Answer:
a) W = 1.63 10⁻²⁸ J, b) W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,
d) W = - 4.93 10⁻²⁸ J
Explanation:
a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m
If we use the law of conservation of energy, work is the change in energy of the system
W = ΔU = U_∞ -U
the potential energy for point charges is
U =k [tex]\sum \frac{q_i q_j}{r_{ij} }[/tex]
in this case we only have two particles
U = k [tex]\frac{q_1q_2}{r_{12} }[/tex]
the distance is
r₁₂ = [tex]\sqrt{(x_2-x_1)^2 + ( y_2-y_1)^2 }[/tex]
r₁₂ =[tex]\sqrt{ 0 + ( 2-0)^2}[/tex]Ra 0 + (2-0)
r₁₂ = √2= 1.4142 m
we substitute
W = k \sum \frac{q_i q_j}{r_{ij} }
let's calculate
W = [tex]\frac{ 9 \ 10^9 (1.6 \ 10^{-19})^2 }{1.4142}[/tex] 9 109 1.6 10-19 1.6 10-19 / 1.4142
W = 1.63 10⁻²⁸ J
b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0
in this case we have two fixed electrons
U = k [tex]( \frac{q_1q_3}{r_{13} } + \frac{q_2q_3}{r_{23} } )[/tex]
in this case all charges are electrons
q₁ = q₂ = q₃ = q
W = U = k q² [tex]( \frac{1}{r_{13} } + \frac{1}{r_{23} } )[/tex]
the distances are
r₁₃ = [tex]\sqrt{(3-0)^2 + 0}[/tex]RA (3.00 -0) 2 + 0
r₁₃ = 3
r₂₃ = [tex]\sqrt{ 3^2 + 2^2}[/tex]Ra (3 0) 2 + (2 0) 2
r₂₃ = √13
r₂₃ = 3.606 m
let's look for the job
W = U
let's calculate
W =[tex]{9 \ 10^3 ( 1.6 10^{-19})^2 }({\frac{1}{3} + \frac{1}{3.606} } )[/tex]
W = 1.407 10⁻²⁷ J
c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,
y₄ = 4.00 m
W = U = k [tex]( \frac{q_1q_4}{r_{14 }} + \frac{q_2q_4}{r_{24} } + \frac{q_3q_4}{r_{34} } )[/tex]
all charges are equal q₁ = q₂ = q₃ = q₄ = q
W = k q² [tex](\frac{1}{r_{14} } + \frac{1}{r_{24} } + \frac{1}{r_{34} } )[/tex]
let's look for the distances
r₁₄ = [tex]\sqrt{3^2 +4^2}[/tex]
r₁₄ = 5 m
r₂₄ = [tex]\sqrt{3^2 + ( 4-2)^2}[/tex]
r₂₄ = √13 = 3.606 m
r₃₄ = [tex]\sqrt{(3-3)^2 + (4-0)^2}[/tex]
r₃₄ = 4 m
we calculate
W = 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{5} + \frac{1}{3.606} + \frac{1}{4} )[/tex]
W = 1.68 10⁻²⁸ J
d) we take the proton to the location x5 = 1m y5 = 1m
W = U = k [tex]( \frac{q_1q_5}{r_{15} } + \frac{q_2q_5}{r_{25} } + \frac{q_3q_5}{r_{35} } + \frac{q_4q_5}{r_{45} } )[/tex]
in this case the charges have the same values but charge 5 is positive and the others negative, so the products of the charges give a negative value
W = - k q² [tex]( \frac{1}{r_{15} } + \frac{1}{r_{25} } + \frac{1}{r_{35} } + \frac{1}{r_{45} } )[/tex]
we look for distances
r₁₅ = [tex]\sqrt{ 1^2 +1^2}[/tex]Ra (1-0) 2 + (1-0) 2
r₁₅ = √ 2 = 1.4142 m
r₂₅ = [tex]\sqrt{ (2-1)^2 +1^2}[/tex]
r₂₅ = √2 = 1.4142 m
r₃₅ = [tex]\sqrt{ ( 3-1)^2 +1^2}[/tex]
r₃₅ = √5 = 2.236 m
r₄₅ = [tex]\sqrt{ (3-1)^2 + (4-1)^2}[/tex]
r₄₅ = √13 = 3.606 m
we calculate
W = - 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{1.4142} +\frac{1}{1.4142} + \frac{1}{2.236} + \frac{1}{3.606} )[/tex]
W = - 4.93 10⁻²⁸ J
An object with a mass of m = 3.85 kg is suspended at rest between the ceiling and the floor by two thin vertical ropes.
The magnitude of the tension in the lower rope is 12.8 N. Calculate the magnitude of the tension in the upper rope.
The tension in the upper rope is 50.53 N.
Tension in the upper rope
The tension in the upper rope is calculated as follows;
T(up) = T(dn) + mg
where;
T(dn) is the tension in the lower ropemg is the weight of the objectT(up) = 12.8 N + (3.85 x 9.8) N
T(up) = 50.53 N
Thus, the tension in the upper rope is 50.53 N.
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Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N. Calculate the magnitude of the tension in the string marked A.
The magnitude of the tension in the string marked A is 39.5 N.
What is the tension in A?The tension in A is determined thus:
The angle at A, θ = tan⁻¹(3/8) = 20.56
When extrapolated below negative x, the angle at B, α = tan⁻¹(5/4) = 51.34
When extrapolated below negative x, the angle at C, β = tan⁻¹(1/6) = 9.46
Taking the horizontal components of tension;
56.3cos(9.46) = A * cos(20.56) + B * cos(51.34)
0.6247B= 55.53 - 0.936A
B = (55.53 - 0.936A)/0.6247 ----(1)
Taking the vertical components of tension;
56.3 * sin(9.46) + A * sin(20.6) = B * sin(51.3)
9.25 + 0.35A = 0.78B ---- (2)
substitute the value (1) in (2)
9.25 + 0.35A = 0.78{(55.53 - 0.936A)/0.6247}
(9.25 + 0.35A) * 0.6247 = 43.31 - 0.73A
0.22A + 0.73A = 43.31 - 5.78
0.93A = 37.53
A = 39.5 N
In conclusion, the tension in A is determined by solving for the vertical and horizontal components of tension.
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A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg initially at rest. What is the initial momentum of the blue and green train combined?
A. 20 kgm/s
B. 50 kgm/s
C. 0 kgm/s
D. 200 kgm/s
The correct option is D. The initial momentum of the blue and green train combined during the collision is 200 kgm/s.
Initial momentum of the blue and green trainApply the principle of conservation of linear momentum as follows;
Pi = m1v1 + m2v2
where;
m1 is mass of blue trainm2 is mass of green trainv1 is velocity of blue trainv2 is velocity green trainPi is the initial momentum of the two trainsPi = (50 x 4) + 30(0)
Pi = 200 kgm/s
Thus, the initial momentum of the blue and green train combined is 200 kgm/s.
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Why is air resistance friction not useful for an airplane?
A. Causes turbulence
B. Speeds it up
C. Slows it down
D. Creates heat
The correct option is C. Air resistance friction is not useful for an airplane because it slows it down.
What is air resistance?Air resistance is the opposition to motion of an object caused by air flow.
High air resistance may cause turbulence during the motion of an air plane. This can lead to engine failure or some mishap during flight.
Air resistance is also a type of friction between air and another material such as airplane.
Effect of air resistance on airplaneFriction between the air and the plane slows the airplane down. This is known as air resistance, or drag.
The faster an object travels through the air, the more it has to fight against drag.
Thus, air resistance friction is not useful for an airplane because it slows it down.
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A 72.5-kg hiker starts at an elevation of 1230 m and climbs to the top of a peak 2660 m high.
What is the hiker's change in potential energy?
Express your answer to three significant figures and include the appropriate units.
What is the minimum work required of the hiker?
Express your answer to three significant figures and include the appropriate units.
Change in potential energy = 1.02 * 10⁶ J
Minimum work required of the hiker = 1.89 * 10⁶ J
What is the change in potential energy of the hiker?The potential energy of a body is calculated as follows:
Potential energy = mghChange in potential energy = Final PE - Initial PE
Change in potential energy = mg(H - h)
Change in potential energy = 72.5 * 9.81 * (2660 - 1230)
Change in potential energy = 1.02 * 10⁶ J
The minimum work required of the hiker is the potential energy at the highest point.
Minimum work = mgH
Minimum work required of the hiker = 1.89 * 10⁶ J
In conclusion, potential energy is energy due to state or position of a body.
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In physics, when a baseball player catches a ball, which one of newtons laws is it an example of?
A. 1st law
B. 2nd law
C. 3rd law
what is the wavelength of a tubas A2(110hz) tone in air at standard conditions/
The wavelength of A2 with frequency (110hz) is 3.01 m.
What is the wavelength of a sound wave?Wavelength of a sound wave is the distance between successive similar points in the wave such as rarefactions or compressions.
A tuba is a musical instrument that produces sound waves.
Wavelength is related to frequency and velocity by the formula below:
Wavelength = velocity/frequencyThe wavelength of A2 with frequency (110hz) is 3.01 m.
In conclusion, the wavelength of a wave is inversely proportional to frequency.
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A 21.0 kg uniform beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 66.0° what is the tension in the cable?
Since upward forces must be equal to the downward forces, the tension in the cable is 225.3 N
What are the two conditions for equilibrium ?The two conditions are;
Sum of the upward forces must be equal to the sum of the downward forces.The sum of the clockwise moment must be equal to the sum of the anticlockwise moment.The given parameters are;
Mass m = 21 kgangle θ = 66°Tsinθ = mg
Tsin 66 = 21 x 9.8
T = 205.8 / 0.914
T = 225.3 N
Therefore, the tension in the cable is 225.3 N
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Radiation from the sun hits Earth unequally and is absorbed by different materials in varying amounts. This is called
A. the Coriolis Effect
B. a thermocline
OC. El Nino Southern Oscillation
OD. differential heating
Radiation from the sun hits earth unequally and is absorbed by different materials in varying amounts. This is called differential heating.
What is differential heating?Differential heating is the property that causes different surfaces to heat up and cool down at different rates. The earth's surface receives different magnitudes of solar radiation and also the earth's surfaces absorb thermal energy in different magnitudes.
The color, shape, texture, surface and presence of constructions can influence the heating or cooling of the earth.
The earth in the equator line is heated more by solar action than that of the poles, since it receives more amount of radiation per unit area.
In general, dry surfaces heat up and cool down faster than wet ones.
Therefore, we can confirm that when radiation from the sun hits earth unequally and is absorbed by different materials in varying amounts is called differential heating.
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Name two units for measuring the diameter of nucleus atom.
The two units for measuring the diameter of nucleus atom are femtometre and metre.
How do you measure the size of the nucleus ?
Nucleus size is expressed in fermi, often known as femtometers. between a lighter and a heavier nucleus. Despite its modest size, the nucleus contains the majority of an atom's mass. The weight or mass of the atom's nucleus and neutrons are determined by neutrons.
femtometre (fm), which equals [tex]10^{-15}[/tex] metre.
A nucleus' diameter largely depends as to how many particles it contains, from about 4 fm for a light nucleus like carbon to 15 fm for a heavy nucleus as lead.
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For light of wavelength 589 nm, calculate the critical angles for the following substances when surrounded by air.
polystyrene
flint glass
(1) The critical angles for the air is 90⁰,
(2) The critical angles for the polystyrene is 39.1 ⁰ and
(3) The critical angles for the flint glass is 37.2 ⁰.
Critical angles for the different mediumθc = sin⁻¹( 1/η)
where;
η is the refractive indexRefractive index of air = 1
Refractive index of polystyrene = 1.5865
Refractive index of flint glass = 1.655
Critical angles for airθc = sin⁻¹( 1/1)
θc = 90⁰
Critical angles for polystyreneθc = sin⁻¹( 1/1.5865)
θc = 39.1 ⁰
Critical angles for flint glassθc = sin⁻¹( 1/1.655)
θc = 37.2 ⁰
Thus, the critical angles for the air is 90⁰, the critical angles for the polystyrene is 39.1 ⁰ and the critical angles for the flint glass is 37.2 ⁰.
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. A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weight to be 7.84 N. She then weighs the crown while it is immersed in water, as shown below in Figure, and now the scale reads 6.86 N. Is the crown made of pure gold? Find the density of the crown and compare it to the density of gold. wag PODECID LE B water
Answer:
Average density of the crown: approximately [tex]8\; {\rm g \cdot mL^{-1}}[/tex].
Hence, if this crown contains no empty space, this crown is not made of pure gold.
Explanation:
Let [tex]m(\text{crown})[/tex] and [tex]V(\text{crown})[/tex] denote the mass and volume of this crown. Let [tex]g[/tex] denote the gravitational field strength.
Since this crown is fully immersed in water, the volume of water displaced [tex]V(\text{water, displaced})[/tex] is equal to the volume of this crown:
[tex]V(\text{water, displaced}) = V(\text{crown})[/tex].
The mass of water displaced would be:
[tex]\begin{aligned}m(\text{water, displaced}) &= \rho(\text{water}) \, V(\text{water, displaced}) \\ &= \rho(\text{water}) \, V(\text{crown})\end{aligned}[/tex].
The weight of water displaced would be [tex]m(\text{water, displaced})\, g = \rho(\text{water}) \, V(\text{crown})\, g\end{aligned}[/tex].
The buoyancy force on this crown is equal to the weight of water that this crown displaced:
[tex]F(\text{buoyancy}) = \rho(\text{water}) \, V(\text{crown})\, g[/tex].
The magnitude of this buoyancy force is [tex]7.84\; {\rm N} - 6.86\; {\rm N} = 0.98\; {\rm N}[/tex]. Rearrange the equation for buoyancy to find [tex]V(\text{crown})[/tex]:
[tex]\begin{aligned} V(\text{crown}) &= \frac{F(\text{buoyancy}) }{\rho(\text{water}) \, g}\end{aligned}[/tex].
Since the weight of this crown is [tex]\text{weight}(\text{crown}) = m(\text{crown})\, g[/tex], the mass of this crown would be [tex]m(\text{crown})= \text{weight}(\text{crown}) / g[/tex].
The average density of this crown would be:
[tex]\begin{aligned}\rho(\text{crown}) &= \frac{m(\text{crown})}{V(\text{crown})} \\ &= \frac{\text{weight}(\text{crown}) / g}{F(\text{buoyancy}) / (\rho(\text{water})\, g)} \\ &= \frac{\text{weight}(\text{crown})}{F(\text{buoyancy})}\, \rho(\text{water}) \\ &= \frac{7.84\; {\rm N}}{0.98\; {\rm N}}\times 1.000 \; {\rm g\cdot mL^{-1}} \\ &= 8.0\; {\rm g \cdot mL^{-1}}\end{aligned}[/tex].
The density of pure gold is significantly higher than [tex]8.0\; {\rm g\cdot mL^{-1}}[/tex]. Hence, if this crown contains no empty space (i.e., no air bubble within the crown), the crown would not be made of pure gold.
A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long as 2.20 m have been recorded. If a frog jumps 2.20 m and the launch angle is 36.5°, find the frog's launch speed and the time the frog spends in the air. Ignore air resistance.
(a)the frog's launch speed (in m/s)
(b)the time the frog spends in the air (in s)
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
What's the expression of range of a projectile motion?Range = U²× sin(2θ)/gU= initial velocity, θ= angle of projectile and g= acceleration due to gravity U=√{Range×g/sin(2θ)}Here, range= 2.20m, = 36.5°U= √{2.20×9.8/sin(73)}U= √{2.20×9.8/sin(73)} = 22.5m/s
What's the expression of time of flight in projectile motion?Time of flight= (2×U×sinθ)/g So, T= (2×22.5×sin36.5°)/9.8= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
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What is the speed of light while traveling through (a) a vacuum, (b) air at 30°C, and (c) air at 0°C?
The speed of light through a transparent substance is 2.00 × 108 m/s. What is the substance?
The wavelength of light from a monochromatic source is measured to be 6.80 × 10−7 m. (a) What is the frequency of this light? (b) What color would you observe?
What is the energy of a photon of red light with a frequency of 4.3 × 1014 Hz?
(1) The speed of light through vacuum is 3 x 10⁸ m/s.
(2) The speed of light in air at 30⁰C is 350 m/s.
(3) The speed of light in air at 0⁰C is 330 m/s.
(4) The transparent substance is glass.
(5) The color of light observed is red.
(6) The energy of a photon of red light is 2.85 x 10⁻¹⁹ J.
Speed of light
The speed of light through vacuum is 3 x 10⁸ m/s.
The speed of light in air at 30⁰C is 350 m/s.
The speed of light in air at 0⁰C is 330 m/s
Refractive index of the substancen = speed of light in vacuum/speed of light in the substance
n = ( 3 x 10⁸ m/s) / ( 2 x 10⁸ m/s)
n = 1.5
The refractive index of glass is 1.5, thus, the transparent substance is glass.
Frequency of lightf = v/λ
f = (3 x 10⁸) / (6.8 x 10⁻⁷)
f = 4.41 x 10¹⁴ Hz
The color of light observed is red based on the frequency and wavelength.
Energy of the photonE = hf
E = (6.626 x 10⁻³⁴)(4.3 x 10¹⁴)
E = 2.85 x 10⁻¹⁹ J
Thus, the speed of light through vacuum is 3 x 10⁸ m/s.
The speed of light in air at 30⁰C is 350 m/s.
The speed of light in air at 0⁰C is 330 m/s.
The transparent substance is glass.
The color of light observed is red.
The energy of a photon of red light is 2.85 x 10⁻¹⁹ J.
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An ideal toroidal solenoid (Figure 1) has inner radius r1 = 15.1 cm and outer radius r2 = 18.3 cm. The solenoid has 180 turns and carries a current of 8.40 A.
a) What is the magnitude of the magnetic field at 12.6 cm from the center of the torus?
b) What is the magnitude of the magnetic field at 16.3 cm from the center of the torus?
c) What is the magnitude of the magnetic field at 20.7 cm from the center of the torus?
The solution for the three questions is mathematically given as
Parts A and C are both zeros.
Part B B = 0.001855Tesla
What is the magnitude of the magnetic field at 12.6 cm from the center of the torus?Parts A and C are both zeros.
For component A, the magnetic field is zero since 12.6 cm is still inside the toroidal solenoid. Part C has no magnetic field since it is 20.7 cm outside of the toroidal solenoid.
Generally, the equation for magnetic field is mathematically given as
B = (mu_0*N*I)/(2*pi*r)
Therefore
B = ((4*pi*10^{-7})*180*8.40)/(2*pi*0.163)
B = 0.001855Tesla
In conclusion, the magnitude of the magnetic field at 16.3 cm from the center of the torus is
B = 0.001855Tesla
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Answer:
Part A & C: 0 T
Part B: 1.855*10^-3 T
Explanation:
The formula that models the magnetic field of a toroid is B=μ0*N*I/2π*r .
Note: Toroids keep their magnetic field rotating within the coils that carry current.
Part A & C: Thus the B field magnitude 12.6 cm & 20.7 cm away from the center is 0 T.
Part B: [tex]B=\frac{(4\pi *10x^{-7} )(180)(8.4 A)}{(2\pi )(16.3*10x^{-2} m)}[/tex]
B=1.855*10^-3 T
What is the approximate uncertainty in the area of a circle of radius 4.3×104cm ?
The approximate uncertainty in the area of a circle is 5.4%.
Area of the circle
A = πr²
where;
r is radius of the circleA = π(4.3 x 10⁴)²
A = 5.81 x 10⁹ cm²
Let the error in measurement = 1 x 10⁴ cm
Error in Area measurementA = π(1 x 10⁴)² = 3.14 x 10⁸ cm²
Uncertainty in measurement% = (error/actual area) x 100%
% = ( 3.14 x 10⁸) / (5.81 x 10⁹) x 100%
% = 5.4 %
Thus, the approximate uncertainty in the area of a circle is 5.4%.
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A proton is moved from the negative to the positive plate of a parallel-plate arrangement. The plates are 1.50 cm apart, and the electric field is uniform with a magnitude of 1 500 N/C.
What is the proton’s potential energy change?
What is the potential difference between the plates?
What is the potential difference between the negative plate and a point midway between the plates?
If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate?
(a) The proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
(b) The potential difference between the negative plate and a point midway between the plates is 11.25 V.
(c) The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
Potential energy of the proton
U = qΔV
where;
q is charge of the protonΔV is potential differenceU = q(Ed)
U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)
U = 3.6 x 10⁻¹⁸ J
Potential difference between the negative plate and a point midwayΔV = E(0.5d)
ΔV = 0.5Ed
ΔV = 0.5 (1500)(1.5 x 10⁻²)
ΔV = 11.25 V
Speed of the protonU = ¹/₂mv²
U = mv²
v² = 2U/m
where;
m is mass of proton = 1.67 x 10⁻²⁷ kgv² = (2 x 3.6 x 10⁻¹⁸) / ( 1.67 x 10⁻²⁷)
v² = 4.311 x 10⁹
v = √(4.311 x 10⁹)
v = 6.57 x 10⁴ m/s
Thus, the proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
The potential difference between the negative plate and a point midway between the plates is 11.25 V.
The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
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A uniform beam of length L = 2.2 m and mass M = 49 kg has its lower end fixed to pivot at a point P on the floor, making an angle θ = 18° as shown in the diagram. A horizontal cable is attached at its upper end B to a point A on a wall. A box of the same mass M as the beam is suspended from a rope that is attached to the beam one-fourth L from its upper end.
What is the y-component Py of the force, in newtons, exerted by the pivot on the beam?
Write an expression for the tension T in the horizontal cable AB.
What is the x-component Px of the force, in newtons, exerted by the pivot on the beam?
I cannot figure part C
Hi there!
Part A.
To solve this part, all we need to do is a summation of vertical forces.
We have the following acting on the beam :
- Force of gravity (Fg, down)
- Force of tension from the rope holding the box (T, down)
- Force exerted by pivot (Py, up)
These sum to zero because the beam is not accelerating vertically.
[tex]\Sigma F = -F_g - T + P_y = 0[/tex]
[tex]P_y = F_g + T[/tex]
The tension force is equal to the box's weight because the forces on the box are balanced. Let's use values and solve.
[tex]P_y = 49(9.8) + 49(9.8) = \boxed{960.4N}[/tex]
Part B.
We must begin by doing a summation of torques. Placing the pivot point at the pivot, we have the following present:
- Force of gravity acting at the center of mass of the rod (CC, at L/2)
- The tension of the horizontal cable acting at the end of the rod (CCW, at L)
- The force of tension in the rope holding the box (CC, at 3L/4)
Since the rod is not rotating, these torques sum to zero.
The equation for torque is:
[tex]\tau = r \times F[/tex]
This is a cross-product, and you must find the lever arm (perpendicular distance between pivot and line of action of force). We will need to use trigonometry for this.
Now, let's find the torque from all three of these forces.
- Force of gravity at center:
The perpendicular distance between the force of gravity and the pivot point is the cosine with respect to the angle made with the floor.
[tex]\tau = Mg\frac{L}{2}cos(\theta) = 49(9.8)*\frac{2.2}{2} cos(18) = 502.367 Nm[/tex]
- Tension of horizontal cable:
The lever arm is the sine with respect to the angle. We will still have to solve for the value of 'T'.
[tex]\tau = TLsin\theta = T(2.2)sin(18) = 0.68T[/tex]
- Tension of rope holding box:
The tension is equal to the weight of the box since the box isn't accelerating. Thus, the torque would be:
[tex]\tau = Mg(\frac{3L}{4}) = 49(9.8)*\frac{3(2.2)}{4}cos(18) = 753.55 Nm[/tex]
Summing with clockwise torques + and counterclockwise -:
[tex]\Sigma \tau = 502.367 + 753.55 - 0.68 T = 0 \\\\1255.917 = 0.68T\\T = \boxed{1847.38 N}[/tex]
Part C.
This part is a lot easier than it seems. All we need to do is a summation of horizontal forces.
We only have two:
- The horizontal tension in the cable to the left (1847.38 N)
- The horizontal force exerted by the pivot on the beam to the right
These two balance out because there is no acceleration of the beam horizontally, so:
[tex]\Sigma F = P_x - T = 0 \\\\P_x = T\\\\P_x = \boxed{1847.38 N}[/tex]
**to the right
Answer:
a) 960.4 N
b) T= 5/4 Mg CotanΘ
c) 1847. 38
Explanation:
a) Py= 2Mg
=2(49 x 9.8)
= 960.4
b) T= (Mg x 1/2 x cos Θ + Mg x 3/4 x cos Θ) / sin Θ
T= 5/4 X Mg cotanΘ
c) T= (5/4) x (49 x 9.8) cotan (18)
T= 1847.37954
= 1847.38
A cart of mass 0.5 kg sits at rest on a table on which it can roll without friction. It is attached to an unstretched spring. You give the mass a push with a constant force over a distance of 5 cm in the direction that compresses the spring, after which the mass starts undergoing simple harmonic motion with a frequency of 0.5 complete oscillations per second and an amplitude of 15 cm.
A) What is the spring constant of the spring?
B) How fast was the cart moving at the instant when you finished pushing it?
C) What force did you exert on the cart?
(A) The spring constant of the spring is 4.94 N.
(B) The speed of the cart after pushing it is 0.47 m/s.
(C) The force applied to the cart is 0.75 N.
Spring constant
ω = √k/m
where;
ω is angular frequencyk is spring constantm is mass0.5 rev/s = 0.5(2π) rad/s = π rad/s = 3.142 rad/s
ω² = k/m
k = mω²
k = 0.5 x (3.142)²
k = 4.94 N/m
Energy stored in the springE = ¹/₂kA²
where;
A is amplitude
E = ¹/₂(4.94)(0.15)²
E = 0.056 J
Speed of the cartE = ¹/₂mv²
2E = mv²
v² = 2E/m
v² = (2 x 0.056)/(0.5)
v² = 0.224
v = √0.224
v = 0.47 m/s
Force exerted on the cartE = ¹/₂FA
2E = FA
F = 2E/A
F = (2 x 0.056)/(0.15)
F = 0.75 N
Thus, the spring constant of the spring is 4.94 N. The speed of the cart after pushing it is 0.47 m/s. The force applied to the cart is 0.75 N.
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For the circuit shown below with a 12.0 V battery, find the total current through the battery and the current through each resistor with the following resistances: R1=5.0 Ohms, R2=10.0 Ohms, R3=15.0 Ohms, R4=33.0 Ohms. Show all your work!
The total current through the battery is 4.77 A.
The current through resistor R1 is 2.4 A.
The current through resistor R2 is 1.2 A
The current through resistor R3 is 0.8 A
The current through resistor R4 is 0.36 A.
Total resistance of the circuitThe total resistance of the circuit is calculated as follows;
1/Rt = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄
1/Rt = 1/5 + 1/10 + 1/15 + 1/33
1/Rt = 0.397
Rt = 1/0.397
Rt = 2.518 ohms
Total current flowing in the circuitI = V/Rt
I = 12/2.518
I = 4.77 A
Current in resistor R₁I₁ = V/R₁
I₁ = 12/5
I₁ = 2.4 A
Current in resistor R₂I₂ = 12/10
I₂ = 1.2 A
Current in resistor R₃I₃ = 12/15
I₃ = 0.8 A
Current in resistor R₄I₄ = 12/33
I₄ = 0.36 A
Thus, the total current through the battery is 4.77 A.
The current through resistor R1 is 2.4 A.
The current through resistor R2 is 1.2 A
The current through resistor R3 is 0.8 A
The current through resistor R4 is 0.36 A.
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The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down. Initial speed of the rock = 1.92×101 m/s. (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10-11 Nm2/kg2.)
A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Calculate the speed of the satellite.
The orbiting speed of the satellite orbiting around the planet Glob is 60.8m/s.
To find the answer, we need to know about the orbital velocity a satellite.
What's the expression of orbital velocity of a satellite?Mathematically, orbital velocity= √(GM/r)G= gravitational constant= 6.67×10^(-11) Nm²/kg², M = mass of sun , r= radius of orbitWhat's the orbital velocity of the satellite in a circular orbit with a radius of 1.45×10⁵ m around the planet Glob of mass 7.88×10¹⁸ kg?Here, M= 7.88×10¹⁸ kg, r= 1.45×10⁵ mOrbital velocity of the orbiting satellite = √(6.67×10^(-11)×7.88×10¹⁸/1.45×10⁵)= 60.8m/s
Thus, we can conclude that the speed of the satellite orbiting the planet Glob is 60.8m/s.
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A rectangular loop of wire with dimensions 1.80 cm by 9.00 cm and resistance 0.800 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude 2.60 T and is directed into the plane of (Figure 1) .
a) At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?
b) What is the direction of the force that the magnetic field exerts on the loop?
(a) The magnitude of the force that the magnetic field exerts on the loop is 0.042 N.
(b) The direction of the force that the magnetic field exerts on the loop will be out of the plane.
Magnetic force exerted on the loop
F = BIL
where;
I is current in the loopL is length of the loopemf = BVb
where;
b is breadth of the loopV is velocityB is magnetic fieldemf = 2.6 x 3 x 0.018 = 0.1404 V
Current in the loop, I = emf/R = 0.1404/0.8 = 0.18 A
Magnetic forceF = BIL
where;
L is length of the loopF = 2.6 x 0.18 x 0.09 = 0.042 N
Direction of the forceThe direction of the force that the magnetic field exerts on the loop will be out of the plane.
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Jane, looking for Tarzan, is running at top speed (6.0 m/s ) and grabs a vine hanging vertically from a tall tree in the jungle.
How high can she swing upward?
Express your answer to two significant figures and include the appropriate units.
Does the length of the vine affect your answer?
(a) The maximum height reached by Jane is 1.8 m.
(b) The length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.
Maximum height Jane can swing
apply the principle of conservation of energy;
P.E = K.E
mgh = ¹/₂mv²
h = v²/2g
where;
v is speed of janeg is acceleration due to gravityh = (6²)/(2 x 9.8)
h = 1.84 m
Time of motion of JaneAssuming Jane to be in simple harmonic motion, the time of motion is calculated as;
T = 2π√(L/g)
where;
L is the length of the vineThus, the length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.
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The head of a hammer with a mass of 1.5 kg is allowed to fall onto a nail from a height of 0.70 m .
What is the maximum amount of work it could do on the nail?
Express your answer to two significant figures and include the appropriate units.
Why do people not just "let it fall" but add their own force to the hammer as it falls?
a)To make the trajectory of the fall more precise.
b)To increase the power of the hammer.
Maximum work done by the hammer on the is 10 J.
Force is added to the hammer in order to increase the power of the hammer; option B.
What is work done?Work done is defined as the product of force and the distance travelled by force.
Energy is used to do work.
The potential energy of the hammer is converted to work done.
Potential energy = mghThe maximum amount of work it could do on the nail is given below:
Maximum work = 1.5 * 9.81 * 0.7
Maximum work = 10 J
Force is added to the hammer in order to increase the power of the hammer.
In conclusion, the work done by the hammer is obtained from the potential energy of the hammer.
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. What is the potential energy of a 0.40 kg ball at a height of 9.2 m?
Explanation:
gravitational potential energy = mgh (must be in S.I. unit)
m= 0.4 kg ; g= 10m/s (gravitational acceleration occurs); h=9.2 m
hence mgh=0.4×10×9.2= 36.8J
unit for energy is joules and since the variables are in S.I. unit, we can use Joules as the final unit for measurement
Answer:
36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s²
Explanation:
Potential Energy = Mass x Gravitational Acceleration x Height
It can be expressed as [tex]\displaystyle{PE = mgh}[/tex] where PE is potential energy, m is mass, g is gravitational acceleration and h is height.
In this case, we know mass is 0.40 kg and height of 9.2 m as well as gravitational acceleration is defined to be 9.8 m/s² (You can also define g = 10 m/s²)
Therefore, substitute given information in formula:
[tex]\displaystyle{PE=0.40 \ \times \ 9.8 \ \times 9.2 }\\\\\displaystyle{PE=36.06 \ J}[/tex]
With g = 10 m/s², you'll get:
[tex]\displaystyle{PE = 0.40 \ \times 10 \ \times 9.2}\\\\\displaystyle{PE = 36.8 \ J}[/tex]
Note that J is for joule unit.
Therefore, the answer is 36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s² - both work.
Find the minimum diameter of an l = 19.3 m long steel wire that will stretch no more than 7.32 mm when a mass of 350 kg is hung on the lower end. (Hint: The Young's modulus of steel is 200.0 GPa.)
The minimum diameter of the steel wire is 7.58 mm.
Stress applied to the steel wireYoung's modulus = stress/strain
strain = e/l = (7.32 x 10⁻³ m) / (19.3 m) = 3.79 x 10⁻⁴
stress = Young's modulus x strain
stress = 200 x 10⁹ N/m² x 3.79 x 10⁻⁴ = 7.59 x 10⁷ N/m²
Area of the wirestress = Force/Area
Area = Force/stress
Area = mg/stress
Area = (350 x 9.8) / (7.59 x 10⁷)
Area = 4.519 x 10⁻⁵ m²
Minimum diameter of the wireArea = πd²/₄
πd²/₄ = 4.519 x 10⁻⁵ m²
πd² = 4(4.519 x 10⁻⁵)
d² = (4 x 4.519 x 10⁻⁵)/π
d² = 5.75 x 10⁻⁵
d = √(5.75 x 10⁻⁵)
d = 7.58 x 10⁻³ m
d = 7.58 mm
Thus, the minimum diameter of the steel wire is 7.58 mm.
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White light is spread out into its spectral components by a diffraction grating. If the grating has 1975 lines per centimeter, at what angle does red light of wavelength 640 nm appear in first-order spectrum? (Assume that the light is incident normally on the grating.)
The angle of the red light is mathematically given as
[tex]\theta = 7.24 \textdegree[/tex]
What angle does red light of wavelength 640 nm appear in the first-order spectrum?Generally, the equation for the grating element is mathematically given as
d= 1 / N
Therefore
d= 1/1965
d= 5.089 * 10^{-6} m
Generally, the equation for the differential formula is mathematically given as
[tex]d sin \theta = m\lambda[/tex]
Therefore
[tex]sin \theta = \lambda / d[/tex]
[tex]sin \theta= (640 * 10 ^ {-9} m)/(5.089 * 10 ^ {-6} m)[/tex]
[tex]\theta = 7.24 \textdegree[/tex]
In conclusion, The angle of the red light
[tex]\theta = 7.24 \textdegree[/tex]
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When a scientific calculator shows the quantity below, what does it mean?
1.5E8
in the absence of friction, the output power of a winding engine is 100kw but thus is reduced by friction to 90kw . how much oil initially at 120° is required per second to to keep the Temperature of the bearing down to 70°C ? specific heat capacity of oil is 2100 j/kg°C.
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Answer:
Explanation:
The angular speed of the rotor is 200 rad/s.
The torque needed to be transmitted by the engine is 180 Nm.
The power of the rotor required to transmit energy to apply a torque τ to rotate a motor with angular speed ω,
P=τω
=180×200W
=36kW
Motorcycle safety helmet extend the time of collision hence decreasing,
a:chance of collision
b:force acting
c: velocity
c:Impulse
Answer:
D. Impulse
Explanation: Hope this helps