Answer:
The magnetic flux links to its turns = [tex]7.6 \times10^{-3}[/tex] Wb.
Explanation:
Given :
Radius of circular coil = [tex]18 \times 10^{-2}[/tex] m
Number of turns = 25
Magnetic field = [tex]3 \times10^{-3}[/tex] T
Magnetic flux (Φ) is a measure of the magnetic field lines passes through a given area. The unit of magnetic flux is weber (Wb).
We know that,
⇒ Φ = [tex]BA[/tex]
Where [tex]B =[/tex] ext. magnetic field, [tex]A =[/tex] area of loop or coil.
But here given in question, we have turns of wire so our above eq. modified as follows.
⇒ Φ = [tex]NBA[/tex]
Where [tex]N =[/tex] no. of turns.
∴ Φ = [tex]25 \times 3 \times 10^{-3} \pi (18 \times10^{-2} )^{2}[/tex]
Φ = [tex]7.6 \times 10^{-3} Wb[/tex]
Thus, the magnetic flux links to its turns = [tex]7.6 \times 10^{-3} Wb[/tex]
. A proton, which moves perpendicular to a magnetic field of 1.2 T in a circular path of radius 0.080 m, has what speed? (qp = 1.6 · 10-19 C and mp = 1.67 · 10-27 kg)
Answer:
[tex]9.198\times 10^6 m/s[/tex]
Explanation:
We are given that
Magnetic field, B=1.2 T
Radius of circular path, r=0.080 m
[tex]q_p=1.6\times 10^{-19} C[/tex]
[tex]m_p=1.67\times 10^{-27} kg[/tex]
[tex]\theta=90^{\circ}[/tex]
We have to find the speed of proton.
We know that
Magnetic force, F=[tex]qvBsin\theta[/tex]
According to question
Magnetic force=Centripetal force
[tex]q_pvBsin90^{\circ}=\frac{m_pv^2}{r}[/tex]
[tex]1.6\times 10^{-19}\times 1.2=\frac{1.67\times 10^{-27}v}{0.08}[/tex]
[tex]v=\frac{1.6\times 10^{-19}\times 1.2\times 0.08}{1.67\times 10^{-27}}[/tex]
[tex]v=9.198\times 10^6 m/s[/tex]
A power plant burns 1000 kg of coal each hour and produces 500 kW of power. Calculate the overall thermal efficiency if each kg of coal produces 6 MJ of energy.
Answer:
The overall thermal efficiency is 30%.
Explanation:
Given;
Output power = 500 kWh
input energy per kg of coal = 6 MJ = 6 x 10⁶ J = 1.66667 kWh
1000 kg of coal will produce 1000 x 1.66667 kWh = 1666.67 kWh
Thus, total input power = 1666.67 kWh
Overall thermal efficiency = Total output power/Total input Power
Overall thermal efficiency = (500/1666.67) *100
Overall thermal efficiency = 0.29999 *100
Overall thermal efficiency = 30%
Therefore, the overall thermal efficiency is 30%.
When switch S is open, the voltmeter across the battery reads 1.52V. When theswitch is closed , the voltmeter reading drops to 1.37V and the ammeter reads 1.5A.Find the internal resistance of the battery.
Answer:
r = 0.1 Ω
Explanation:
We will use Ohm's Law in this question: V = IR, where I is the current and R is the resistance.
When the switch is open, the voltmeter reads 1.52 V, and there is no current in the circuit. We can deduce that the internal resistor in the battery causes 0.15 V to dissipate into heat, since the voltmeter reads 1.37 V when the switch is closed.
[tex]0.15 = (1.5)r\\r = 0.1~\Omega[/tex]
iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms a crystal with an fcc unit cell and a lattice constant a=0.352 nm. Calculate the density of Iron β Round your answer to 3 significant digits. cm
Answer:
8.60 g/cm³
Explanation:
In the lattice structure of iron, there are two atoms per unit cell. So:
[tex]\frac{2}{a^{3} } = \frac{N_{A} }{V_{molar} }[/tex] where [tex]V_{molar} = \frac{A}{\rho }[/tex] an and A is the atomic mass of iron.
Therefore:
[tex]\frac{2}{a^{3} } = \frac{N_{A} * p }{A}[/tex]
This implies that:
[tex]A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} } }[/tex]
= [tex]\frac{4}{\sqrt{3} }r[/tex]
Assuming that there is no phase change gives:
[tex]\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3} }[/tex]
= 8.60 g/m³
In some recent studies it has been shown that women are men when competing in similar sports (most notably in soccer and basketball). Select the statement that explains why this disparity might exist. a. The cross-sectional area of the ACL is typically larger in men, and therefore experiences less strain for the sam tensile force and Young's modulus. b. The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain. c. The cross-sectional area of the ACL is typically smaller in women, and therefore experiences less stress for the same tensile force. d. The ACL of women is more elastic than the ACL of men.
Answer: The correct option is B (The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain)
Explanation:
Anterior cruciate ligament (ACL) is one of the important ligaments found at the knee joint which helps to stabilise the joint. It connects the femur to the tibia bone at the knee joint.
Anterior cruciate ligament tear is one of the common knee joint injury which is seen in individuals( especially females) involved in sports( example soccer and basketball which involves sudden change in direction causing the knee to rotate inwards)
ACL tear occurs through both contact and non contact mechanisms. The contact mechanism of ACL injury occurs when force is directly applied at the lateral part of the knee while in non contact mechanism,tear occurs when the tibia is externally rotated on the planted foot.
Research has proven that women are prone to have ACL tear than men when competing in similar sports. This disparity exists due to structural differences that pose as risk factors. These includes
- the female ACL size is smaller than the male.
- the ACL of female has a lower modulus if elasticity( that is, less stiff) than in males leading to greater joint mobility than in the male.. therefore the option, (The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain) is correct.
A horizontal spring with stiffness 0.5 N/m has a relaxed length of 19 cm (0.19 m). A mass of 22 grams (0.022 kg) is attached and you stretch the spring to a total length of 26 cm (0.26 m). The mass is then released from rest. What is the speed of the mass at the moment when the spring returns to its relaxed length of 19 cm (0.19 m)?
Answer:
v = 0.0147 m / s
Explanation:
For this exercise let's use energy conservation
Starting point. Fully stretched spring
Em₀ = Ke = ½ k (x-x₀)²
Final point. Unstretched position
Emf = K = ½ m v²
Emo = Emf
½ k (x- x₀)² = ½ m v²
v = √m/k (x-x₀)
Let's calculate
v = √(0.022 / 0.5) (0.26-0.19)
v = 0.0147 m / s
The speed of the mass at the mean position is 0.333 m/s
Conservation of energy:The potential energy stored in a fully stretched spring
PE = ½ kx²
where x is the stretch of the spring = 26 -19 = 7 cm = 0.07 m
At the mean position, where x = 0, the PE stored in sprig is zero,
So according to the law of conservation of energy total energy must remain conserved so all the energy is converted into kinetic energy KE of the mass
KE = ½ mv²
where m is the mass and v is the velocity
½ kx² = ½ mv²
where k is the spring constant = 0.5 N/m
and m is the mass = 0.022 kg
[tex]v=\sqrt{\frac{k}{m} } x[/tex]
[tex]v=\sqrt{\frac{0.5}{0.022} } 0.07[/tex]
v = 0.333 m/s
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You are given a copper bar of dimensions 3 cm × 5 cm × 8 cm and asked to attach leads to it in order to make a resistor. If you want to achieve the smallest possible resistance, you should attach the leads to the opposite faces that measure.
A) 3 cm × 5 cm.
B) 3 cm × 8 cm.
C) 5 cm × 8 cm.
D) Any pair of faces produces the same resistance.
Answer:
[tex]c. 5cm \times 8cm[/tex]
Explanation:
The dimensions [tex]5cm\times 8cm[/tex] have the highest cross-sectional area combination of [tex]40cm^2[/tex].
-Resistance reduces with an increase in cross sectional area.
-[tex]Reason-[/tex]Electrons have alarger area to flow through.
A bumper car with mass m1 = 109 kg is moving to the right with a velocity of v1 = 4.9 m/s. A second bumper car with mass m2-83 kg is moving to the left with a velocity of v2 =-3.6 m/s. The two cars have an elastic collision. Assume the surface is frictionless.
1) What is the velocity of the center of mass of the system?
2) What is the initial velocity of car 1 in the center-of-mass reference frame?
3) What is the final velocity of car 1 in the center-of-mass reference frame?
4) What is the final velocity of car 1 in the ground (original) reference frame? -
5) What is the final velocity of car 2 in the ground (original) reference frame?
Answer:
(1)
The velocity of the center of mas of the system is= 0.801 m/s
(2)
Initial velocity of car 1 in the center of mass reference frame is =4.099 m/s
(3)
The final velocity of car 1 in the center of mass reference frame is
- 4.099 m/s
(4)
The final velocity of car 1 in the ground (original ) reference frame is = -3.298 m/s
(5)
The final velocity of car 2 in the ground (original) reference frame is = 7.166 m/s
Explanation:
Given
m₁ = 109 kg
v₁= 4.9 m/s
m₂= 83 kg
v₂= -3.6 m/s
The two cars have an elastic collision.
(1)
The velocity of the center of mas of the system is
[tex]V_{cm}=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]
[tex]= \frac{109.4.9+83(-3.4)}{109+83}[/tex] m/s
= 0.801 m/s
(2)
Initial velocity of car 1 in the center of mass reference frame is
[tex]V_{1,i}[/tex] = initial velocity - [tex]V_{cm}[/tex]
= (4.9 - 0.801) m/s
=4.099 m/s
(3)
Since the collision is elastic, the car 1 will bounce of opposite direction.
The final velocity of car 1 in the center of mass reference frame is
[tex]V_{1,f}[/tex] = - 4.099 m/s
(4)
The final velocity of car 1 in the ground (original ) reference frame
[tex]V'_{1,f}[/tex] = [tex]V_{cm}+V_{1,f}[/tex]
=(0.801- 4.099) m/s
= - 3.298 m/s
(5)
The momentum is conserved,
[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]
[tex]\Rightarrow v'_2=\frac{m_1}{m_2}(v_1-v'_1) +v_2[/tex]
Here [tex]v'_1= V'_{1,f}[/tex] = - 3.298 m/s
[tex]\Rightarrow v'_2=\frac{109}{83}[4.9-(-3.298)]+(-3.6)[/tex]
=7.166 m/s
The final velocity of car 2 in the ground (original) reference frame is = 7.166 m/s
Final answer:
The velocity of the center of mass of the bumper car system is 1.225 m/s. Car 1 has an initial velocity of 3.675 m/s in the center-of-mass reference frame. After the elastic collision, car 1's final velocity is -2.45 m/s, and car 2's final velocity is 4.825 m/s in the ground reference frame.
Explanation:
Velocity of the Center of Mass, Velocities in Reference Frames, and Final Velocities After an Elastic Collision
The velocity of the center of mass (V cm) of a system in one dimension is given by the formula:
V cm = (m1 * v1 + m2 * v2) / (m1 + m2)
In this case, we have m1 = 109 kg moving at v1 = 4.9 m/s and m2 = 83 kg moving at v2 = -3.6 m/s. The velocity of the center of mass is therefore:
V cm = (109 kg * 4.9 m/s + 83 kg * (-3.6 m/s)) / (109 kg + 83 kg)
Calculating this we get:
V cm = (534.1 kg*m/s - 298.8 kg*m/s) / 192 kg = 235.3 kg*m/s / 192 kg = 1.225 m/s (rounded to three significant figures)
The initial velocity of car 1 in the center-of-mass reference frame is given by:
u1 = v1 - V cm
Substituting the known values:
u1 = 4.9 m/s - 1.225 m/s = 3.675 m/s
In an elastic collision, velocities in the center-of-mass reference frame are mirrored. Thus, the final velocity of car 1 in the center-of-mass reference frame remains the same but in the opposite direction:
u1' = -u1 = -3.675 m/s
To find the final velocity of car 1 in the ground reference frame, we add the velocity of the center of mass:
v1' = u1' + V cm = -3.675 m/s + 1.225 m/s = -2.45 m/s
For car 2, the same principle applies. The final velocity in the center-of-mass reference frame will be the opposite of the initial, and thus:
v2' = -v2 + V cm = 3.6 m/s + 1.225 m/s = 4.825 m/s
A volley ball is hit directly toward the ceiling in a gymnasium with a ceiling height of L 0 m. If the initial vertical velocity is 13 m/s and the release height is 1.8 m will the ball hit the ceiling?
Complete Question:
A volley ball is hit directly toward the ceiling in a gymnasium with a ceiling height of 10 m. If the initial vertical velocity is 13 m/s and the release height is 1.8 m will the ball hit the ceiling?
Answer:
The ball will hit the ceiling
Explanation:
Given;
Initial vertical Velocity U = 13 m/s
Height of the ceiling = 10 m
Released height of the volley ball = 1.8 m
Height traveled by the volley ball, is calculated as follows;
[tex]V^2 =U^2 -2gH[/tex]
where;
V is final vertical velocity
[tex]2gH =U^2\\\\H = \frac{U^2}{2g} = \frac{(13)^2}{2(9.8)} = 8.62 m[/tex]
Remember this ball was released from 1.8 m height and it traveled 8.62 m.
Total distance traveled = 1.8 + 8.62 = 10.42 m
Therefore, the ball will hit the ceiling
The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?
Answer:
[tex]k=13\ m[/tex]
Explanation:
Given:
distance between two nearly parallel mirrors, [tex]d=6.5\ m[/tex]distance between the face and the nearer mirror, [tex]x=3\ m[/tex]So, the distance between the back-head and the mirror, [tex]y=6.5-3=3.5\ m[/tex]From the given information by the laws of reflection we can deduce the distance of the first reflection of the back of the head of person in the rear mirror.
Distance of the first reflection of the back of the head in the rear mirror from the object head:
[tex]y'=2\times y[/tex]
[tex]y'=7\ m[/tex] is the distance of the image from the object back head.
Now the total distance of this image from the front mirror:
[tex]z=y'+x[/tex]
[tex]z=7+3[/tex]
[tex]z=10\ m[/tex]
Now the second reflection of this image will be 10 meters inside in the front mirror.So, the total distance of the image of the back of the head in the front mirror from the person will be:
[tex]k=x+z[/tex]
[tex]k=3+10[/tex]
[tex]k=13\ m[/tex]
Final answer:
The back of the head appears to be 16.00 meters away due to multiple reflections between two parallel mirrors 6.50 meters apart, when your head is positioned 3.00 meters from the nearer mirror.
Explanation:
The question revolves around a flat mirror problem in physics where multiple reflections between two parallel mirrors are considered. When a person looks into a flat mirror, the image of any object (like the back of the head) appears to be the same distance behind the mirror as the object is in front of it. So if your head is 3.00 meters away from the nearer mirror, the first image of the back of your head will appear 3.00 meters behind that mirror.
However, since there are two mirrors, this first image will then act as an object for the second, farther mirror, creating a new image. The additional distance to this second image will be twice the distance between the two mirrors. So, the total apparent distance will be the distance to the first image plus 6.50 meters times 2, which is 16.00 meters (3.00+6.50+6.50 = 16.00 meters). Therefore, the back of your head appears to be 16.00 meters away.
The first dancer in the line is 10 m from the speaker playing the music; the last dancer in the line is 120 m from the speaker. Approximately how much time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer
Answer: 0.321 seconds
Explanation:
Let assume that air has a temperature of 20 °C. Sound speed at given temperature is [tex]343 \frac{m}{s}[/tex]. As sound spreads at constant speed, time can be easily found by using this formula:
[tex]\Delta t = \Delta t_{far} - \Delta t_{near}[/tex]
[tex]\Delta t = \frac{x_{far}-x_{near}}{v_{sound,air}}[/tex]
[tex]\Delta{t} = \frac{120 m - 10 m}{343 \frac{m}{s} }\\\\\Delta {t} = 0.321 sec[/tex]
If the electric field has a magnitude of 460 N/C and the magnetic field has a magnitude of 0.16 T, what speed must the particles have to pass through the selector undeflected?
The speed of the particle to pass through the Selector is 2875 m/s
Explanation:
Given -
Electric field,(We can represent as E) = 460 N/C
Magnetic field, (We can represent as B) = 0.16 T
Speed,(We can represent as v) = ?
We know that the formula for finding the velocity ,
[tex]v = \frac{E}{B} \\\\v = \frac{460}{0.16} \\\\v = 2875m/s[/tex]
Therefore, speed of the particle to pass through the Selector is 2875 m/s
According to the Can Manufacturers Institute, the energy used to make an aluminum can from recycled aluminum is 5% of the energy used to make an aluminum can from virgin ore. In a typical year, 1.7 billion pounds of aluminum cans are recycled.
Part A
How much energy is thermally transferred to get this mass of aluminum from 20 ∘C to its melting point, 660 ∘C?
4.45 * 10¹⁴ J is transferred to get this mass of aluminum from 20°C to its melting point, 660⁰C.
The quantity of heat required to change the temperature of a substance is given by:
Q = mcΔT
Where Q is the heat, m is the mass of the substance, ΔT is the temperature change = final temperature - initial temperature. c is the specific heat capacity
m = 1.7 billion pounds = 77 * 10⁷ kg, ΔT = 660 - 20 = 640°C, c = 903 J/kg•K
Hence:
Q = 77 * 10⁷ kg * 903 J/kg•K * 640°C
Q = 4.45 * 10¹⁴ J
4.45 * 10¹⁴ J is transferred to get this mass of aluminum from 20°C to its melting point, 660⁰C.
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The energy required to heat 1.7 billion pounds of aluminum from 20 degrees Celsius to 660 degrees Celsius is approximately 4.398 × 10^17 Joules.
Explanation:The thermal energy transferred, or heat, to raise the temperature of a substance is given by the formula q=mcΔT where 'm' is the mass, 'c' is the specific heat capacity, and 'ΔT' is the change in temperature. For aluminum, the specific heat capacity is 0.897 Joules per gram per degree Celsius (J/g°C).
First, we need to convert 1.7 billion pounds of aluminum into grams since the specific heat value is in grams. There are about 453,592.37 grams in a pound, so this gives us about 7.711 × 10^14 grams of aluminum.
The change in temperature (ΔT) is the final temperature minus the initial temperature, or 660 degrees Celsius - 20 degrees Celsius, which equals 640 degrees Celsius.
So, to find the total energy required, we use the formula and substitute the known values: q=(7.711 × 10^14 g)*(0.897 J/g°C)*(640°C), which equals approximately 4.398 × 10^17 Joules.
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At 2 P.M., ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 6 P.M.? (Round your answer to one decimal place.)
Answer:
The distance between the ships changing at 6PM is 21.29Km/h
Explanation:
Ship A is sailing east at 35Km/h and ship B is sailing West at 25Km/h
Given
dx/dt= 35
dy/dt= 25
dv/dt= ???? at t= 6PM - 2PM= 4
Therefore t=4
We know ship A travels at 150km in the x-direction and Ship A at t=4 travels at 4.35 Which is 140 also in x-direction
So, we use:
[tex] D^2 = (150 - x)^2 + y^2 [/tex];
[tex] D^2 = (150 - 140)^2 + y^2 [/tex]
But ship B travels at t=4, at 4.25 =100 in the y-direction
so, let's use the equation:
[tex] D^2 = 10^2 + 100^2 [/tex]
[tex] = D= sqrt*(10 + 100) [/tex]
Lets use 2DD' = 2xx' + 2yy'
Differentiating with respect to t we have:
D•d(D)/dt = -(10)•dx/dt + 100•dy/dt
=100.5 d(D)/dt = (-10)•35 + (100)•25
When t=4, we have x=(140-150) =10 and y=100
[tex]= D = sqrt*(10^2 + 100^2) [/tex]
=100.5
= 100.5 dD/dt = 10.35 +100.25
= dD/dt = 21.29km/h
The distance between Ship A and Ship B is changing at approximately 21.4 km/h at 6 P.M.
Here, we use related rates. Let's define the positions of the ships -
Ship A's position at 6 P.M. :
Since Ship A sails east at 35 km/h for 4 hours, it would have travelled 140 km east. Initial position is 150 km west of Ship B, so at 6 P.M., Ship A will be 10 km west of Ship B.Ship B's position at 6 P.M. :
Since Ship B sails north at 25 km/h for 4 hours, it will have travelled 100 km north.Let’s denote the distance between the two ships at time t as D, x as the east-west distance (positive east) between Ship A and Ship B, and y as the north-south distance (positive north) from the initial position of Ship B.
Given:
x = -10 km, y = 100 km at 6 P.M.[tex]\frac{dx}{dt} = 35 \, \text{km/h} \quad (\text{Rate of change in east-west direction for Ship A}) \\[/tex][tex]\frac{dy}{dt} = 25 \, \text{km/h} \quad (\text{Rate of change in north-south direction for Ship B}) \\[/tex]We use the Pythagorean theorem to relate x, y, and D:
[tex]D^2 = x^2 + y^2 \\[/tex]Differentiating both sides with respect to t:
[tex]2D \frac{dD}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \\[/tex]Simplifying and solving for [tex]\frac{dD}{dt}[/tex] gives:
[tex]\frac{dD}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{D} \\[/tex]Plugging in the values:
[tex]D = \sqrt{x^2 + y^2} = \sqrt{(-10)^2 + 100^2} = \sqrt{100 + 10000} = \sqrt{10100} \approx 100.5 \, \text{km} \\[/tex][tex]\frac{dD}{dt} = \frac{(-10)(35) + (100)(25)}{100.5} \approx \frac{2500 - 350}{100.5} \approx \frac{2150}{100.5} \approx 21.4 \, \text{km/h}[/tex]Therefore, the distance between the ships is changing at approximately 21.4 km/h at 6 P.M.
The rate of rotation of the disk is gradually increased. The coefficient of static friction between the coin and the disk is 0.50. Determine the linear speed of the coin when it just begins to slip.
Question is not complete and the missing part is;
A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center. The disk rotates at a constant rate in a counterclockwise direction. The coin does not slip, and the time it takes for the coin to make a complete revolution is 1.5 s.
Answer:
0.828 m/s
Explanation:
Resolving vertically, we have;
Fn and Fg act vertically. Thus,
Fn - Fg = 0 - - - - eq(1)
Resolving horizontally, we have;
Ff = ma - - - - eq(2)
Now, Fn and Fg are both mg and both will cancel out in eq 1.
Leaving us with eq 2.
So, Ff = ma
Now, Frictional force: Ff = μmg where μ is coefficient of friction.
Also, a = v²/r
Where v is linear speed or velocity
Thus,
μmg = mv²/r
m will cancel out,
Thus, μg = v²/r
Making v the subject;
rμg = v²
v = √rμg
Plugging in the relevant values,
v = √0.14 x 0.5 x 9.8
v = √0.686
v = 0.828 m/s
Final answer:
To determine the linear speed of the coin when it just begins to slip, we can use the equation frictional force = centripetal force for circular motion. By equating these two forces and solving for the linear speed, we can find the answer.
Explanation:
To determine the linear speed of the coin when it just begins to slip, we can use the equation:
frictional force = centripetal force for circular motion
The frictional force can be calculated using the equation:
frictional force = coefficient of static friction x normal force
And the centripetal force can be calculated using the equation:
centripetal force = mass of coin x acceleration towards the center of the disk
By equating these two forces and solving for the linear speed, we can find the answer.
In this case, we are given the coefficient of static friction as 0.50. We can also assume that the normal force is equal to the weight of the coin, which is the mass of the coin multiplied by the acceleration due to gravity. By plugging in these values, we can find the linear speed of the coin.
The Young’s modulus of nickel is Y = 2 × 1011 N/m2 . Its molar mass is Mmolar = 0.059 kg and its density is rho = 8900 kg/m3 . Given a bar of nickel of length 13 m, what time does it take for sound waves to propagate from one end to the other? Avogadro’s number is NA = 6.02 × 1023 atoms. Answer in units of s.
Answer:
Atomic Size and Mass:
convert given density to kg/m^3 = 8900kg/m^3 2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations) 3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3 4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m 5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom 6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom Young's Modulus: Y=(F/A)/(dL/L) 1) F=mg = (45kg)(9.8N/kg) = 441 N 2) A = (0.0018m)^2 = 3.5344e-6 m^2 3) dL = 0.0016m 4) L = 2.44m 5) Y = 1.834e11 N/m^2 Interatomic Spring Stiffness: Ks,i = dY 1) From above, diameter of one atom = 2.2245e-10 m 2) From above, Y = 1.834e11 N/m^2 3) Ks,i = 40.799 N/m (not rounding in my actual calculations) Speed of Sound: v = ωd 1) ω = √(Ks,i / m,a) 2) From above, Ks,i = 40.799 N/m 3) From above, m,a = 9.7974e-26 kg 4) ω=2.0406e13 N/m*kg 5) From above, d=2.2245e-10 m 6) v=ωd = 4539 m/s (not rounding in actual calculations) Time Elapsed: 1) length sound traveled = L+dL = 2.44166 m 2) From above, speed of sound = 4539 m/s 3) T = (L+dL)/v = 0.000537505 s
One wire possesses a solid core of copper, with a circular cross-section of radius 3.78 mm. The other wire is composed of 19 strands of thin copper wire bundled together. Each strand has a circular cross-section of radius 0.756 mm. The current density J in each wire is the same, J=2950 A/m².
1. How much current does each wire carry?
2. The resistivity of copper is 1.69 x 10⁻⁸ ohm m. What is the resistance of a 1.00 m length of each wire?
Answer:
a) Current in wire 1 = 0.132 A
Current in wire 2 = 0.101 A
b) Resistance of wire 1 = R₁ = 0.000376 Ω = (3.76 × 10⁻⁴) Ω = 0.376 mΩ
Resistance of wire 2 = R₂ = 0.000495 Ω = (4.95 × 10⁻⁴) Ω = 0.495 mΩ
Explanation:
Current density, J = (current) × (cross sectional area)
Current density for both wires = J = 2950 A/m²
For wire 1,
Cross sectional Area = πr² = π(0.00378²)
A₁ = 0.00004491 m²
For wire 2,
With the assumption that the strands are well banded together with no spaces in btw.
Cross sectional Area = 19 × πr² = π(0.000756)²
A₂ = 0.00003413 m²
Current in wire 1 = I₁ = J × A₁ = 2950 × 0.00004491 = 0.132 A
Current in wire 2 = I₂ = J × A₂ = 2950 × 0.00003413 = 0.101 A
b) Resistance = ρL/A
ρ = resistivity for both wires = (1.69 x 10⁻⁸) Ω.m
L = length of wire = 1.00 m for each of the two wires
A₁ = 0.00004491 m²
A₂ = 0.00003413 m²
R₁ = ρL/A₁ = (1.69 x 10⁻⁸ × 1)/0.00004491
R₁ = 0.000376 Ω = (3.76 × 10⁻⁴) Ω = 0.376 mΩ
R₂ = ρL/A₂ = (1.69 x 10⁻⁸ × 1)/0.00003413
R₂ = 0.000495 Ω = (4.95 × 10⁻⁴) Ω = 0.495 mΩ
Hope this helps!!
3/122 The collar has a mass of 2 kg and is attached to the light spring, which has a stiffness of 30 N/m and an unstretched length of 1.5 m. The collar is released from rest at A and slides up the smooth rod under the action of the constant 50‐N force. Calculate the velocity v of the collar as it passes position B.
Explanation:
According to the law of conservation of energy, work done by the force is as follows.
[tex]W_{F} = F Cos (30^{o}) \times 1.5[/tex]
= 64.95 J
Now, gain in potential energy is as follows.
P.E = mgh
= [tex]2 \times 9.8 \times 1.5[/tex]
= 29.4 J
Gain in potential energy will be as follows.
= [tex]\frac{1}{2}kx^{2}_{2} - \frac{1}{2}kx^{2}_{1}[/tex]
= [tex]\frac{1}{2} \times 30 N/m \times [(2.5 - 1.5)^{2} - (2 - 1.5)^{2}][/tex]
= 11.25
As,
[tex]W_{f} = u_{1} + u_{2} + \frac{1}{2}mv^{2}[/tex]
[tex]\frac{1}{2}mv^{2} = W_{f} - u_{1} - u_{2}[/tex]
= 64.95 J - 29.4 - 11.25
= 24.3
[tex]v^{2} = \frac{24.3 \times 2}{2}[/tex]
v = 4.92 m/s
Therefore, we can conclude that relative velocity at point B is 4.92 m/s.
To calculate the velocity of the collar as it passes position B, we need to consider the forces acting on the collar and use Newton's second law of motion. The collar is attached to a light spring, so the force exerted by the spring can be calculated using Hooke's law. The net force acting on the collar is the sum of the force exerted by the spring and the constant 50-N force.
Explanation:To calculate the velocity of the collar as it passes position B, we need to consider the forces acting on the collar and use Newton's second law of motion. The collar is attached to a light spring, so the force exerted by the spring can be calculated using Hooke's law. The net force acting on the collar is the sum of the force exerted by the spring and the constant 50-N force. We can equate this net force to the mass of the collar times its acceleration. Solving for the acceleration gives us the magnitude of the velocity as it passes position B.
Using Hooke's law, the force exerted by the spring can be calculated as:
F = k * x
where F is the force, k is the stiffness of the spring, and x is the displacement of the collar from its equilibrium position. Since the collar is attached to a light spring, we can assume that the displacement of the spring is negligible. Therefore, the force exerted by the spring is zero.
The net force is then equal to the constant 50-N force:
Net force = 50 N
Applying Newton's second law, we have:
Net force = mass * acceleration
Solving for acceleration gives us:
Acceleration = Net force / mass = 50 N / 2 kg = 25 m/s^2
Since velocity is the derivative of displacement, we can integrate the acceleration with respect to time to find the velocity:
v = a * t
where v is the velocity, a is the acceleration, and t is the time.
Since the collar is released from rest at position A, the time taken to reach position B can be determined using the equation of motion:
s = u*t + (1/2)*a*t^2
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
Since the collar starts from rest, the initial velocity is zero. Solving for t gives us:
t = sqrt((2*s) / a) = sqrt((2*1.5 m) / 25 m/s^2) = 0.7746 s
Finally, substituting the values of acceleration and time into the equation for velocity gives us:
v = a * t = 25 m/s^2 * 0.7746 s = 19.365 m/s
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A concave mirror with a radius of curvature of 10 cm is used in a flashlight to produce a beam of near-parallel light rays.
The distance between the light bulb and the mirror is most nearly _____.
Answer: 5cm
Explanation: Since the radius of curvature is 10cm, the focal length of the mirror (f) is
f = r/2
Where r is the radius of curvature.
Since r = 10cm, f = 10/2 = 5cm.
To produce a parallel light rays of a flash light, it means the the image will be at infinity this making the image distance to be infinite.
From the mirror formulae
1/u + 1/v = 1/f
Where u = object distance =?, v = image distance = infinity and f = focal length = 5cm.
Let us substitute the parameters, we have that
1/u + 1/∞ = 1/5
1/∞ = 0
Hence 1/u = 1/5
u = 5cm.
Hence the object needs to be placed nearly 5cm to the mirror
Final answer:
The light bulb should be placed at the focal length of the concave mirror, which is 5 cm, to produce near-parallel light rays in a flashlight.
Explanation:
The distance between the light bulb and a concave mirror to produce a beam of near-parallel light rays in a flashlight is most nearly the focal length of the mirror. Since the mirror's radius of curvature (R) is 10 cm, using the formula R = 2f, we can calculate the focal length (f). Dividing the radius of curvature by 2, we get f = R/2 = 10 cm / 2 = 5 cm. Therefore, the light bulb should be placed approximately 5 cm from the mirror to achieve near-parallel rays.
A particle of mass 2.37 kg is subject to a force that is always pointed towards East or West but whose magnitude changes sinusoidally with time. With the positive x-axis pointed towards the East, the x-component of the force is given as follows:
Fx = F₀cos(ωt), where F₀ = 2 N and ω = 1.1 rad/s.
At t = 0 the particle is at x₀ = 0 and has the x-component of the velocity, vₓ = 0.
What is the x-component of velocity (vₓ) in meters per second at t= 1.5 seconds?
Answer:
[tex]v(1.5)=0.7648\ m/s[/tex]
Explanation:
Dynamics
When a particle of mass m is subject to a net force F, it moves at an acceleration given by
[tex]\displaystyle a=\frac{F}{m}[/tex]
The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by
[tex]F=2cos1.1t[/tex]
The variable acceleration is calculated by:
[tex]\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}[/tex]
[tex]a=0.8439cos1.1t[/tex]
The instant velocity is the integral of the acceleration:
[tex]\displaystyle v(t)=\int_{t_o}^{t_1}a.dt[/tex]
[tex]\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt[/tex]
Integrating
[tex]\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}[/tex]
[tex]\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)[/tex]
[tex]\boxed{v(1.5)=0.7648\ m/s}[/tex]
Final answer:
To find the x-component of velocity at t= 1.5 seconds for a particle under a sinusoidally varying force, we derive the equation of motion, integrate the acceleration, and calculate the velocity, resulting in [tex]v_x[/tex] = 0.567 m/s.
Explanation:
The question pertains to finding the x-component of velocity (vx) of a particle at t= 1.5 seconds, given the mass and the sinusoidally varying force. Since the force applied on the particle varies as Fx = F0cos(ωt), where F0 = 2 N and ω = 1.1 rad/s, we can find the acceleration and then integrate it with respect to time to find velocity. The acceleration ax is given by Fx/m = (2cos(1.1t))/2.37. To find the change in velocity, we integrate ax with respect to time, giving us vx = ∫ ax dt = ∫ (2cos(1.1t))/2.37 dt. Evaluating this integral from 0 to 1.5 s, we use the definite integral which simplifies to vx(t) = (2/2.37)(sin(1.1(1.5))-sin(1.1(0)))/1.1. The calculation yields vx = 0.567 m/s, indicating the particle's velocity in the x-direction at 1.5 seconds.
Car A is accelerating in the direction of its motion at the rate of 3 ft /sec2. Car B is rounding a curve of 440-ft radius at a constant speed of 30 mi /hr. Determine the velocity and acceleration which car B appears to have to an observer in car A if car A has reached a speed of 45 mi /hr for the positions represented.
Answer:
Incomplete question
Check attachment for the diagram of the problem.
Explanation:
The acceleration of the car A is given as
a=3ft/s²
Car B is rounding a curve of radius
r=440ft
Car B is moving at constant speed of Vb=30mi/hr.
Car A reach a speed of 45mi/hr
Note, 1 mile = 5280ft
And 1 hour= 3600s
Then
Va=45mi/hr=45×5280/3600
Va=66ft/s
Also,
Vb=30mi/hour=30×5280/3600
Vb=44ft/s
Now,
a. Let write the relative velocity of car B, relative to car A
Vb = Va + Vb/a
Then,
Using triangle rule, because vectors cannot be added automatically
Vb/a²= Vb²+Va²-2Va•VbCosθ
From the given graphical question the angle between Va and Vb is 60°.
Vb/a²=44²+66² - 2•44•66Cos60
Vb/a²=1936+ 4356 - 5808Cos60
Vb/a² = 3388
Vb/a = √3388
Vb/a = 58.21 ft/s
The direction is given as
Using Sine Rule
a/SinA = b/SinB = c/SinC
i.e.
Va/SinA = Vb/SinB = (Vb/a)/SinC
66/SinA = 44/SinB = 58.21/Sin60
Then, to get B
44/SinB = 58.21/Sin60
44Sin60/58.21 = SinB
0.6546 = SinB
B=arcsin(0.6546)
B=40.89°
b. The acceleration of Car B due to Car A.
Let write the relative acceleration of car B, relative to car A.
Let Aa be acceleration of car A
Ab be the acceleration of car B.
Ab = Aa + Ab/a
Given the acceleration of car A
Aa=3ft/s²
Then to get the acceleration of car B, using the tangential acceleration formular
a = v²/r
Ab = Vb²/r
Ab = 44²/440
Ab = 4.4ft/s²
Using cosine rule again as above
Ab/a²= Aa²+Ab² - 2•Aa•Ab•Cosθ
Ab/a²= 3²+4.4²- 2•3•4.4•Cos30
Ab/a²= 9+19.36 - 22.863
Ab/a² = 5.497
Ab/a = √5.497
Ab/a = 2.34ft/s²
To get the direction using Sine rule again, as done above
Using Sine Rule
a/SinA = b/SinB = c/SinC
i.e.
Aa/SinA = Ab/SinB = (Ab/a)/SinC
3/SinA = 4.4/SinB = 2.34/Sin30
Then, to get B
4.4/SinB = 2.34/Sin30
4.4Sin30/2.34 = SinB
0.9402 = SinB
B=arcsin(0.9402)
B=70.1°
Since B is obtuse, the other solution for Sine is given as
B= nπ - θ. , when n=1
B=180-70.1
B=109.92°
To determine the velocity and acceleration which car B appears to have to an observer in car A, we need to consider the relative motion between the two cars. The velocity of car B as observed by the observer in car A is approximately 29955/176 ft/sec. The acceleration of car B as observed by the observer in car A is approximately 1/23966164627200 mi^2/s^2.
Explanation:To determine the velocity and acceleration which car B appears to have to an observer in car A, we need to consider the relative motion between the two cars. Car B is rounding a curve at a constant speed, so its velocity remains constant. However, the observer in car A will perceive car B as having a different velocity and acceleration. The velocity of car B to the observer in car A will depend on the relative motion between the two cars, while the acceleration of car B to the observer in car A will depend on the change in direction of car B's motion.
Let's calculate the velocity and acceleration of car B as observed by an observer in car A:
Velocity: Since car B is rounding a curve with a radius of 440 ft and a constant speed of 30 mi/hr, we can use the formula v = rω to find the angular velocity ω. The angular velocity ω is equal to the speed divided by the radius, so ω = (30 mi/hr) / (440 ft) = (30 mi/hr) / (5280 ft/mi) / (440 ft) = 1/1760 rad/sec. The observer in car A will perceive car B's velocity as the vector sum of its actual velocity in the curve (tangent to the curve) and the observer's velocity in the direction of the curve (opposite to the centripetal force). Since car A has reached a speed of 45 mi/hr, its velocity can be converted to ft/sec as (45 mi/hr) / (5280 ft/mi) = 15/176 ft/sec. Therefore, the velocity of car B as observed by the observer in car A will be (30 mi/hr) + (15/176 ft/sec) = (660/22 ft/sec) + (15/176 ft/sec) = (660/22 + 15/176) ft/sec = (29955/176) ft/sec.
Acceleration: Since car B is rounding a curve at a constant speed, its acceleration is directed towards the center of the curve and has a magnitude of v^2 / r, where v is the velocity and r is the radius. Substituting the values, we get the acceleration as (30 mi/hr)^2 / (440 ft) = ((30 mi/hr)^2) / ((5280 ft/mi) / (440 ft)) = (900 mi^2/hr^2) / (5280 ft/mi) * (440 ft) = (900 mi^2 * ft^2) / (5280 hr^2) * (440) ft = (900 * 5280 * 440) ft^2 / hr^2 = (2119680000/5280) ft^2 / hr^2 = (400800 ft^2/hr^2) = (400800 ft^2/hr^2) * (1/3600 hr^2/s^2) * (1 mi^2 / (5280 ft)^2) = (400800 / 3600) * (1/5280)^2 mi^2/s^2 = (111/9900) * (1/5280)^2 mi^2/s^2 = (11/990) * (1/5280)^2 mi^2/s^2 = (11/990) * (1/5280)^2 mi^2/s^2 = (1/266611200) mi^2/s^2 = (1/266611200) * (5280 ft/mi)^2 = (1/266611200) * 5280^2 ft^2/s^2 = (1/266611200) * 13939200 ft^2/s^2 = (1/266611200) * 13939200 ft^2/s^2 = (1/19) ft^2/s^2 = (1/19) * (1/5280)^2 mi^2/s^2 = (1/19) * (1/13939200) mi^2/s^2 = (1/19) * (1/13939200) mi^2/s^2 = (1/26268580800) mi^2/s^2 = (1/26268580800) * (5280 ft/mi)^2 = (1/26268580800) * 5280^2 ft^2/s^2 = (1/26268580800) * 13939200 ft^2/s^2 = (1/26268580800) * 13939200 ft^2/s^2 = (1/237896) ft^2/s^2 = (1/237896) * (1/5280)^2 mi^2/s^2 = (1/237896) * (1/13939200) mi^2/s^2 = (1/237896) * (1/13939200) mi^2/s^2 = (1/23966164627200) mi^2/s^2.
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Suppose you have a coffee mug with a circular cross section and vertical sides (uniform radius). What is its inside radius if it holds 375 g of coffee when filled to a depth of 7.50 cm
Answer:
0.0399 m
Explanation:
We are given that
Mass of coffee=375g=[tex]\frac{375}{1000}=0.375 kg[/tex]
1kg=1000g
Depth=h=7.5 cm=[tex]7.5\times 10^{-2} m[/tex]
[tex]1 cm=10^{-2} m[/tex]
Density of coffee=[tex]\rho=1000kg/m^3[/tex]
We have to find the inside radius of coffee mug.
We know that
[tex]\rho=\frac{m}{V}[/tex]
Substitute the values
[tex]1000=\frac{0.375}{\pi r^2h}[/tex]
[tex]r^2=\frac{0.375}{1000\times 7.5\times 10^{-2}\times 3.14}[/tex]
By using [tex]\pi=3.14[/tex]
[tex]r=\sqrt{\frac{0.375}{1000\times 7.5\times 10^{-2}\times 3.14}}[/tex]
[tex]r=0.0399 m[/tex]
Hence, the inside radius=0.0399 m
You're driving a vehicle of mass 1350 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 71 m. The maximum horizontal component of the force that the road can exert on the tires is only 0.23 times the vertical component of the force of the road on the tires (in this case the vertical component of the force of the road on the tires is mg, the weight of the car, where as usual g = +9.8 N/kg, the magnitude of the gravitational field near the surface of the Earth). The factor 0.23 is called the "coefficient of friction" (usually written "", Greek "mu") and is large for surfaces with high friction, small for surfaces with low friction.
(a) What is the fastest speed you can drive and still make it around the turn? Invent symbols for the various quantities and solve algebraically before plugging in numbers.
maximum speed =_______________ m/s
Answer:
[tex]v=12.65\ m.s^{-1}[/tex]
Explanation:
Given:
mass of vehicle, [tex]m=1350\ kg[/tex]radius of curvature, [tex]r=71\ m[/tex]coefficient of friction, [tex]\mu=0.23[/tex]During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:
[tex]m.\frac{v^2}{r} =\mu.N[/tex]
where:
[tex]\mu=[/tex] coefficient of friction
[tex]N=[/tex] normal reaction force due to weight of the car
[tex]v=[/tex] velocity of the car
[tex]1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)[/tex]
[tex]v=12.65\ m.s^{-1}[/tex] is the maximum velocity at which the vehicle can turn without skidding.
Final answer:
The maximum speed at which a car can safely make a turn on a flat road, given the mass of the car, the turn's radius of curvature, and the coefficient of friction, is approximately 25 m/s. This calculation demonstrates that the car's load does not influence its ability to negotiate the turn safely on a flat surface.
Explanation:
The student is asking how to calculate the maximum speed at which a car can safely make a turn on a flat road, given the car's mass, the turn's radius of curvature, and the coefficient of friction between the tires and the road. First, let's denote the mass of the car as m, gravitational acceleration as g, the radius of curvature as R, and the coefficient of friction as μ. To find the maximum speed v, we use the fact that the centripetal force needed to make the turn must be less than or equal to the maximum static friction force, which is μmg. This gives the condition mv²/R ≤ μmg. Solving for v, we find v = √(μgR).
By plugging in the numbers: m = 1350 kg, R = 71 m, g = 9.8 m/s², and μ = 0.23, we get v = √(0.23 * 9.8 * 71) which calculates to be approximately 25 m/s. Note, because coefficients of friction are approximate, the answer is given to only two digits.
This result is quite significant as it shows that the maximum safe speed is independent of the car's mass due to the proportional relationship between friction and normal force, which in turn is proportional to mass. This implies that how heavily loaded the car is does not affect its ability to negotiate the turn, assuming a flat surface.
A solid metal sphere with radius 0.430 m carries a net charge of 0.270 nC . Part A Find the magnitude of the electric field at a point 0.106 m outside the surface of the sphere. Express your answer using three significant figures. E
Answer:
8.46 N/C
Explanation:
Using Gauss law
[tex]E=\frac {kQ}{r^{2}}[/tex]
Gauss's Law states that the electric flux through a surface is proportional to the net charge in the surface, and that the electric field E of a point charge Q at a distance r from the charge
Here, K is Coulomb's constant whose value is [tex]9\times 10^{9} Nm^{2}/C^{2}[/tex]
r = 0.43 + 0.106 = 0.536 m
[tex]E=\frac {9\times 10^{9}\times 0.270\times 10^{-9}}{0.536^{2}}=8.4581755402094007\approx 8.46 N/C[/tex]
The magnitude of the electric field at a point 0.106 m outside a solid metal sphere with a radius of 0.430 m and a net charge of 0.270 nC is approximately 892 N/C, to three significant figures.
Explanation:The subject of your question is Physics, specifically focussing on electrostatics and the calculation of the electric field outside a charged sphere. The formula for the electric field (E) due to a point charge is given by Coulomb's Law, which is E = kQ/r^2, where 'Q' is the charge, 'r' is the distance from the charge, and 'k' is Coulomb's constant, approximately 8.99 x 10^9 N.m^2/C^2. Since the electric field due to a uniformly distributed spherical charge behaves as if all the charge is concentrated at the center, you can use this formula for the magnitude of the electric field outside the sphere.
Substituting the provided values (converted to appropriate units), we find E = (8.99 x 10^9 N.m^2/C^2 x 0.270 x 10^-9 C)/(0.536 m)^2, which gives E approximately equal to 892 N/C to three significant figures.
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What is the speed of a point on the earth's surface located at 3/43/4 of the length of the arc between the equator and the pole, measured from equator
Answer:
[tex]v=177.95m/s[/tex]
Explanation:
First, determine circle's radius between Earth's pole and the location. This can be calculated as:
[tex]r=R_e_a_r_t_hCos(90\frac{3}{4})\\R_e_a_r_t_h=6.37\times10^6m\\r=6.37\times10^6\times Cos67.5\textdegree\\r=2,437,693.46m\\[/tex]
The angular speed of earth is constant and is :
[tex]w=\frac{2\pi}{T}=\frac{2\pi}{24\times 3600}\\=7.3\times10^{-5}rad/s[/tex]
Velocity is:
[tex]v=wr\\=7.3\times10^{-5}\times 2,437,693.46\\v=177.95m/s[/tex]
A factory worker moves a 30.0 kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25.
1. What magnitude of force must the worker apply?
2. How much work is done on the crate by the worker's push?
3. How much work is done on the crate by friction?
4. How much work is done by normal force? By gravity?
5. What is the net work done on the crate?
To solve this problem we will apply the concepts related to the Friction force and work. The friction force can be defined as the product between the Normal Force (Mass by gravity) and the dynamic friction constant. In the case of Work this is defined as the product of the distance traveled by the applied force. Then we will solve the points sequentially to find the answer to each point,
PART A) The friction force with the given data is,
[tex]F_f = \mu_k mg[/tex]
Here,
[tex]\mu_k[/tex] = Kinetic coefficient
m = Mass
g = Gravitational acceleration
[tex]F_f = (0.25)(30kg)(9.8m/s^2)[/tex]
[tex]F_f = 73.5N[/tex]
PART B) The work done by the worker is the distance traveled for the previously force found, then
[tex]W_w = rF_f[/tex]
[tex]W_w = (4.5m)(73.5N)[/tex]
[tex]W_w = 330.75J[/tex]
PART C) The work done by the friction force would be the distance traveled with the previously calculated force, therefore
[tex]W_f = -rF_f[/tex]
[tex]W_f = -(4.5m)(73.5N)[/tex]
[tex]W_f = -330.75J[/tex]
[tex]W_f = -331J[/tex]
PART D) The work done by the normal force is,
[tex]W_N = N (r) Cos(90)[/tex]
[tex]W_N = 0J[/tex]
The work done by gravitational force is
[tex]W_g = rF_gcos(90)[/tex]
[tex]W_g = 0J[/tex]
PART E) The expression for the total work done is,
[tex]W_{net} = W_f +W_w +W_N+W_g[/tex]
[tex]W_{net} = 330.75-330.75+0+0[/tex]
[tex]W_{net} = 0J[/tex]
Therefore the net work done by the system is 0J
A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the final number of microstates is 0.599 times that of the initial number of microstates?
Answer:
Entropy Change = 0.559 Times
Explanation:
Entropy change is determined by the change in the micro-states of a system. As we know that the micro-states are the same as measure of disorderness between initial and final states, that's the the amount of change in micro-states determine how much of entropy has changed in the system.
Thin Layer Chromatography consists of three parts: The analyte, the stationary phase, and mobile phase. Match each of these terms to what it was in our experiment. Stationary Phase ____ a) The solvent
Mobile Phase ____ b) Silica
Analyte ____ c) One of the analgesiscs
Answer:
Analyte⇒ one of analgesics
stationery phase⇒ silica
mobile phase⇒ solvent
Explanation:
during the thin layer chromatography non volatile mixtures are separated.The technique is performed on the plastic or aluminum foil that is coated with a thin layer.
You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrical wire, using all the metal, with a resistance of 1.3 Ω. How long will your wire be? What will be its diameter? The resistivity of copper is 1.7 x 10-8 Ωm. The mass density of copper is 8.96 g/cm3.
Answer:
Length = 2.92 m
Diameter = 0.11 mm
Explanation:
We have [tex]m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}[/tex] , where:
[tex]l[/tex] is the length
[tex]m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3[/tex]
We divide the first equation by the second equation to get:
[tex]\frac{m}{R} = \frac{d A^2}{\rho}[/tex]
[tex]A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2[/tex]
Using this Area, we find the diameter of the wire:
[tex]D = \sqrt{\frac{4A}{\pi}}[/tex]
[tex]D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}[/tex]
[tex]D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm[/tex]
To find the length, we multiply the two equations stated initially:
[tex]mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}[/tex]
[tex]l^2 = 8.534\\l = 2.92 \ m[/tex]
While sliding a couch across a floor, Andrea and Jennifer exert forces FA and F on the couch. Andrea's force is due north with a magnitude of 140.0 N and Jennifer's force is 260 east of north with a magnitude of 220.0 N (a) Find the net force (in N) in component form. net : (b) Find the magnitude (in N) and direction (in degrees counterclockwise from the east axis) of the net force magnitude direction X o counterclockwise from the east axis (c) If Andrea and Jennifer's housemates, David and Stephanie, disagree with the move and want to prevent its relocation, with what combined force Fos (in N) should they push so that the couch does not move? (Express your answer in vector form.) Fos
Answer:
See the answers and the explanation below.
Explanation:
To solve this problem we must make a free body diagram with the forces applied as well as the direction. In the attached images we can see the nomenclature of the direction of the forces and the free body diagram.
a)
Sum of forces in y-axis
Fy = 140 - (220*sin(10))
Fy = 101.8 [N]
Sum of forces in x-axis
Fx = - (220*cos(10))
Fx = - 216.65 [N]
b)
For the above result, for there to be balance we realize that we need one equal to the resulting y-axis but in the opposite direction and another opposite force in direction but equal in magnitude on the x-axis.
Fy = - 101.8 [N]
Fx = 216.65 [N]
c )
Now we need to use the Pythagorean theorem to find the result of these forces.
[tex]F = \sqrt{(101.8)^{2} +(216.65)^{2} } \\F=239.4[N][/tex]
And the direction will be as follows.
α = tan^(-1) (101.8 / 216.65)
α = 25.16° (south to the east)
F = 216.65 i - 101.8 j [N]