The direction of rotation of a proton can be determined using the right-hand rule in physics when considering the direction of its magnetic moment and the uniform magnetic field it's immersed in. The resultant rotation direction works to facilitate energy minimization in the system.
Explanation:In the presence of an external magnetic field, the principle of physics concerning the interaction between magnetic fields and magnetic moments state that the magnetic moment of an object, in this case a proton, will align itself along the direction of the external magnetic field. This means, if the magnetic moment vector mu of the proton is directed along the positive z-axis, and the magnetic field is also along the positive z-axis, the proton's rotation will be in the direction opposite to that of the magnetic field following the right-hand rule. The proton's rotation and the magnetic field directions facilitate energy minimization in the system. Hence, the direction of rotation of a proton, based on the direction of its magnetic moment and the direction of the uniform magnetic field it is immersed in, can be determined by the right-hand rule; if your thumb points in the direction of the proton's magnetic moment (as opposed to the magnetic field), your fingers will curl in the direction of its rotation.
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Determine whether the following actions cause the fission reaction in the reactor to speed up or slow down.
a. speeds up fission: Adding the moderator to the reactor
b. speeds up fission: Removing the control rods from the reactor
c. slows down fission: Inserting the control rods into the reactor
d. slows down fission: Removing the moderator from the reactor
e. slows down fission: A sudden loss of primary coolant water in a pressurized water reactor
Answer:
option B and C
Explanation:
Control rod are used to regulate the nuclear reactor.
When you insert control rod in the reactor it slows down the nuclear fission inside the reactor and the energy produced in the reactor will be less.
When you remove control road from the reactor the nuclear fission increase inside the reactor and the energy production is high.
Control rod consist of boron, boron absorb the neutron which help to control the nuclear fission.
Hence, the correct answer is option B and C
A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2.000 s.) The length of a seconds pendulum is 0.9923 m at Tokyo and 0.9941 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?
Explanation:
We know the equation
[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]
where l is length of pendulum, g is acceleration due to gravity and T is period.
Rearranging
[tex]g= \frac{4\pi^2l}{T^2}[/tex]
Length of pendulum in Tokyo = 0.9923 m
Length of pendulum in Cambridge = 0.9941 m
Period of pendulum in Tokyo = Period of pendulum in Cambridge = 2s
We have
[tex]\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}= \frac{\frac{4\pi^2 l_{\texttt{Tokyo}}}{ T_{\texttt{Tokyo}}^2}}{\frac{4\pi^2 l_{\texttt{Cambridge}}}{ T_{\texttt{Cambridge}}^2}}\\\\\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}=\frac{\frac{0.9923}{2^2}}{\frac{0.9941}{2^2}}=0.998[/tex]
Ratio of free fall acceleration of Tokyo to Cambridge = 0.998
Electromagnetic radiation of 5.16Ă—1016 Hz frequency is applied on a metal surface and caused electron emission. Determine the work function of the metal if the maximum kinetic energy (Ek) of the emitted electron is 4.04Ă—10-19 J.
Answer:
Work function of the metal, [tex]W_o=3.38\times 10^{-17}\ J[/tex]
Explanation:
We are given that
Frequency of the electromagnetic radiation, [tex]f=5.16\times 10^{16}[/tex] Hz
The maximum kinetic energy of the emitted electron, [tex]K=4.04\times 10^{-19}\ J[/tex]
We need to find the work function of the metal.
We know that the maximum kinetic energy of ejected electron
[tex]K=h\nu-w_o[/tex]
Where h=Plank's constant=[tex]6.63\times 10^{-34} J.s[/tex]
[tex]\nu[/tex] =Frequency of light source
[tex]w_o[/tex]=Work function
Substitute the values in the given formula
Then, the work function of the metal is given by :
[tex]W_o=h\nu -K[/tex]
[tex]W_o=6.63\times 10^{-34}\times 5.16\times 10^{16}-4.04\times 10^{-19}[/tex]
[tex]W_o=3.38\times 10^{-17}\ J[/tex]
So, the work function of the metal is [tex]3.38\times 10^{-17}\ J[/tex]. Hence, this is the required solution.
Carbon dioxide makes up approximately 0.04% of Earth's atmosphere. If you collect a 2.9 L sample from the atmosphere at sea level (1.00 atm) on a warm day (27 ∘C), how many CO2 molecules are in your sample?
Answer:
number of molecules= 2.83 x 10^19
Explanation:
0.04 % means 0.04 L CO_2 in 100 L atmosphere
so for 2.9 L atmosphere CO2 vol = ( 0.04/100) x 2.9 = 0.00116
T = 27°C = 300 K , P = 1 atm ,
n = PV/RT = ( 1 x 0.00116) ÷ ( 0.08206x300) = 4.71×10^{-5}
number of molecules = 6.023 x 10^23 x 4.71×10^{-5} = 2.83 x 10^19
The purpose of studying a system is to identify the strengths and weaknesses of the existing system and examine current inputs, outputs, processes, security and controls, and system performance.
The purpose of system is to make system more reliable and efficient for desired work completion
Explanation:
In most of the times, the system analysts operate in a dynamic environment where change is the necessary process . A business application, a business firm or a computer system may be the required environment. In order To rebuild a system, the key elements must be considered which we need to examine are as follows:
1. Outputs and inputs.
2. Security and Control.
3. processor
4. Environment.
5. Feedback.
6. Boundaries and interface.
A 4.9 kg block slides down an inclined plane that makes an angle of 27◦ with the horizontal. Starting from rest, the block slides a distance of 2.7 m in 5.4 s. The acceleration of gravity is 9.81 m/s 2 . Find the coefficient of kinetic friction between the block and plane.
Answer:
μk = 0.488
Explanation:
Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
We define the x-axis in the direction parallel to the movement of the block on the inclined plane and the y-axis in the direction perpendicular to it.
Forces acting on the block
W: Weight of the block : In vertical direction
FN : Normal force : perpendicular to the inclined plane
fk : kinetic Friction force: parallel to the inclined plane
Calculated of the W
W= m*g
W= 4.9 kg* 9.8 m/s² = 48.02 N
x-y weight components
Wx = Wsin θ = 48.02*sin27° = 21.8 N
Wy = Wcos θ = 48.02*cos27° = 42.786 N
Calculated of the FN
We apply the formula (1)
∑Fy = m*ay ay = 0
FN - Wy = 0
FN = Wy
FN = 42.786 N
Calculated of the fk
fk = μk* FN= μk*42.786 Equation (1)
Kinematics of the block
Because the block moves with uniformly accelerated movement we apply the following formula to calculate the acceleration of the block :
d = v₀*t+(1/2)*a*t² Formula (2)
Where:
d:displacement (m)
v₀: initial speed (m/s)
t: time interval (m/s)
a: acceleration ( m/s²)
Data:
d= 2.7 m
v₀ = 0
t= 5.4 s
We replace data in the formula (2)
d = v₀*t+(1/2)*a*t²
2.7 = 0+(1/2)*a*( 5.4)²
2.7 = (14.58)*a
a = 2.7 / (14.58)
a= 0.185 m/s²
We apply the formula (1) to calculated μk:
∑Fx = m*ax , ax= a : acceleration of the block
Wx-fk= m*a , fk=μk*42.786 of the Equation (1)
21.8 - (42.786)*μk = (4.9)*(0.185)
21.8 -0.907= (42.786)*μk
20.89 = (42.786)*μk
μk = (20.89) / (42.786)
μk = 0.488
A Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia In one set of experiments, they studied the maximum speed that quolls could run around a curved path without slipping. One quoll was running at 2.4 m/s around a curve with a radius of 1.6 m when it started to slip.
What was the coefficient of static friction between the quoll's feet and the ground in this trial?
Answer:
Coefficient of static friction = 0.37
Explanation:
At the point the the quoll slides, quoll attains its maximum velocity.
So Ne = (mv^2)/r ....equa 1
And N =mg....equ 2
Where N vertical force of qoull acting on the surface, e = coefficient of friction, m=mass, g=9.8m/s^2, r =radius =1.6m, v= max velocity of quill = 2.4m/s
Sub equ 2 into equ 1
Mge= (mv^2)/r ...equa3
Simplfy equ3
e = v^2/(gr)...equ 4
Sub figures above
e = 5.76/(9.8*1.6)
e = 0.37
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth.
What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if
(a) its initial speed is 0.351 of the escape speed from Earth and
(b) its initial kinetic energy is 0.351 of the kinetic energy required to escape Earth? (Give your answers as unitless numbers.)
(c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?
Answer:
Explanation:
We shall apply law of conservation of mechanical energy for projectile being thrown .
Total energy on the surface = total energy at height h required
a ) At height h , velocity = .351 x ( 2 GM/R x h )
[tex]\frac{-GM}{R} + \frac{m\times(.351\times\sqrt{2GM})^2 }{2R } = \frac{-GMm}{R+h} + 0[/tex]
[tex]\frac{-GMm}{R} +\frac{1}{2}\times \frac{-2GMm}{R} \times0.123=\frac{-GMm}{R+h}[/tex]
[tex]\frac{0.877GMm}{R} =\frac{-GMm}{R+h}[/tex]
h = .14 R
b )
[tex]\frac{-GM}{R} + \frac{m\times(.351\times2GM) }{2R } = \frac{-GMm}{R+h} + 0[/tex]
[tex]\frac{-0.649GMm}{R} = \frac{-GMm}{R+h}[/tex]
h = .54 R
c ) least initial mechanical energy required at launch if the projectile is to escape Earth
= GMm / R + 1/2 m (2GM/R)
= 0
A tennis ball is thrown upward from the top of a 680 foot high building at a speed of 56 feet per second. The tennis ball's height above ground can be modeled by the equation . When does the tennis ball hit the ground?
Answer:
t = 8.5 s
Explanation:
Kinematic equation of the movement of the tennis ball that is thrown upwards :
y = y₀ + v₀*t -½ g*t² Equation (1)
Where :
y : position of the ball as a function of time
y₀ : Initial position of the ball
t: time
g: acceleration due to gravity in m/s²
Known data
g = 32 ft/s²
y₀ = 680 ft
v₀ = 56 ft/s
Calculation of the time it takes for the ball to thit the ground
We replace data en the equation (1)
y = y₀ + v₀*t -½ g*t²
0 = 680+(56)*t -½( 32) *t²
16*t²-(56)*t- 680 = 0 equation (2)
solving equation (2) quadratic:
t₁ = 8.5 s
t₁ = -5 s
Time cannot be negative so the time it takes for the ball to hit the ground is t = 8.5 s
Suppose we consider the system of the three capacitors as a single "equivalent" capacitor. Given the charges of the three individual capacitors calculated in the previous part, find the total charge Qtot for this equivalent capacitor.
Answer:
Qtot = 6C * deltaV
Explanation:
you can find the total capacitance from adding 1C+2C+3C=6C. and the total voltage is 1V. Capacitance = charge/voltage--> C = Q / V--> 6C = Q / deltaV. this makes Qtot = 6C* deltaV
The Total charge for the equivalent circuit is = [tex]Q_{tot}[/tex] = 6c * ΔV
Although your question is incomplete I found the missing part online and used it to resolve the question
Given data :
Total capacitance ( C ) = 6C ( 1 + 2 + 3 )
voltage = 1 V
Three capacitors having values of ; 1 C, 2 C, 3 C
Determine the total charge ( Qtot )
Applying the formula ; Q = CV ---- ( 1 )
where; Q = charge
C = capacitance
change in V = ΔV
∴ [tex]Q_{tot}[/tex] = 6c * ΔV
Hence the total charge Qtot for the equivalent capacitor = 6c * ΔV
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A car slams on its brakes creating an acceleration of -3.2 m/s2 it comes to a rest after traveling a distance of 210 m what was it's velocity before it began to accelerate
The initial velocity of the car is 36.6 m/s
Explanation:
The motion of the car is a uniformly accelerated motion (=constant acceleration), therefore we can apply suvat equations:
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
For the car in this problem, we have:
v = 0 is the final velocity (the car comes to a stop)
[tex]a=-3.2 m/s^2[/tex] is the acceleration
s = 210 m is the displacement of the car
Solving for u, we find the initial velocity:
[tex]u=\sqrt{v^2-2as}=\sqrt{0-(2)(-3.2)(210)}=36.6 m/s[/tex]
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An astronaut drops a hammer on the moon . It takes 1 second to hit the ground after being dropped, and it is going 1.6m/s when it lands. What is the acceleration due to gravity on thr moon?
Answer:
the value of acceleration due to gravity in moon is 1.6m/[tex]s^{2}[/tex] along downward direction
Explanation:
Here, the acceleration is constant and it is equal to acceleration due to gravity in moon. Therefore the question depicts a situation of uniformly accelerated motion in a straight line. So, let us refresh the three equations of uniformly accelerated straight line motion.
v = u + at
[tex]s = ut + \frac{1}{2}at^{2}[/tex]
[tex]v^{2} = u^{2} +2as[/tex]
where,
u = initial velocity
v = final velocity
s = displacement
a = acceleration
t = time
Since we are dealing with vectors (velocity, acceleration and displacement), we have to take their directions in to account. So we must adopt a coordinate system according to our convenience. Here, we are taking point of throwing as origin, vertically upward direction as positive y axis and vertically downward direction as negative y axis.
t = 1s
u = 0 (since the hammer is dropped)
v = -1.6m/s (since its direction is downward)
a = ?
The only equation that connects all the above quantities is
v = u + at
therefore,
a = [tex]\frac{v - u}{t}[/tex]
substituting the values
a = [tex]\frac{-1.6 - 0}{1}[/tex]
a = -1.6m/[tex]s^{2}[/tex]
Thus, the value of acceleration due to gravity in moon is 1.6m/[tex]s^{2}[/tex]. The negative sign indicates that it is along downward direction.
An 19-cm-long bicycle crank arm, with a pedal at one end, is attached to a 23-cm-diameter sprocket, the toothed disk around which the chain moves. A cyclist riding this bike increases her pedaling rate from 65 rpm to 90 rpm in 10 s .
Answer:
The tangential acceleration of the pedal is 0.0301 m/s².
Explanation:
Given that,
Length = 19 cm
Diameter = 23 cm
Time = 10 sec
Initial angular velocity = 65 rpm
Final velocity = 90 rpm
Suppose we need to find the tangential acceleration of the pedal
We need to calculate the tangential acceleration of the pedal
Using formula of tangential acceleration
[tex]a_{t}=r\alpha[/tex]
[tex]a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{\omega_{2}-\omega_{1}}{t}[/tex]
[tex]a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{90\times\dfrac{2]pi}{60}-65\times\dfrac{2\pi}{60}}{10}[/tex]
[tex]a_{t}=0.0301\ m/s^2[/tex]
Hence, The tangential acceleration of the pedal is 0.0301 m/s².
Two identical strings, of identical lengths of 2.00 m and linear mass density of μ=0.0065kg/m, are fixed on both ends. String A is under a tension of 120.00 N. String B is under a tension of 130.00 N. They are each plucked and produce sound at the n=10 mode. What is the beat frequency?
Answer:
beat frequency = 13.87 Hz
Explanation:
given data
lengths l = 2.00 m
linear mass density μ = 0.0065 kg/m
String A is under a tension T1 = 120.00 N
String B is under a tension T2 = 130.00 N
n = 10 mode
to find out
beat frequency
solution
we know here that length L is
L = n × [tex]\frac{ \lambda }{2}[/tex] ........1
so λ = [tex]\frac{2L}{10}[/tex]
and velocity is express as
V = [tex]\sqrt{\frac{T}{\mu } }[/tex] .................2
so
frequency for string A = f1 = [tex]\frac{V1}{\lambda}[/tex]
f1 = [tex]\frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}[/tex]
f1 = [tex]\frac{10}{2L} \sqrt{\frac{T1}{\mu } }[/tex]
and
f2 = [tex]\frac{10}{2L} \sqrt{\frac{T2}{\mu } }[/tex]
so
beat frequency is = f2 - f1
put here value
beat frequency = [tex]\frac{10}{2*2} \sqrt{\frac{130}{0.0065}}[/tex] - [tex]\frac{10}{2*2} \sqrt{\frac{120}{0.0065} }[/tex]
beat frequency = 13.87 Hz
Six baseball throws are shown below. In each case the baseball is thrown at the same initial speed and from the same height h above the ground. Assume that the effects of air resistance are negligible. Rank these throws according to the speed of the baseball the instant before it hits the ground.
Answer:
The final velocities of all the six balls will be same.
Explanation:
According to law of conservation of energy:
Gain in K.E = Loss in potential energy
½ mv^2 = mgh
Where “m” and “g” are constant. The interchange in energies will occur only with the change in velocity and height. Since, balls are thrown from the same hight with the same initial velocity so, their final velocities will also be same just before striking the ground.
The six balls will reach the ground at the same time, hence the final velocity of the balls will be the same.
During a downward motion of an object, the speed of the object increases as the object moves downwards and becomes maximum before the object hits the ground.
The equation for estimating the final velocity of the six balls is given as;
[tex]v_f = v_i + gt[/tex]
If air resistance is negligible, the six balls will reach the ground at the same time, hence the final velocity of the balls will be the same.
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Sound wave A delivers 2J of energy in 2s. Sound wave length B delivers 10J of energy in 5s. Sound wave C delivers 2mJ of energy in 1ms. Rank in order, from largest to smallest, the sound powers of Pa, Pb, Pc of these three waves.Explain. What equation would you use to determine this?
Answer:
[tex]P_c=P_b>P_a[/tex]
Explanation:
E = Energy
T = Time
Power is given by the equation
[tex]P=\frac{E}{T}[/tex]
For first case
[tex]P_a=\frac{2}{2}\\\Rightarrow P_a=1\ W[/tex]
For second case
[tex]P_b=\frac{10}{5}\\\Rightarrow P_b=2\ J[/tex]
For third case
[tex]P_c=\frac{2\times 10^{-3}}{1\times 10^{-3}}\\\Rightarrow P_b=2\ J[/tex]
The rank of power would be [tex]P_c=P_b>P_a[/tex]
In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter contained 75.0 g H2O initially at 23.2°C. The final temperature of the solution was 31.8°C. What was the change in enthalpy for the dissolution of this compound?
Answer:
The enthalpy for dissolution is - 305.558 J/g
Solution:
Mass of the ionic compound, m = 10.00 g
Mass of water, m' = 75.0 g
Initial temperature, T = [tex]23.2^{\circ}C[/tex]
Final Temperature, T' = [tex]31.8^{\circ}C[/tex]
Now,
To calculate the change in enthalpy:
We know that the specific heat of water is 4.18 [tex]J/g^{\circ}C[/tex]
Total mass of the solution, M = m + m' = 10.00 + 75.0 = 85.0 g
Temperature, difference, [tex]\Delta T = T' - T = 31.8 - 23.2 = 8.6^{\circ}C[/tex]
Thus
The heat absorbed by the solution is given by:
[tex]Q = MC_{w}\Delta T = 85.0\times 4.18\times 8.6 = 3055.58\ J[/tex]
Enthalpy, [tex]\Delta H = -\frac{Q}{m} = - \frac{3055.58}{10} = - 305.558\ J/g[/tex]
What causes the temperature to increase with height through the stratosphere
Answer: The temperature increases on the stratosphere with the altitude, given that absorption of the ultraviolet rays by the ozone.
Explanation: On the stratosphere, the water vapor and the umidity are almost nonexistents and, in view of the absorption of ultraviolet rays by the ozon, the temperature increases, reaching 35,6º Fahrenheit.
The ozone is a unusual type of oxygen molecule. In the stratosphere, the ozone appears on a large scale and warms it up by the absorption of the ultraviolet rays energy.
The temperature in the stratosphere increases with altitude due to the absorption of UV radiation by ozone, which leads to exothermic reactions that generate heat, especially at higher altitudes where ozone concentration is greater.
Explanation:The temperature increases with height in the stratosphere primarily because of the absorption of ultraviolet (UV) radiation by ozone. Ozone, a form of oxygen molecule, is very good at absorbing UV radiation from the Sun. When UV radiation is absorbed by ozone, it leads to exothermic chemical reactions which generate heat, thus raising the temperature of the surrounding atmosphere.
At higher altitudes in the stratosphere, the concentration of ozone is greater and, consequently, more UV radiation is absorbed, leading to a higher temperature in this region compared to the lower part of the stratosphere. It is this temperature inversion that distinguishes the stratosphere from the troposphere below, where the temperature normally decreases with altitude. The absorption of UV radiation and its conversion into heat makes the top of the stratosphere hotter as it is closer to the source of shortwave radiation, and the amount of absorbed shortwave radiation diminishes towards the bottom of the stratosphere.
The cheetah is one of the fastest-accelerating animals, because it can go from rest to 19.6 m/s (about 44 mi/h) in 2.9 s. If its mass is 108 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in the following units.
(a) watts(b) horsepower.
Answer:
a)P =14288.4 W
b)P = 19.16 horsepower
Explanation:
Given that
m= 108 kg
Initial velocity ,u= 0 m/s
Final velocity ,v= 19.6 m/s
t= 2.9 s
Lets take acceleration of Cheetah is a m/s²
We know that
v= u + a t
19.6 = 0 + a x 2.9
a= 6.75 m/s²
Now force F
F= m a
F= 108 x 6.75 N
F= 729 N
Now the power P
P = F.v
P = 729 x 19.6 W
P =14288.4 W
We know that
1 W= 0.0013 horsepower
P = 19.16 horsepower
P =14288.4 W
Listed following are four models for the long-term expansion (and possible contraction) of the universe. Rank the models from left to right based on their predictions for the average distance between galaxies five billion years from now, from smallest to largest.
a. recollapsing universe
b. accelerating universe
c. coasting universe
d. critical universe
Answer:
gdsz
Explanation:
dsgzz cxvzdgctfgdsvftgdsftrdsfdtsardtgasfd5t6sgftsfdrstfdtgsv6cr5vsd5rw5
The masses of the blocks are m1 = 16.0 kg and m2 = 12.0 kg, the mass of the pulley is M = 5.00 kg, and the radius of the pulley is R = 0.300 m. Block m2 is initially on the floor, and block m1 is initially 4.60 m above the floor when it is released from rest. The pulley's axis has negligible friction. The mass of the string is small enough to be ignored, and the string does not slip on the pulley, nor does it stretch.
a. How much time (in s) does it take block m1 to hit the floor after being released?
b. How would your answer to part (a) change if the mass of the pulley were neglected? (Enter the time, in seconds, it takes block m1 to hit the floor if the mass of the pulley were neglected.)
Answer:
a)Time taken will be 2.783 s
b)Time taken will be 2.564 s
Explanation:
a)Since the pulley has mass ,
It will have a MOMENT OF INERTIA . in other terms, whn the string slides upon it, it will produce a torque ( due to the tension) and thus it will make the pulley roll.The string soesn't get slackened. Thus the acceleration along the string must be constant - which is the string constraint. the FBD's of the bodies are attached. from them ,[tex]m_{1}g-T_{1}= m_{1}a//T_{2}-m_{2}g=m_{2}a[/tex] ------3
Since the string doesn't slip, the acceleration of the pulley at the end point of contact of the string must be equal to [tex]a[/tex]or,αR = a ; ------1
writing the torque equation about COM of the pulley , we get[tex](T_{1} - T_{2})*R=m*R^{2}*(alpha)=5*R^{2}*\frac{a}{R}[/tex] ------2
solving these we get ,
[tex]a=\frac{4g}{33}[/tex]
∴
a)time taken :
[tex]s=ut+\frac{1}{2} at^{2}\\u=0\\a=\frac{4g}{33} \\s=4.6\\4.6=\frac{1}{2} * \frac{4*9.8}{33} *t^{2}\\t= 2.783 sec[/tex]
ANS : 2.783 sec
b)
In case of B, mass is zero.
Thus, there is no rotation of the pulley. this is equivalent to a normal 1 Dimension motion question
equations are:[tex]m_{1}g-T_{1}=m_{1}a\\T_{1}-m_{2}g=m_{2}a\\a= \frac{m_{1}-m_{2}}{m_{1}+m_{2}}g\\a=\frac{4g}{28} \\a=\frac{g}{7} \\a=1.4ms^{-2}[/tex]
Thus time t will be ,
[tex]s=\frac{1}{2} at^{2}\\4.6=\frac{1}{2}*1.4*t^{2}\\ t=2.564 sec[/tex]
ANS : 2.564 sec
The time it takes the block to reach the ground can be found by making
use of the law of conservation of energy.
a. The time it takes the the block m₁ to hit the floor is approximately 2.596 seconds.b. The time it takes the the block m₁ to hit the floor, if the mass of the pulley were neglected is approximately 2.562 seconds.Reasons:
Given parameters are;
Mass of block m₁ = 16.0 kg
Mass of block m₂ = 12.0 kg
Mass of the pulley, M = 5.00 kg
By conservation of energy, we have;
m₁g·h - m₂·g·h = 0.5×(m₁ + m₂)·v² + 0.5·I·ω²
[tex]\omega = \dfrac{v}{R}[/tex]
[tex]Moment \ of \ inertia\ of \ pulley, I = \dfrac{1}{2} \cdot M \cdot R^2[/tex]
Therefore;
[tex]m_1 \cdot g \cdot h - m_2 \cdot g \cdot h = 0.5 \cdot (m_1 + m_2) \cdot v^2 + 0.5 \cdot I \cdot \dfrac{v}{R}[/tex]
Which gives;
(16 - 12)×9.81×4.6 = 0.5×(16+12)×v² + 0.5×(0.5×5×0.3²)× [tex]\left(\dfrac{v}{0.3} \right)^2[/tex]
Solving gives, v ≈ 21.93 m/s
We have;
v ≈ 3.544
v² = 2·a·h
[tex]a = \dfrac{v^2}{2 \times h}[/tex]
Which gives;
[tex]a = \dfrac{3.544^2}{2 \times 4.6} \approx 1.365[/tex]
v = a×t
[tex]t = \dfrac{v}{a} = \dfrac{3.544}{1.365} \approx 2.596[/tex]
The time it takes the the block m₁ to hit the floor, t ≈ 2.596 seconds
b. When the mass of the pulley is neglected, we have;
[tex]m_1 \cdot g \cdot h - m_2 \cdot g \cdot h = 0.5 \cdot (m_1 + m_2) \cdot v^2[/tex]
(16 - 12)×9.81×4.6 = 0.5×(16+12)×v²
180.504 = 14·v²
[tex]v = \sqrt{\dfrac{180.504}{14} } \approx 3.591[/tex]
[tex]a = \dfrac{3.591^2}{2 \times 4.6} \approx 1.401[/tex]
[tex]t = \dfrac{v}{a} = \dfrac{3.591}{1.401} \approx 2.562[/tex]
The time it takes the the block m₁ to hit the floor, if the mass of the pulley were neglected, t ≈ 2.562 seconds.
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A 0.150-kg metal rod carrying a current of 15.0 A glides on two horizontal rails 0.510 m apart. If the coefficient of kinetic friction between the rod and rails is 0.160, what vertical magnetic field is required to keep the rod moving at a constant speed?
Answer:
B = 0.0307 T = 30.74 mT
Explanation:
Given
m = 0.150 kg
I = 15.0 A
d = 0.510 m
μk = 0.16
B = ?
Balancing the forces on the rod in the j direction
N - m*g = 0 ⇒ N = m*g
and in the i direction
I*d*B - μk*N = 0 ⇒ B = μk*N / (I*d)
⇒ B = μk*m*g / (I*d)
⇒ B = (0.16)*(0.150 kg)*(9.8 m/s²) / (15.0 A*0.510 m)
⇒ B = 0.0307 T = 30.74 mT
You apply forces of magnitude 4.2×104N to the top and bottom surfaces of a brass cube. The forces are tangent to each surface and parallel to the sides of each surface. If the cube is 2.5 cm on a side, what is the resulting shear displacement? The shear modulus for brass is 3.5×1010Pa.
Answer: [tex]4.8(10)^{-5} m[/tex]
Explanation:
We can solve this problem by the following equation:
[tex]\eta=\frac{F.h}{A \Delta x}[/tex]
Where:
[tex]\eta=3.5(10)^{10}Pa[/tex] is the shear modulus for brass
[tex]F=4.2(10)^{4}N[/tex] is the applied force
[tex]h=2.5 cm=0.025 m[/tex] is the height of the cube
[tex]A=h^{2}=(0.025 m)^{2}=625(10)^{-6} m^{2}[/tex] is the area of each surface of the cube
[tex]\Delta x[/tex] is the shear displacement
Finding [tex]\Delta x[/tex]:
[tex]\Delta x=\frac{F.h}{A \eta}[/tex]
[tex]\Delta x=\frac{(4.2(10)^{4}N)(0.025 m)}{(625(10)^{-6} m^{2})(3.5(10)^{10}Pa)}[/tex]
Finally:
[tex]\Delta x= 4.8(10)^{-5} m[/tex]
You throw a 50.0g blob of clay directly at the wall with an initial velocity of -5.00 m/s i. The clay sticks to the wall, and the collision takes about 20.0 ms (2.00 x 10^-2 s). a) What is the change in momentum for the blob of clay?
Answer:0.25 kg-m/s
Explanation:
Given
mass of blob [tex]m=50 gm [/tex]
initial velocity [tex]u=-5 m/s\ \hat{i}[/tex]
time of collision [tex]t=20 ms[/tex]
we know Impulse is equal to change in momentum
initial momentum [tex]P_i=mu[/tex]
[tex]P_i=50\times 10^{-3}\times (-5)=-0.25 kg-m/s[/tex]
Final momentum [tex]P_f=50\times 10^{-3}v[/tex]
[tex]P_f=0[/tex] as final velocity is zero
Impulse [tex]J=P_f-P_i[/tex]
[tex]J=0-(-0.25)[/tex]
[tex]J=0.25 kg-m/s[/tex]
A molecule moves down its concentration gradient using a transport protein in the plasma membrane. This is an example of
Final answer:
Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane.
Explanation:
Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane. This process does not require the input of energy and allows substances to diffuse across the membrane more easily. For example, glucose is transported into cells using glucose transporters that utilize facilitated transport. This process is important for the movement of larger or charged molecules that cannot freely diffuse across the cell membrane.
Consider the following electron configurations to answer the question:
(i) 1s2 2s2 2p6 3s1
(ii) 1s2 2s2 2p6 3s2
(iii) 1s2 2s2 2p6 3s2 3p1
(iv) 1s2 2s2 2p6 3s2 3p4
(v) 1s2 2s2 2p6 3s2 3p5
The electron configuration of the atom that is expected to have a positive electron affinity is ________.
Answer:
(ii) 1s2 2s2 2p6 3s2
Explanation:
Electron Affinity is the energy change that occur when an atom gains an electron.
X₍₉₎ + e⁻ → X⁻ ΔE = Eea
ΔE is change in energy
Eea is electron affinity
Often, electron affinity has negative energy values. The more negative the electron affinity, the easier it is to add an electron to a particular atom. Electron affinity increases across the period in the periodic table. However, there are few exceptions:
1. The electron affinities of group 18 (8A) elements are greater than zero. This is because the atom has a filled valence shell, an addition of electron causes the electron to move to a higher energy shell.
2. The electron affinities of group 2 (2A) elements are more positive because addition of an electron requires it to reside in the previously unoccupied p sub-shell.
3. The electron affinities of group 15 (5A) elements are more positive because addition of an electron requires it to be put in an already occupied orbital.
Applying these consideration to the elements given in the question:
(i) The sum of the electron in 1s²2s² 2p⁶ 3s¹ = 2+2+6+1 = 11 Sodium (Na)
(ii) The sum of the electron in 1s² 2s² 2p⁶ 3s² = 2+2+6+2 = 12 Magnesium (Mg)
(iii) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p¹ = 2+2+6+2+1 = 13 Aluminium (Al)
(iv) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁴ = 2+2+6+2+4 = 16 Sulfur (S)
(v) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁵ = 2+2+6+2+5 = 17 Chlorine (Cl)
The atom that is expected to have a positive electron affinity is Magnesium which is a group 2A element with electron configuration of 1s² 2s² 2p⁶ 3s².
An atom with a positive electron affinity attracts electrons more than one with negative electron affinity. Typically, these atoms tend to have half-filled or filled subshell configurations. In the given examples, the atom with electron configuration 1s2 2s2 2p6 3s2 3p1 is likely to have a positive electron affinity.
Explanation:The electron affinity of an atom is a measure of the energy change when an electron is added to a neutral atom to form a negative ion. An atom with a positive electron affinity attracts electrons more than one with negative electron affinity. Usually, the atom would tend to have a half-filled (e.g. 3p3) or filled (e.g. 3p6) subshell configuration.
If we consider the given electron configurations: (i) 1s2 2s2 2p6 3s1 (ii) 1s2 2s2 2p6 3s2 (iii) 1s2 2s2 2p6 3s2 3p1 (iv) 1s2 2s2 2p6 3s2 3p4 (v) 1s2 2s2 2p6 3s2 3p5, it appears that (iii) 1s2 2s2 2p6 3s2 3p1 should have a positive electron affinity. This atom is a Potassium atom (K). It has a positive electron affinity because it is just one electron away from obtaining a full orbit shell (in the 3s and 3p orbitals) which would provide a more stable electron configuration.
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As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an axis through its center. Since you were a good physics student, you know what to do. You measure the diameter of the wheel to be 0.88 m and find that it weighs 280 N . You mount the wheel, using frictionless bearings, on a horizontal axis through the wheel's center. You wrap a light rope around the wheel and hang a 6.32 kg mass from the free end of the rope. You release the mass from rest; the mass descends and the wheel turns as the rope unwinds. You find that the mass has speed 4.0 m/s after it has descended 2.5 m .(a) What is the moment of inertia of the wheel for an axis perpendicular to the wheel at its center?
Final answer:
The moment of inertia of the wheel for an axis perpendicular to the wheel at its center is 0.964 kg * m^2.
Explanation:
To calculate the moment of inertia of the wheel, we can use the principle of conservation of energy. The initial gravitational potential energy of the mass is equal to the final rotational kinetic energy of the wheel. This can be represented by the equation:
mg * h = 1/2 * Iω^2
Where m is the mass, g is the acceleration due to gravity, h is the distance the mass has descended, I is the moment of inertia of the wheel, and ω is the angular velocity of the wheel. Rearranging the equation:
I = 2mg * h / ω^2
Substituting the given values:
I = 2 * 6.32 kg * 9.8 m/s^2 * 2.5 m / (4.0 m/s)^2
I = 0.964 kg * m^2
Therefore, the moment of inertia of the wheel for an axis perpendicular to the wheel at its center is 0.964 kg * m^2.
Calculate the work required to move a planet’s satellite of mass 571 kg from a circular orbit of radius 2R to one of radius 3R, where 8.8 × 106 m is the radius of the planet. The mass of the planet is 7.76 × 1024 kg. Answer in units of J]
Final answer:
The work required to move a satellite from an orbit of radius 2R to 3R around a planet is calculated using the gravitational potential energy formula and is found to be 3.897×1010 J.
Explanation:
To calculate the work required to move a satellite from one circular orbit to another around a planet, we must consider the gravitational potential energy differences in the two orbits.
The gravitational potential energy (U) of an object of mass m in orbit around a planet of mass M at a distance r is given by U = -GmM/r, where G is the gravitational constant (6.67×10-11 N m2/kg2).
For the initial orbit at radius 2R, the potential energy is U1 = -GmM/(2R), and for the final orbit at radius 3R, the potential energy is U2 = -GmM/(3R). The work done (W) in moving the satellite is the difference in gravitational potential energy, W = U2 - U1. Substituting the values, we get:
W = (-GmM/3R) - (-GmM/2R) = (GmM/6R)
Let's calculate the work required using the given values: G = 6.67×10-11 N m2/kg2, m = 571 kg, M = 7.76×1024 kg, R = 8.8×106 m.
W = (6.67×10-11 N m2/kg2 × 571 kg × 7.76×1024 kg) / (6 × 8.8×106 m)
W = 3.897×1010 J
Therefore, the work required to move the satellite from a circular orbit of radius 2R to one of radius 3R is 3.897×1010 J.
Four particles, one at each of the four corners of a square with 2.0-m-long edges, are connected by massless rods. The masses are m1=m3=3.0 kg and m2=m4=4.0 kg. Find the moment of inertia of the system about the z axis. (the z axis runs through m2, which is at the origin, m1 is on the y axis, and m3 is on the x axis. Use the parallel-axis theorem and the result for Problem 41 to find them moment of inertia of the four-particle system about an axis that passes through the center of mass and is parallel with the z axis. Check your result by direct computation.
Answer:
Explanation:
The particles are in x-y plane with coordinates of masses as follows
m₂ at (0,0 ) m₁ at ( 0,2 ), m₄ at ( 2,2 ) and m₃ at (2,0 )
Moment of inertia about z axis
I_z = 0 + 3 x 2² + 4 x (2√2)² + 3 x 2²
= 12 + 32 + 12
= 56 kgm²
Now let us find out moment of inertia about axis through CM
According to theorem of parallel axis
I_z = I_g + m x r²
Here m is total mass that is 14 kg and r is distance between two axis which is √2 m
56 = I_g + 14 x (√2)²
I_g = 56 - 28
= 28 kgm²
We can directly compute I_g as follows
I_g = 4 x (√2)² +3 x (√2)² +4 x (√2)²+3 x (√2)²
= 8 +6 +8 +6
= 28 kgm²
So the result obtained earlier is correct.
A car traveling 6.0 m/s is uninformly accelerating at a rate of 3.0 m/s^2 for 15 seconds. What is it’s final velocity?
Answer:
The answer to your question is 11.2 m/s
Explanation:
Data
Initial speed (vo) = 6.0 m/s
Acceleration (a) = 3.0 m/s²
time = 15 s
Final speed = ?
Formula
d = vot + [tex]\frac{1}{2} at^{2}[/tex]
vf² = vo² + 2ad
Process
d = (6)(15) + [tex]\frac{1}{2} (3)(15)^{2}[/tex]
d = 90 + 337.5
d = 427.5 m
vf² = (6)² + 2(3)(15)
vf² = 36 + 90
vf² = 126
vf = 11.2 m/s