car, the cop immediately accelerates his vehicle from 0 to 90 mph in 4.5 seconds. The cop car has a maximum velocity of 90 mph. At what time does the cop car meet the speeding car and at what distance?
Answer:
Distance= 4 miles
Time = 36.3 seconds
Explanation:
80 mph = 178.95 m/s
90 mph = 201.32 m/s
V = u +at
201.32= 0+a(4.5)
201.32/4.5= a
44.738 m/s² = a
Acceleration of the cop car
= 44.738 m/s²
Distance traveled at 4.5seconds
For the cop car
S= ut + ½at²
S= 0(4.5) + ½*44.738*4.5
S= 100.66 meters
Distance traveled at 4.5seconds
For the speeding car
4.5*178.95=805.275
The cop car will still cover 704.675 +x distance while the speeding car covers for their distance to be equal
X/178.95= (704.675+x)/201.32
X-0.89x= 626.37
0.11x= 626.37
X= 5694.3 meters
The time = 5694.3/178.95
Time =31.8 seconds
So the distance they meet
= 5694.3+805.275
= 6499.575 meters
= 4.0 miles
The Time = 4.5+31.8
Time = 36.3 seconds
an electron travels at 0.3037 times the speed of light through a magnetic field and feels a force of 1.2498 pN. What is the magnetic field in teslas
Answer:
Explanation:
Charge on an electron (q) = 1.6 * 10 ^ -19 C
Velocity of electron (v) = 0.3037 * 300,000,000 = 91,110,000 m/sec
We know that, Force exerted on moving particle moving through a magnetic field :
[tex]F= q * v * B ( q,v\ and\ B\ are\ mutually\ perpendicular)[/tex]
1.2498 * 10 ^ -12 = 1.6 * 10^ -19 * 91110000 * B
B = 0.08573 T
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?
Answer:
6000 counts per secondExplanation:
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;
2000 counts per second = 1 meter ... 1
In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;
x count per second = 3 meter ... 2
Solving the two expressions simultaneously for x we will have;
2000 counts per second = 1 meter
x counts per second = 3 meter
Cross multiply to get x
2000 * 3 = 1* x
6000 = x
This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample
Determine the next possible thickness of the film (in nm) that will provide the proper destructive interference. The index of refraction of the glass is 1.58 and the index of refraction of the film material is 1.48.
Answer:
I know the answer
Explanation:
We want to choose the film thickness such that destructive interference occurs between the light reflected from the air-film interface (call it wave 1) and from the film-lens interface (call it wave 2). For destructive interference to occur, the phase difference between the two waves must be an odd multiple of half-wavelengths.
You can think of the phases of the two waves as second hands on a clock; as the light travels, the hands tick-tock around the clock. Consider the clocks on the two waves in question. As both waves travel to the air-film interface, their clocks both tick-tock the same time-no phase difference. When wave 1 is reflected from the air-film boundary, its clock is set forward 30 seconds; i.e., if the hand was pointing toward 12, it's now pointing toward 6. It's set forward because the index of refraction of air is smaller than that of the film.
Now wave 1 pauses while wave two goes into and out of the film. The clock on wave 2 continues to tick as it travels in the film-tick, tock, tick, tock.... Clock 2 is set forward 30 seconds when it hits the film-lens interface because the index of refraction of the film is smaller than that of the lens. Then as it travels back through the film, its clock still continues ticking. When wave 2 gets back to the air-film interface, the two waves continue side by side, both their clocks ticking; there is no change in phase as they continue on their merry way.
So, to recap, since both clocks were shifted forward at the two different interfaces, there was no net phase shift due to reflection. There was also no phase shift as the waves travelled into and out from the air-film interface. The only phase shift occured as clock 2 ticked inside the film.
Call the thickness of the film t. Then the total distance travelled by wave 2 inside the film is 2t, if we assume the light entered pretty much normal to the interface. This total distance should equal to half the wavelength of the light in the film (for the minimum condition; it could also be 3/2, 5/2, etc., but that wouldn't be the minimum thickness) since the hand of the clock makes one revolution for each distance of one wavelength the wave travels (right?).
Convert 76.2 kilometers to meters?
Answer
76200meters
Explanation:
we know that 1km=1000meters
to convert km into meters we we divide km by meters
=76.2/1000
=76200meters
Two 1.0 nF capacitors are connected in series to a 1.5 V battery. Calculate the total energy stored by the capacitors.
Answer:
1.125×10⁻⁹ J
Explanation:
Applying,
E = 1/2CV²................... Equation 1
Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.
Given; C = 1.0 nF, = 1.0×10⁻⁹ F, V = 1.5 V
Substitute into equation 1
E = 1/2(1.0×10⁻⁹×1.5²)
E = 1.125×10⁻⁹ J
Hence the energy stored by the capacitor is 1.125×10⁻⁹ J
If a disk rolls on a rough surface without slipping, the acceleration of the center of gravity (G) will _ and the friction force will b
Answer:
Will be equal to alpha x r; less than UsN
The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the ground. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held?a. The tube should be held horizontally, parallel to the ground b. The potential difference between the ends of the tube does not depend on the tube's orientation. c. The tube should be held vertically perpendicular to the ground
Answer:
b) True. potencial diferencie does not depend on orientation
Explanation:
In this exercise we are asked to show which statements are true.
The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.
It does not refer to the height of the system.
We can now review the claims
a) False. Potential not to be refers to height
b) True. Does not depend on orientation
c) False The potential does not refer to the altitude but to the Earth's charge
A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 5.0 μg dust particle is suspended in midair just above the center of the carpet.
Required:
What is the charge on the dust particle?
Answer:
The charge on the dust particle is [tex]q_d = 6.94 *10^{-13} \ C[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 2.0 \ m[/tex]
The width is [tex]w = 4.0 \ m[/tex]
The charge is [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]
The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]
Generally the electric field on the carpet is mathematically represented as
[tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]
Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]
[tex]E = -70621.5 \ N/C[/tex]
Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles
[tex]F__{E}} = F__{G}}[/tex]
=> [tex]q_d * E = m * g[/tex]
=> [tex]q_d = \frac{m * g}{E}[/tex]
=> [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]
=> [tex]q_d = 6.94 *10^{-13} \ C[/tex]
A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly above the fish
Answer:
Apparent depth (Da) = 60.15 cm (Approx)
Explanation:
Given:
Distance from fish (D) = 80 cm
Find:
Apparent depth (Da)
Computation:
We know that,
Refractive index of water (n2) = 1.33
So,
Apparent depth (Da) = D(n1/n2)
Apparent depth (Da) = 80 (1/1.33)
Apparent depth (Da) = 60.15 cm (Approx)
The apparent depth of the fish is 60 cm.
To calculate the apparent depth of the fish, we use the formula below.
Formula:
R.F(water) = Real depth(D)/Apparent depth(D')R.F = D/D'.................... Equation 1Where:
R.F = Refractive index of waterMake D' The subject of the equation.
D' = D/R.F................... Equation 2From the question,
Given:
D = 80 cmR.F = 1.333Substitute these values into equation 2
D' = 80/1.33D' = 60.01D' = 60 cmHence, the apparent depth of the fish is 60 cm
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A circular conducting loop of radius 31.0 cm is located in a region of homogeneous magnetic field of magnitude 0.700 T pointing perpendicular to the plane of the loop. the loop is connected in series with a resistor of 265 ohms. The magnetic field is now increased at a constant rate by a factor of 2.30 in 29.0 s.
Calculate the magnitude of induced emf in the loop while the magnetic field is increasing.
With the magnetic field held constant a ts its new value of 1.61 T, calculate the magnitude of its induced voltage in the loop while it is pulled horizontally out of the magnetic field region during a time interval of 3.90s.
Answer:
(a) The magnitude of induced emf in the loop while the magnetic field is increasing is 9.5 mV
(b) The magnitude of the induced voltage at a constant magnetic field is 124.7 mV
Explanation:
Given;
radius of the circular loop, r = 31.0 cm = 0.31 m
initial magnetic field, B₁ = 0.7 T
final magnetic field, B₂ = 2.3B₁ = 2.3 X 0.7 T = 1.61 T
duration of change in the field, t = 29
(a) The magnitude of induced emf in the loop while the magnetic field is increasing.
[tex]E = A*\frac{\delta B}{\delta t} \\\\[/tex]
[tex]E = A*\frac{B_2 -B_1}{\delta t}[/tex]
Where;
A is the area of the circular loop
A = πr²
A = π(0.31)² = 0.302 m²
[tex]E = A*\frac{B_2 -B_1}{\delta t} \\\\E = 0.302*\frac{1.61-0.7}{29} \\\\E = 0.0095 \ V\\\\E = 9.5 \ mV[/tex]
(b) the magnitude of the induced voltage at a constant magnetic field
E = A x B/t
E = (0.302 x 1.61) / 3.9
E = 0.1247 V
E = 124.7 mV
Therefore, the magnitude of the induced voltage at a constant magnetic field is 124.7 mV
Without actually calculating any logarithms, determine which of the following intervals the sound intensity level of a sound with intensity 3.66×10^−4W/m^2 falls within?
a. 30 and 40
b. 40 and 50
c. 50 and 60
d. 60 and 70
e. 70 and 80
f. 80 and 90
g. 90 and 100
Answer:
f. 80 and 90
Explanation:
1 x 10⁻¹² W/m² sound intensity falls within 0 sound level
1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level
1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level
1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level
1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level
1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level
1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level
1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level
1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level
1 x 10⁻³ W/m² sound intensity falls within 90 sound level
Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.
f. 80 and 90
front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the rear tire
Answer:
The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.
Explanation:
This question suggests that the car is accelerating forward. Thus, the easiest way for us to know what friction is doing is for us to know what happens when we turn friction off.
Now, if there is no friction and the car is stopped, if we push down on the accelerator, it will make the front wheels to spin in a clockwise manner. This spin occurs on the frictionless surface with the rear wheels doing nothing while the car doesn't move.
Now, if we apply friction to just the front wheels, the car will accelerate forward while the back wheels would be dragging along the road and not be spinning. Thus, friction opposes the motion and as such, it must act im a direction opposite to where the car is going. This must be static friction.
The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.
Which of these cannot be a resistor in a series or parallel circuit?
A)switch
B) battery
C) light bulb
D) all of these are resistors
Answer:
it is going to D. all of these are resistors
A baseball (m=145g) traveling 35 m/s moves a fielder's glove backward 23 cm when the ball is caught. What was the average force exerted by the ball on the glove?
Answer:
386.13 N
Explanation:
The kinetic energy of the baseball is converted into workdone in moving the glove backward( work energy theorem).
Therefore, KE of the ball
[tex]\frac{1}{2} mv^2 =\frac{1}{2}(0.145)35^2\\ = 88.81 \text{J}[/tex]
Now, workdone in moving the glove
W= Fd
where F = Force applied, d = displacement of the glove= 0.23 cm.
88.81 = F×0.23
F= 88.81/0.23 = 386.13 N
Astronomers think planets formed from interstellar dust and gases that clumped together in a process called? A. stellar evolution B. nebular aggregation C. planetary accretion D. nuclear fusion
Answer:
C. planetary accretion
Explanation:
Astronomers think planets formed from interstellar dust gases that clumped together in a process called planetary accretion.
Answer:
[tex]\boxed{\sf C. \ planetary \ accretion }[/tex]
Explanation:
Astronomers think planets formed from interstellar dust and gases that clumped together in a process called planetary accretion.
Planetary accretion is a process in which huge masses of solid rock or metal clump together to produce planets.
A golfer hits a 42 g ball, which comes down on a tree root and bounces straight up with an initial speed of 15.6 m/s. Determine the height the ball will rise after the bounce. Show all your work.
Answer:
12.2 m
Explanation:
Given:
v₀ = 15.6 m/s
v = 0 m/s
a = -10 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (15.6 m/s)² + 2 (-10 m/s²) Δy
Δy = 12.2 m
[tex] \LARGE{ \boxed{ \rm{ \green{Answer:}}}}[/tex]
Given,
The initial speed is 15.6 m/s The mass of the ball is 42g = 0.042kgFinding the initial kinetic energy,
[tex]\large{ \boxed{ \rm{K.E. = \frac{1}{2}m {v}^{2}}}}[/tex]
⇛ KE = (1/2)mv²
⇛ KE = (1/2)(0.042)(15.6)²
⇛ KE = 5.11 J
|| ⚡By conservation of energy, the potential energy at the highest point will also be 5.11 J, since there is no kinetic energy at the highest point because the ball is not moving (we neglect energy lost due to air resistance, heat, sound, etc.) ⚡||
So, we have:
[tex] \large{ \boxed{ \rm{P.E. = mgh}}}[/tex]
⇛ h = PE/(mg)
⇛ h = 5.11 J /(0.042 × 9.8)
⇛ h = 12.41 m
✏The ball will rise upto a height of 12.41 m
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If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself.
Required:
a. What is its frequency?
b. What type of electromagnetic radiation might this be?
Answer:
a) 5.5×10^17 Hz
b) visible light
Explanation:
Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;
λ= 5.5 × 10^-10 m
Since;
c= λ f and c= 3×10^8 ms-1
f= c/λ
f= 3×10^8/5.5 × 10^-10
f= 5.5×10^17 Hz
The electromagnetic wave is visible light
Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 2.2% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of the galaxy relative to the earth
Answer:
Vrel= 0.75c
Explanation:
See attached file
Which of the following explains why a “control” is important in a case-control study of a disease? The researchers need to control the bias that those who contracted the disease may create when they talk to others. The researchers need to compare those who contracted the disease to those who did not. The researchers need to compare those who contracted the disease to those who contracted previous diseases. The researchers need to control the disease so that it is not spread further.
The researchers need to compare those who contracted the disease to those who did not.
Which is one criterion that materials of a technological design should meet? They must be imported. They must be affordable. They must be naturally made. They must be locally produced.
Answer:
they must be affordable because they have to pay for it or they wont get the stuff they are bying.
Explanation:
need a brainliest please.
Answer: B, they must be affordable.
Explanation:
The ancient Greek Eratosthenes found that the Sun casts different lengths of shadow at different points on Earth. There were no shadows at midday in Aswan as the Sun was directly overhead. 800 kilometers north, in Alexandria, shadow lengths were found to show the Sun at 7.2 degrees from overhead at midday. Use these measurements to calculate the radius of Earth.
Answer:
The radius of the earth is [tex]r = 6365.4 \ km[/tex]
Explanation:
From the question we are told that
The distance at Alexandria is [tex]d_a = 800 \ km = 800 *10^{3} \ m[/tex]
The angle of the sun is [tex]\theta = 7.2 ^o[/tex]
So we want to first obtain the circumference of the earth
So let assume that the earth is circular ([tex]360 ^o[/tex])
Now from question we know that the sun made an angle of [tex]7.2 ^o[/tex] so with this we will obtain how many [tex](7.2 ^o)[/tex] are in [tex]360^o[/tex]
i.e [tex]N = \frac{360}{7.2}[/tex]
=> [tex]N = 50[/tex]
With this value we can evaluate the circumference as
[tex]c = 50 * 800[/tex]
[tex]c = 40000 \ km[/tex]
Generally circumference is mathematically represented as
[tex]c = 2\pi r[/tex]
[tex]40000 = 2 * 3.142 * r[/tex]
=> [tex]r = 6365.4 \ km[/tex]
An electric heater draws 13 amperes of current when connected to 120 volts. If the price of electricity is $0.10/kWh, what would be the approximate cost of running the heater for 8 hours?
(A) $0.19
(B) $0.29
(C) $0.75
(D) $1.25
(E) $1.55
Answer:
C $0.75 my friend I wish it is right answer
