Answer:
B.
Explanation:
The water collects in the ocean; it is then evaporated by the sun. After evaporation the water turns into water vapor, it then condenses to form clouds.
The ocean water prior to the part of the water cycle should be option B.
Ocean water:The ocean water should be collected back in the ocean prior to the part of the water cycle.
Because this should be done when it is evaporated by the sun. When the evaporation is done so the water should be transformed into water vapor.
Find out more information about the Water here:brainly.com/question/4381433?referrer=searchResults
Suppose that you release a small ball from rest at a depth of 0.700 m below the surface in a pool of water. If the density of the ball is 0.200 that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? (Neglect any transfer of energy to the splashing and waves produced by the emerging ball.)
Number________
Units ________
Answer:
photo
Explanation:
B. CO
A wave has frequency of 2 Hz and a wave length of 30 cm. the velocity of the wave is
A. 60.0 ms
B. 6.0 ms
D. 0.6 ms
Answer:
0.6 m/s
Explanation:
2Hz = 2^-1 = 2 /s
30cm = .3m
Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.
(.3m)*(2 /s) = 0.6 m/s
If a diode at 300°K with a constant bias current of 100μA has a forward voltage of 700mV across it, what will the voltage drop across this same diode be if the bias current is increased to 1mA? g
Answer:
the voltage drop across this same diode will be 760 mV
Explanation:
Given that:
Temperature T = 300°K
current [tex]I_1[/tex] = 100 μA
current [tex]I_2[/tex] = 1 mA
forward voltage [tex]V_r[/tex] = 700 mV = 0.7 V
To objective is to find the voltage drop across this same diode if the bias current is increased to 1mA.
Using the formula:
[tex]I = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]
[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]
where;
[tex]V_r[/tex] = 0.7
[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix}[/tex]
[tex]I_2 = I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix}[/tex]
[tex]\dfrac{I_1}{I_2} = \dfrac{ I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]
[tex]\dfrac{100 \ \mu A}{1 \ mA} = \dfrac{ \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]
Suppose n = 1
[tex]V_T = \dfrac{T}{11600} \\ \\ V_T = \dfrac{300}{11600} \\ \\ V_T = 25. 86 \ mV[/tex]
Then;
[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { nV_T} -1} \end {pmatrix}[/tex]
[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { 25.86} -1} \end {pmatrix}[/tex]
[tex]e^{\dfrac{V_r'}{nv_T}-1} = 5.699 \times 10^{12}[/tex]
[tex]{e^\dfrac{V_r'}{nv_T}} = 5.7 \times 10^{12}[/tex]
[tex]{\dfrac{V_r'}{nv_T}} =log_{e ^{5.7 \times 10^{12}}}[/tex]
[tex]{\dfrac{V_r'}{nv_T}} =29.37[/tex]
[tex]V_r'=29.37 \times nV_T[/tex]
[tex]V_r'=29.37 \times 25.86[/tex]
[tex]V_r'=759.5 \ mV[/tex]
[tex]Vr' \simeq[/tex] 760 mV
Thus, the voltage drop across this same diode will be 760 mV
Assume that Mars and Earth are in the same plane and that their orbits around the Sun are circles (Mars is ≈230×10^6km from the Sun and Earth is ≈150×10^6km from the Sun).
a) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is nearest Earth?
b) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is farthest from Earth?
Answer:
Explanation:
speed of light ≈ 300 000 000 m/s
a) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is nearest Earth?
(230e9 - 150e9) / 3e8 = 277 s or 4 minutes 27 seconds.
b) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is farthest from Earth?
(230e9 + 150e9) / 3e8 = 1267 s or 21 minutes 6 seconds.
A spark is generated in an automobile spark plug when there is an electric potential of 3000 V across the electrode gap. If 60 W of power is generated in a single spark that delivers a total charge of 3 nC, how long does it take for the spark to travel across the gap?
A. 50 ns
B. 75 ns
C. 125 ns
D. 150 ns
E. 225 ns 5
Answer:
The correct option is d
Explanation:
From the question we are told that
The electric potential is [tex]V = 3000 \ V[/tex]
The power is [tex]P = 60 \ W[/tex]
The charge delivered is [tex]q = 3nC = 3.0 *10^{-9} \ C[/tex]
Generally the power generated is mathematically represented as
[tex]P = I V[/tex]
=> [tex]I = \frac{P}{V }[/tex]
=> [tex]I = \frac{60 }{3000 }[/tex]
=> [tex]I = 0.02 \ A[/tex]
This current flow is mathematically represented as
[tex]I = \frac{q -q_o}{\Delta t }[/tex]
Where [tex]q_o[/tex] is the charge delivered at t=0 s which is 0s
So
[tex]0.02 = \frac{ (3.0 *10^{-9}) -0 }{t - 0 }[/tex]
[tex]t = 1.50 *10^{-7 } \ s[/tex]
[tex]t = 150 *10^{-9 } \ s[/tex]
A uniform narrow tube 1.90 m long is open at both ends. It resonates at two successive harmonics of frequencies 280 Hz and 294 Hz.(a) What is the fundamental frequency?_____Hz(b) What is the speed of sound in the gas in the tube?________ m/s
Answer:
a)14Hz
b)26.6m/s
Explanation:
a)we were given
the first harmonics frequencies as 280 Hz
The second harmonic frequency as 294 Hz.
The fundamental frequency is equal to the gap which means the distance that exist between the harmonics, then
the fundamental frequency=(294 - 280 = 10 Hz)
= 14Hz
b) We know the frequency and the wavelength of the sound wave (
We were told that the wavelength must be twice the length of the tube then, velocity can be calculated as
And fundamental frequency= 14Hz, and distance of 1.90 m then
v = f*2L = (14Hz)*2*(1.90 m) = 26.6m/s
Therefore, the speed of sound in the gas in the tubes is 26.6m/s
A major strike-slip earthquake on the San Andreas fault in California will cause a catastrophic tsunami affecting residents of San Francisco.
a) true
b) false
Answer:
a) true
Explanation:
The san andres is a transform fault that forms boundary between the Pacific and the North Atlantic plate and this slip strike is characterized by the latex motion the fault runs in the length of the California state. This plate is widely estimated for the high magnitude of earthquakes and varies from 7.7 to 8.3 magnitude. They are capable of producing a deadly tsunami that can devastate the pacific northwest.Magnetic resonance imaging needs a magnetic field strength of 1.5 T. The solenoid is 1.8 m long and 75 cm in diameter. It is tightly wound with a single layer of 1.50-mm-diameter superconducting wire.
What current is needed?
Answer:
The current needed is 1790.26 A
Explanation:
Given;
magnitude of magnetic field, B = 1.5 T
length of the solenoid, L = 1.8 m
diameter of the solenoid, d = 75 cm = 0.75 m
The magnetic field is given by;
[tex]B = \frac{\mu_o NI }{L}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
I is current in the solenoid
N is the number of turns, calculated as;
[tex]N = \frac{Length \ of\ solenoid}{diameter \ of \ wire} \\\\N = \frac{1.8}{1.5*10^{-3}} =1200 \ turns[/tex]
The current needed is calculated as;
[tex]I = \frac{BL}{\mu_o N} \\\\I = \frac{1.5 *1.8}{4\pi *10^{-7} *1200} \\\\I = 1790.26 \ A[/tex]
Therefore, the current needed is 1790.26 A.
Answer:
I = 1790.5 A
Explanation:
The magnetic field due to a solenoid is given by the following formula:
B = μ₀NI/L
where,
B = Magnetic Field Required = 1.5 T
μ₀ = 4π x 10⁻⁷ T/A.m
L = length of Solenoid = 1.8 m
I = Current needed = ?
N = No. of turns = L/diameter of wire = 1.8 m/1.5 x 10⁻³ m = 1200
Therefore,
1.5 T = (4π x 10⁻⁷ T/A.m)(1200)(I)/1.8 m
I = (1.5 T)(1.8 m)/(1200)(4π x 10⁻⁷ T/A.m)
I = 1790.5 A
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m2. What is the maximum value of the magnetic field in the wave
Answer:
7.78x10^-8T
Explanation:
The Pointing Vector S is
S = (1/μ0) E × B
at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,
S = (1/μ0) E B
where S, E and B are magnitudes. The average value of the Pointing Vector is
<S> = [1/(2 μ0)] E0 B0
where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)
Also at any instant,
E = c B
where E and B are magnitudes, so it must also be true at the instant of peak values
E0 = c B0
Substituting for E0,
<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²
Solve for B0.
Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)
= 7.79 x10 ^-8 T
If you weigh 685 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 25.0 km
CHECK COMPLETE QUESTION BELOW
you weigh 685 NN on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 25.0 kmkm ? Take the mass of the sun to be msmsm_s = 1.99×1030 kgkg , the gravitational constant to be GGG = 6.67×10−11 N⋅m2/kg2N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be ggg = 9.8 m/s2m/s2 .
Answer:
5.94×10^15N
Explanation:
the weight on the surface of a neutron star can be calculated by below expresion
W= Mg
W= weight of the person
m= mass of the person
g=gravity of the neutron star
But we need the mass which can be calculated as
m= W/g
m= 685/9.81
m= 69.83kg
From the gravitational law equation we have
F= GMm/r^2
G= gravitational constant = 6.67x10⁻¹¹
M= mass of the neutron star = 1.99x10³⁰ kg
r = distance between the person and the surface
Then r can be calculated as = 25/2 = 12.5 km , we divide by two because it's the distance between the person and the surface
g=gravity of the neutron star can be calculated as
g=(6.67×10^-11 ×1.99×10^30)/(12.5×10^3)^2
= 8.50×10^13m/s^2
Then from W= mg we can find our weight
W= 8.50×10^13m/s^2 × 69.83
= 5.94×10^15N
Therefore, weight on the surface of a neutron star is 5.94×10^15N
Why are the meters squared in the formula to calculate acceleration?
Answer:
During acceleration, you are moving across a distance over a time, but also increasing how fast we are doing it. Therefore, it means by how many meters per second the velocity changes every second
Explanation:
Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r = 0.5 m and r = 1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?
The direction for each field vector is perpendicular to equipotential lines.
Take a snapshot of the simulation showing equipotential lines and paste to a word document.
....................
15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and direction of the magnetic field. Now change to 2 loops, then to 1 loop. What do you observe the relationship to be between the magnitude of the magnetic field and the number of loops for the same current
Answer:
we see it is a linear relationship.
Explanation:
The magnetic flux is u solenoid is
B = μ₀ N/L I
where N is the number of loops, L the length and I the current
By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops
B = (μ₀ I / L) N
the amount between paracentesis constant, in the case of 4 loop the field is worth
B = cte 4
N B
4 4 cte
3 3 cte
2 2 cte
1 1 cte
as we see it is a linear relationship.
In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,
you check the weather and find that the winds are coming from the west at 15 milers per hour. this information describes the winds
Answer:
Velocity
Explanation:
We finds that the winds are coming from the west at 15 miles per hour. This information shows the velocity of the wind. Since, velocity is a vector quantity. It has both magnitude and direction. 15 miles per hour shows the speed of wind and west shows the direction of wind motion.
Hence, the given information describes wind velocity.
A swimmer is treading water with their head above the surface of a pool and sees a penny at the bottom of the pool 5.0 mm below. How deep does the coin appear to be? (Index of refraction of water = 1.33) [Conceptual note: Does the coin appear to be shallower or deeper?]
Answer:
The apparent depth is [tex]D' = 0.00376 \ m[/tex]
Explanation:
From the question we are told that
The depth of the water is [tex]D = 5.0 \ mm = 5.0 *10^{-3} \ m[/tex]
The refractive index of water is [tex]n = 1.33[/tex]
Generally the apparent depth of the coin is mathematically represented as
[tex]D' = D * [\frac{ n_a}{n} ][/tex]
Here [tex]n_a[/tex] is the refractive index of air the value is [tex]n_a = 1[/tex]
So
[tex]D' = 5.0 *10^{-3} * [\frac{1}{1.33} ][/tex]
[tex]D' = 0.00376 \ m[/tex]
The apparent depth will be 0.00376 m.
What is an index of refraction?
The index of refraction of a substance also known as the refraction index is a dimensionless quantity that specifies how quickly light passes through it in optics.
d is the depth of the water =5.0 mm =5.0 ×10⁻³
n is the refractive index of water =1.33
[tex]\rm n_a[/tex] is the refractive index of wire=1
The apparent depth of the coin is given as;
[tex]\rm D'=D \times \frac{n_a}{n} \\\\ \rm D'=5.0 \times 10^{-3} \times \frac{1}{1.33} \\\\ \rm D'=0.00376 \ m[/tex]
Hence the apparent depth will be 0.00376 m.
To learn more about the index of refraction refer to the link;
https://brainly.com/question/23750645
Can someone please help!!!
Answer:
W = F • ∆x
so for work to be done, a force and displacement has to be in the same direction. (Ex: a box is being pushed forward and it's also moving forward.)
If the solenoid is 45.0 cm long and each winding has a radius of 8.0 cm , how many windings are in the solenoid
Answer:
The number of windings is 1.
Explanation:
The radius of the solenoid = 8.0 cm = 0.08 m
Length of the solenoid = 45.0 cm = 0.45 m
number of turn = ?
circumference of each winding = 2πr = 2 x 3.142 x 0.08 = 0.503 m
The number of windings = (Length of the solenoid)/(circumference of each winding)
==> 0.45/0.503 = 0.89 ≅ 1
A long, horizontal hose of diameter 5.4 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1.2 cm. Water squirts from the nozzle at velocity 20 m/sec. Assume that the water has no viscosity or other form of energy dissipation.
a) What is the velocity of the water in the hose?
b) What is the pressure differential between the water in the hose and water in the nozzle?
c) How long will it take to fill a tub of volume 120 liters with the hose?
Answer:
a) 0.988 m/s
b) 199512 Pa
c) 57.52 s
Explanation:
given that
A
A1 v1 = A2 v2
d1² v1 = d2² v2
v2 = [d1/d2]² v1
v2 = (1.2/5.4)² * 20
v2 = 0.049 * 20
v2 = 0.988 m/s
B
P + 1/2 ρ v² = K.
[p2 - p1] = 1/2 ρ [v1² - v2²]
[p2 - p1] = 1/2 * 1000 [20² - 0.988²]
[p2 - p1] = 500 * (400 - 0.976)
[p2 - p1] = 500 * 399
[p2 - p1] = 199512 Pa
C
Flow rate = AV = π [d²/ 4 ] * v
= π [0.012² / 4 ] * 20 = 0.00226 m³ /s
= π [0.054² / 4 ] * 0.988 = 0.00226 m³ /s
130 liters = 0.13 m³
t = 0.13/ 0.00226 = 57.52 s
The capacitor is originally charged. How does the current I in the ammeter behave as a function of time after the switch is closed?
1. I = 0 always.
2. I = constant, not equal to 0.
3. I increases, then is constant.
4. I instantly jumps up, then slowly decreases.
5. None of the above.
Answer:
The current in the ammeter is zero.
(1) is correct option.
Explanation:
Given that,
The capacitor is charged.
We need find the current after closed switched
We know that,
When switch is closed then the capacitor behave as a short circuit, and the all current flows through it. the current is zero.
Then, the ammeter reads zero.
Hence, The current in the ammeter is zero.
(1) is correct option.
A donkey is attached by a rope to a wooden cart at an angle of 23° to the horizontal. the tension in the rope is 210 n. if the cart is dragged horizontally along the floor with a constant speed of 7 km/h, calculate how much work the donkey does in 35 minutes.
Answer:
787528.7 J
Explanation:
Work done: This can be defined as the product of force and distance along the direction of force. The S.I unit of work is Joules (J).
From the question,
W = Tcos∅(d)............. Equation 1
Where W = work done, T = tension in the rope, ∅ = the angle of the rope to the horizontal, d = distance.
But,
d = v(t)..................... equation 2
Where v = velocity, t = time
Substitute equation 2 into equation 1
W = Tcos∅(vt)............. Equation 3
Given: T = 210 N, ∅ = 23°, v = 7 km/h = 1.94 m/s, t = 35 min = 2100 s
Substitute into equation 3
W = 210(cos23°)(1.94×2100)
W = 787528.7 J
What is the pathway of sound through fluids starting at the oval window through to dissipation of the sound waves at the round window
a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleration does it experience
Answer:
The angular acceleration is [tex]\alpha = 3.235 \ rad/s ^2[/tex]
Explanation:
From the question we are told that
The moment of inertia is [tex]I = 0.034\ kg \cdot m^2[/tex]
The net torque is [tex]\tau = 0.11\ N \cdot m[/tex]
Generally the net torque is mathematically represented as
[tex]\tau = I * \alpha[/tex]
Where [tex]\alpha[/tex] is the angular acceleration so
[tex]\alpha = \frac{\tau }{I}[/tex]
substituting values
[tex]\alpha = \frac{0.1 1}{ 0.034}[/tex]
[tex]\alpha = 3.235 \ rad/s ^2[/tex]
If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y-direction, the wave is traveling in the
Answer:
The wave is travelling in the ±z-axis direction.
Explanation:
An electromagnetic wave has an oscillating magnetic and electric field. The electric and magnetic field both oscillate perpendicularly one to the other, and the wave travels perpendicularly to the direction of oscillation of the electric and magnetic field.
In this case, if the magnetic field is in the +x-axis direction, and the electric field is in the +y-axis direction, we can say with all assurance that the wave will be travelling in the ±z-axis direction.
A sinusoidal sound wave moves through a medium and is described by the displacement wave function s(x, t) = 1.99 cos(15.2x − 869t) where s is in micrometers, x is in meters, and t is in seconds. (a) Find the amplitude of this wave. µm (b) Find the wavelength of this wave. cm (c) Find the speed of this wave. m/s (d) Determine the instantaneous displacement from equilibrium of the elements of the medium at the position x = 0.050 9 m at t = 2.94 ms. µm (e) Determine the maximum speed of a element's oscillatory motion. mm/s
Answer:
a) A = 1.99 μm , b) λ = 0.4134 m , c) v = 57.2 m / s , d) s = - 1,946 nm ,
e) v_max = 1,739 mm / s
Explanation:
A sound wave has the general expression
s = s₀ sin (kx - wt)
where s is the displacement, s₀ the amplitude of the wave, k the wave vector and w the angular velocity, in this exercise the expression given is
s = 1.99 sin (15.2 x - 869 t)
a) the amplitude of the wave is
A = s₀
A = 1.99 μm
b) wave spectrum is
k = 2π /λ
in the equation k = 15.2 m⁻¹
λ = 2π / k
λ = 2π / 15.2
λ = 0.4134 m
c) the speed of the wave is given by the relation
v = λ f
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 869 / 2π
f = 138.3 Hz
v = 0.4134 138.3
v = 57.2 m / s
d) To find the instantaneous velocity, we substitute the given distance and time into the equation
s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)
s = 1.99 sin (0.77368 - 2.55486)
remember that trigonometry functions must be in radians
s = 1.99 (-0.98895)
s = - 1,946 nm
The negative sign indicates that it shifts to the left
e) the speed of the oscillating part is
v = ds / dt)
v = - s₀(-w) cos (kx -wt)
the maximum speed occurs when the cosines is 1
v_maximo = s₀w
v_maximum = 1.99 869
v_maximo = 1739.31 μm / s
let's reduce to mm / s
v_maxio = 1739.31 miuy / s (1 mm / 103 mu)
v_max = 1,739 mm / s
a) A is = 1.99 μm , b) λ is = 0.4134 m , c) v is = 57.2 m / s , d) s is = - 1,946 nm, e) v_max is = 1,739 mm / s
Calculation of Wavelength
When A sound wave has the general expression is:
Then, s = s₀ sin (kx - wt)
Now, where s is the displacement, Then, s₀ is the amplitude of the wave, k the wave vector, and w the angular velocity, Now, in this exercise the expression given is
s is = 1.99 sin (15.2 x - 869 t)
a) When the amplitude of the wave is
A is = s₀
Thus, A = 1.99 μm
b) When the wave spectrum is
k is = 2π /λ
Now, in the equation k = 15.2 m⁻¹
Then, λ = 2π / k
After that, λ = 2π / 15.2
Thus, λ = 0.4134 m
c) When the speed of the wave is given by the relation is:
Then, v = λ f
Now, the angular velocity and frequency are related is:
w is = 2π f
Then, f = w / 2π
After that, f = 869 / 2π
Now, f = 138.3 Hz
Then, v = 0.4134 138.3
Thus, v = 57.2 m / s
d) Now, To find the instantaneous velocity, When we substitute the given distance and time into the equation
Then, s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)
After that, s = 1.99 sin (0.77368 - 2.55486)
Then remember that trigonometry functions must be in radians
After that, s = 1.99 (-0.98895)
Thus, s = - 1,946 nm
When The negative sign indicates that it shifts to the left
e) When the speed of the oscillating part is
Then, v = ds / dt)
Now, v = - s₀(-w) cos (kx -wt)
When the maximum speed occurs when the cosines is 1
Then, v_maximo = s₀w
After that, v_maximum = 1.99 869
v_maximo = 1739.31 μm / s
Now, let's reduce to mm / s
Then, v_maxio = 1739.31 miuy / s (1 mm / 103 mu)
Therefore, v_max = 1,739 mm / s
Finf more informmation about Wavelength here:
https://brainly.com/question/6352445
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.
Transistor circuits are sometimes referred to as switching circuits - Why is this?
Explanation:
One of the most common uses for transistors in an electronic circuit is as simple switches. In short, a transistor conducts current across the collector-emitter path only when a voltage is applied to the base. When no base voltage is present, the switch is off. When base voltage is present, the switch is on.
An oil film (n = 1.48) of thickness 290 nm floating on water is illuminated with white light at normal incidence. What is the wavelength of the dominant color in the reflected light? A. Blue (470 nm) B. Green (541 nm) C. Violet (404 nm) D. Yellow (572 nm)
Answer:
The correct option is D
Explanation:
From the question we are told that
The refractive index of oil film is [tex]k = 1.48[/tex]
The thickness is [tex]t = 290 \ nm = 290*10^{-9} \ m[/tex]
Generally for constructive interference
[tex]2t = [m + \frac{1}{2} ]* \frac{\lambda}{k}[/tex]
For reflection of a bright fringe m = 1
=> [tex]2 * (290*10^{-9}) = [1 + \frac{1}{2} ]* \frac{\lambda}{1.48}[/tex]
=> [tex]\lambda = 5.723 *10^{-7} \ m[/tex]
This wavelength fall in the range of a yellow light
3 QUESTIONS PLEASE ANSWER!
Answer:
1. A
2. C
3. D
Explanation:
An isolated system consists of two masses. The first, m1, has a mass of 1.90 kg, and is initially traveling to the east with a speed of 6.71 m/s. The second, m2, has a mass of 2.94 kg, and is initially traveling to the west with an unknown initial speed. The two masses collide head-on in a completely inelastic collision that stops them both. Calculate the initial kinetic energy of m2.
Answer:
m1v1=m2v2, v2=4.3m/s KE=(0.5)(2.94)(4.3)=6.2J
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s ). If a particular disk is spun at 998.0 rad/s while it is being read, and then is allowed to come to rest over 0.502 seconds , what is the magnitude of the average angular acceleration of the disk?
Answer:
1988.05 rad/s^2
Explanation:
The angular speed of the optical disk ω = 998.0 rad/s
the time taken to come to rest t = 0.502 s
The magnitude of the average angular acceleration ∝ = ω/t
∝ = 998.0/0.502 = 1988.05 rad/s^2