Explanation:
Mary had 21 plants when she went on vacation.
When she got back, she only had 14 left alive.
We need to find the percent decrease in the number of plants.
Decrease in plants = 21 - 14 = 7
Percent decrease is given by :
[tex]\%=\dfrac{7}{21}\times 100\\\\\%=33.33\%[/tex]
So, there is 33% pf decrease in the number of plants.
When light travels from one medium to another with a different index of refraction, how is the light's frequency and wavelength affected
Answer:
The frequency does not change, but the wavelength does
Explanation:
Here are the options
A. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, the frequency changes and the wavelength does not.
B. The frequency does change, but the wavelength remains unchanged.
C. Both the frequency and wavelength change.
D. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, neither the wavelength nor the frequency changes.
E. The frequency does not change, but its wavelength does.
When light goes through one medium to the next, the frequency doesn't really change seeing as frequency is dependent on wavelength and light wave velocity. And when the wavelength shifts from one medium to the next.
[tex]n= \frac{C}{V} \ and\ \frac{\lambda_o}{\lambda_m}[/tex]
where [tex]\lambda_o[/tex] indicates wavelength in vacuum
[tex]\lambda_m[/tex] indicates wavelength in medium
n indicates refractive index
v indicates velocity of light wave
c indicates velocity of light
And wavelength is medium-dependent. Frequency Here = v[tex]\lambda[/tex] and shift in wavelength and velocity, not shifts in overall frequency.
Therefore the correct option is E
which objects would have a greater gravitational force between them, Objects A and B, or Objects B and C
Answer:
Objects that are closer together have a stronger force of gravity between them.
Explanation:
For example, the moon is closer to Earth than it is to the more massive sun, so the force of gravity is greater between the moon and Earth than between the moon and the sun.
During the new moon phase, why is the Moon not visible in the sky?
Answer:
Explanation:
The moon gets the light from the sun. When the moon lies between the sun and the earth, only the back portion of the moon gets the light from the sun. So the side facing the sun does not get any light and appears to be dark or does not appear at all.
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plz mark as brainliest!!!!!!!
Answer:
The moon is between the sun, and Earth and reflects light back towards the sun.
Explanation:
A P E X test answer. Just took the test and this is the correct answer.
if an object weighs 550 n and the area is 1 cube
A woman was told in 2020 that she had exactly 15 years to live. If she travels away from the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:
Answer:
2035
Explanation:
The doctor does not travel with the woman, and therefore, he won't experience any relativistic effect on his time. The doctor will judge time by the time here on earth. Technically, the last new year's day the doctor, who is here on earth, would expect the woman to celebrate will be in 2020 + 15 years = 2035
A mass weighing 64 pounds stretches a spring 6 inches. The mass moves through a medium offering a damping force that is numerically equal to β times the instantaneous velocity. Determine the values of β > 0 for which the system will exhibit oscillatory motion. (Enter your answer as a single inequality.)
Answer:
chk photo
Explanation:
The values of β for which the system will exhibit oscillatory motion are those greater than 0.
What is oscillatory motion?To determine the values of β for which the system will exhibit oscillatory motion, you need to perform a damped oscillation analysis on the system.
The equation of motion for a damped oscillator is given by:
F = ma = -kx - βv
where F is the force, m is the mass, a is the acceleration, k is the spring constant, x is the displacement of the mass from its equilibrium position, β is the damping coefficient, and v is the velocity of the mass.
To determine the values of β for which the system will exhibit oscillatory motion, you need to solve for the roots of the characteristic equation, which is obtained by setting the acceleration (a) equal to zero:
0 = -kx - βv
βv + kx = 0
x(β + k) = 0
The roots of this equation are x = 0 and x = -(β + k). If x = 0 is a root of the characteristic equation, then the system will exhibit steady-state behavior (i.e., the mass will come to rest at its equilibrium position and not oscillate). If x = -(β + k) is a root of the characteristic equation, then the system will exhibit oscillatory behavior (i.e., the mass will oscillate around its equilibrium position).
Therefore, the values of β for which the system will exhibit oscillatory motion are those for which x = -(β + k) is a root of the characteristic equation. This means that β must be such that:
-(β + k) < 0
β > -k
Since k is a positive value (the spring constant is always positive), the inequality simplifies to:
β > 0
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Water flows at 0.00027 m3/s through a 10-m long garden hose lying on the ground, with a radius of 0.01 m. Water has a viscosity of 1 mPa.s What is the magnitude of gauge pressure in Pa of the water entering the hose
Answer:
The gauge pressure is [tex]P = 687.4 \ Pa[/tex]
Explanation:
From the question we are told that
The rate of flow is [tex]Q = 0.00027 m^3 /s[/tex]
The height is h = 10 m
The radius is r = 0.01 m
The viscosity is [tex]\eta = 1mPa \cdot s = 1 *10^{-3} \ Pa\cdot s[/tex]
Generally the gauge pressure according to Poiseuille's equation is mathematically represented as
[tex]P = 8 \pi \eta * \frac{L * v }{ A}[/tex]
Here v is the velocity of the water which is mathematically represented according to continuity equation as
[tex]v = \frac{Q}{A }[/tex]
Where A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 *(0.01)^2[/tex]
[tex]A = 3.142 *10^{-4} \ m^2[/tex]
So
[tex]v = \frac{ 0.00027}{3.142*10^{-4}}[/tex]
[tex]v = 0.8593 \ m/s[/tex]
So
[tex]P = 8 * 3.142 * 1.0*10^{-3}* \frac{10 * 0.8593 }{ 3.142*10^{-4}}[/tex]
[tex]P = 687.4 \ Pa[/tex]
To protect her new two-wheeler, Iroda Bike
buys a length of chain. She finds that its
linear density is 0.65 lb/ft.
If she wants to keep its weight below 1.4 lb,
what length of chain is she allowed?
Answer in units of ft.
Answer:
2.2 ft
Explanation:
0.65 lb / 1 ft = 1.4 lb / x
x ≈ 2.2 ft
At what frequency should a 200-turn, flat coil of cross sectional area of 300 cm2 be rotated in a uniform 30-mT magnetic field to have a maximum value of the induced emf equal to 8.0 V
Answer:
The frequency of the coil is 7.07 Hz
Explanation:
Given;
number of turns of the coil, 200 turn
cross sectional area of the coil, A = 300 cm² = 0.03 m²
magnitude of the magnetic field, B = 30 mT = 0.03 T
Maximum value of the induced emf, E = 8 V
The maximum induced emf in the coil is given by;
E = NBAω
Where;
ω is angular frequency = 2πf
E = NBA(2πf)
f = E / 2πNBA
f = (8) / (2π x 200 x 0.03 x 0.03)
f = 7.07 Hz
Therefore, the frequency of the coil is 7.07 Hz
14 A particle of mass m strikes a vertical rigid wall perpendicularly from the left with velocity v.
rigid wall
If the collision is perfectly elastic, the total change in momentum of the particle that occurs as a
result of the collision is
A. 2mv to the right.
B. 2my to the left.
C. my to the right.
D. my to the left.
Answer:
C. mv to the right
Explanation:
momentum of thr particle=m1v1
momentum of the wall=m2v2
m1v1+ m2v2 =m1u1+ m2u2 since the wall doesn't move it's momentum is zero.
m1v1 =m1u1
therefore change in that occurs as result of the collision is C. mv to the right
The total change in momentum of the particle of mass m that collides elastically with a vertical rigid wall perpendicularly from the left with velocity v is 2mv to the left (option B).
The total change in momentum is given by:
[tex] \Delta p = p_{f} - p_{i} [/tex]
In the initial state, the particle is moving to the right until it collides with the rigid wall, so:
[tex] p_{i} = mv [/tex]
In the final state, the particle moves backward after the collision with the wall, so:
[tex] p_{f} = -mv [/tex]
The minus sign is because it is moving in the negative x-direction (to the left)
Hence, the total change in momentum is:
[tex] \Delta p = -mv - mv = -2mv [/tex]
Therefore, the total change in momentum of the particle is 2mv to the left (option B).
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PLEASE HELP FAST Five-gram samples of brick and glass are at room temperature. Both samples receive equal amounts of energy due to heat flow. The specific heat capacity of brick is 0.22 cal/g°C and the specific heat capacity of glass is 0.22 cal/g°C. Which of the following statements is true? 1.The temperature of each sample will increase by the same amount. 2.The temperature of each sample will decrease by the same amount. 3.The brick will get hotter than the glass. 4.The glass will get hotter than the brick.
Answer:
1.The temperature of each sample will increase by the same amount
Explanation:
This is because, since their specific heat capacities are the same and we have the same mass of each substance, and the same amount of energy due to heat flow is supplied to both the glass and brick at room temperature, their temperatures would thereby increase by the same amount.
This is shown by the calculation below
Q = mcΔT
ΔT = Q/mc where ΔT = temperature change, Q = amount of heat, m = mass of substance and c = specific heat capacity of substance.
Since Q, m and c are the same for both substances, thus ΔT will be the same.
So, the temperature of each sample will increase by the same amount
Sammy is 5 feet and 5.3 inches tall.tall.what is sammy's height in metres?
Answer:
65.3
Explanation:
1 foot = 12 inches
Sammy is 5 feet tall.
5 feet = ? inches
Multiply the feet value by 12 to find in inches.
5 × 12
= 60
Add 5.3 inches to 60 inches.
60 + 5.3
= 65.3
Answer:
It will be 》》》》1.664716m
The velocity function (in meters per second) is given for a particle moving along a line. Find the total distance traveled by the particle during the given interval
Answer:
s=((vf+vi)/2)t vf is final velocity and vi is initial velocity
An object is placed in a room where the temperature is 20 degrees C. The temperature of the object drops by 5 degrees C in 4 minutes and by 7 degrees C in 8 minutes. What was the temperature of the object when it was initially placed in the room
Answer:
28.3°C
Explanation:
Using
T(t) = (T(0) - 20)*(e^(-k*t)) + 20
for some positive number k, and some initial temperature T(0).
Boundary conditions:
T(4) = T(0) - 5 _______ (i)
T(8) = T(0) - 7 _______ (ii)
==> solving for T(0) and k :
(i):
(T(0) - 20)*(e^(-k*4)) + 20 = T(0) - 5 ==>
(T(0) - 20)*(e^(-k*4)) = T(0) - 20 - 5
(T(0) - 20)*(e^(-k*4)) = (T(0) - 20) - 5
5 = (T(0) - 20) - (T(0) - 20)*(e^(-k*4))
5 = (T(0) - 20) * ( 1 - e^(-k*4) )
(ii):
(T(0) - 20)*(e^(-k*8)) + 20 = T(0) - 7
(T(0) - 20)*(e^(-k*8)) = (T(0) - 20) - 7
7 = (T(0) - 20) - (T(0) - 20)*(e^(-k*8))
7 = (T(0) - 20) * (1 - e^(-k*8))
In both results, subsitute x = e^(-4k) and C = (T(0) - 20)
(i): 5 = C * (1 - x)
(ii): 7 = C * (1 - x^2) = C * (1-x)*(1+x)
Substitute C*(1-x) from (i) into (ii):
(ii): 7 = 5*(1+x) ==> (1+x) = 7/5 ==> x = 2/5
back into (i):
(i): 5 = C * (1 - 2/5) ==> 5 = C * 3/5 ==> C = 25/3
C = T(0) - 20 ==>
T(0) = C + 20 = 25/3 + 20 = 25/3 + 60/3 = 85/3
= 28.3°C
A certain digital camera having a lens with focal length 7.50 cmcm focuses on an object 1.70 mm tall that is 4.70 mm from the lens.
Part A. How far must the lens be from the photocells?
s = cm
Part B. Is the image on the photocells erect or inverted? Real or virtual?
a. The image is erect and real.
b. The image is inverted and real.
c. The image is erect and virtual.
d. The image is inverted and virtual.
Part C. How tall is the image on the photocells?
|h?| = cm
Part D. A SLR digital camera often has pixels measuring 8.00?m
Answer:
a. 7.62cm
b. Real and inverted
c. 2.76 cm
d. 3450
Explanation:
We proceed as follows;
a. the lens equation that relates the object distance to the image distance with the focal length is given as follows;
1/f = 1/p + 1/q
making q the subject of the formula;
q = pf/p-f
From the question;
p = 4.70m
f = 7.5cm = 0.075m
Substituting these, we have ;
q = (4.7)(0.075)/(4.7-0.075) = 0.3525/4.625 = 0.0762 = 7.62 cm
b. The image is real and inverted since the image distance is positive
c. We want to calculate how tall the image is
Mathematically;
h1 = (q/p)h0
h1 = (7.62/4.70)* 1.7
h1 = 2.76 cm
d. We want to calculate the number of pixels that fit into this image
Mathematically:
n = h1/8 micro meter
n = 2.76cm/8 micro meter = 2.76 * 10^-2/8 * 10^-6 = 3450
Polarized sunglasses:
a. block most sunlight because sunlight is polarized
b. are better but work the same way as non-polarized sunglasses
c. are polarized to filter out certain wavelengths of light
d. block reflected light because reflected light is partially polarized.
Polarized sunglasses creates filter of vertical openings for light. The light rays will reach the eyes of human vertically only.
The sun rays will not reach human eye directly which will create a shield for sun light burden on human eye.
Polarized sunglasses are best used for blocking and eliminating certain wavelengths of light.
Therefore the correct answer is option C. Polarizes Sunglasses are polarized and it filter out certain wavelengths of light.
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An electrostatic paint sprayer contains a metal sphere at an electric potential of 25.0 kV with respect to an electrically grounded object. Positively charged paint droplets are repelled away from the paint sprayer's positively charged sphere and towards the grounded object. What charge must a 0.168-mg drop of paint have so that it will arrive at the object with a speed of 18.8 m/s
Answer:
The charge is [tex]Q = 2.177 *10^{-9} \ C[/tex]
Explanation:
From the question we are told that
The electric potential is [tex]V = 25.0 \ kV = 25.0 *10^{3}\ V[/tex]
The mass of the drop is [tex]m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg[/tex]
The speed is [tex]v = 18.8 \ m/s[/tex]
Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question is mathematically represented as
[tex]Q = \frac{m v^2 }{ 2 * V }[/tex]
Substituting values
[tex]Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }[/tex]
[tex]Q = 2.177 *10^{-9} \ C[/tex]
A microwave oven operates at 2.4 GHz with an intensity inside the oven of 2300 W/m2 . Part A What is the amplitude of the oscillating electric field
Answer:
The amplitude of the oscillating electric field is 1316.96 N/C
Explanation:
Given;
frequency of the wave, f = 2.4 Hz
intensity of the wave, I = 2300 W/m²
Amplitude of oscillating magnetic field is given by;
[tex]B_o = \sqrt{\frac{2\mu_o I}{c} }[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
I is intensity of wave
c is speed of light = 3 x 10⁸ m/s
[tex]B_o = \sqrt{\frac{2*4\pi *10^{-7}*2300}{3*10^8} } \\\\B_o = 4.3899 *10^{-6} \ T[/tex]
The amplitude of the oscillating electric field is given by;
E₀ = cB₀
E₀ = 3 x 10⁸ x 4.3899 x 10⁻⁶
E₀ = 1316.96 N/C
Therefore, the amplitude of the oscillating electric field is 1316.96 N/C
A double-convex thin lens is made of glass with an index of refraction of 1.52. The radii of curvature of the faces of the lens are 60 cm and 72 cm. What is the focal length of the lens
Answer:
63 cm
Explanation:
Mathematically;
The focal length of a double convex lens is given as;
1/f = (n-1)[1/R1 + 1/R2]
where n is the refractive index of the medium given as 1.52
R1 and R2 represents radius of curvature which are given as 60cm and 72cm respectively.
Plugging these values into the equation, we have:
1/f = (1.52-1)[1/60 + 1/72)
1/f = 0.0158
f = 1/0.0158
f = 63.29cm which is approximately 63cm
Proposed Exercises: Strength and Acceleration in Circular Movement In the situation illustrated below, a 7kg sphere is connected to a rope so that it can rotate in a vertical plane around an O axis perpendicular to the plane of the figure. When the sphere is in position A, it has a speed of 3m/s. Determine for this position the modulus of tension on the string and the rate at which the tangential velocity is increased.
Answer:
81 N
7.1 m/s²
Explanation:
Draw a free body diagram of the sphere. There are two forces:
Weight force mg pulling straight down,
and tension force T pulling up along the rope.
Sum of forces in the centripetal direction:
∑F = ma
T − mg sin 45° = m v² / r
T = m (g sin 45° + v² / r)
T = (7 kg) (10 m/s² sin 45° + (3 m/s)² / 2 m)
T = 81 N
Sum of forces in the tangential direction:
mg cos 45° = ma
a = g cos 45°
a = (10 m/s²) cos 45°
a = 7.1 m/s²
A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x axis at 37 m/s in the positive direction. The second, with mass 22 g , moves along the y axis at 34 m/s in the positive direction. Find the velocity of third piece.
Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
A student holds a bike wheel and starts it spinning with an initial angular speed of 7.0 rotations per second. The wheel is subject to some friction, so it gradually slows down.
In the 10.0 s period following the inital spin, the bike wheel undergoes 60.0 complete rotations. Assuming the frictional torque remains constant, how much more time Δ????s will it take the bike wheel to come to a complete stop?
The bike wheel has a mass of 0.625 kg0.625 kg and a radius of 0.315 m0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque ????fτf that was acting on the spinning wheel.
Answer:
a) Δt = 24.96 s , b) τ = 0.078 N m
Explanation:
This is a rotational kinematics exercise
θ = w₀ t - ½ α t²
Let's reduce the magnitudes the SI system
θ = 60 rev (2π rad / 1 rev) = 376.99 rad
w₀ = 7.0 rot / s (2π rad / 1 rpt) = 43.98 rad / s
α = (w₀ t - θ) 2 / t²
let's calculate the annular acceleration
α = (43.98 10 - 376.99) 2/10²
α = 1,258 rad / s²
Let's find the time it takes to reach zero angular velocity (w = 0)
w = w₀ - alf t
t = (w₀ - 0) / α
t = 43.98 / 1.258
t = 34.96 s
this is the total time, the time remaining is
Δt = t-10
Δt = 24.96 s
To find the braking torque, we use Newton's law for angular motion
τ = I α
the moment of inertia of a circular ring is
I = M r²
we substitute
τ = M r² α
we calculate
τ = 0.625 0.315² 1.258
τ = 0.078 N m
The total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.
Given data:
The initial angular speed of wheel is, [tex]\omega = 7.0 \;\rm rps[/tex] (rps means rotation per second).
The time interval is, t' = 10.0 s.
The number of rotations made by wheel is, n = 60.0.
The mass of bike wheel is, m = 0.625 kg.
The radius of wheel is, r = 0.315 m.
The problem is based on rotational kinematics. So, apply the second rotational equation of motion as,
[tex]\theta = \omega t-\dfrac{1}{2} \alpha t'^{2}[/tex]
Here, [tex]\theta[/tex] is the angular displacement, and its value is,
[tex]\theta =2\pi \times 60\\\\\theta = 376.99 \;\rm rad[/tex]
And, angular speed is,
[tex]\omega = 2\pi n\\\omega = 2\pi \times 7\\\omega = 43.98 \;\rm rad/s[/tex]
Solving as,
[tex]376.99 = 43.98 \times 10-\dfrac{1}{2} \alpha \times 10^{2}\\\\\alpha = 1.25 \;\rm rad/s^{2}[/tex]
Apply the first rotational equation of motion to obtain the value of time to reach zero final velocity.
[tex]\omega' = \omega - \alpha t\\\\0 = 43.98 - 1.25 \times t\\\\t = 35.18 \;\rm s[/tex]
Then total time is,
T = t - t'
T = 35.18 - 10
T = 25.18 s
Now, use the standard formula to obtain the value of braking torque as,
[tex]T = m r^{2} \alpha\\\\T = 0.625 \times (0.315)^{2} \times 1.25\\\\T = 0.0775 \;\rm Nm[/tex]
Thus, we can conclude that the total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.
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How much time will elapse if a radioisotope with a half-life of 88 seconds decays to one-sixteenth of its original mass?
Answer:
352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.
Explanation:
The decay of radioisotopes are represented by the following ordinary differential equation:
[tex]\frac{dm}{dt} = -\frac{t}{\tau}[/tex]
Where:
[tex]t[/tex] - Time, measured in seconds.
[tex]\tau[/tex] - Time constant, measured in seconds.
[tex]m[/tex] - Mass of the radioisotope, measured in grams.
The solution of this expression is:
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]
Where [tex]m_{o}[/tex] is the initial mass of the radioisotope, measured in kilograms.
The ratio of current mass to initial mass is:
[tex]\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }[/tex]
The time constant is now calculated in terms of half-life:
[tex]\tau = \frac{t_{1/2}}{\ln2}[/tex]
Where [tex]t_{1/2}[/tex] is the half-life of the radioisotope, measured in seconds.
Given that [tex]t_{1/2} = 88\,s[/tex], the time constant of the radioisotope is:
[tex]\tau = \frac{88\,s}{\ln 2}[/tex]
[tex]\tau \approx 126.957\,s[/tex]
Now, if [tex]\frac{m(t)}{m_{o}(t)} = \frac{1}{16}[/tex] and [tex]\tau \approx 126.957\,s[/tex], the time is:
[tex]t = -\tau \cdot \ln\frac{m(t)}{m_{o}}[/tex]
[tex]t = -(126.957\,s)\cdot \ln \frac{1}{16}[/tex]
[tex]t \approx 352\,s[/tex]
352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.
PLEASE HELP Which of the following describes the efficiency of real machines? 1.Efficiency is 100% 2.IMA is always larger than the AMA 3.IMA is always smaller than the AMA 4.IMA is equal to the AMA
Answer:
IMA is always larger than the AMA
Explanation:
IMA is Ideal Mechanical Advantage and it equals the length of effort that is divided by the length of resistance which is given by the formula
IMA= Fr/Fe
Where Fr is the resistance force
Fe is the effort force.
IM= de/dr
Where de is the distance of the applied effort
dr is the distance traveled by the load.
In any real machine, the effort is needed to overcome friction and because of this, the ideal mechanical advantage(IMA) is always larger than the actual mechanical advantage (AMA)
To protect her new two-wheeler, Iroda Bike
buys a length of chain. She finds that its
linear density is 0.68 lb/ft.
If she wants to keep its weight below 1.4 lb,
what length of chain is she allowed?
Answer in units of ft.
Answer:
1.8/0.61 =2.95 ft
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Two instruments produce a beat frequency of 5 Hz. If one has a frequency of 264 Hz, what could be the frequency of the other instrument
Answer:
259 Hz or 269 Hz
Explanation:
Beat: This is the phenomenon obtained when two notes of nearly equal frequency are sounded together. The S.I unit of beat is Hertz (Hz).
From the question,
Beat = f₂-f₁................ Equation 1
Note: The frequency of the other instrument is either f₁ or f₂.
If the unknown instrument's frequency is f₁,
Then,
f₁ = f₂-beat............ equation 2
Given: f₂ = 264 Hz, Beat = 5 Hz
Substitute into equation 2
f₁ = 264-5
f₁ = 259 Hz.
But if the unknown frequency is f₂,
Then,
f₂ = f₁+Beat................. Equation 3
f₂ = 264+5
f₂ = 269 Hz.
Hence the beat could be 259 Hz or 269 Hz
Convierta 164 decimetros a hectometros
Answer:
sinco
Explanation:
When light of wavelength 233 nm shines on a metal surface the maximum kinetic energy of the photoelectrons is 1.98 eV. What is the maximum wavelength (in nm) of light that will produce photoelectrons from this surface
Answer:
λmax = 372 nm
Explanation:
First we find the energy of photon:
E = hc/λ
where,
E = Energy of Photon = ?
λ = Wavelength of Light = 233 nm = 2.33 x 10⁻⁷ m
c = speed of light = 3 x 10⁸ m/s
h = Planks Constant = 6.626 x 10⁻³⁴ J.s
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.33 x 10⁻⁷ m)
E = 8.5 x 10⁻¹⁹ J
Now, from Einstein's Photoelectric Equation:
E = Work Function + Kinetic Energy
8.5 x 10⁻¹⁹ J = Work Function + (1.98 eV)(1.6 x 10⁻¹⁹ J/1 eV)
Work Function = 8.5 x 10⁻¹⁹ J - 3.168 x 10⁻¹⁹ J
Work Function = 5.332 x 10⁻¹⁹ J
Since, work function is the minimum amount of energy required to emit electron. Therefore:
Work Function = hc/λmax
λmax = hc/Work Function
where,
λmax = maximum wavelength of light that will produce photoelectrons = ?
Therefore,
λmax = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5.332 x 10⁻¹⁹ J)
λmax = 3.72 x 10⁻⁷ m
λmax = 372 nm
A neutron star has a mass of between 1.4-2.8 solar masses compressed to the size of:
A. Earth
B. The state of Oregon
C. North America
D. An average city
The correct answer is D. An average city
Explanation:
A neutron star differs from others due to its massive density, this means a lot of matter is compressed in a small area. Indeed, neutron stars have a mass of around 1.4 to 2.8 times the mass of the sun. But these are considerably small as they only measure around 20 kilometers, which is the size of an average city. Additionally, neutron stars are this dense because they are the result of a regular star exploding, which leads to a super-dense core, or neutron star. In this context, the mass of a neutron star is compressed to the size of an average city.
Warm blooded animals are homeothermic; that is, they maintain an approximately constant body temperature. (Forhumans it's about 37 oC.) When they are in an environment that is below their optimum temperature, they use energy derived from chemical reactions within their bodies to warm them up. One of the ways that animals lose energy to their environment is through radiation. Every object emits electromagnetic radiation that depends on its temperature. For very hot objects like the sun, that radiation is visible light. For cooler objects, like a house or a person, that radiation is in the infrared and is invisible. Nonetheless, it still carries energy. Other ways that energy is lost by a warm animal to a cool environment includes conduction (direct touching of a cooler object) and convection (cooler air moving and carrying thermal energy away). See Heat Transfer for a discussion of all three.
For this problem, we'll just consider how much energy an animal needs to burn (obtain from internal chemical reactions) in order to stay warm just from radiation losses. The rate at which an object loses energy through radiation is given by the Stefan-Boltzmann equation:
Rate of energy loss = AεσT4
where T is the absolute (Kelvin) temperature, A is the area of the object, ε is the emissivity (unitless and =1 for a perfect emitter, less for anything else), and σ is the Stefan-Boltzmann constant:
σ = 5.67 x 10-8 J/(s m2 K4)
Consider a patient trying to sleep naked in a cool room (55 oF = 13 oC). Assume that the person being considered is a perfect emitter and absorber of radiation (ε = 1), has a surface area of about 2.5 m2, and a mass of 80 kg.
a. A person emits thermal radiation at a rate corresponding to a temperature of 37 oC and absorbs radiation at a rate (from the air and walls) corresponding to a temperature of 13 oC. Calculate the individual's net rate of energy loss due to radiation (in Watts = Joules/second).
net rate of energy loss = Watts
b. Assume the patient produces no energy to keep warm. If they have a specific heat about equal to that of water (1 Cal/kg-oC) how much would their temperature fall in one hour? (1 Cal = 1kcal = 103 cal)
ΔT = oC
c. Given that the energy density of fat is about 9 Cal/g, how many grams of fat would the person have to utilize to maintain their body temperature in that environment for one hour?
amount of fat needed = g
Answer:
a) 360.7 J/s
b) 16.23 °C
c) 34.48 g
Explanation:
The mass of the person = 80 kg
The person is a perfect emitter, ε = 1
surface area of the person = 2.5 m^2
a) If he emits radiation at 37 °C, [tex]T_{out}[/tex] = 37 + 273 = 310 K
and receives radiation at 13 °C, [tex]T_{in}[/tex] = 13 + 273 = 286 K
Rate of energy loss E = Aεσ([tex]T^{4} _{out}[/tex] - [tex]T^{4} _{in}[/tex] )
where σ = 5.67 x 10^-8 J/(s m^2 K^4)
substituting values, we have
E = 2.5 x 1 x 5.67 x 10^-8 x ([tex]310^{4}[/tex] - [tex]286^{4}[/tex]) = 360.7 J/s
b) If they have specific heat about equal to that of water = 1 Cal/kg-°C
but 1 Cal = 1 kcal = 10^3 cal
specific heat of person is therefore = 10^3 cal/kg-°C
heat loss = 360.7 J/s = 360.7 x 3600 = 1298520 J/hr
heat lost in 1 hour = 1 x 1298520 = 1298520 J
This heat lost = mcΔT
where ΔT is the temperature fall
m is the mass
c is the specific heat equivalent to that of water
the specific heat is then = 10^3 cal/kg-°C
equating, we have
1298520 = 80 x 10^3 x ΔT
1298520 = 80000ΔT
ΔT = 1298520/80000 = 16.23 °C
c) 1298520 J = 1298520/4184 = 310.35 Cal
density of fat = 9 Cal/g
gram of fat = 310.35/9 = 34.48 g