In a mixture of helium and neon gases at thermal equilibrium, the average speed of helium atoms will be greater than the average speed of neon atoms.
Explanation:In a mixture of two gases at thermal equilibrium, the average speed of the atoms will depend on their respective molar masses. The average speed of an atom can be calculated using the root mean square speed formula, v = sqrt(3kT/m), where v is the average speed, k is the Boltzmann constant, T is the temperature in Kelvin, and m is the molar mass of the atom. In this case, helium has a smaller molar mass than neon, so according to the formula, the average speed of helium atoms will be greater than the average speed of neon atoms.
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The average speed of helium atoms is higher than that of neon atoms because the molar mass of helium is much smaller than that of neon. At the same temperature, lighter molecules move faster. Thus, the correct answer is C.
To determine the average speeds of helium and neon atoms in a mixture at thermal equilibrium at temperature T, we need to consider the relationship between temperature, mass, and molecular speed.
The average kinetic energy per atom at a given temperature T is the same for both helium and neon. This can be expressed as:
[tex]\frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k_B T[/tex]
where,
[tex]k_B[/tex]= Boltzmann constant
[tex]T[/tex]= temperature
[tex]m[/tex]= molar mass of the gas
Since both gases have the same average kinetic energy, the differences in their masses will influence their average speeds. Specifically, the equation for the root-mean-square speed (v_{rms}) of a gas is:
[tex]v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}[/tex]
Given:
Molar mass of helium (He) =[tex]4.0 g/mol[/tex]
Molar mass of neon (Ne) = [tex]20.2 g/mol[/tex]
We see that helium has a much smaller molar mass compared to neon. Therefore, for the same temperature T:
[tex]v_{rms, He} > v_{rms, Ne}[/tex]
This means the average speed of helium atoms is greater than that of neon atoms. Thus, the correct answer is:
C. the average speed of the helium atoms is greater than the average speed of the neon atoms.
Angular oscillation of the slotted link is achieved by the crank OA, which rotates clockwise at the steady speed N = 118 rev/min. Find the angular velocity of the slotted link when θ = 61°
Answer:
The angular velocity is 4.939rad/sec
Explanation:
θ = 61°
N = 118 rev/min
The formula for angular velocity is given as;
w = θ/t
where;
w = angular velocity
θ = angle
t = time in rad/sec
1 rev/ min = 0.1047 radian/second.
Therefore 118 rev/min = 118*0.1047rad/sec
=12.35rad/sec
Substituting into the formula, we have
w = 61/12.35
w = 4.939rad/sec
Once Kate's kite reaches a height of 52 ft (above her hands), it rises no higher but drifts due east in a wind blowing 6 ft divided by s. How fast is the string running through Kate's hands at the moment that she has released 105 ft of string?
Answer:
5.213ft
Explanation:
Z² = x² + y²
x = √(z² - y²)
y = 52ft, dx = 6ft, z = 105ft, dz = ?
d(z² = x² + y²)
2zdz = 2xdx
dz = xdx/z
But x = √(z² - y²)
dz = √(z² - y²)/z * dx
dz = [√(105² - 52²)/105] * 6
dz = √(8521)/ 17.5
dz = 5.213ft
2. On January 21 in 1918, Granville, North Dakota, had a surprising change in temperature. Within 12 hours, the temperature changed from 237 K to 283 K. What is this change in temperature in the Celsius and Fahrenheit scales?
Answer:
The temperature change in Celsius is 46°C.
The temperature change in Fahrenheit is 82.8°F.
Explanation:
A degree of Celsius scale is equal to that of kelvin scale; therefore,
[tex]\Delta C = \Delta K = 283K-237K\\\\ \boxed{\Delta C = 46^oC}[/tex]
A degree in Fahrenheit is 1.8 times the Celsius degree; therefore
[tex]\Delta F = 1.8(46^o)[/tex]
[tex]\boxed{\Delta F = 82.8^oF}[/tex]
Hence, the temperature change in Celsius is 46°C, and the temperature change in Fahrenheit is 82.8°F.
Final answer:
The temperature change of 46 degrees from Kelvin to Celsius in Granville, North Dakota, is equivalent to a change of 82.8 degrees Fahrenheit.
Explanation:
On January 21, 1918, the temperature in Granville, North Dakota, changed from 237 K to 283 K within 12 hours. To convert the change in temperature to the Celsius scale, we first recognize that 0 degrees Celsius is equivalent to 273.15 K. Therefore, the initial temperature in Celsius would have been 237 K - 273.15 K = -36.15 extdegree C, and the final temperature would have been 283 K - 273.15 K = 9.85 extdegree C. The change in the Celsius scale is then 9.85 extdegree C - (-36.15 extdegree C) = 46 extdegree C
To convert the change in temperature to the Fahrenheit scale, we start with the change in Celsius and use the conversion formula (F = C imes \frac{9}{5} + 32). Since we are looking for the change, we only need to convert the difference in Celsius to Fahrenheit. Therefore, the change in Fahrenheit is 46 extdegree C imes \frac{9}{5} = 82.8 extdegree F.
Two negative charges that are both - 3.8 C push each other apart with a force of 19.0 N. How far apart are the two charges?
Answer:
The separation distance between the two charges must be 82704.2925 m
Explanation:
Given:
Two negative charges that are both q = -3.8 C
Force of 19 N
Question: How far apart are the two charges, s = ?
First, you need to get the electrostatic force of this two negative charges:
[tex]F=\frac{kq}{s^{2} } \\s=\sqrt{\frac{kq}{F} }[/tex]
Here
k = electric constant of the medium = 9x10⁹N m²/C²
Substituting values:
[tex]s=\sqrt{\frac{9x10^{9}*(-3.8)^{2} }{19} } =82704.2925m[/tex]
A cyclist coasts up a 9.00° slope, traveling 12.0 m along the road to the top of the hill. If the cyclist’s initial speed is 9.00 m/s, what is the final speed? Ignore friction and air resistance.
Answer:
[tex]v_{f} \approx 6.647\,\frac{m}{s}[/tex]
Explanation:
The final speed of the cyclist is determined by applying the Principle of Energy Conservation:
[tex]\frac{1}{2}\cdot m\cdot v_{o}^{2} + m\cdot g\cdot h_{o} = \frac{1}{2}\cdot m\cdot v_{f}^{2} + m\cdot g\cdot h_{f}[/tex]
[tex]\frac{1}{2} \cdot v_{o}^{2} + g\cdot (h_{o}-h_{f}) = \frac{1}{2}\cdot v_{f}^{2}[/tex]
[tex]v_{f}^{2}=v_{o}^{2} + 2\cdot g \cdot (h_{o}-h_{f})[/tex]
[tex]v_{f} = \sqrt{v_{o}^{2}+2\cdot g \cdot (h_{o}-h_{f})}[/tex]
[tex]v_{f} = \sqrt{v_{o}^{2}-2\cdot g \cdot \Delta s \cdot \sin \theta}[/tex]
[tex]v_{f} = \sqrt{(9\,\frac{m}{s} )^{2}-2\cdot (9.807\,\frac{m}{s} )\cdot (12\,m)\cdot \sin 9^{\textdegree}}[/tex]
[tex]v_{f} \approx 6.647\,\frac{m}{s}[/tex]
Answer:
the final speed of the 10.85 m/s.
Explanation:
Given that,
Slope with respect to horizontal, [tex]\theta=9^{\circ}[/tex]
Distance travelled, d = 12 m
Initial speed of the cyclist, u = 9 m/s
We need to find the final speed of the cyclist. Let h is the height of the sloping surface such that,
[tex]h=d\times sin\thetah\\\\=12\times sin(9) \\\\h = 1.877 m[/tex]
v is the final speed of the cyclist. It can be calculated using work energy theorem as
[tex]\dfrac{1}{2}m(v^2-u^2)\\\\=mgh\dfrac{1}{2}(v^2-u^2)\\\\=gh\dfrac{1}{2}\times (v^2-(9.0)^2)\\\\=9.8\times 1.87\\\\v = 10.85 m/s[/tex]
Thus,the final speed of the 10.85 m/s.
Sheet metal of two different alloys is produced by rolling to 0.1 mm thickness. You must select one of these alloys for a sheet forming operation. To make the selection, tensile samples are cut from each alloy at 0o , 45o , and 90o to the rolling direction. The initial gauge section of each tensile sample is 0.75 mm width. After subjecting all samples to the same tensile load, th
Answer: Alloy A is best chosen for the sheet forming operation
Explanation:
taking both alloy A and B into consideration;
For alloy A;
the true strain ration, R₀ = width strain / thickness strain
R₀ = in (0.65/0.75) / in (0.092/0.1) = 1.716
at 45⁰ and 90⁰, the true strain ratio becomes;
R45⁰ = in (0.63/0.75) / in (0.093/0.1) = 2.4025
R90⁰ = in (0.67/0.75) / in (0.087/0.1) = 0.8099
where R(avg) = (R₀+2Ras+Rao) / A = (1.716+2(2.4025)+0.8099) = 3.0340
R(avg) = 3.0340
For alloy B;
R₀ = in (0.7/0.75) / in (0.082/0.1) = 0.3477
R45⁰ = in (0.69/0.75) / in (0.083/0.1) = 0.4475
R90⁰ = in (0.71/0.75) / in (0.078/0.1) = 0.2206
R(avg) = (R₀+2Ras+Rao) / A = (0.3477+2(0.4475)+0.2206) = 0.365825
R(avg) = 0.365825
comparing both we have that,
R(avg) for allow A > R(avg) for allow B
∴ Alloy A is the best to be selected for sheet formation operation.
cheers i hope this helps!!!1
The uniform slender bar AB has a mass of 7 kg and swings in a vertical plane about the pivot at A. If angular velocity of the bar is 3 rad/s when θ = 35o , compute the force supported by the pin at A at that instant
Answer:
49 N
Explanation:
The diagram of the bar is obtained online and attached to this solution.
The free body diagram is also attached.
Since the weight of the bar acts at the middle of the bar, the torque due to the weight of the bar is given by
τ = mgx
where x = (L/2) cos 35° = 0.45 × cos 35° = 0.3686 m
τ = (7)(9.8)(0.3686) = 25.29 Nm
The force acting on pin A = torque ÷ (length × sin 35°) = 25.29 ÷ (0.9 × sin 35°)
= 25.29 ÷ 0.5162 = 48.99 N = 49 N
Hope this Helps!!!
Answer:
The force supported by the pin at A is 69.081 N
Explanation:
The diagram is in the figure attach. The angular acceleration using the moment expression is:
[tex]-mg(\frac{Lcos\theta }{2} )=I\alpha \\\alpha =\frac{-3g}{2L} cos\theta[/tex]
Where
L = length of the bar = 900 mm = 0.9 m
[tex]\alpha =\frac{-3*9.8cos35}{2*0.9} =-13.38rad/s^{2}[/tex]
The acceleration in point G is equal to:
[tex]a_{G} =a_{A} +\alpha kr_{G/A} -w^{2} r_{G/A}[/tex]
Where
aA = acceleration at A = 0
w = angular velocity of the bar = 3 rad/s
rG/A = position vector of G respect to A = [tex]\frac{L}{2} cos\theta i-\frac{L}{2} cos\theta j[/tex]
[tex]a_{G} =(\frac{L}{2}\alpha sin\theta -\frac{w^{2}Lcos\theta }{2} )i+(\frac{L}{2}\alpha cos\theta +\frac{w^{2}Lsin\theta }{2} )j=(\frac{0.9*(-13.38)*sin35}{2} -(\frac{3^{2}*0.9*cos35 }{2} )i+(\frac{0.9*(-13.38)*cos35}{2} +\frac{3^{2} *0.9*sin35)}{2} )j=-6.76i-2.61jm/s^{2}[/tex]
The force at A in x is equal to:
[tex]A_{x} =ma_{G} =7*(-6.76)=-47.32N[/tex]
The force at A in y is:
[tex]A_{y} =ma_{G} +mg=(7*(-2.61))+(7*9.8)=50.33N[/tex]
The magnitude of force A is equal to:
[tex]A=\sqrt{A_{x}^{2}+A_{y}^{2} } =\sqrt{(-47.32^{2})+50.33^{2} } =69.081N[/tex]
The compressive strength of our bones is important in everyday life. Young’s modulus for bone is about 1.4 * 1010 Pa. Bone can take only about a 1.0% change in its length before fracturing. (a) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0 cm2 ? (This is approximately the cross-sectional area of a tibia, or shin bone, at its narrowest point.) (b) Estimate the maximum height from which a 70 kg man could jump and not fracture his tibia. Take the time between when he first touches the floor and when he has stopped to be 0.030 s, and assume that the stress on his two legs is distributed equally.
Answer:
Explanation:
Young modulus ε = 1.4 × 10¹⁰ Pa
ΔL = 1% of the original length = 0.01 x where x is the original length
cross sectional area = 3.0 cm² =( 3 .0 / 10000) m²= 0.0003 m²
ε = Stress / strain
stress = ε × strain
stress = F /A
F force = ε × A × ( ΔL / L) = 1.4 × 10¹⁰ Pa × 0.0003 m² × 0.01 = 4.2 × 10⁴ N
b) F net = F max - mg ( weight) = 84000 - ( 70 × 9.8 m/s² ) ( F is double since the stress on the two leg is equally distributed)
f net = ma = 84000 - ( 70 × 9.8 m/s² )
a = (84000 - ( 70 × 9.8 m/s² )) = 1190.2 m/s²
v = u + at
where final velocity equal zero
- u = -at since it coming downwards
u = at = 1190.2 m/s² × 0.03s = 35.706 m/s
using conservation of energy
1/2 mv² = mgh
1/2v²/ g = h
h = 0.5 × (35.706 m/s )² / 9.8 = 65.04 m
A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.827 m and 2.89 kg, respectively. When the propeller rotates at 507 rpm (revolutions per minute), what is its rotational kinetic energy K
Answer:
3241.35J
Explanation:
No. Of rods = 5
Mass = 2.89kg
Length (L) = 0.827m
W = 507rpm
Kinetic energy of rotation = ½I*ω²
For each rod, the moment of inertia (I) = ML² / 3
I = ML² / 3
I = [2.89*(0.827)²] / 3
I = 1.367 / 3 = 0.46kgm²
ω = 507 rev/min. Convert rev/min to rev/sec.
507 * 2Πrads/60s = 53.09rad/s
ω = 53.09rad/s
k.e = ½ I * ω²
K.E = ½ * 0.46 * (53.09)²
K.E = 648.27.
But there five (5) rods, so kinetic energy is equal to
K.E = 5 * 648.27 = 3241.35J
The rotational kinetic energy of a propeller with five blades, each modeled as a uniform rod, can be determined using formulas for moment of inertia and rotational kinetic energy.
Explanation:The kinetic energy of rotating objects depends on their moment of inertia and angular velocity. Given that each propeller blade is modeled as a uniform rod rotating about its end, we can calculate the moment of inertia (I) of all five blades using the formula I=5*(1/3*m*L^2), where m is the mass and L is the length of each rod. After we find the moment of inertia, we can determine the angular velocity (ω) in radian per second, given that the propeller rotates at 507 rpm, by using the conversion factor ω=(507*2*pi)/60. Finally, we calculate the rotational kinetic energy (K) using the formula K=1/2*I*ω^2.
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20 POINTS TRUE OR FALSE:
An object with constant velocity will change its motion only if a balanced force acts on it.
A) TRUE
B) FALSE
Hello :)
True :D
Gülün selamı vaar <3
Cerenin de selamı vaar <3
-Your Welcome--Merve-
The magnetic dipole moment of Earth has magnitude 8.00 1022 J/T.Assume that this is produced by charges flowing in Earth’s molten outer core. If the radius of their circular path is 3500 km, calculate the current they produce.
Answer:
[tex]2.08\cdot 10^9 A[/tex]
Explanation:
The magnetic dipole moment of a circular coil with a current is given by
[tex]\mu = IA[/tex]
where
I is the current in the coil
[tex]A=\pi r^2[/tex] is the area enclosed by the coil, where
[tex]r[/tex] is the radius of the coil
So the magnetic dipole moment can be rewritten as
[tex]\mu = I\pi r^2[/tex] (1)
Here we can assume that the magnetic dipole moment of Earth is produced by charges flowing in Earth’s molten outer core, so by a current flowing in a circular path of radius
[tex]r=3500 km = 3.5\cdot 10^6 m[/tex]
Here we also know that the Earth's magnetic dipole moment is
[tex]\mu = 8.0\cdot 10^{22} J/T[/tex]
Therefore, we can re-arrange eq (1) to find the current that the charges produced:
[tex]I=\frac{\mu}{\pi r^2}=\frac{8.00\cdot 10^{22}}{\pi (3.5\cdot 10^6)^2}=2.08\cdot 10^9 A[/tex]
A horizontal uniform bar of mass 2.5 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.74 m from the other end. A monkey of mass 1.25 kg walks from one end of the bar to the other. Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar.
Answer:
The tension T₁ in string 1 at the moment that the monkey is halfway between the ends of the bar is 34.68 N
Explanation:
Given;
mass of the uniform horizontal bar, m₁ = 2.5 kg
length of the bar, L = 3.0 m
mass of the monkey, m₂ = 1.25 kg
distance from the right end of the second string, d = 0.74 m
For a body to remain in rotational equilibrium, the net external torque acting on it due to applied external forces must be equal.
∑τ = 0
For vertical equilibrium of bar-string system, in which T₁ and T₂ are the tension on both ends of the string;
T₁ + T₂ = (m₁ + m₂)g
T₁ + T₂ = (2.5 + 1.25) 9.8
T₁ + T₂ = 36.75 N
For rotational equilibrium when the monkey is halfway between the ends of the bar, take moment about the left end of the string.
(m₁ + m₂) L /2 = T₂(L - d)
(m₁ + m₂)0.5L = T₂( L - d)
(2.5 + 1.25)0.5 x 3 = T₂ ( 3 - 0.74)
4.6875 = T₂ (2.26)
T₂ = (4.6875) / (2.26)
T₂ = 2.074 N
Thus, T₁ = 36.75 N - T₂
T₁ = 36.75 N - 2.074 N
T₁ = 34.68 N
You have a summer job at a company that developed systems to safely lower large loads down ramps. Your team is investigating a magnetic system by modeling it inthe laboratory. The safety system is a conducting bar that slides on two parallel conducting rails thatrun down the ramp (similar to the one in the previous problem). The bar is perpendicular to the railsand is in contact with them. At the bottom of the ramp, the two rails are connected together. The box slides down the rails through a uniform vertical magnetic field. The magnetic field is supposed to causethe bar to slide down the ramp at a constant velocity even when friction between the bar and the rails negligible.Before setting up the laboratory model, your task is to calculate the constant velocity of the bar (sliding down the ramp on rails in a vertical magnetic field) as a function of the mass of the bar, the strength ofthe magnetic field, the angle of the ramp from the horizontal, the length of the bar (which is the same asthe distance between the rails), and the resistance of the bar. Assume that all of the other conductors inthe system have a much smaller resistance than the bar.a) If the force due to the changing flux exactly cancells out the net force due to the combination of gravity and normal force, then the bare will cease to accelerate and instead move at a constant velocity. Please solve for this velocity algebraically.b)Write out the units for each of your variables and show (by cancellation and substitution) that the units for your veloctiy will be m/s on both left and right side of your equation.
Answer:
Note that the emf induced is
emf = B d v cos (A)
---> v = emf / [B d cos (A)]
where
B = magnetic field
d = distance of two rails
v = constant speed
A = angle of rails with respect to the horizontal
Also, note that
I = emf/R
where R = resistance of the bar
Thus,
I = B d v cos (A) / R
Thus, the bar experiences a magnetic force of
F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.
Thus, the component of this parallel to the incline is
F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R
As this is equal to the component of the weight parallel to the incline,
B^2 d^2 v cos^2 (A) / R = m g sin (A)
where m = the mass of the bar.
Solving for v,
v = [R m g sin (A) / B^2 d^2 cos^2 (A)] [ANSWER, the constant speed, PART A]
******************************
v = [R m g sin (A) / B^2 d^2 cos^2 (A)]
Plugging in the units,
m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]
Note that T = kg / (s * C), and ohm = J * s/C^2
Thus,
m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]
= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]
As J = kg*m^2/s^2, cancelling C^2,,
= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]
Cancelling kg^2,
= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]
Cancelling m^2/s^2,
= [s * m/s^2]
Cancelling s,
=m/s [DONE! WE SHOWED THE UNITS ARE CORRECT! ]
A 0.450-kg hockey puck, moving east with a speed of 5.25 m/s , has a head-on collision with a 0.850-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed (magnitude of the velocity) of each object after the collision
Explanation:
Given that,
Mass of the hockey puck, m₁ = 0.45 kg
Initial peed of the hockey puck, u₁ = 5.25 m/s (east)
Mass of other puck, m₁ = 0.85 kg
Initial speed of other puck, u₂ = 0 (at rest)
Let v₁ and v₂ are the final speeds of both pucks after the collision respectively. Using the conservation of momentum as :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1v_1+m_2v_2=0.45\times 5.25+0.85\times 0\\\\m_1v_1+m_2v_2=2.36\\\\0.45v_1+0.85v_2=2.36.........(1)[/tex]
The coefficient of restitution for elastic collision is equal to 1.
[tex]C=\dfrac{v_2-v_1}{u_1-u_2}\\\\1=\dfrac{v_2-v_1}{u_1-u_2}\\\\1=\dfrac{v_2-v_1}{5.25-0}\\\\v_2-v_1=5.25.......(2)[/tex]
On solving equation (1) and (2) we get :
[tex]v_1=-1.611\ m/s\\\\v_2=3.63 m/s[/tex]
Hence, this is the required solution.
n the melt atomization process to make fine metal powders, there usually is a variation in powder size because not all atomized droplets are of the same size. In an experiment, it was found that titanium droplets of 10 μm dia reached their solidification temperature (1660 C) in 2 milliseconds when the surrounding gas temperature was 1100 C and the initial droplet temperature was 1860 C. How long would it take for droplets of 5 μm, 20 μm, 30 μm and 50 μm diameter to cool by the same amount if they all had the same initial temperature? Given: density of Ti = 4510 kg.m-3.
Answer:
Check the explanation
Explanation:
Atomization, also called the spraying technique or procedure, refers to a process in which molten metals are broken down into little or tiny drops of liquid through the use of high-speed fluids (liquid as water, gas as air or inert gas) or fluids with centrifugal force, and then solidified into a powdered state.
Kindly check the attached image below to get the step by step solution to the above question.
The speed of light in a vacuum is 300,000.0 kmls. How long does it take light to travel
150,000,000 km from the Sun to Earth?
Answer:
8 minutes
Explanation:
The time it takes for light to travel from the Sun to Earth can be found using the formula time = distance/speed. Given the speed of light is known to be 300,000 km/s and the distance is 150,000,000 km, it will take light about 500 seconds to travel from the Sun to the Earth.
Explanation:To calculate the time it takes for light to travel from the Sun to Earth, you need to use the formula for time which is distance/speed. The speed of light is 300,000 km/s and the distance from the Sun to Earth is 150,000,000 km.
So, Time = Distance / Speed = 150,000,000 km / 300,000 km/s = 500 seconds
Therefore, it takes approximately 500 seconds for light to travel from the Sun to Earth.
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4. Analyze: What can you say about the acceleration of dividers when the pressure increases
from left to right?
Velocity changes with time and this is called acceleration.
What is acceleration?The question is incomplete but I will try to explain the meaning of acceleration to you. The term acceleration refers to the change of velocity with time.
a = Δv/t
a = accelerationΔv = change in velocityt = time intervalHence, velocity changes with time and this is called acceleration.
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The increase of the pressure of the system would lead to an increase of acceleration.
What happens when the pressure equates to the force of action?As pressure equals to the force upon action. Here F is the force and the A the area. As Newton's second law F = ma We substitute P A = m a
Hence the pressure is directly proportional to acceleration.
Find out more information about acceleration.
brainly.com/question/4279093.
What are potential impacts of pollution on a watershed? Check all that apply.
excess runoff
loss of farmland
loss of areas for tourism
contaminated drinking water
deaths of sea creatures that are used as a food source
limits to potential economic activities such as a fishing
Answer:
contaminated drinking water
deaths of sea creatures that are used as a food source
limits to potential economic activities such as a fishing
Explanation:
A watershed is a large area that comprises of drainage area of all the surrounding water bodies meeting at a common affluence point before draining into sea or ocean or any other large water body. Pollution in this area can pollute the small water streams flowing through it, thereby polluting the larger water body into which it drains.
Thus, the water extracted for drinking from such area will be contaminated. Pollution in larger water body can cause death of water creature and hence pose a threat to fishing.
Answer:
loss of areas for tourism
contaminated drinking water
deaths of sea creatures that are used as a food source
limits to potential economic activities such as fishing
The ultracentrifuge is an important tool for separating and analyzing proteins. Because of the enormous centripetal accelerations, the centrifuge must be carefully balanced, with each sample matched by a sample of identical mass on the opposite side. Any difference in the masses of opposing samples creates a net force on the shaft of the rotor, potentially leading to a catastrophic failure of the apparatus. Suppose a scientist makes a slight error in sample preparation and one sample has a mass 10 mg larger than the opposing sample.
Complete Question
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Answer:
The net force is [tex]F_{net}= 6.44 *10^{-4} N[/tex]
Explanation:
Generally the net force is a force that come up due to the unequal centripetal force(A difference in centripetal force ) and it is mathematically represented as
[tex]F_{net} = \Delta F_{cen}[/tex]
and the difference in centripetal force [tex]\Delta F_{cen}[/tex] is mathematically represented as
[tex]\Delta F_{cen} = \Delta m* rw^2[/tex]
Which the difference in mass multiplied by the centripetal acceleration
Substituting 10 mg = [tex]10 *10^{-3}g[/tex] for [tex]\Delta m[/tex] , 12 cm = [tex]\frac{12}{100} = 0.12m[/tex] for radius
and 70,000 rpm = [tex]70,000 *[\frac{2 \pi rad}{1 rev}][\frac{1 min}{60s} ] = 7326.7 rad/s[/tex]
[tex]F_{net} = \Delta F_{cen} = 10*10^{-3} * 0.12 * 7326.7[/tex]
[tex]F_{net}= 6.44 *10^{-4} N[/tex]
How much current is in a circuit that includes a 9.0-volt battery and a bulb with a resistance of 4.0 ohms? A. 0.44 amps B. 36 amps C. 2.3 amps D. 13 amps
Answer:
C. 2.3A
Explanation:
V/Ω=A
9V / 4Ω = 2.25 ≅ 2.3A
The current flowing through the circuit is 2.3 A. So, option C is correct.
What is Ohm's law ?Ohm's law states that, the voltage applied across a circuit is directly proportional to the steady current flowing through the circuit and also proportional to the resistance of the circuit.
Here,
Voltage applied in the circuit, V = 9 V
Resistance of the bulb, R = 4 Ω
According to Ohm's law,
V [tex]\alpha[/tex] I
V [tex]\alpha[/tex] R
So, V = IR
Therefore, current flowing through the circuit,
I = V/R
I = 9/4
I = 2.25 A approximately, 2.3 A
Hence,
The current flowing through the circuit is 2.3 A.
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A helicopter lifts a 81 kg astronaut 19 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/15. How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her
Answer:
a) The work done on the astronaut by the force from the helicopter is [tex]W_{h}=16087.68\ J[/tex].
b) The work done on the astronaut by the gravitational force is [tex]W_{g}=-15082.2\ J[/tex] .
Explanation:
We are told that the mass of the astronaut is [tex]m=81\ kg[/tex], the displacement is [tex]\Delta x=19\ m[/tex], the acceleration of the astronaut is [tex]|\vec{a}|=\frac{g}{15}[/tex] and the acceleration of gravity is [tex]g=9.8\ \frac{m}{s^{2}}[/tex] .
We suppose that in the vertical direction the force from the helicopter [tex]F_{h}[/tex] is upwards and the gravitational force [tex]F_{g}[/tex] is downwards. From the sum of forces we can get the value of [tex]F_{h}[/tex]:
[tex]F_{h}-F_{g}=m.a[/tex]
[tex]F_{h}-mg=m.\frac{g}{15}[/tex]
[tex]F_{h}=mg(1+\frac{1}{15})[/tex]
[tex]F_{h}=(\frac{16}{15}).81\ kg.\ 9.8\ \frac{m}{s^{2}}\ \Longrightarrow\ F_{h}=846.72\ N[/tex]
We define work as the product of the force, the displacement of the body and the cosine of the angle [tex]\theta[/tex] between the direction of the force and the displacement of the body:
[tex]W=F.\Delta x.\ cos(\theta)[/tex]
a) The work done on the astronaut by the force from the helicopter
[tex]W_{h}=F_{h}.\Delta x[/tex]
[tex]W_{h}=846.72\ N.\ 19\ m[/tex]
[tex]W_{h}=16087.68\ J[/tex]
b) The work done on the astronaut by the gravitational force
[tex]W_{g}=-F_{g}.\Delta x[/tex]
[tex]W_{g}=-mg\Delta x[/tex]
[tex]W_{g}=-81\ kg.\ 9.8\ \frac{m}{s^{2}}.\ 19\ m[/tex]
[tex]W_{g}=-15082.2\ J[/tex]
Answer:
a) Work done on the astronaut by the force from the helicopter = 16.104 kJ
b) Work done on the astronaut by the gravitational force = -15.082 kJ
Explanation:
mass of the astronaut, m = 81 kg
height, h = 19 m
acceleration of the astronaut, a = g/15
Since the astronaut is lifted up, using the third law of motion:
T - mg = ma
T = mg + ma
T = (81*9.81) + 81*(9.81/15)
T = 847.584 N
Work done on the astronaut by the helicopter
Work done = Tension * height
W = T* h
W = 847.584 * 19
Work done, W = 16104.096 Joules
W = 16.104 kJ
b) Work done on the astronaut by the gravitational force on her
[tex]W = -f_{g} h[/tex]
[tex]f_{g} = mg = 81 * 9.8\\f_{g} = 793.8 N[/tex]
[tex]W = -793.8 * 19\\W =- 15082.2 J[/tex]
W = -15.082 kJ
A magnetic dipole with a dipole moment of magnitude 0.0243 J/T is released from rest in a uniform magnetic field of magnitude 57.5 mT. The rotation of the dipole due to the magnetic force on it is unimpeded. When the dipole rotates through the orientation where its dipole moment is aligned with the magnetic field, its kinetic energy is 0.458 mJ. (a) What is the initial angle between the dipole moment and the magnetic field
Answer:
47.76°
Explanation:
Magnitude of dipole moment = 0.0243J/T
Magnetic Field = 57.5mT
kinetic energy = 0.458mJ
∇U = -∇K
Uf - Ui = -0.458mJ
Ui - Uf = 0.458mJ
(-μBcosθi) - (-μBcosθf) = 0.458mJ
rearranging the equation,
(μBcosθf) - (μBcosθi) = 0.458mJ
μB * (cosθf - cosθi) = 0.458mJ
θf is at 0° because the dipole moment is aligned with the magnetic field.
μB * (cos 0 - cos θi) = 0.458mJ
but cos 0 = 1
(0.0243 * 0.0575) (1 - cos θi) = 0.458*10⁻³
1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³
1 - cos θi = 0.3278
collect like terms
cosθi = 0.6722
θ = cos⁻ 0.6722
θ = 47.76°
Answer:
The initial angle between the dipole moment and the magnetic field is 47.76⁰
Explanation:
Given;
magnitude of dipole moment, μ = 0.0243 J/T
magnitude of magnetic field, B = 57.5 mT
change in kinetic energy, ΔKE = 0.458 mJ
ΔKE = - ΔU
ΔKE = - (U₂ -U₁)
ΔKE = U₁ - U₂
U₁ -U₂ = 0.458 mJ
[tex](-\mu Bcos \theta_i )- (-\mu Bcos \theta_f) = 0.458 mJ\\\\-\mu Bcos \theta_i + \mu Bcos \theta_f = 0.458 mJ\\\\\mu Bcos \theta_f -\mu Bcos \theta_i = 0.458 mJ\\\\\mu B(cos \theta_f - cos \theta_i ) = 0.458 mJ[/tex]
where;
θ₁ is the initial angle between the dipole moment and the magnetic field
[tex]\theta_f[/tex] is the final angle which is zero (0) since the dipole moment is aligned with the magnetic field
μB(cos0 - cosθ₁) = 0.458 mJ
Substitute the given values of μ and B
0.0243 x 0.0575 (1 - cosθ₁) = 0.000458
0.00139725 (1 - cosθ₁) = 0.000458
(1 - cosθ₁) = 0.000458 / 0.00139725
(1 - cosθ₁) = 0.327787
cosθ₁ = 1 - 0.327787
cosθ₁ = 0.672213
θ₁ = cos⁻¹ (0.672213)
θ₁ = 47.76⁰
Thus, the initial angle between the dipole moment and the magnetic field is 47.76⁰
which description best tells how the arrangement of the sun, moon, and earth affect the range of the tides during a spring tide
Answer:in a spring tide the sun , moon and the earth are in a straight line which causes regular higher tide and low tide to increase.
Explanation:
In a spring tide, the Moon, Earth, and Sun are all in allignment, which causes the greatest gravitational pull on the water of Earth. Basically, when the Moon and Sun align towards and in the same direction at the Earth, their gravitational pulls join together making the pull stronger. (see image below)
What this very strong pull does is make high tides higher than usual, and low tides lower than usual. This is because the pull on tides is very strong and more water is being pulled to form higher high tides, making there less water for there to be in low tides.
(image is from oceanservice.gov)
A uniform meter stick (with a length of 1 m) is pivoted to rotate about a horizontal axis through the 25 cm mark on the stick. The stick is released from rest in a horizontal position. The moment of inertia of a uniform rod about an axis perpendicular to the rod and through the center of mass of the rod is given by 1 12 ML2. Determine the magnitude of the initial angular acceleration of the stick
Answer:
The initial angular acceleration is [tex]16.8 s^{-2}[/tex]
Explanation:
From the parallel axis theorem the moment of inertia about 25cm mark is
[tex]I = \dfrac{1}{12}ML^2+MD^2[/tex]
since [tex]L = 1m[/tex] & [tex]D = 0.25m[/tex], we have
[tex]I = \dfrac{1}{12}M(1)^2+M(0.25m)^2[/tex]
[tex]I = \dfrac{7}{48} M[/tex]
Now, the gravitational force (equal to the weight of the object) acts as a torque [tex]\tau[/tex] on the center of mass of the rod, which induces angular acceleration [tex]\alpha[/tex]according to
[tex]\tau = I\alpha[/tex]
since [tex]\tau = Mg D[/tex]
[tex]MgD =I \alpha[/tex]
[tex]MgD = \dfrac{7}{48} M \alpha[/tex]
solving for [tex]\alpha[/tex] we get
[tex]\boxed{\alpha = \dfrac{48}{7} gD}[/tex]
putting in [tex]g= 9.8m/s^2[/tex] and [tex]D = 0.25m[/tex] we get:
[tex]\boxed{\alpha = 16.8\: s^{-2}.}[/tex]
which is the initial angular acceleration.
The initial angular acceleration of the stick is 16.812 radians per square second.
First, we calculate the resulting moment of inertia by Steiner theorem:
[tex]I_{O} = I_{g} + M\cdot r^{2}[/tex] (1)
Where:
[tex]I_{g}[/tex] - Moment of inertia with respect to center of mass, in kilogram-square meters.[tex]M[/tex] - Mass of the stick, in meters. [tex]r[/tex] - Distance between the point of rotation and the center of mass, in meters.[tex]I_{O}[/tex] - Moment of inertia with respect to the point of rotation, in meters.If we know that [tex]I_{g} = \frac{1}{12} \cdot M\cdot L^{2}[/tex], [tex]L = 1\,m[/tex] and [tex]r = 0.25\,m[/tex], then the formula for the moment of inertia is:
[tex]I_{O} = \frac{1}{12}\cdot M + \frac{1}{16}\cdot M[/tex]
[tex]I_{O} = \frac{7}{48}\cdot M[/tex] (2)
We know that initial angular acceleration ([tex]\alpha[/tex]), in radians per square second, is solely due to gravity. By the second Newton's law and the D'Alembert principle, we derive an expression for the initial angular acceleration of the stick:
[tex]\Sigma M = M\cdot g\cdot r = I_{O}\cdot \alpha[/tex] (3)
Where:
[tex]g[/tex] - Gravitational acceleration, in meters per square second.[tex]r[/tex] - Distance between the point of rotation and the center of mass, in meters.
By (2) and (3) we have the following simplified expression:
[tex]M\cdot g \cdot r = \frac{7}{48} \cdot M\cdot \alpha[/tex]
[tex]\alpha = \frac{48}{7}\cdot g \cdot r[/tex] (4)
If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]r = 0.25\,m[/tex], then the initial angular acceleration of the stick is:
[tex]\alpha = \frac{48}{7}\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.25\,m)[/tex]
[tex]\alpha = 16.812\,\frac{rad}{s^{2}}[/tex]
The initial angular acceleration of the stick is 16.812 radians per square second.
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2. A 3.5 m long string is fixed at both ends and vibrates in its 7th harmonic with an amplitude (at an antinode) of 2.4 cm. If the speed of waves on the string is 150 m/s, what is the maximum speed for a point on the string at an antinode?
Answer:
Maximum speed for a point on the string at anti node will be 22.6 m/sec
Explanation:
We have given length of string L = 3.5 m
For 7th harmonic length of the string [tex]L=\frac{7\lambda }{2}[/tex]
So [tex]\lambda =\frac{2L}{7}[/tex]
Speed of the wave in the string is 150 m/sec
Frequency corresponding to this wavelength [tex]f=\frac{v}{\lambda }=\frac{7v}{2L}[/tex]
So angular frequency will be equal to [tex]\omega =2\pi f=2\pi \times \frac{7v}{2L}=2\times 3.14\times \frac{7\times 150}{2\times 3.5}=942rad/sec[/tex]
Maximum speed is equal to [tex]v_m=A\omega =0.024\times 942=22.60m/sec[/tex]
So maximum speed for a point on the string at anti node will be 22.6 m/sec
If 4.1 × 1021 electrons pass through a 40 Ω resistor in 5 min, what is the potential difference across the resistor? The fundamental charge is 1.602 × 10−19 C . Answer in units of V
Answer:
Potential difference across resistor will be 87.66 volt
Explanation:
We have given number of electrons [tex]n=4.1\times 10^{21}[/tex]
Charge on one electron [tex]e=1.6\times 10^{-19}C[/tex]
So total charge [tex]Q=4.1\times 10^{21}\times 1.6\times 10^{-19}=656C[/tex]
Time is given t = 5 min
1 minute = 60 sec
So 5 minute = 5×60 = 300 sec
So current [tex]i=\frac{Q}{t}=\frac{656}{300}=2.1866A[/tex]
Resistance is given R = 40 ohm
Sp from ohm's law potential difference across resistor v = iR = 2.1866×40 = 87.466 volt
The potential difference across the 40 Ω resistor when 4.1 × 1021 electrons have passed through it over a time period of 5 minutes is calculated using Ohm's law, which results in 87.6 Volts.
Explanation:The subject of this question is based on Ohm's law, an important concept in Physics. Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.
First, calculate the total charge (Q) passed through the resistor. This can be done by multiplying the number of electrons by the fundamental charge of one electron (Q = electrons x electron charge). Thus, Q = 4.1 × 1021 electrons x 1.602 × 10-19 C = 656.82 Coulombs.
Next, use the formula to obtain the electric current (I) passed through the resistor. Since current (in Amperes) is equal to the total charge (in Coulombs) divided by the time (in seconds) and time here is given in minutes, we first need to convert minutes to seconds. So, 5 min = 5 x 60 = 300 seconds. Hence, I = Q / t = 656.82 C / 300 s = 2.19 Ampere.
Lastly, we can now find the potential difference (V) across the resistor. According to Ohm's law, V = I * R. Thus, V = 2.19 Ampere * 40 Ω = 87.6 Volts.
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Charge is distributed uniformly along a long straight wire. The electric field 2.00 cm from the wire is 20.0 NC/ directed radially inward towards the axis of symmetry. The linear charge density on the wire is
The linear charge density on the wire is approximately 0.002 N/Cm.
Explanation:The linear charge density on the wire can be calculated using the formula:
λ = E × r / 2πk
where λ is the linear charge density, E is the electric field, r is the distance from the wire, and k is the Coulomb's constant. Plugging in the known values of E = 20.0 N/C and r = 2.00 cm = 0.02 m, we can solve for λ:
λ = (20.0 N/C) × (0.02 m) / (2π × 8.99 × 10^9 Nm^2/C^2) ≈ 0.002 N/Cm
Therefore, the linear charge density on the wire is approximately 0.002 N/Cm.
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a 500 pound metal star is hanging on two cables which are attached to the ceiling. the left hand cable makes a 13 degree angle with the ceiling while the rigth hand cables makes a 77 degree angle with the ceiling. What is the tension on each of the cables
Answer:
T1 = 112.07[lb]
T2 = 487.3 [lb]
Explanation:
To solve this problem we must perform a static balance analysis, for this we perform a free body diagram. In this free body diagram we use the angles mentioned in the description of the problem.
Performing a sum of forces on the X-axis equal to zero, we can find an equation that relates the tension of the T1 & T2 cables.
Then we perform a summation of forces on the Y-axis, in which we can find another equation. In this new equation, we replace the previous one and we can find the tension T2.
T1 = 112.07[lb]
T2 = 487.3 [lb]
An excited hydrogen atom releases an electromagnetic wave to return to its normal state. You use your futuristic dual electric/magnetic field tester on the electromagnetic wave to find the directions of the electric field and magnetic field. Your device tells you that the electric field is pointing in the negative z direction and the magnetic field is pointing in the positive y direction. In which direction does the released electromagnetic wave travel
Answer:
The answer is -x direction.
Explanation:
According to the right hand rule for electromagnetism, when the thumb points up, index finger points forward and the middle finger is perpendicular to the index finger, the thumb points in the direction of the magnetic force, the index finger in the direction of the charge movement and the middle finger in the direction of the electromagnetic lines. If the electric field is pointing in the -z direction which is into the screen and the magnetic field is in the +y direction which is upwards, then the magnetic lines are in the -x direction.
I hope this answer helps.
A block of mass m attached to a horizontally mounted spring with spring constant k undergoes simple harmonic motion on a frictionless surface. How would the maximum speed of the block be affected if the spring constant was increased by a factor of 4 while holding the amplitude of oscillation constant? It would decrease by a factor of 14 . It would increase by a factor of 4. It would remain unchanged. It would increase by a factor of 2. It would decrease by a factor of 12 .
Question:
a. It would decrease by a factor of 14 .
b. It would increase by a factor of 4.
c. It would remain unchanged.
d. It would increase by a factor of 2.
e. It would decrease by a factor of 12 .
Answer:
The correct option is;
d. It would increase by a factor of 2
Explanation:
Here we have
[tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2 + F_kx[/tex]
The formula for maximum mass is given by the following relation;
[tex]v_{max1} =A\sqrt{\frac{k}{m} }[/tex]
Where:
[tex]v_{max}[/tex] = Maximum speed of mass on spring
k = Spring constant
m = Mass attached to the spring
A = Amplitude of the oscillation
Therefore, when k is increased by a factor of 4 we have;
[tex]v_{max2} =A\sqrt{\frac{4 \times k}{m} } = 2 \times A\sqrt{\frac{ k}{m} }[/tex]
Therefore, [tex]v_{max2} = 2 \times v_{max1}[/tex], that is the velocity will increase by a factor of 2.