Answer:
a) 800N × 20 cm = 1100N × x cm
16000= 1100x
x= 14.5
therefore it must be placed on the (50 + 14.5)cm mark
= 64.5 cm mark
b) 800N × 25 cm = (1100N × 10 cm)+(450N × x cm)
20000 = 11000 + 450x
450x = 9000
x = 20 cm
therefore it must be placed on the (50 + 20)cm mark
= 70 cm mark
a) The distance at which the mass m₂(110 g) should be suspended is 64.54 cm.
b) The distance at which mass m₃(45 g) should be suspended is 70 cm.
What is meant by principle of moments?According to the Principle of Moments, when a body is balanced or is at equilibrium, the total clockwise and anticlockwise moments about a given point are equal.
a) m₁ = 80 g
m₂ = 110 g
r₁ = 30 cm
According to the Principle of Moments,
m₁r₁ = m₂r₂
Therefore, the distance,
r₂ = m₁r₁/m₂
r₂ = 80 x 20/110
r₂ = 14.54 cm
So, the distance at which mass m₂ should be suspended is,
r' = 50 + 14.54
r' = 64.54 cm
b) m₁ = 80 g
m₂ = 110 g
m₃ = 45 g
r₁ = 25 cm
r₂ = 60 cm
According to the Principle of Moments,
m₁r₁ = m₂r₂ + m₃r₃
80 x 25 = (110 x 10) + (45 x r₃)
45 x r₃ = 2000 - 1100
r₃ = 900/45
r₃ = 20 cm
So, the distance at which mass m₃ should be suspended is,
r' = 50 + 20
r' = 70 cm.
Hence,
a) The distance at which the mass m₂(110 g) should be suspended is 64.54 cm.
b) The distance at which mass m₃(45 g) should be suspended is 70 cm.
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The following represents a mass attached to a spring oscillating in simple harmonic motion. X(t) = 4.0 cos(3.0t +0.10) units of meters, radians and seconds
a. What is the amplitude? (1)
b. What is the angular frequency? (1)
c. What is the frequency? (2)
d. What is the period? (2)
e. What is the phase constant? (1)
f. What is the maximum speed? (2)
g. If the mass m= 1.2 kg, what is the spring constant? (2)
h. If the mass m= 1.2 kg, what is the total energy of the oscillator? (3)
i. What is the potential energy of the oscillator at t=0 s? (3)
j. What is the kinetic energy of the oscillator at t=0 s? (3)
Answer:
a) A = 4.0 m , b) w = 3.0 rad / s , c) f = 0.477 Hz , d) T = 20.94 s
Explanation:
The equation that describes the oscillatory motion is
x = A cos (wt + fi)
In the exercise we are told that the expression is
x = 4.0 cos (3.0 t + 0.10)
let's answer the different questions
a) the amplitude is
A = 4.0 m
b) the frequency or angular velocity
w = 3.0 rad / s
c) angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 3 / 2π
f = 0.477 Hz
d) the period
frequency and period are related
T = 1 / f
T = 1 / 0.477
T = 20.94 s
e) the phase constant
Ф = 0.10 rad
f) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф)
speed is maximum when sine is + -1
v = A w
v = 4 3
v = 12 m / s
g) the angular velocity is
w² = k / m
k = m w²
k = 1.2 3²
k = 10.8 N / m
h) the total energy of the oscillator is
Em = ½ k A²
Em = ½ 10.8 4²
Em = 43.2 J
i) the potential energy is
Ke = ½ k x²
for t = 0 x = 4 cos (0 + 0.1)
x = 3.98 m
j) kinetic energy
K = ½ m v²
for t = 00.1 ²
v = A w sin 0.10
v = 4 3 sin 0.10
v = 1.98 m / s
the coefficient of static friction between mass mA
and the table is 0.40, whereas the coefficient of kinetic friction
is 0.20.
(a) What minimum value of mA will keep the system from
starting to move?
(b) What value(s) of mA will keep the system moving at
constant speed?
[Ignore masses of the cord and the (frictionless) pulley.]
Answer:
(a) 5.0 kg
(b) 10 kg
Explanation:
Draw a free body diagram for each block. There are 4 forces on block A:
Weight force mAg pulling down,
Normal force N pushing up,
Tension force T pulling right,
and friction force Nμ pushing left.
There are 2 forces on block B:
Weight force mBg pulling down,
and tension force T pulling up.
Whether the system is just starting to move, or moving at constant speed, the acceleration is 0.
Sum of forces on B in the -y direction:
∑F = ma
mBg − T = 0
mBg = T
Sum of forces on A in the +y direction:
∑F = ma
N − mAg = 0
N = mAg
Sum of forces on A in the +x direction:
∑F = ma
T − Nμ = 0
T = Nμ
Substitute:
mBg = mAg μ
mA = mB / μ
(a) When the system is just starting to move, μ = 0.40.
mA = 2.0 kg / 0.40
mA = 5.0 kg
(b) When the system is moving at constant speed, μ = 0.20.
mA = 2.0 kg / 0.20
mA = 10 kg
m_1=5kg
The value(s) of mA will keep the system moving at constant speed is
m=10kg
From the question we are told
the coefficient of static friction between mass mA and the table is 0.40, where as the coefficient of kinetic friction is 0.20.
a)
Generally the equation for the Tension is mathematically given as
T=mg
Where
[tex]m_1g=m_2g[/tex]
Therefore
[tex]m_1=\frac{2.0}{0.4}\\\\m_1=5kg[/tex]
b
Generally the equation for the Tension is mathematically given as
[tex]T=f\\\\T=u_km_1g\\\\\m_1=\frac{m_2}{u}\\\\m_1=\frac{2}{0.2}[/tex]
m=10kg
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When an auditorium has a solid wall, sound waves will tend to perfectly reflect off the wall (i.e. with a 180o phase change). If listening to music, as from an orchestra, the incoming and reflected waves will interfere with each other. For a listener sitting 0.5 m from the wall, what is the lowest frequency which gets suppressed by this interference
This question is incomplete, the complete question is;
When an auditorium has a solid wall, sound waves will tend to perfectly reflect off the wall (i.e. with a 180o phase change). If listening to music, as from an orchestra, the incoming and reflected waves will interfere with each other. For a listener sitting 0.5 m from the wall, what is the lowest frequency which gets suppressed by this interference? Use vsound=330 m/s.
Answer: f = 165 Hz
the lowest frequency which gets suppressed by this interference is 165 Hz
Explanation:
For a reflected wave (out of phase), the path difference between the incoming and reflected wave should be equal to the half integral multiple of wavelength.
r₂ - r₁ = ( m + 1/2) λ/2
r₂ is the distance from the source to observer via reflection
r₁ is distance from source to observer
here r₂ would travel an additional distance of 0.5 m due to reflection that straight approaching wave.
Therefor to have minimum/lowest possible frequency, we say m = 0
we substitute
0.5 = ( 0 + 1/2 ) λ/2
λ = 2m
The frequency would be
f = Vsound / λ
f = 330 / 2
f = 165 Hz
Therefore the lowest frequency which gets suppressed by this interference is 165 Hz
A student attaches a length of nylon fishing line to a fence post. She stretches it out and shakes the end of the rope in her hand back and forth to produce waves on the line. The most efficient way for her to increase the wavelength is to
Answer:
increase the tension on the hose and shake the end fewer times per second.
Explanation:
This is because Tension of the rope is directly proportional to wavelength and inversely proportional to the frequency ( shakes per second) so increasing tension increases wavelength and vice versa
Wavelength= Tension/ frequency
The most efficient way for her to increase the wavelength is to increase the rate of shaking the rope per second.
What is wavelength?Wavelength measures the distance of wave that occur between one crest and another crest.
To increase the wavelength the rope end should be shaked in a number of time per second this will help build up tension in the rope.
Therefore, she should increase the rate of shaking the rope per second.
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A 1.70 mm string of weight 0.0135 NN is tied to the ceiling at its upper end, and the lower end supports a weight WW. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation
y(x,t) = (8.50mm)cos(172rad?m?1x?2730rad?s?1t)
Assume that the tension of the string is constant and equal to W.
1) How much time does it take a pulse to travel the full length of the string?
2) What is the weight W?
3) How many wavelengths are on the string at any instant of time?
4) What is the equation for waves traveling down the string?
a) y(x,t) = (8.50 mm)cos(172rad?m?1 x ?2730rad?s?1t)
b) y(x,t) = (8.50 mm)cos(172rad?m?1 x +2730rad?s?1t)
c) y(x,t) = (10.5 mm)cos(172rad?m?1 x +2730rad?s?1t)
d) y(x,t) = (10.5 mm)cos(172rad?m?1 x ?2730rad?s?1t)
Answer:
d) y(x,t) = (10.5 mm)cos(172rad?m?1 x ?2730rad?s?1t)
Using the differential equation modeling Newton's Law of Cooling dTdt=k(T−Te)dTdt=k(T−Te), Answer the following. Brewing Coffee: The brewing temperature of the water used is very important. It should be between 195 F and 205 F. The closer to 205 F the better. Boiling water (212 F) should never be used, as it will burn the coffee. Water that is less than 195 F will not extract properly. On the other hand, coffee that has a temperature of 205 F is too hot to drink. Coffee is best when it is served at a temperature of 140 F to 155 F (the Goldilocks range). Suppose coffee is initially brewed at 205 F and the room temperature is 70 F. Determine the value of kk if the temperature of the coffee drops from 205 F to 200 F in the first two minutes after brewing. Round answer to 4 decimal places.
Answer:
k = -3.1450 10⁻⁴ s⁻¹
Explanation:
In this exercise we are given the equation that describes the cooling process
dT / dt = k (T -)
Let's solve is this equation,
dT / (T-T_ {e}) = k dt
change of variable for integration
T -T_{e} = T ’
dT = dT '
∫ dT ’/ T’ = k ∫ dt
we integrate
ln T ’= k t
we change to the initial variables
ln (T - T_{e}) = k t
Let's evaluate from the lower limit T = T for t = 0 to the upper limit T = T₀ for time t
ln (T₀ -T_{e}) - ln (T -T_{e}) = k (t-0)
we simplify
ln (T₀ -T_{e} / T -T_{e}) = k t
k = ln (T₀ -T_{e}) / (T-Te) / t
In the exercise they indicate that the temperature T = 205 F, the ambient temperature is T_{e} = 70F, the temperature to which T₀ = 200 F falls in a time t = 2 min = 120 s
Let's calculate
k = ln [(200- 70) / (205 -70)] / 120
k = -0.0377403 / 120
k = -3.1450 10⁻⁴ s⁻¹
Suppose an electron and a proton move at the same speed. Which particle has a longer de Broglie wavelength
Answer:
Therefore, electron will have a longer de Broglie Wavelength.
Explanation:
The de Broglie wavelength is given by the following formula:
λ = h/mv
where.
λ = de Broglie wavelength
h = Plank's Constant
m = mass of the particle.
v = speed of the particle
Since, the speed of both electron and proton is same and Plank's constant is also a constant. Therefore, the de Broglie wavelength depends solely upon the mass of electron and proton, as follows:
λ ∝ 1/m
It shows that wavelength is inversely proportional to the mass of particle.
Since, the mass of electron is less than the mass of proton.
Therefore, electron will have a longer de Broglie Wavelength.
A gas is compressed by an isothermal process that decreases its volume by a factor of 2. In this process, the pressure:________
a. Does not change
b. Increases by a factor of less than 2.
c. Increases by a factor of 2.
d. Increases by a factor of more than 2.
Answer:
In this process, the pressure increases by a factor of 2.
Explanation:
For an isothermal process, the temperature remains constant throughout the process.
Also, PV = constant
It means that there is an inverse relationship between pressure and volume of the gas.
If the volume is decreased by a factor of 2, the pressure increases by a factor of 2. Hence, the correct option is (c).
Two long straight wires are parallel and 9.5 cm apart. They are to carry equal currents such that the magnetic field at a point halfway between them has magnitude 280 T.
(a) Should the currents be in the same or opposite directions?
(b) How much current is needed?
Answer:
(a) the current will flow in opposite direction
(b) the current needed is 33.25 A
Explanation:
(a) At the center of the two parallel wires, the two wires will have the same magnitude of magnetic field. In order to have a non a zero value of magnetic field at the center, the field must be in the same direction and the current will flow in opposite direction according to right hand rule.
(b) How much current is needed
Given;
distance between the two parallel wires, d = 9.5 cm = 0.095 m
magnitude of magnetic field at a point halfway between the wires, [tex]B_c[/tex] = 280 μT (This unit was corrected to obtain feasible current)
The magnetic field at distance R due to an infinite wire is given by;
[tex]B = \frac{\mu_o I}{2\pi R}[/tex]
At the center of the wire, [tex]B_c = 2B[/tex]
[tex]B_c = 2(\frac{\mu_o I}{2\pi R} )\\\\B_c = \frac{\mu_o I}{\pi R}[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
R is the center point between the wires, R = d/2 = 0.095m / 2 = 0.0475 m
I is the current needed
[tex]B_c = \frac{\mu_o I}{\pi R} \\\\I = \frac{B_c \pi R}{\mu_o} \\\\I = \frac{280* 10^{-6}*\pi *0.0475}{4\pi *10^{-7}} \\\\I = 33.25 \ A[/tex]
Tectonic plates are large segments of the Earth's crust that moves slowly. Suppose that one such plate has an average speed of 4.0 cm/per year. a.) what distance does it move in 1 second at this speed. b.) What is the speed per kilometer per million years
Answer:
a. [tex]3.04\times 10^{-8}m[/tex]
b. = 40 km/ million years
Explanation:
The computation is shown below;
According to the question, Data provided in the question
Average speed = 4.0 cm / per year
Distance move in 1 second at this speed
Based on the above information
a. For distance move in 1 second is
As we know that
[tex]d_1 = v_g \times t\\\\ = 4\ cm \times \frac{1}{100\times 365.25\times 3,600} \times 1\s\\\\= 3.04\times 10^{-8}m[/tex]
b. For speed per kilometer per million years is
[tex]v_1 = 4\times \frac{10^6}{10^5} \\\\[/tex]
= 40 km/ million years
How is the work done to hoist the counterweight related to the potential energy of the counterweight at its specified height?
Answer:
The work done to lift the counterweight equals the potential energy acquired
Explanation:
since this is vertically applied force on the counterweight, and the distance the force is displacing the counterweight is in the same direction as the applied force, it equals the gained potential energy
When monochromatic light illuminates a grating with 7000 lines per centimeter, its second order maximum is at 62.4°. What is the wavelength of the lig
Answer:
633nm
Explanation:
Given the following :
Number of lines per centimeter(N) = 7000
Angle θ = 62.4°
Order (n) = 2
If grating element = d
Wavelength (λ) = (d* SinΘ) / 2
If number of lines = 7000 per cm
Converting to metre :
100 cm = 1m
7000 lines per 1 cm
Number of lines per m:
7000 lines * 100 = 700,000 lines per meter
Recall :
d = reciprocal of N
d = 1 / 700,000
d = 0.00000142857
Substituting into (λ) = (d* SinΘ) / 2
λ = (0.00000142857 * Sin 62.4°) / 2
λ = 0.00000126600 / 2
λ = 0.000000633002
λ = 0.000000633
λ = 633 × 10^-9 m = 633nm
In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. Let's say we now have
3C+4D=5
2C+5D=2
None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. Which of the following expressions would result from that process?
C= 53−43D
C= 1−52D
D= 25−25C
D= 54−34C
SubmitMy AnswersGive Up
Correct
Part E - Solving for Two Variables
Now that you have one of the two variables in Part D isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying an Expression Primers.
Enter the answer as two numbers (either fraction or decimal), separated by a comma, with C first.
Need the answer with work shown for Part E.
Answer:
Explanation:
Given the simultaneous equation,
3C+4D=5 .............. 1
2C+5D=2 ............... 2
Solving for the value of C and D using substitution method.
From equation 1;
3C = 5-4D
Divide both sides by 3
3C/3 = (5-4D)/3
C = (5-4D)/3 .... 3
From equation 2:
2C+5D=2
5D = 2-2C
Divide both sides by 5;
5D/5 = 2-2C/5
D = (2-2C)/5 ..... 4
Substitute equation 4 into 3;
C = 5-4{(2-2C)/5}/3
C = [5 - (8-8C/5)]/3
C = [25-(8-8C)/5]/3
C = (17+8C)/15
15C = 17+8C
15C-8C = 17
7C = 17
C = 17/7
Substitute C = 17/7 into equation 4 to get the value of D
D = (2-2(17/7))/5
D = (2-34/7)/5
D = 14-34/35
D = -20/35
D = -4/7
Hence the value of C = 17/7, D = -4/7
A pendulum oscillates 50 times in 6 seconds. Find its time period and frequency?
Explanation:
time taken fir 50 oscillations is 6 seconds
time taken for 1 oscillation is 6/50
convert it into a decimal
A 1.10kg block is attached to a spring with spring constant 18 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 40cm/s .What is the block's speed at the point where x = 0.45 A?
Answer:
The velocity is [tex]v_x = 0.356 \ m/s[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 1.10 \ kg[/tex]
The spring constant is [tex]k = 18 \ N/m[/tex]
The speed is [tex]v = 40 \ cm / s = 0.4 m/s[/tex]
The position considered is x = 0.45 A
Here A is the amplitude which is mathematically represented as
[tex]A = v * \sqrt{\frac{m}{k} }[/tex]
=> [tex]A = 0.4 * \sqrt{\frac{1.10}{18 } }[/tex]
=> [tex]A = 0.0989 \ m[/tex]
So [tex]x = 0.45 * 0.0989[/tex]
=> [tex]x = 0.045 \ m[/tex]
Generally the speed at x is mathematically represented as
[tex]v_x = \sqrt{ \frac{k}{m} * [A^2 - x^2 ]}[/tex]
=> [tex]v_x = \sqrt{ \frac{18}{ 1.10} * [0.0989^2 - 0.045^2 ]}[/tex]
=> [tex]v_x = 0.356 \ m/s[/tex]
An helicopter lowers a probe into lake Chad which is suspended on a cable. the probe has a mass of 500kg and its average density is 1400kg/m³. what is the tension in the cable?
Answer:
1,401.85NExplanation:
If the mass of the probe is 500kg, its weight W = mass acceleration due to gravity.
Weight of the probe = 500*9.81
Weight of the probe = 4,905N
If its average density = 1400kg/m³
Volume = Mass/Density
Volume = 500/1400
Volume = 0.3571m³
According to the floatation principle, the volume of the probe is equal to the volume of liquid displaced. Hence the volume of water displaced is 0.357m³.
Since density of water is 1000kg/m³, we can find the mass of the water using the formula;
Mass of water = Density of water * Volume of water
Mass of water = 1000*0.3571
Mass of water = 357.1kg
Weight of water displaced = 3571 * 9.81 = 3503.15N
The tension in the cable will be the difference between the weight of the probe and weight of the displaced fluid.
Tension in the cable = 4,905N - 3503.15N
Tension in the cable = 1,401.85N
Hence the tension in the cable is 1,401.85N
In coming to a stop, a car leaves skid marks 80 m long on the highway. Assuming a deceleration of 3.5 m/s2 , estimate the speed of the car just before braking.
Answer:
The speed of the car just before braking is 23.66 m/s.
Explanation:
Given;
mark of the skid, d = 80 m
deceleration of the car, a = 3.5 m/s²
To determine the speed of the car just before braking, we apply the following kinematic equation;
[tex]v^2 = u^2 + 2ad\\\\v^2 = 0 + 2(3.5)(80)\\\\v^2 = 560\\\\v= \sqrt{560}\\\\v = 23.66 \ m/s[/tex]
Therefore, the speed of the car just before braking is 23.66 m/s.
A 121-kg astronaut (including space suit) acquires a speed of 2.90 m/s by pushing off with her legs from a 1600-kg space capsule. Use the reference frame in which the capsule is at rest before the push.
A) What is the velocity of the space capsule after the push in the reference frame?
B) If the push lasts 0.660 s , what is the magnitude of the average force exerted by each on the other?
C) What is the kinetic energy of the astronaut after the push in the reference frame?
D) What is the kinetic energy of the capsule after the push in the reference frame?
Answer:
a) 0.22 m/s
b) 531.67 N
c) 508.81 J
d) 38.72 J
Explanation:
the mass of the astronaut = 121 kg
astronaut's push of speed = 2.9 m/s
mass of the space capsule = 1600 kg
a) according to the conservation of momentum, the summation of the total momentum in a system must be equal to zero.
let us take the direction of the astronaut as positive.
Astronaut's momentum p = mv
where
m is the mass
v is the velocity
momentum p = 121 x 2.9 = 350.9 kg-m/s
The space capsules momentum = mv
==> 1600 x (-v) = -1600v this is because the space capsule moves in the opposite direction to the astronaut.
according to conservation of momentum
350.9 + (-1600v) = 0
350.9 = 1600v
v = 350.9/1600 = 0.22 m/s
b) magnitude of the force F is the rate of change of momentum.
The astronaut and the space capsule both change momentum from 0 to 350.9 kg-m/s. In 0.66 seconds, the force will be
F = [tex]\frac{m(v - u)}{t}[/tex]
where
u is their initial velocity = 0 m/s
where v = 2.9
t = 0.66
substituting, we have
F = [tex]\frac{121(2.9 - 0)}{0.66}[/tex] = 350.9/0.66 = 531.67 N this same force is experienced by the space capsule
c) Kinetic energy of the astronaut = [tex]\frac{1}{2} mv^{2}[/tex]
m is the mass = 121 kg
v is the velocity = 2.9 m/s
KE = [tex]\frac{1}{2}*121*2.9^{2}[/tex] = 508.81 J
d) Kinetic energy of the space capsule = [tex]\frac{1}{2} mv^{2}[/tex]
KE = [tex]\frac{1}{2}* 1600* 0.22^{2}[/tex] = 38.72 J
A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform electric field of increasing strength also passes through the ring, parallel to the wire. The magnetic field through the ring is zero.
a. At what rate is the electric field strength increasing?
b. is the electric field in the direction of the current or opposite to the current?
Answer:
a
[tex]\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s[/tex]
b
The direction of the electric field is opposite that of the current
Explanation:
From the question we are told that
The current is [tex]I = 17\ A[/tex]
The diameter of the ring is [tex]d = 3.0 \ cm = 0.03 \ m[/tex]
Generally the radius is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
[tex]r = \frac{0.03}{2}[/tex]
[tex]r = 0.015 \ m[/tex]
The cross-sectional area is mathematically represented as
[tex]A = \pi r^2[/tex]
=> [tex]A = 3.142 * (0.015^2)[/tex]
=> [tex]A = 7.07 *10^{-4 } \ m^ 2[/tex]
Generally according to ampere -Maxwell equation we have that
[tex]\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt }[/tex]
Now given that [tex]\= B = 0[/tex] it implies that
[tex]\oint \= B \cdot \= ds = 0[/tex]
So
[tex]\mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt } = 0[/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]
[tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
[tex]\phi[/tex] is magnetic flux which is mathematically represented as
[tex]\phi = E * A[/tex]
Where E is the electric field strength
So
[tex]\mu_o I + \epsilon_o \mu _o \frac{ d [EA] }{dt } = 0[/tex]
=> [tex]\frac{dE}{dt} =- \frac{I}{\epsilon_o * A }[/tex]
=> [tex]\frac{dE}{dt} =- \frac{17}{8.85*10^{-12} * 7.07*10^{-4} }[/tex]
=> [tex]\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s[/tex]
The negative sign shows that the direction of the electric field is opposite that of the current
Which option gives an
object's temperature in Sl units?
A. 0°C
B. 273 K
C. 273 kg
D. 32°F
Answer:B
Explanation: I just did it on a p e x
273 K gives the object's temperature in the SI unit therefore the correct answer is option B
What is a unit of measurement?A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation. Any additional quantity of that type can be stated as a multiple of the measurement unit.
The International System of Units, sometimes known as the SI system of units, is the most frequently used and acknowledged system of units in use nowadays. There are three additional units and 7 SI basic units in this system of SI units.
The three supplemental SI units are radian, steradian, and becquerel, whereas the base SI units are meter, kilogram, second, kelvin, ampere, candela, and mole. These base units can be used to create all other SI units.
Thus,273 K gives the object's temperature in the SI unit therefore the correct answer is option B
Learn more about the unit of measurement from here
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Select the correct answer.
Based on the law of conservation of energy, which statement is false?
O A. Energy is lost when machines don't work right.
OB.
We can't add more energy to the universe.
Ос.
We can't destroy energy that exists in the universe.
OD. Energy changes from one form to another.
Answer:A is the correct answer
Explanation:
In a simple model of a potassium iodide (KI) molecule, we assume the K and I atoms bond ionically by the transfer of one electron from K to I.(a) The ionization energy of K is 4.34 eV, and the electron affinity of I is 3.06 eV. What energy is needed to transfer an electron from K to I, to form K+ and I? ions from neutral atoms? This quantity is sometimes called the activation energy Ea.eV(b) A model potential energy function for the KI molecule is the Lennard
This question is incomplete, the complete question is;
In a simple model of a potassium iodide (KI) molecule, we assume the K and I atoms bond ionically by the transfer of one electron from K to I.
(a) The ionization energy of K is 4.34 eV, and the electron affinity of I is 3.06 eV. What energy is needed to transfer an electron from K to I, to form K+ and I- ions from neutral atoms? This quantity is sometimes called the activation energy Ea.eV
(b) A model potential energy function for the KI molecule is the Lennard - jones potential:
U(r) = 4∈[ (α/r)¹² - (α/r)⁶ ] + Ea
where r is the internuclear separation distance and α and ∈ are adjustable parameters (constants) . The Ea term is added to ensure the correct asymptotic behavior at large r and is activation energy calculated in a. At the equilibrium separation distance, r=r₀=0.305 nm, U(r) is a minimum, and dU/dr=0. In addition, U(r₀)=-3.37 eV.
Us the experimental values for the equilibrium sepeartion and dissociation energy of KI to determine/find 'α' and '∈'.
(c) calculate the force needed to break the KI molecule in nN
Answer:
a) energy is needed to transfer an electron from K to I, to form K+ and I- ions from neutral atoms is 1.28 eV
b) α = 0.272, ∈ = 4.65 eV
c) the force needed to break the KI molecule in nN 65.6 nN
Explanation:
a) The ionization energy of K is 4.34 ev ( energy needed to remove the outer most electrons)
And the electron affinity of I is 3.06 ev ( which is energy released when electron is added)
Now the energy that is need to transfer an electron from K to I,
i.e the ionization energy of K(4.34 ev) and the electron affinity of I (3.06 ev)
RE = 4.34 - 3.06 = 1.28 eV
b)
from the question we have
U(r) = 4∈[ (α/r)¹² - (α/r)⁶ ] + Ea
now taking d/drU(r₀)=0 (at r = r₀)
= 4∈d/dr [ (α/r)¹² - (α/r)⁶ ] = 0
= ( -12(α¹²/r¹³)) - (-6 (α⁶/r⁷)) = 0
12(α¹²/r¹³) = 6 (α⁶/r⁷)
α⁶ = r⁶/2
α = r/(2)^1/6
at equilibrium r = r₀ = 0.305 nm
α = 0.305 nm / (2)^1/6
C = 0.0305/1.1246
α = 0.272
Now substituting the values of U(r₀), α, Eₐ in the initial expression
U(r) = 4∈[ (α/r)¹² - (α/r)⁶ ] + Ea
we have
- 3.37eV = 4∈ [ (0.272 nm / 0.305 nm)¹² - (0.272 nm / 0.305 nm )⁶ ] + 1.28
- 1.65 eV = ∈(0.25 - 0.5)
∈ = 4.65 eV
c)
Now to break the molecule then the potential energy should be zero(0)
and we know r = 0.272 nm
therefore force needed to break the molecule is
F = -dU/dR_r-α
F = -4∈ (-12/α + 6/α)
F = -4(4.65eV) ( -12/0.272nm + 6/0.272nm)
F = 65.6 nN
According to the chart, one gram of copper and
gram(s) of gold
would change their temperatures by approximately the same amount by
adding heat to them.
A) one
B) two
C) three
D) four
Answer:
C) three
Explanation:
Let gram of gold required be m . Let temperature change in both be Δ t .
heat absorbed = mass x specific heat x change in temperature
for copper
heat absorbed = 1 x .385 x Δt
for gold
heat absorbed = m x .129 x Δt
So
m x .129 x Δt = 1 x .385 x Δt
m = 2.98
= 3 g approximately .
A beam of light in air enters a glass slab with an index of refraction of 1.40 at an angle of incidence of 30.0°. What is the angle of refraction? (index of refraction of air=1)
Answer:
[tex] \boxed{\sf Angle \: of \: refraction \: (r) = {sin}^{ - 1} ( \frac{1}{2.8} )} [/tex]
Given:
Refractive index of air ( [tex] \sf \mu_{air} [/tex] )= 1
Refractive index of glass slab ( [tex] \sf \mu_{glass} [/tex]) = 1.40
Angle of incidence (i) = 30.0°
To Find:
Angle of refraction (r)
Explanation:
From Snell's Law:
[tex] \boxed{ \bold{ \sf \mu_{air}sin \ i = \mu_{glass}sin \: r}}[/tex]
[tex] \sf \implies 1 \times sin \: 30 ^ \circ = 1.4sin \:r[/tex]
[tex] \sf sin \:30^ \circ = \frac{1}{2} : [/tex]
[tex] \sf \implies \frac{1}{2} = 1.4 sin \: r[/tex]
[tex] \sf \frac{1}{2} = 1.4 sin \: r \: is \: equivalent \: to \: 1.4 sin \: r = \frac{1}{2} : [/tex]
[tex] \sf \implies 1.4 sin \: r = \frac{1}{2} [/tex]
Dividing both sides by 1.4:
[tex] \sf \implies \frac{\cancel{1.4} sin \: r}{\cancel{1.4}} = \frac{1}{2 \times 1.4} [/tex]
[tex] \sf \implies sin \: r = \frac{1}{2 \times 1.4} [/tex]
[tex] \sf \implies sin \: r = \frac{1}{2.8} [/tex]
[tex] \sf \implies r = {sin}^{ - 1} ( \frac{1}{2.8} )[/tex]
[tex] \therefore[/tex]
[tex] \sf Angle \: of \: refraction \: (r) = {sin}^{ - 1} ( \frac{1}{2.8} )[/tex]
You toss a ball straight up in the air. Immediately after you let go of it, what force or forces are acting on the ball
Answer and Explanation: When tossing a ball up in the air, the forces acting on the ball are due to Gravity, which is defined by gravitational acceleration on that location on Earth (approximately 9.8 m/s²) multiplied by mass of the ball; Force of thrown, i.e., the force you threw the ball and air resistance force, which is proportional to the square of the ball's through the air and the ball's cross section area. To facilite calculations, air resistance force is normally ignored.
Answer:
weight and drag
Explanation:
In a photoelectric experiment, a metal is irradiated with light of energy 3.56 eV. If a stopping potential of 1.10 V is required, what is the work function of the metal?
Answer:
The work function is [tex]\phi = 2.46 \ eV[/tex]
Explanation:
From the question we are told that
The light energy is [tex]E = 3.56 eV[/tex]
The stopping voltage is [tex]V = 1.10 \ V[/tex]
Generally work function is mathematically represented as
[tex]\phi = E - KE[/tex]
Where KE is the kinetic energy of the ejected electron and it is mathematically represented as
[tex]KE = V * e[/tex]
Where e is the charge on the electron
So
[tex]KE = 1.10eV[/tex]
Thus
[tex]\phi = 3.56eV - 1.10 eV[/tex]
=> [tex]\phi = 2.46 \ eV[/tex]
a sled is moving with a velocity of 8m/s. The sled slows to a stop over a time of 4s, covering a distance of 16m. What is the sleds acceleration
Calculate the acceleration ,a .
Formula used:-s = ut + 1/2 at²
Solution:-According to the second equation of motion, we have
s = ut + 1/2 at²
★Substituting the values in the above formula,we get:
⇒ 16 = 8 × 4 + 1/2 × a × 4
⇒ 16 = 32 + 2a
⇒ 2a = 16 - 32
⇒ 2a = -16
⇒ a = -16/2
⇒ a = -8 m/s²
Hence,the acceleration is -8 m/s² .
Find the minimum value of n in the Balmer series for which the predicted wavelength is in the ultraviolet region of the spectrum. View Available Hint(s)
Answer:
λ =365.4 nm
Explanation:
Boh's atomic model of the Hydrogen atom the energy of each level is
Eₙ = - 13.606 / n²
where the synergy is in electonvotes and the value of E₀ = 13.606 eV is the energy of the base state of hydrogen.
An atomic transition occurs when an electron goes from an excited state and joins everything of lower energy.
ED = 13.606 (1 / n₀² - 1 /[tex]n_{f}^{2}[/tex])
we are going to apply this relationship to answer slash.
At the beginning of the studies of atomic transitions, each group did not consider having a different name
name Initial state
Lymman 1
Balmer 2
the final state is any other state sta the continuum that corresponds to n = inf
Let's look for the highest energy of the Balmer series
ΔE = 13.606 (1/2² - 1 /∞)
ΔE = 3.4015 eV
Let's use the Planck relation for the energy
E = h f = h c /λ
λ = h c / E
Let's reduce the energy to J
E = 3.4015 eV (1.6 10⁻¹⁹ J / 1 eV) = 5.4424 10⁻¹⁹
λ = 6.63 10⁻³⁴ 3 10⁸ / 5.4424 10⁻¹⁹
λ = 3.654 10⁻⁷ m
λ = 3,654 10⁻⁷ m (10⁹ nm / 1m)
λ =365.4 nm
this eta radiation in the ultraviolet range
A cylinder contains 3.5 L of oxygen at 350 K and 2.7 atm . The gas is heated, causing a piston in the cylinder to move outward. The heating causes the temperature to rise to 620 K and the volume of the cylinder to increase to 9.1 L.What is the gas pressure? P= _____atm
Answer:
The pressure is [tex]P_2 = 1.84 \ a.t.m[/tex]
Explanation:
From the question we are told that
The first volume of is [tex]v_1 = 3.5 \ L[/tex]
The first pressure is [tex]P_1 = 2.7 \ a.t.m[/tex]
The first temperature is [tex]T_1 = 350 \ K[/tex]
The new temperature is [tex]T_2 = 620 \ K[/tex]
The new volume is [tex]V_2 = 9.1 \ a.t.m[/tex]
Generally according to the combined gas law we have that
[tex]\frac{P_1 V_1 }{T_1 } = \frac{P_2 V_2 }{T_2 }[/tex]
=> [tex]P_2 = \frac{P_1 * V_1 * T_2 }{T_1 * V_2 }[/tex]
=> [tex]P_2 = \frac{ 2.7 * 3.5 * 620 }{ 350 * 9.1 }[/tex]
=> [tex]P_2 = 1.84 \ a.t.m[/tex]
3.60×10^23 nitrogen molecules collide with a 16.0 cm^2 wall each second. Assume that the molecules all travel with a speed of 450 m/s and strike the wall head on. What is the pressure on the wall?
Answer:
The pressure on the wall is 9416.25 N/m²
Explanation:
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