The question is incomplete, the complete question is;
Light of frequency f falls on a metal surface and ejects electrons of maximum kinetic energy K by the photoelectric effect.
Part A If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be
K/2.
K.
2K.
greater than 2K.
Answer:
2K
Explanation:
Given that the kinetic energy of photo electrons is given by;
K= E -Wo
Where;
K = kinetic energy
E= energy of incident photon
Wo = work function
But;
E= hf
Wo = fo
h= Plank's constant
f= frequency of incident photon
fo= Threshold frequency
So:
K= hf - hfo
Where the frequency of incident light is doubled;
K= 2hf - hfo
Hence, maximum kinetic energy of the emitted electrons in this case will be 2K
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is
The question is incomplete. The complete question is :
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s) and the dissolution rate (kg/s).
Solution :
From flow over sphere, the mass transfer equation can be written as :
[tex]$Sh = 2 + 0.6 Re^{1/2} Sc^{1/3}$[/tex]
where, Sherood number, [tex]$Sh = \frac{K_L d}{D_{eff}}$[/tex]
Reynolds number, [tex]$Re=\frac{Vd\rho}{\mu}$[/tex]
Schmid number, [tex]$Sc= \frac{\mu}{\rho D_{eff}}$[/tex]
So,
[tex]$\frac{K_L d}{D_{eff}}=2+0.6 \left( \frac{V d \rho}{\mu} \right)^{1/2} \ \left( \frac{\mu}{\rho D_{eff}} \right)^{1/3}$[/tex]
Diameter, d = 1 cm = [tex]$1 \times 10^{-2}$[/tex] m
V = 1 m/s
[tex]$\rho = 1000 \ kg/m^3$[/tex]
[tex]$\mu = 10^{-3} \ kg/m/s$[/tex]
[tex]$D_{eff} = 2 \times 10^{-9} \ m^2/s$[/tex]
[tex]$\frac{K_L \times 10^{-2}}{2 \times 10^{-9}}=2+0.6 \left( \frac{1 \times 10^{-2} \times 10^3}{10^{-3}} \right)^{1/2} \ \left( \frac{10^{-3}}{10^3 \times 2 \times 10^{-9}} \right)^{1/3}$[/tex]
[tex]$K_L \times 5 \times 10^6=478.22$[/tex]
[tex]$K_L=9.5644 \times 10^{-5}$[/tex] m/s
So the mass transfer coefficient is 9.5644 [tex]$\times 10^{-5}$[/tex] m/s. It is given solubility,
[tex]$\Delta C = 2 \ kg/m^3$[/tex]
[tex]$N = Md^2 \times \Delta C \times K_L$[/tex]
[tex]$N= M \times (10^{-2})^2 \times 2 \times 9.5644 \times 10^{-5}$[/tex]
[tex]$N= 6 \times 10^{-8}$[/tex] kg/s (dissolution rate)
A proton moves perpendicular to a uniform magnetic field at a speed of 1.75 107 m/s and experiences an acceleration of 2.25 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.
Answer:
B = 0.013(-j) T
Explanation:
Given that,
The speed of a proton, [tex]v=1.75\times 10^7\ m/s[/tex]
Acceleration experienced by the proton,[tex]a=2.25\times 10^3\ m/s[/tex]
We need to find the magnitude and the direction of the magnetic field. At equilibrium,
[tex]ma=qvB\\\\B=\dfrac{ma}{qv}\\\\B=\dfrac{1.67\times 10^{-27}\times 2.25\times 10^{13}}{1.6\times 10^{-19}\times 1.75\times 10^{7}}\\\\B=0.013\ T[/tex]
The velocity is in +z direction, force in +x direction, then the field must be in -y direction.
If a boy lifts a mass of 6kg to a height of 10m and travels horizontally with a constant velocity of 4.2m/s, calculate the work done? Explain your answer.
Answer:
W = 641.52 J
Explanation:
The work done here will be the sum of potential energy and the kinetic energy of the boy. Here potential energy accounts for vertical motion part while the kinetic energy accounts for the horizontal motion part:
[tex]Work\ Done = Kinetic\ Energy + Potential\ Energy\\\\W = K.E +P.E\\\\W = \frac{1}{2}mv^2+mgh\\\\[/tex]
where,
W = Work Done = ?
m = mass = 6 kg
v = speed = 4.2 m/s
g = acceleration dueto gravity = 9.81 m/s²
h = height = 10 m
Therefore,
[tex]W = \frac{1}{2}(6\ kg)(4.2\ m/s)^2+(6\ kg)(9.81\ m/s^2)(10\ m)[/tex]
W = 52.92 J + 588.6 J
W = 641.52 J
point charges q1=50 uc and q2=-25 uc are placed 1 m apart. what is the force on a third chare q3=2 uc placed midway between q1 and q2? where must q3 of the preceding problem be placed so that the net force on it is zero?
Answer:
d = -1 m
The negative sign indicates that the charge from that force of the space of the two spheres.
Explanation:
That is a problem of electric forces, given by Coulomb's law
F = [tex]k \frac{ q1q2}{r^2}[/tex]
We use that charges of the same sign repel and charges of different signs do not attract, so the net force is
∑ = F₁₃ + F₂₃
F_ {net} = [tex]k \frac{q_1q_3}{r_{13}^2} + k \frac{q_2q_3}{ r_{23}^}[/tex]
a) the charge is placed at the midpoint between the other two
r₁₃ = r₁₂ = R = ½ m = 0.5mF_ {net} =[tex]\frac{k}{R^2 } \ q3 ( q1+q2)[/tex]
calculate us
F_ {net} = 9 10⁹ / 0.5² 2 10⁻⁶ (50 -25) 10⁻⁶
F_ {net} = 1,800 N
b) where must be placed q3 so that the force is zero
for this case the charge q3 is outside the spheres
∑ F = 0
F₁₁₃ = F₂₃
k q_1 / r_{13}² = k q₂ q₃ / r₂₃²
q₁/ r₁₂² = q₂ / r₂₃²
suppose the distance
r₁₂ = d
the he other sphere is
r₂₃ = d + 1
we substitute
q₃ / d² = q₂ / (d + 1) ²
(d + 1) ² = q₂ / q₃ d²
d² (1 - q₂/ q₃) + 2d + 1 = 0
we solve the equation of a second
d = [-2 + [tex]\sqrt{2^2 - 4 1 ( 1+25/50}[/tex] ] / 2
d = -2 /2
d = -1 m
The negative sign indicates that the charge from that force of the space of the two spheres.
A 3.1-mole sample of an ideal gas is gently heated at constant temperature 320 K. It expands from initial volume 23 L to final volume V2. A total of 1.7 kJ of heat is added during the expansion process. What is V2? Let the ideal-gas constant R = 8.314 J/(mol • K).
From the ideal gas law,
PV = nRT ==> P = nRT/V
where P is the pressure exerted by the gas on the container. The work W done by this pressure as the volume of the gas changes from V₁ to V₂ is given by the integral,
[tex]W = \displaystyle \int_{V_1}^{V_2}P\,\mathrm dV \implies W = nRT \ln\left(\dfrac{V_2}{V_1}\right)[/tex]
and solving for V₂ gives
[tex]V_2 = V_1\exp\left(\dfrac{W}{nRT}\right)[/tex]
If you add 1.7 kJ of heat to the system, which does the aforementioned work, the gas will expand to a volume of
[tex]V_2 = (23\,\mathrm L)\exp\left(\dfrac{1.7\,\mathrm{kJ}}{(3.1\,\mathrm{mol})\left(8.314\frac{\rm J}{\mathrm{mol}\cdot\mathrm K}\right)(320\,\mathrm K)}\right) \approx \boxed{28 \,\mathrm L}[/tex]
Define hydropower or hydroelectric power ?
No Spam..
[tex]\:[/tex]
Hydroelectric power, also called hydropower is the electricity produced from generators driven by turbines that convert the potential energy of falling or fast-flowing water into mechanical energy.
Answer:
Hydroelectric power/hydropower - electricity produced by a hydraulic source, specifically energy generated falling or flowing water
A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500m/s strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 300m/s.
A) Compute the magnitude of the velocity of the stone after it is struck.
B) Compute the direction of the velocity of the stone after it is struck. (degrees from the initial direction of the bullet)
Answer:
Explanation:
Given that:
mass of stone (M) = 0.100 kg
mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg
initial velocity of stone ([tex]u_{stone}[/tex]) = 0 m/s
Initial velocity of bullet ([tex]u_{bullet}[/tex]) = (500 m/s)i
Speed of the bullet after collision ([tex]v_{bullet}[/tex]) = (300 m/s) j
Suppose we represent [tex](v_{stone})[/tex] to be the velocity of the stone after the truck, then:
From linear momentum, the law of conservation can be applied which is expressed as:
[tex]m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}[/tex]
[tex](2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}[/tex]
[tex](2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j= (0.100 \ kg)v_{stone}[/tex]
[tex]v_{stone}= (1.25\ kg.m/s)i-(0.75\ kg m/s)j[/tex]
[tex]v_{stone}= (12.5\ m/s)i-(7.5\ m/s)j[/tex]
∴
The magnitude now is:
[tex]v_{stone}=\sqrt{ (12.5\ m/s)^2-(7.5\ m/s)^2}[/tex]
[tex]\mathbf{v_{stone}= 14.6 \ m/s}[/tex]
Using the tangent of an angle to determine the direction of the velocity after the struck;
Let θ represent the direction:
[tex]\theta = tan^{-1} (\dfrac{-7.5}{12.5})[/tex]
[tex]\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}[/tex]
1. A 2.7-kg copper block is given an initial speed of 4.0 m/s on a rough horizontal surface. Because of friction, the block finally comes to rest. (a) If the block absorbs 85% of its initial kinetic energy as internal energy, calculate its increase in temperature.
Answer:
ΔT = 0.017 °C
Explanation:
According to the given condition, the change in internal energy of the block must be equal to 85% of its kinetic energy:
Change in Internal Energy = (0.85)(Kinetic Energy)
[tex]mC\Delta T = (0.85)\frac{1}{2}mv^2\\\\C\Delta T = (0.425)v^2\\\\\Delta T = \frac{0.425v^2}{C}[/tex]
where,
ΔT = increase in temperature = ?
v = speed of block = 4 m/s
C = specific heat capacity of copper = 389 J/kg.°C
Therefore,
[tex]\Delta T = \frac{(0.425)(4\ m/s)^2}{389}\\\\[/tex]
ΔT = 0.017 °C
At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magnetic B=B k and electric E=−E k fields with the initial velocity v=v0 i. The magnitudes of the fields: B=0.18 T, E=278 V/m, and the initial speed v0=2.1 m/s are given. Find at what time t, the particle's speed would become equal to v(t)=3.78·v0:
Answer:
10.78 s
Explanation:
The force on the charge is computed by using the equation:
[tex]F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j[/tex]
F = ma
∴
[tex]a ^{\to}= \dfrac{F^{\to}}{m}[/tex]
[tex]a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)[/tex]
[tex]a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j[/tex]
At time t(sec; the partiCle velocity becomes [tex]v(t) = 3.78 v_o[/tex]
The velocity of the charge after the time t(sec) is expressed by using the formula:
[tex]v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}[/tex]
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 24 ft/s2. What is the distance (in ft) traveled before the car comes to a stop? (Round your answer to one decimal place.)
The car has initial speed 50 mi/h ≈ 73 ft/s, so it covers a distance x such that
0² - (73 ft/s)² = 2 (-24 ft/s²) x
==> x ≈ 111.0 ft
An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to 109 rad/sec in 1.87 seconds. The blades of the fan have a diameter of 6.7 meters and their deceleration rate is 4.7 rad/sec2.
What was the initial angular speed of the fan in rev/sec?
ωi =
Answer:
wo = 18.75 rev / s
Explanation:
This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²
w_f = w₀ - a t
w₀ = w_f + a t
w₀ = 109 + 4.7 1.87
w₀ = 117.8 rad / s
let's reduce to revolutions / s
w₀ = 117.8 rad / s (1 rev / 2pi rad)
w₀ = 18.75 rev / s
Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.
why meter cube is called derived unit
Answer:
Because it is the result of two more fundamental units, a derived unit is termed that. For volume, the cubic meter (m³) is the fundamental unit of area. Any number that cannot be measured directly with any equipment is referred to as a derived unit. For example, we can't quantify a substance's density using a rule, scale, or bucket.
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An object of mass 80 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object down, a buoyancy force of 1/50 times the weight of the object is pushing the object up (weight=mg). If we assume that water resistance exerts a force on the abject that is proportional to the velocity of the object, with proportionality constant 10 N-sec/m, find the equation of motion of the object. After how many seconds will the velocity of the object be 40 m/s? Assume that the acceleration due to gravity is 9.81 m/sec^2.
Answer:
a) Fnet = mg - Fb - Fr
b) 8.67 secs
Explanation:
mass of object = 80 kg
Buoyancy force = 1/50 * weight ( 80 * 9.81 ) = 15.696
Proportionality constant = 10 N-sec/m
a) Calculate equation of motion of the object
Force of resistance on object due to water = Fr ∝ V
= Fr = Kv = 10 V
Given that : Fb( due to buoyancy ) , Fr ( Force of resistance ) acts in the positive y-direction on the object while mg ( weight ) acts in the negative y - direction on the object.
Fnet = mg - Fb - Fr
∴ Equation of motion of the object ( Ma = mg - Fb - Fr )
b) Calculate how long before velocity of the object hits 40 m/s
Ma = mg - Fb - Fr
a = 9.81 - 0.1962 - 0.125 V = 9.6138 - 0.125 V
V = u + at ---- ( 1 )
u = 0
V = 40 m/s
a = 9.6138 - 0.125 V
back to equation 1
40 = 0 + ( 9.6138 - 0.125 (40) ) t
40 = 4.6138 t
∴ t = 40 / 4.6138 = 8.67 secs
any four difference between speed and acceleration
Speed:
Measured in distance / timeIt tells the speed of a bodyIt is a scalar quantitySpeed is always positiveIt's unit is meter per secondAcceleration:
Measured in velocity / timeIt tells the change in the velocityImplied on change in a velocityIt is a vector quantityIt's unit is meter per second square
Which of the following groups is the largest ?
population
community
ecosystem
biome
Answer:
B. Community
Took science classes for 6 years now
Two friends, Al and Jo, have a combined mass of 194 kg. At the ice skating rink, they stand close together on skates, at rest and facing each other. Using their arms, they push on each other for 1 second and move off in opposite directions. Al moves off with a speed of 7.9 m/sec in one direction and Jo moves off with a speed of 6.7 m/sec in the other. You can assume friction is negligible.
What is Al's mass? 110.58 What is Jo's mass? If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Al on Jo? If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Jo on Al?
Answer:
The mass of Al is 89.027 kilograms.
The mass of Jo is 104.973 kilograms.
The magnitude of the force of Jo on Al is 596.481 newtons.
Explanation:
Given the absence of external forces, this situation can be described will by Principle of Linear Momentum Conservation and Impact Theorem on each skater:
Al:
[tex]m_{1}\cdot (v_{1, f}-v_{1, o}) = -F \cdot \Delta t[/tex] (1)
Jo:
[tex]m_{2}\cdot (v_{2,f}-v_{2,o}) = F\cdot \Delta t[/tex] (2)
Total mass:
[tex]m_{1} + m_{2} = 194\,kg[/tex]
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex] - Masses of the skaters, in kilograms.
[tex]v_{1,o}[/tex], [tex]v_{1,f}[/tex] - Initial and final velocities of Al, in meters per second.
[tex]v_{2,o}[/tex], [tex]v_{2,f}[/tex] - Initial and final velocities of Jo, in meters per second.
[tex]F[/tex] - Impact force between skaters, in newtons.
[tex]\Delta t[/tex] - Impact time, in seconds.
If we know that [tex]v_{1,o} = 0\,\frac{m}{s}[/tex], [tex]v_{1,f} = -7.9\,\frac{m}{s}[/tex], [tex]\Delta t = 1\,s[/tex], [tex]v_{2,o} = 0\,\frac{m}{s}[/tex] and [tex]v_{2,f} = 6.7\,\frac{m}{s}[/tex], then the masses of the skaters are, respectively:
[tex](194-m_{2})\cdot (-7.9) = -F[/tex] (1b)
[tex]m_{2} \cdot 6.7 = F[/tex] (2b)
(2b) in (1b):
[tex](194-m_{2})\cdot (-7.9) = -m_{2}\cdot 6.7[/tex]
[tex]-1532.6 +7.9\cdot m_{2} = -6.7\cdot m_{2}[/tex]
[tex]14.6\cdot m_{2} = 1532.6[/tex]
[tex]m_{2} = 104.973\,kg[/tex]
[tex]m_{1} = 194\,kg - 104.973\,kg[/tex]
[tex]m_{1} = 89.027\,kg[/tex]
And the magnitude of the force is:
[tex]F = 6.7\cdot m_{2}[/tex]
[tex]F = 596.481\,N[/tex]
The mass of Al is 89.027 kilograms.
The mass of Jo is 104.973 kilograms.
The magnitude of the force of Jo on Al is 596.481 newtons.
Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to view distant objects with her eye relaxed. With her glasses on, what is the closest object on which she can focus?
Answer:
[tex]u=34cm[/tex]
Explanation:
From the question we are told that:
Far point is [tex]V=34 cm[/tex]
Near point is [tex]u=17 cm[/tex]
Therefore
Focal Length
[tex]f=-34cm[/tex]
Generally the equation for the Lens is mathematically given by
[tex]\frac{1}{u}=\frac{1}{f}-\frac{1}{v}[/tex]
[tex]\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}[/tex]
[tex]u=34cm[/tex]
HELP MEEEEEEE PLEASEEEEEEEEE
Answer:
D) Q = 80 C
Explanation:
Given;
current flowing in the light bulb, I = 2A
time of current flow, t = 40,000 ms = 40,000 x 10⁻³ s = 40 s
The quantity of the charge is calculated as;
Q = It
where;
Q is the quantity of the charge (Coulombs)
Q = (2 ) x (40)
Q = 80 C
Therefore, the quantity of charge flowing in the circuit is 80 C
D) Q = 80 C
The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 6 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 64,000 m/s. What are the masses of the two stars
Answer:
the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
Explanation:
Given the data in the question;
Time period = 6 months = 1.577 × 10⁷ s
orbital speed v = 64000 m/s
since its a circular orbit,
v = 2πr / T
we solve for r
r = vT/ 2π
r = ( 64000 × 1.577 × 10⁷ ) / 2π
r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU
Now, from Kepler's law
T² = r³ / ( m₁ + m₂ )
T = 6 months = 0.5 years
we substitute
(0.5)² = (1.0737)³ / ( m₁ + m₂ )
0.25 = 1.2378 / ( m₁ + m₂ )
( m₁ + m₂ ) = 1.2378 / 0.25
( m₁ + m₂ ) = 4.9512
m₁ = m₂ = 4.9512 / 2 = 2.4756 solar mass
we know that solar mass = 1.989 × 10³⁰ kg
so
m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg
m₁ = m₂ = 4.92 × 10³⁰ kg
Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
Choir members are traditionally grouped into four main vocal registers: bass,
tenor, alto, and soprano. Men with low-pitched voices tend to sing bass, while
those with slightly higher-pitched voices sing in the tenor range. Similarly,
women with low-pitched voices sing alto, while women with high-pitched
voices sing soprano. Which vocal register sings at the lowest frequency?
A. Bass
B. Alto
C. Soprano
D. Tenor
Answer:
The bass has the lowest frequency ,it only causes vibrations but at a lower frequency
calculate the pressure of water having density 1000 kilo per metre square at a depth of 20 m inside the water
Answer:
the pressure of the water at the given depth is 196,200 N/m².
Explanation:
Given;
density of the water, ρ = 1000 kg/m³
depth of the water, h = 20 m
acceleration due to gravity, g = 9.81 m/s²
The pressure at the given depth of the water is calculated as;
P = ρgh
P = 1000 x 9.81 x 20
P = 196,200 N/m²
Therefore, the pressure of the water at the given depth is 196,200 N/m².
I’m a photoelectric effect, which property of the incident light determines how much kinetic energy the ejected electrons have ?
A) brightness
B) frequency
C) size of the beam
D) none of the above
Answer:
b = frequency
Joule is a SI unit of power
Measuring cylinder is used to measure the volume of a liquid
Answer:
The SI unit of power is watt
a stone is thrown vertically upwards with a velocity of 20 m per second what will be its velocity when it reaches a height of 10.2 m
Answer:
Explanation:
Here's the info we have:
initial velocity is 20 m/s;
final velocity is our unknown;
displacement is -10.2 m; and
acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation
v² = v₀² + 2aΔx and filling in accordingly to solve for v:
[tex]v=\sqrt{(20)^2+2(-9.8)(-10.2)}[/tex] Rounding to the correct number of sig fig's to simplify:
[tex]v=\sqrt{400+2.0*10^2}[/tex] to get
v = [tex]\sqrt{600}=20\frac{m}{s}[/tex] If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.
So either 20 m/s or 24 m/s
Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k = 8.99 × 1012N/C. where, d= 11 cm Q= 12.5 C
Answer:
The electric field is 9.3 x 10^12 N/C and the direction is away from the charge.
Explanation:
charge, Q = 12.5 C
distance, d = 11 cm = 0.11 m
Let the electric field is E.
[tex]E =\frac{K Q}{d^2}\\\\E = \frac{9\times 10^9\times 12.5}{0.11\times 0.11}\\\\E = 9.3\times 10^{12} N/C[/tex]
The direction of electric filed is away from the charge.
Flapping flight is very energy intensive. A wind tunnel test
on an 89 g starling showed that the bird used 12 W of
metabolic power to fly at 11 m/s. What is its metabolic power for starting flight?
Answer:
The metabolic power for starting flight=134.8W/kg
Explanation:
We are given that
Mass of starling, m=89 g=89/1000=0.089 kg
1 kg=1000 g
Power, P=12 W
Speed, v=11 m/s
We have to find the metabolic power for starting flight.
We know that
Metabolic power for starting flight=[tex]\frac{P}{m}[/tex]
Using the formula
Metabolic power for starting flight=[tex]\frac{12}{0.089}[/tex]
Metabolic power for starting flight=134.8W/kg
Hence, the metabolic power for starting flight=134.8W/kg
A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth. Biologically speaking, at the end of the mission, the astronaut's age would be:_______.
a) exactly 50 years.
b) exactly 25 years.
c) exactly 30 years.
d) less than 50 years.
e) more than 50 years.
Answer:
I think D) less than 50 years
Biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years. The correct option is d.
Who is an astronaut?An astronaut observes and performs the experiments based on the universe.
A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth.
Due to special relativity, between space and Earth, both moving with different speeds.
The total age will be less than 30 +20 =50 years. In space, he is moving with speed of light. So, time will move slowly. As measured with respect to Earth, exact time spent in space 20 years will be less on Earth.
So, biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years.
Thus, the correct option is d.
Learn more about astronaut.
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What computer measures physical quantities?
Answer:
Three Types of Computer The Computer are classified into three main types:: • Analog Computers • Digital Computers • Hybrid Computers (Analog + Digital) Analog Computers:: Analog Computer measures “Physical Quantities” for example Temperature, Voltage, Pressure, and Electric Current.
A velocity of ship A relative to ship B is 10m/s in the direction N45E . If the velocity of B is 20m/s in the direction N60W . Find the velocity of ship A and direction.
Answer:
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Explanation:
