Answer:
Power, P = 307.31 W
Explanation:
It is given that,
Kasek climb at an angle of 6° with the horizontal at a steady speed of 4.0 m/s.
The total mass of bicycle and Kasek is 75 kg
We need to find the Kasek's power output to climb the same hill at the same speed. The angle is made with the horizontal. It means that,
F = F sinθ
So,
Power output is given by :
[tex]P=mg\sin\theta\times v\\\\P=75\times 9.8\times \sin(6)\times 4\\\\P=307.31\ W[/tex]
So, Kasek's power output to climb the same hill is 307.31 W.
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]
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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The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rev/min in 8.00s, starting from rest?
Answer:
Explanation:
From the question we are told that
The moment of inertia is [tex]I = 2.50 \ kg \cdot m^2[/tex]
The final angular speed is [tex]w_f = 400 rev/min = \frac{400 * 2\pi}{60} = 41.89 \ rad/s[/tex]
The time taken is [tex]t = 8.0 s[/tex]
The initial angular speed is [tex]w_i = 0\ rad/s[/tex]
Generally the average angular acceleration is mathematically represented as
[tex]\alpha = \frac{w_f - w_i }{t}[/tex]
=> [tex]\alpha = \frac{41.89}{8}[/tex]
=> [tex]\alpha = 5.24 \ rad/s^2[/tex]
Generally the torque is mathematically represented as
[tex]\tau = I * \alpha[/tex]
=> [tex]\tau = 5.24 * 2.50[/tex]
=> [tex]\tau = 13.09 \ N \cdot m[/tex]
Read the passage about the pygmy shrew.
The pygmy shrew is the smallest mammal in North America. However, when comparing the amount of food eaten to its body weight, the pygmy shrew eats more food than any other mammal. It will consume two to three times its own body weight in food daily. One explanation is that the pygmy shrew uses energy at a high rate. In fact, its heart beats over one thousand times per minute.
What is the best explanation for what happens to the food's mass and energy when it is consumed by the pygmy shrew?
Answer:
A very high metabolism and a very small size.
Explanation:
The pygmy shrew is a very small mammal, that forages day and night. The metabolism of the Pygmy shrew is so high that it must eat at least every 30 minutes or it might die. The best explanation for what happens to the food's mass and energy is that most of the food mass is rapidly used fro building up of the shrew due to its very high metabolism, and a bigger portion of the food is lost from the surface of the body of the shrew, due to its very small size. The combination of these two factors; a very high metabolism (rapidly uses up food material, and generates a large amount of heat in a very short time) and the very small size (makes heat loss due to surface area to volume ratio high) explains what happens to the food mass and energy.
How are period and frequency related to each other?
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.
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]
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.
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)
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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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 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]
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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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
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?
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
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]
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:
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]
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
A ranger in a national park is driving at 15.0 m/s when a deer jumps into the road 60 m ahead of the vehicle. After a reaction time, t, the ranger applies the brakes to produce an acceleration of a = -3.00 m/s2. What is the maximum reaction time allowed if she is to avoid hitting the deer?
Answer:
t = 5 s
Explanation:
In uniform rectilinear movement, the equation for final speed is:
vf = v₀ + a*t
In this case we need that the car stops just before 60 m after applied the brakes, then
vf = v₀ - a*t
vf = final speed = 0
v₀ = initial speed = 15 m/s
And negative acceleration is 3 m/s²
0 = 15 (m/s) - 3 ( m/s²)*t
t = 15 / 3 (m/s /m/s²)
t = 5 s
The point is that with that value ranger will hit the deer so in order to not to hit the deer that time should be smaller than 5 seconds
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An aircraft travelling at 600 km/h accelerates steadily at 10 km/h per second. Taking the
speed of sound as 1100 km/h at the aircraft's altitude, how long will it take to reach the
Sound barrier'?
Answer:
50 seconds
Explanation:
Acceleration = change in velocity / change in time
a = Δv / Δt
10 km/h/s = (1100 km/h − 600 km/h) / t
t = 50 s
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
An ideal gas is at a temperature of 320 K. What is the average translational kinetic energy of one of its molecules
Answer:
6.624 x 10^-21 J
Explanation:
The temperature of the ideal gas = 320 K
The average translational energy of an ideal gas is gotten as
[tex]K_{ave}[/tex] = [tex]\frac{3}{2}K_{b}T[/tex]
where
[tex]K_{ave}[/tex] is the average translational energy of the molecules
[tex]K_{b}[/tex] = Boltzmann constant = 1.38 × 10^-23 m^2 kg s^-2 K^-1
T is the temperature of the gas = 320 K
substituting value, we have
[tex]K_{ave}[/tex] = [tex]\frac{3}{2} * 1.38*10^{-23} * 320[/tex] = 6.624 x 10^-21 J
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]
Find the work done by the gas during the following stages. (a) A gas is expanded from a volume of 1.000 L to 4.000 L at a constant pressure of 2.000 atm. (b) The gas is then cooled at constant volume until the pressure falls to 1.500 atm
Answer:
a) 607.95 J
b) 0 J
Explanation:
a) Initial volume = 1 L = 0.001 m^3
final volume = 4 L = 0.004 m^3
pressure = 2 atm = 202650 Pa (1 atm = 101325 Pa)
work done by the gas on the environment = PΔV
P is the pressure = 101325 Pa
ΔV is the change in volume from the initial volume to the final volume
ΔV = 0.004 m^3 - 0.001 m^3 = 0.003 m^3
work done by the gas = 202650 x 0.003 = 607.95 J
b) If the gas is cooled at constant volume, then the gas does no work. For a gas to do work, there must be a change in its volume.
Therefore the work done in cooling at constant volume until pressure falls to 1.5 atm = 0 J
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
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
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
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].