Answer:
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Explanation:bcz idk
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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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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 .
On a day that the temperature is 10.0°C, a concrete walk is poured in such a way that the ends of the walk are unable to move. Take Young's modulus for concrete to be 7.00 109 N/m2 and the compressive strength to be 2.00 109 N/m2. (The coefficient of linear expansion of concrete is 1.2 10-5(°C−1).)
What is the stress in the cement on a hot day of 42.0°C? N/m2
Answer:
The stress is [tex]stress = 2688000 \ N[/tex]
Explanation:
From the question we are told that
The first temperature is [tex]T_1 = 10 ^o \ C[/tex]
The young modulus is [tex]Y = 7.00 *10^9\ N/m^2[/tex]
The compressive strength is [tex]\sigma = 2.00 *10^{9} \ N/m^2[/tex]
The coefficient of linear expansion is [tex]\alpha = 1.2 *10^{-5} \ ^o C ^{-1}[/tex]
The second temperature is [tex]T_2 = 42.0^o \ C[/tex]
Generally the change in length of the concrete is mathematically represented as
[tex]\Delta L = \alpha * L * [T_2 - T_1 ][/tex]
=> [tex]\frac{\Delta L}{L} = \alpha * [T_2 - T_1 ][/tex]
=> [tex]strain = \alpha * [T_2 - T_1 ][/tex]
Now the young modulus is mathematically represented as
[tex]Y = \frac{stress}{strain}[/tex]
=> [tex]7.00 *10^9 = \frac{stress}{\alpha(T_2 - T_1 ) }[/tex]
=> [tex]stress = \alpha (T_2 - T_1 ) * 7.00 *10^9[/tex]
=> [tex]stress = 1.2* 10^{-5} (42 - 10 ) * 7.00 *10^9[/tex]
=> [tex]stress = 2688000 \ N[/tex]
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]
If the prism is surrounded by a fluid, what is the maximum index of refraction of the fluid that will still cause total internal reflection
Answer:
n₁ > n₂.
prisms are made of glass with refractive index n₂ = 1.50, so the fluid that surrounds the prism must have an index n₁> 1.50
Explanation:
Total internal reflection occurs when the refractive index of the incident medium the light is greater than the medium to which the light is refracted, let's use the refraction equation
n₁ sin θ₁ = n₂ sin θ₂
the incident medium is 1, at the limit point where refraction occurs is when the angle in the refracted medium is 90º, so sin θ₂ = 1
n₁ sin θ₁ = n₂
sin θ₁ = n₂ / n₁
We mean that this equation is defined only for n₁ > n₂.
In our case, for the total internal reflection to occur, the refractive incidence of the medium must be greater than the index of refraction of the prism.
In general, prisms are made of glass with refractive index n₂ = 1.50, so the fluid that surrounds the prism must have an index n₁> 1.50
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
A double-slit experiment is performed with light of wavelength 640 nm. The bright interference fringes are spaced 1.6 mm apart on the viewing screen.What will the fringe spacing be if the light is changed to a wavelength of 360nm?
Answer:
1.44*10^-3m
Explanation:
Given that distance BTW two bright fringes is
DetaY = lambda* L/d
So for second wavelength
Deta Y2= Lambda 2* L/d
=lambda 2 x deta y1/ lambda1
So substituting
= 360 x 10^-9 x (1.6*10^-3/640*10^-9)
1.44*10^ -3m
Bailey wants to find out which frozen solid melts the fastest: soda, ice, or orange juice. She pours each of the three liquids into the empty cubes of an ice tray, and then places the ice tray in the freezer overnight. The next day, she pulls the ice tray out and sets each cube on its own plate. She then waits and watches for them to melt. When the last part of the frozen liquid melts, she records the time.
Answer:
its 45 over 6
Explanation:the answer is in the question
Answer: Only the melted cube's shape changed.
Explanation:
Can someone tell me a very very simple physics experiment topic that links to biology?
Explanation:
One idea would be to investigate the correlation between your pulse pressure and your pulse rate. To do this, you'll need a blood pressure monitor.
First, measure your resting pressure and rate. Then exercise for 30 seconds. Measure your new blood pressure and pulse rate. Wait for your pressure and rate to return to normal, then repeat the trial for 1 minute, 1.5 minutes, 2 minutes, etc.
List the results in a table. This should include the amount of exercise time, your pulse rate, your systolic pressure (the high number, which is your blood pressure during contraction of your heart muscle), and your diastolic pressure (the low number, which is your blood pressure between heartbeats). Calculate your pulse pressure (systolic minus diastolic) for each trial. Graph the pulse pressure on the x-axis, and your pulse rate (beats per minute) on the y-axis.
What do you hypothesize will be the shape of the graph? Consider Bernoulli's formula, which relates fluid pressure and flow. How close do the results match your hypothesis? What might explain any differences?
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.
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
How many electrons would have to be removed from a coin to leave it with a charge of +1.0 10-7 C?
Answer:
[tex]n=6.25\times 10^{11}[/tex]
Explanation:
We need to find the number of electrons that would have to be removed from a coin to leave it with a charge of [tex]+10^{-7}\ C[/tex]. Then the number of electrons be n. Using quantization of electric charge as :
q = ne
e is charge on an electron
[tex]n=\dfrac{q}{e}\\\\n=\dfrac{10^{-7}}{1.6\times 10^{-19}}\\\\n=6.25\times 10^{11}[/tex]
So, the number of electrons are [tex]6.25\times 10^{11}[/tex].
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
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
A person is lying on a diving board 3.00 m above the surface of the water in a swimming pool. She looks at a penny that is on the bottom of the pool directly below her. To her, the penny appears to be a distance of 8.00 m from her.
Required:
What is the depth of the water at this point?
Answer:
The depth of water at the point is [tex]d_A = 6.55 \ m[/tex]
Explanation:
From the question we are told that
The height of the person above water is [tex]d = 3.00 \ m[/tex]
The distance of the coin as seen by the person is [tex]d' = 8.00 \ m[/tex]
Generally the apparent depth is mathematically represented as
[tex]d_a = \frac{d_A}{n}[/tex]
Here [tex]d_A[/tex] is the actual depth of water while n is the refractive index of water with a constant value [tex]n = 1.33[/tex]
Now from the point the person is the apparent depth is evaluated as
[tex]d_a = d'-d[/tex]
=> [tex]d_a = 8 - 3[/tex]
=> [tex]d_a = 5 \ m[/tex]
So
[tex]5 = \frac{d_A}{1.33}[/tex]
=> [tex]d_A = 5 * 1.33[/tex]
=> [tex]d_A = 6.55 \ m[/tex]
The center of the galaxy is filled with low-density hydrogen gas that scatters light rays. An astronomer wants to take a picture of the center of the galaxy. Will the view be better using ultra violet light, visible light, or infrared light? Explain.
Answer:
Infrared light
Explanation:
Infrared light is the spectrum of electromagnetic wave given off by a body possessing thermal energy. Infrared light is preferred over visible light in this region of space because visible light is easily scattered in the presence of fine particles. Infrared ray makes it easy for us to observe Cold, dark molecular clouds of gas and dust in our galaxy that glows when irradiated by the stars . Infrared can also be used to detect young forming stars, even before they begin to emit visible light. Stars emit a smaller portion of their energy in the infrared spectrum, so nearby cool objects such as planets can be more readily detected with infrared light which won't be possible with an ultraviolet or visible light.
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
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Find the position of the center of mass of two bodies points of masses m1 and m2 joined by a rod of mass negligible in length d. Find the acceleration of the center of mass for the case that m1 = 1 [kg] and m2 = 3 [kg] and the applied forces of the figure with d = 2 [m]
Explanation:
If m₁ is at the origin, then the center of mass is at:
x = (m₁ × 0 m + m₂ × d m) / (m₁ + m₂)
x = m₂ d / (m₁ + m₂)
If m₁ = 1 kg, m₂ = 3 kg, and d = 2 m:
x = (3 kg) (2 m) / (1 kg + 3 kg)
x = 1.5 m
Sum of forces in the x direction:
∑F = ma
16 N = (1 kg + 3 kg) aₓ
aₓ = 4 m/s²
Sum of forces in the y direction:
∑F = ma
20 N = (1 kg + 3 kg) aᵧ
aᵧ = 5 m/s²
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
Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.
If she runs fhe same course again, what constant speed would let her finish in the same time as in the first race?
Answer:
The velocity is [tex]v = 4.76 \ m/s[/tex]
Explanation:
From the question we are told that
The first distance is [tex]d_1 = 4.0 \ km = 4000 \ m[/tex]
The first speed is [tex]v_1 = 5.0 \ m/s[/tex]
The second distance is [tex]d_2 = 1.0 \ km = 1000 \ m[/tex]
The second speed is [tex]v_2 = 4.0 \ m/s[/tex]
Generally the time taken for first distance is
[tex]t_1 = \frac{d_1 }{v_1 }[/tex]
[tex]t_1 = \frac{4000}{5}[/tex]
[tex]t_1 = 800 \ s[/tex]
The time taken for second distance is
[tex]t_1 = \frac{d_2 }{v_2 }[/tex]
[tex]t_1 = \frac{1000}{4}[/tex]
[tex]t_1 = 250 \ s[/tex]
The total time is mathematically represented as
[tex]t = t_1 + t_2[/tex]
=> [tex]t = 800 + 250[/tex]
=> [tex]t = 1050 \ s[/tex]
Generally the constant velocity that would let her finish at the same time is mathematically represented as
[tex]v = \frac{d_1 + d_2}{t }[/tex]
=> [tex]v = \frac{4000 + 1000}{1050 }[/tex]
=> [tex]v = 4.76 \ m/s[/tex]
The constant speed that will let her finish in the same time as in the first race is 4.76 m/s
Determination of the time taken for first 4 KmDistance = 4 Km = 4 × 1000 = 4000 mSpeed = 5 m/sTime 1 =?
Time 1 = distance / speed
Time 1 = 4000 / 5
Time 1 = 800 s
Determination of the time taken for the last 1 KmDistance = 1 Km = 1 × 1000 = 1000 mSpeed = 4 m/sTime 2 =?Time 2 = distance / speed
Time 2 = 1000 / 4
Time 2 = 250 s
Determination of the constant speedTotal distance = 4000 + 1000 = 5000 mTotal time = 800 + 250 = 1050 sConstant speed =?Constant speed = Total distance / total time
Constant speed = 5000 / 1050
Constant speed = 4.76 m/s
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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)
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
Light of wavelength 575 nm passes through a double-slit and the third order bright fringe is seen at an angle of 6.5^degree away from the central fringe. What is the separation between the double slits? a) 5.0 mu m b) 10 mu m c) 15 mu m d) 20 mu m e) 25 mu m
Answer:
The correct option is C
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 575 *10^{-9} \ m[/tex]
The angle is [tex]\theta = 6.5^o[/tex]
The order of maxima is n = 3
Generally for constructive interference
[tex]dsin \theta = n * \lambda[/tex]
=> [tex]d = \frac{n * \lambda }{ sin \theta }[/tex]
=> [tex]d = \frac{3 * 575 *10^{-9} }{ sin 6.5 }[/tex]
=> [tex]d = 15.24 *10^{-6} \ m[/tex]
=> [tex]d = 15 \mu m[/tex]
A 1.70 kg block slides on a horizontal, frictionless surface until it encounters a spring with a force constant of 955 N/m. The block comes to rest after compressing the spring by a distance of 4.60 cm. The other end of the spring is attached to a wall. Find the initial speed of the block.
Answer:
The initial speed of the block is 1.09 m/s
Explanation:
Given;
mass of block, m = 1.7 kg
force constant of the spring, k = 955 N/m
compression of the spring, x = 4.6 cm = 0.046 m
From principle of conservation of energy
kinetic energy of the block = elastic potential energy of the spring
¹/₂mv² = ¹/₂kx²
mv² = kx²
[tex]v = \sqrt{\frac{kx^2}{m} }[/tex]
where;
v is the initial speed of the block
x is the compression of the spring
[tex]v = \sqrt{\frac{955*(0.046)^2}{1.7} } \\\\v = 1.09 \ m/s[/tex]
Therefore, the initial speed of the block is 1.09 m/s
In a container of negligible mass, 020 kg of ice at an initial temperature of - 40.0 oC is mixed with a mass m of water that has an initial temperature of 80.0 oC. No heat is lost to the surroundings. If the final temperature of the system is 20.0 oC, what is the mass m of the water that was initially at 80.0 oC
Answer:
The mass is [tex]m_w = 0.599 \ kg[/tex]
Explanation:
From the question we are told that
The mass of ice is [tex]m_c = 0.20 \ kg[/tex]
The initial temperature of the ice is [tex]T_i = -40.0 ^oC[/tex]
The initial temperature of the water is [tex]T_{iw} = 80^o C[/tex]
The final temperature of the system is [tex]T_f = 20^oC[/tex]
Generally according to the law of energy conservation,
The total heat loss is = total heat gained
Now the total heat gain is mathematically represented as
[tex]H = H_1 + H_2 + H_3[/tex]
Here [tex]H_1[/tex] is the energy required to move the ice from [tex]-40^oC \to 0^oC[/tex]
And it mathematically evaluated as
[tex]H_1 = m_c * c_c * \Delta T[/tex]
Here the specific heat of ice is [tex]c_c = 2100 \ J \cdot kg^{-1} \cdot ^oC^{-1}[/tex]
So
[tex]H_1 = 0.20 * 2100 * (0-(-40))[/tex]
[tex]H_1 = 16800\ J[/tex]
[tex]H_2[/tex] is the energy to melt the ice
And it mathematically evaluated as
[tex]H_2 = m * H_L[/tex]
The latent heat of fusion of ice is [tex]H_L = 334 J/g = 334 *10^{3} J /kg[/tex]
So
[tex]H_2 = 0.20 * 334 *10^{3}[/tex]
[tex]H_2 = 66800 \ J[/tex]
[tex]H_3[/tex] is the energy to raise the melted ice to [tex]20^oC[/tex]
And it mathematically evaluated as
[tex]H_3 = m_c * c_w * \Delta T[/tex]
Here the specific heat of water is [tex]c_w= 4190\ J \cdot kg^{-1} \cdot ^oC^{-1}[/tex]
[tex]H_3 = 0.20 * 4190* (20-0))[/tex]
[tex]H_3 = 16744 \ J[/tex]
So
[tex]H = 16800 + 66800 + 16744[/tex]
[tex]H = 100344\ J[/tex]
The heat loss is mathematically evaluated as
[tex]H_d = m * c_h ( 80 - 20 )[/tex]
[tex]H_d = m_w * 4190 * ( 80 - 20 )[/tex]
[tex]H_d = 167600 m_w[/tex]
So
[tex]167600 m_w = 100344[/tex]
=> [tex]m_w = 0.599 \ kg[/tex]
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
Light from a 600 nm source goes through two slits 0.080 mm apart. What is the angular separation of the two first order maxima occurring on a screen 2.0 m from the slits
Answer:
The angular separation is [tex]k = 0.8594^o[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 600 \ nm = 600*10^{-9} \ m[/tex]
The distance of separation between the slit is [tex]d = 0.080 \ mm = 0.080 *10^{-3} \ m[/tex]
The distance from the screen is
Generally the condition for constructive interference is mathematically represented as
[tex]d \ sin(\theta) = n \lambda[/tex]
=> [tex]\theta = sin ^{-1} [ \frac{n * \lambda }{ d } ][/tex]
here [tex]\theta[/tex] is the angular separation between the central maxima and one side of the first order maxima
given that we are considering the first order of maxima n = 1
=> [tex]\theta = sin ^{-1} [ \frac{1 * 600*10^{-9} }{ 2.0 } ][/tex]
=> [tex]\theta = sin ^{-1} [ 0.0075 ][/tex]
=> [tex]\theta = 0.4297^o[/tex]
So the angular separation of the two first order maxima is
[tex]k = 2 * \theta[/tex]
[tex]k = 2 * 0.4297[/tex]
[tex]k = 0.8594^o[/tex]
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]
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 0.145 kg baseball pitched at 33.m/s is hit on a horizontal line drive straight back at the pitcher at 46.0 m/s. If the contact time between bat and ball is 5.70×10−3 s, calculate the magnitude of the force (assumed to be constant) exerted on the ball by the bat.
Answer:
F = 2009.64 N
Explanation:
It is given that,
Mass of a baseball, m = 0.145 kg
Initial speed if the baseball, u = 33 m/s
It hit on a horizontal line drive straight back at the pitcher at 46.0 m/s, final velocity, v = -46 m/s
Time of contact between the bat and the ball is [tex]t=5.7\times 10^{-3}\ s[/tex]
We need to find the magnitude of the force exerted by the ball on the bat. It can be calculated using impulse-momentum theorem. So,
[tex]Ft=m(v-u)\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.145\times (-46-33)}{5.7\times 10^{-3}}\\\\F=-2009.64\ N[/tex]
So, the magnitude of force exerted on the ball by the bat is 2009.64 N.