Answers
Answer:
K = 227.04 W/m.°C
Explanation:
First we need to find the heat required to melt the ice:
q = m H
where,
q = heat required = ?
m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg
H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg
Therefore,
q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)
q = 2839 J
Now, we find the heat transfer rate through rod:
Q = q/t
where,
t = time = (10 min)(60 s/1 min) = 600 s
Q = Heat Transfer Rate = ?
Therefore,
Q = 2839 J/600 s
Q = 4.73 W
From Fourier's Law of Heat Conduction:
Q = KA ΔT/L
where,
K = Thermal Conductivity = ?
A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²
L = Length of rod = 60 cm = 0.6 m
ΔT = Difference in temperature = 100°C - 0°C = 100°C
Therefore,
4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m
K = (4.73 W)/(0.0208 m.°C)
K = 227.04 W/m.°C
Related Questions
The block moves up an incline with constant speed. What is the total work WtotalWtotalW_total done on the block by all forces as the block moves a distance LLL
Answers
Answer:
External force W₁ = F L
Friction force W₂ = - fr L
weight component W₃ = - mg sin θ L
Y Axis Force W=0
Explanation:
When the block rises up the plane with constant velocity, it implies that the sum of the forces is zero.
For these exercises it is indicated to create a reference system with the x axis parallel to the plane and the y axis perpendicular
let's write the equations of translational equilibrium in given exercise
X axis
F - fr -Wₓ = 0
F = fr + Wₓ
the components of the weight can be found using trigonometry
Wₓ = W sin θ
[tex]W_{y}[/tex] = W cos θ
let's look for the work of these three forces
W = F x cos θ
External force
W₁ = F L
since the displacement and the force have the same direction
Friction force
W₂ = - fr L
since the friction force is in the opposite direction to the displacement
For the weight component
W₃ = - mg sin θ L
because the weight component is contrary to displacement
Y Axis
N- Wy = 0
in this case the forces are perpendicular to the displacement, the angle is 90º and the cosine 90 = 0
therefore work is worth zero
Rod cells in the retina of the eye detect light using a photopigment called rhodopsin. 1.8 eV is the lowest photon energy that can trigger a response in rhodopsin. Part A What is the maximum wavelength of electromagnetic radiation that can cause a transition
Answers
Answer:
The maximum wavelength of the e-m wave is 6.9 x 10^-7 m
Explanation:
Energy required to trigger a response = 1.8 eV
we convert to energy in Joules.
1 eV = 1.602 x 10^-19 J
1.8 eV = [tex]x[/tex] J
[tex]x[/tex] = 1.8 x 1.602 x 10^-19 = 2.88 x 10^-19 J
The energy of an electromagnetic wave is gotten as
E = hf
where
h is the Planck's constant = 6.63 x 10^-34 J-s
and f is the frequency of the wave.
substituting values, we have
2.88 x 10^-19 = 6.63 x 10^-34 x f
f = (2.88 x 10^-19)/(6.63 x 10^-34)
f = 4.34 x 10^14 Hz
We know that the frequency of an e-m wave is given as
f = c/λ
where
c is the speed of light = 3 x 10^8 m/s
λ is the wavelength of the e-m wave
From this we can say that
λ = c/f
λ = (3 x 10^8)/(4.34 x 10^14)
λ = 6.9 x 10^-7 m
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
Answers
Answer:
C $0.75 my friend I wish it is right answer
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.
Answers
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:
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.
Answers
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?).
If a convex lens were made out of very thin clear plastic filled with air, and were then placed underwater where n = 1.33 and where the lens would have an effective index of refraction n = 1, the lens would act in the same way
a. as a flat refracting surface between water and air as seen from the water side.
b. as a concave mirror in air.
c. as a concave lens in air.
d. as the glasses worn by a farsighted person.
e. as a convex lens in air.
Answers
Answer:
D. A convex lens in air
Explanation:
This is because the air tight plastic under water will reflect light rays in the same manner as a convex lens
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?
Answers
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]
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
Answers
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
A 137 kg horizontal platform is a uniform disk of radius 1.53 m and can rotate about the vertical axis through its center. A 68.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 25.9 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.
Answers
Answer:
The moment of inertia is [tex]I= 312.09 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The mass of the platform is m = 137 kg
The radius is r = 1.53 m
The mass of the person is [tex]m_p = 68.7 \ kg[/tex]
The distance of the person from the center is [tex]d_c =1.19 \ m[/tex]
The mass of the dog is [tex]m_d = 25.9 \ kg[/tex]
The distance of the dog from the person [tex]d_d = 1.45 \ m[/tex]
Generally the moment of inertia of the system is mathematically represented as
[tex]I = I_1 + I_2 + I_3[/tex]
Where [tex]I_1[/tex] is the moment of inertia of the platform which mathematically represented as
[tex]I_1 = \frac{m * r^2}{2}[/tex]
substituting values
[tex]I_1 = \frac{ 137 * (1.53)^2}{2}[/tex]
[tex]I_1 = 160.35 \ kg\cdot m^2[/tex]
Also [tex]I_2[/tex] is the moment of inertia of the person about the axis which is mathematically represented as
[tex]I_2 = m_p * d_c^2[/tex]
substituting values
[tex]I_2 = 68.7 * 1.19^2[/tex]
[tex]I_2 = 97.29 \ kg \cdot m^2[/tex]
Also [tex]I_3[/tex] is the moment of inertia of the dog about the axis which is mathematically represented as
[tex]I_3 = m_d * d_d^2[/tex]
substituting values
[tex]I_3 = 25.9 * 1.45^2[/tex]
[tex]I_3 = 54.45 \ kg \cdot m^2[/tex]
Thus
[tex]I= 160.35 + 97.29 + 54.45[/tex]
[tex]I= 312.09 \ kg \cdot m^2[/tex]
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?
Answers
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
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?
Answers
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) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
pF
(b) Determine the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
kV
Answers
Explanation:
(a) Given that,
Area of a parallel plate capacitor, [tex]A=1.8\ cm^2=1.8\times 10^{-4}\ m^2[/tex]
The separation between the plates of a capacitor, [tex]d=0.01\ mm = 10^{-5}\ m[/tex]
The dielectric constant of, k = 2.1
When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :
[tex]C=\dfrac{k\epsilon_o A}{d}[/tex]
Putting all the values we get :
[tex]C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF[/tex]
(b) We know that the Teflon has dielectric strength of 60 MV/m, [tex]E=60\times 10^6\ V/m[/tex]
The voltage difference between the plates at this critical voltage is given by :
[tex]V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V[/tex]
or
V = 0.6 kV
We have that the Capacitance and potential difference is mathematically given as
[tex]Vmax=\frac{Q}{334.68pF}[/tex]C=334.68pF
Capacitance &potential differenceQuestion Parameters:
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
a)
Generally the equation for the Capacitance is mathematically given as
[tex]C=\frac{ke_0A}{d}\\\\Therefore\\\\C=\frac{2.1*1.80e-4*8.85e12}{0.01e-3}\\\\[/tex]
C=334.68pF
b)
Generally the equation for the Capacitance is mathematically given as
[tex]Vmax=\frac{Q}{C}[/tex]
Where
Q is the charge on the plates, and hence not given
Therefore, maximum potential difference is
[tex]Vmax=\frac{Q}{334.68pF}[/tex]
For more information on potential difference visit
https://brainly.com/question/14883923
A radiation worker is subject to a dose of 200 mrad/h of maximum QF neutrons for one 40 h work week. How many times the yearly allowable effective dose did she receive?
Answers
Answer:
16 times.
Explanation:
The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr
Time consumed, t = 40 hr
The magnitude of Q.F for the neutrons, Q.F = 2
Thus the effective radiation dose is:
[tex]R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad[/tex]
Thus, the effective dose allowable yearly = 16 times
Two protons moving with same speed in same direction repel each other but what about two protons moving with different speed in the same direction?
Answers
Answer:In the case of two proton beams the protons repel one another because they have the same sign of electrical charge. There is also an attractive magnetic force between the protons, but in the proton frame of reference this force must be zero! Clearly then the attractive magnetic force that reduces the net force between protons in the two beams as seen in our frame of reference is relativistic. In particular the apparent magnetic forces or fields are relativistic modifications of the electrical forces or fields. As such modifications, they cannot be stronger than the electrical forces and fields that produce them. This follows from the fact that switching frames of reference can reduce forces, but it can’t turn what is attractive in one frame into a repulsive force in another frame.
In the case of wires the net charges in two wires are zero everywhere along the wires. That makes the net electrical forces between the wires very nearly zero. Yet the relativistic magnetic forces and fields will be of the same sort as in the case of two beams of charges of a single sign. This is true even in the frame of reference of what we think as the moving charges, that is, the electrons. In the frame of reference moving at the drift velocity of these current-carrying electrons, it is the protons or positively charged ions that are moving in the other direction. Consequently in any frame of reference for current-carrying wires in parallel, the net electrical force will be essentially zero, and there will be a net attractive magnetic force
Explanation:
Explanation:
Particles with similar charges (both positive or both negative) will always repel each other, regardless of their speed or direction.
In the lab , you have an electric field with a strength of 1,860 N/C. If the force on a particle with an unknown charge is 0.02796 N, what is the value of the charge on this particle.
Answers
Answer:
The charge is [tex]q = 1.50 *10^{-5} \ C[/tex]
Explanation:
From the question we are told that
The electric field strength is [tex]E = 1860 \ N/C[/tex]
The force is [tex]F = 0.02796 \ N[/tex]
Generally the charge on this particle is mathematically represented as
[tex]q = \frac{F}{E}[/tex]
=> [tex]q = \frac{0.02796}{ 1860}[/tex]
=> [tex]q = 1.50 *10^{-5} \ C[/tex]
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
Answers
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.
Two 1.0 nF capacitors are connected in series to a 1.5 V battery. Calculate the total energy stored by the capacitors.
Answers
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
Terms to describe the opposition by a material.to being magnetised is
Answers
Answer:
Repulsion
Explanation:
6. What is the bulk modulus of oxygen if 32.0 g of oxygen occupies 22.4 L and the speed of sound in the oxygen is 317 m/s?
Answers
Answer:
[tex] \boxed{\sf Bulk \ modulus \ of \ oxygen \approx 143.5 \ kPa} [/tex]
Given:
Mass of oxygen (m) = 32.0 g = 0.032 kg
Volume occupied by oxygen (V) = 22.4 L = 0.0224 m³
Speed of sound in oxygen (v) = 317 m/s
To Find:
Bulk modulus of oxygen
Explanation:
[tex]\sf Density \ of \ oxygen \ (\rho) = \frac{m}{V}[/tex]
[tex]\sf \implies Bulk \ modulus \ of \ oxygen \ (B) = v^{2} \rho[/tex]
[tex]\sf \implies B = v^{2} \times\frac{m}{V}[/tex]
[tex]\sf \implies B = {(317)}^{2} \times \frac{0.032}{0.0224} [/tex]
[tex]\sf \implies B = {(317)}^{2} \times 1.428[/tex]
[tex]\sf \implies B = 100489 \times 1.428[/tex]
[tex]\sf \implies B = 143498.292 \: Pa[/tex]
[tex]\sf \implies B \approx 143.5 \: kPa[/tex]
NASA is doing research on the concept of solar sailing. A solar sailing craft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion.
A) Should the sail be absorptive or reflective? Why?
B)The total power output of the sun is 3.90 × 1026 W . How large a sail is necessary to propel a 1.06 × 104 kg spacecraft against the gravitational force of the sun?
Answers
Answer:
A = 6.8 km²
Explanation:
A) The sail should be reflective. This is so that, it can produce the maximum radiation pressure.
B) let's begin with the formula used to calculate the average solar sail in orbit around the sun. Thus;
F_rad = 2IA/c
I is given by the formula;
I = P/(4πr²)
Thus;
F_rad = (2A/c) × (P/(4πr²)) = PA/2cπr²
Where;
A is the area of the sail
r is the distance of the sail from the sun
c is the speed of light = 3 × 10^(8) m/s
P is total power output of the sun = 3.90 × 10^(26) W
Now,F_rad = F_g
Where F_g is gravitational force.
Thus;
PA/2cπr² = G•m•M_sun/r²
r² will cancel out to givw;
PA/2cπ = G•m•M_sun
Making A the subject, we have;
A = (2•c•π•G•m•M_sun)/P
Now, m = 1.06 × 10⁴ kg and M_sun has a standard value of 1.99 × 10^(30) kg
G is gravitational constant and has a value of 6.67 × 10^(-11) Nm²/kg²
Thus;
A = (2 × 3 × 10^(8) × π × 6.67 × 10^(-11) × 1.06 × 10^(4) × 1.99 × 10^(30))/(3.90 × 10^(26))
A = 6.8 × 10^(6) m² = 6.8 km²
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.
Answers
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
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
Answers
Answer:
Vrel= 0.75c
Explanation:
See attached file
The Bohr model pictures a hydrogen atom in its ground state as a proton and an electron separated by the distance a0 = 0.529 × 10−10 m. The electric potential created by the electron at the position of the proton is
Answers
Answer:
E = -8.23 10⁻¹⁷ N / C
Explanation:
In the Bohr model, the electric potential for the ground state corresponding to the Bohr orbit is
E = k q₁ q₂ / r²
in this case
q₁ is the charge of the proton and q₂ the charge of the electron
E = - k e² / a₀²
let's calculate
E = - 9 10⁹ (1.6 10⁻¹⁹)² / (0.529 10⁻¹⁰)²
E = -8.23 10⁻¹⁷ N / C
2. The nuclear model of the atom held that
a. electrons were randomly spread through "a sphere of uniform positive
electrification."
b. matter was made of tiny electrically charged particles that were smaller than the
atom
C. matter was made of tiny, indivisible particles.
d. the atom had a dense, positively charged nucleus.
Answers
Answer:
the atom had a dense, positively charged nucleus.
Explanation:
Ernest Rutherford, based on the experiment carried out by two of his graduate students, established the authenticity of the nuclear model of the atom.
According to the nuclear model, an atom is made up of a dense positive core called the nucleus. Electrons are found to move round this nucleus in orbits. This is akin to the movement of the planets round the sun in the solar system.
The mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal. (1) Write an equation that represents this direct proportion, in which D is the proportionality constant. The density of lead metal is 11.3 g/cm3. (2) What is the mass of a piece of lead metal that has a volume of 17.3 cm3
Answers
Answer:
1) M = 11.3V2) 195.49 gramsExplanation:
1) If the mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal, this is expressed mathematically as shown;
M ∝ V
M = kV
For every proportionality sign, there will always be a proportionality constant 'k'
Since the proportionality constant is the density (D) of the metal, the equation will become;
M = DV
Given the density to be 11.3 g/cm3, the equation will become;
M = 11.3V
Hence, the equation that represents this direct proportion, in which D is the proportionality constant with metal density of 11.3g/cm³ is M = 11.3V
2) If the volume of the metal is 17.3cm³, on substituting this values into the equation in (1) to get the mass of the metal, we will have;
M = 11.3V
M = 11.3 * 17.3
M = 195.49 grams
Hence, the mass of a piece of lead metal that has a volume of 17.3 cm³ is 195.49 grams.
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.
Answers
The researchers need to compare those who contracted the disease to those who did not.
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.
Answers
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:
https://brainly.com/question/372577?referrer=searchResults
"A satellite requires 88.5 min to orbit Earth once. Assume a circular orbit. 1) What is the circumference of the satellites orbit
Answers
Answer:
circumference of the satellite orbit = 4.13 × 10⁷ m
Explanation:
Given that:
the time period T = 88.5 min = 88.5 × 60 = 5310 sec
The mass of the earth [tex]M_e[/tex] = 5.98 × 10²⁴ kg
if the radius of orbit is r,
Then,
[tex]\dfrac{V^2}{r} = \dfrac{GM_e}{r^2}[/tex]
[tex]{V^2} = \dfrac{GM_e r}{r^2}[/tex]
[tex]{V^2} = \dfrac{GM_e }{r}[/tex]
[tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]
Similarly :
[tex]T = \sqrt{\dfrac{ 2 \pi r} {V} }[/tex]
where; [tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]
Then:
[tex]T = {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {GM_e }} }[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {6.674\times 10^{-11} \times 5.98 \times 10^{24} }} }[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ 3.991052 \times 10^{14} }}[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {19977617.48}[/tex]
[tex]5310 \times 19977617.48= 2 \pi r^{3/2}}[/tex]
[tex]1.06081149 \times 10^{11}= 2 \pi r^{3/2}}[/tex]
[tex]\dfrac{1.06081149 \times 10^{11}}{2 \pi}= r^{3/2}}[/tex]
[tex]r^{3/2}} = \dfrac{1.06081149 \times 10^{11}}{2 \pi}[/tex]
[tex]r^{3/2}} = 1.68833392 \times 10^{10}[/tex]
[tex]r= (1.68833392 \times 10^{10})^{2/3}}[/tex]
[tex]r= 2565.38^2[/tex]
r = 6579225 m
The circumference of the satellites orbit can now be determined by using the formula:
circumference = 2π r
circumference = 2π × 6579225 m
circumference = 41338489.85 m
circumference of the satellite orbit = 4.13 × 10⁷ m
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
Answers
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
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
Answers
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
