Final answer:
If the width of both slits in a double-slit interference experiment is doubled without changing the distance between their centers, the spacing of the fringes decreases and the intensity of the bright fringes decreases.
Explanation:
In a double-slit interference experiment, if the width of both slits is doubled without changing the distance between their centers:
a. The spacing of the fringes will decrease.
b. The intensity of the bright fringes will decrease.
When the width of the slits is increased, the interference fringes become wider and less compact. This means that the spacing between the fringes becomes smaller. Additionally, the intensity of the bright fringes decreases because spreading out the light over a wider area results in less light reaching a specific point on the screen.
A single water molecule were oriented such that its dipole moment (magnitude 6.186 × 10 − 30 Cm is along a z axis. How much torque does an 8500 N/C electric field exert on the dipole if the field lies in x z -plane, 42 ∘ above the x -axis?
Answer:
Torque=6.261×10^-43Nm
Explanation:
Torque is given by÷
Torque=dqEsin(phi)
Where E is a constant=1.6022×10^-19
d= distance = 6.186×10^-30cm
Changing to metres=6.186×10^-28m
q= the charge=8500NC
Phi =42°
Torque= (6.186×10^-28)×8500×(1.6022×10^-19)sin42°
Torque= 8.425×10^-43 × 0.7431
Torque= 6.261 ×10^-43
The torque exerted by an 8500 N/C electric field exerts on the dipole is 6.261 ×10⁻⁴³ Nm.
What is torque?The force which causes the object to rotate about any axis is called perpendicular distance.
Torque is a twisting or turning force that frequently results in rotation around an axis, which may be a fixed point or the center of mass. The ability of something rotating, such as a gear or a shaft, to overcome turning resistance is another way to think of torque.
Given:
A single water molecule was oriented such that its dipole moment of magnitude 6.186 × 10 − 30 Cm is along a z-axis,
The electric field = 8500 N/C,
∅ = 42°
Calculate the torque as shown below,
Torque = d × q × E × sin(∅)
Where E is a constant = 1.6022×10⁻¹⁹
distance = 6.186×10⁻³⁰ cm = 6.186×10⁻²⁸ m
Torque = 6.186×10⁻²⁸ × 8500 × (1.6022 × 10⁻¹⁹) sin42°
Torque = 8.425×10⁻⁴³ × 0.7431
Torque = 6.261 ×10⁻⁴³
Thus, the torque is 6.261 ×10⁻⁴³ Nm.
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A rigid tank contains air at a pressure of 70 psia and a temperature of 55 ˚F. By how much will the pressure increase as the temperature is increased to 115 ˚F?
Answer:
P_2 = 62.69 psi
Explanation:
given,
P₁ = 70 psia T₁ = 55° F = (55 + 459.67) R
P₂ = ? T₂ = 115° F = (115 + 459.67) R
we know,
p = ρ RT
ρ is the density which is constant
R is also constant
now,
[tex]\dfrac{P_2}{P_1} =\dfrac{T_2}{T_1}[/tex]
[tex]\dfrac{P_2}{70} =\dfrac{55+459.67}{115+459.67}[/tex]
P_2 = 62.69 psi
Hence, the increase in Pressure is equal to P_2 = 62.69 psi
A 5.55 L cylinder contains 1.21 mol of gas A and 4.93 mol of gas B, at a temperature of 27.3 °C. Calculate the partial pressure of each gas in the cylinder. Assume ideal gas behavior.
Answer:
Pressure of gas A = 544.324 kPa
Pressure of gas B = 2217.784 kPa
Explanation:
Data provided in the question:
Total volume of the cylinder, V = 5.55 L = 0.00555 m³ [1 m³ = 1000 L]
Moles on gas A, [tex]n_a[/tex] = 1.21 mol
Moles on gas A, [tex]n_b[/tex] = 4.93 mol
Temperature, T = 27.3°C = 27.3 + 273 = 300.3 K
Now,
Pressure = [tex]\frac{nRT}{V}[/tex]
here,
R is the ideal gas constant = 8.314 J/mol.K
Therefore,
Pressure of gas A = [tex]\frac{n_aRT}{V}[/tex]
= [tex]\frac{1.21\times8.314\times300.3}{0.00555}[/tex]
= 544324.32 Pa
= 544.324 kPa
Pressure of gas B = [tex]\frac{n_bRT}{V}[/tex]
= [tex]\frac{4.93\times8.314\times300.3}{0.00555}[/tex]
= 2217784.22 Pa
= 2217.784 kPa
A toy helicopter is flying in a straight line at a constant speed of 4.5 m/s. If a projectile is launched vertically with an initial speed of v0 = 28 m/s, what horizontal distance d should the helicopter be from the launch site S if the projectile is to be traveling downward when it strikes the helicopter? Assume that the projectile travels only in the vertical direction.
Final answer:
To find the horizontal distance, we use the equation d = horizontal velocity * T, where T is the time of flight. Using the given values, the horizontal distance is 25.71 m.
Explanation:
To determine the horizontal distance the helicopter should be from the launch site, we need to find the time it takes for the projectile to reach the helicopter. Since the projectile only travels in the vertical direction, we can use the equation:
T = (2 * v0) / g
where T is the time of flight, v0 is the initial vertical velocity, and g is the acceleration due to gravity. Plugging in the values, we get:
T = (2 * 28 m/s) / 9.8 m/s^2 = 5.71 s
Now, we can find the horizontal distance using the equation:
d = horizontal velocity * T
The horizontal velocity of the helicopter is its constant speed, which is 4.5 m/s. Plugging in the values, we get:
d = 4.5 m/s * 5.71 s = 25.71 m
Consider a horizontal semi-transparent plate at a temperature 360 K that is irradiated from above and below. Air at [infinity] 310 K flows over both sides of the plate with a convective coefficient, 45 W/m2·K. The absorptivity of the plate is 0.4. If the radiosity is 4500 W/m2, find the irradiation, in W/m2, and emissivity of the plate. Is the plate diffuse-gray?
Answer:
[tex]G=6750\ W.m^2[/tex] is the irradiation
[tex]\epsilon=0.4725[/tex]
No, it is not a grey body because we don't have its transmissivity is not zero.
Explanation:
Given that:
Temperature of air, [tex]T_{\infty}=310\ K[/tex]
temperature of plate, [tex]T_s=360\ K[/tex]
convective heat transfer coefficient, [tex]h=45\ W.m^{-2}.K^{-1}[/tex]
absorptivity of plate, [tex]\alpha=0.4[/tex]
radiosity of the plate, [tex]J=4500\ W.m^{-2}[/tex]
From the energy balance eq. :
[tex]E_{in}=E_{out}[/tex]
since two surfaces are involved
[tex]2G=2J+2q_{conv}[/tex]
[tex]G=J+q_{conv}[/tex]
where
[tex]q_{conv}=[/tex] heat transfer per unit area due to convection
[tex]G=[/tex] irradiation per unit area
[tex]G=J+h.\Delta T[/tex]
[tex]G=4500+45\times (360-310)[/tex]
[tex]G=6750\ W.m^2[/tex] is the irradiation
Since radiosity includes transmissivity, reflectivity and emissivity.
[tex]J=G.\tau+G.\rho+E[/tex]
[tex]J=G.\tau+G.\rho+\epsilon\times \sigma\times T^{4}[/tex]
where:
[tex]\tau=[/tex] transmissivity
[tex]\rho=[/tex] reflectivity
[tex]\epsilon=[/tex] emissivity
Since the absorptivity of the plate is 0.4 and the plate is semitransparent so the transmissivity and reflectivity = 0.3 each.
[tex]4500=0.3\times 6750+0.3\times 6750+\epsilon\times 5.67\times 10^{-8}\times 360^4[/tex]
[tex]\epsilon=0.4725[/tex]
No, it is not a grey body because we don't have its transmissivity is not zero.
The Stefan-Boltzmann law of radiation can be used to find the irradiation and emissivity of a plate. The irradiation can be found using the net rate of heat transfer equation, and the emissivity can be found by rearranging the equation. The plate is diffuse-gray if its emissivity is between 0 and 1.
Explanation:The irradiation and emissivity of the plate can be found using the Stefan-Boltzmann law of radiation. The rate of heat transfer by emitted radiation is given by P = 6 AeT^4, where o = 5.67 × 10^-8 J/s. m². K^4 is the Stefan-Boltzmann constant, A is the surface area of the object, and T is its temperature in kelvins.
To find the irradiation, we can use the equation Qnet = σe A (T₂^4 – T₁^4), where Qnet is the net rate of heat transfer, σ is the Stefan-Boltzmann constant, e is the emissivity of the body, A is the surface area of the object, T₂ is the temperature of the surrounding air, and T₁ is the temperature of the plate.
To find the emissivity of the plate, we can rearrange the equation Qnet = σe A (T₂^4 – T₁^4) to solve for e. The plate is diffuse-gray if its emissivity is between 0 and 1. If the emissivity is 0, the plate is a perfect reflector, and if the emissivity is 1, the plate is a perfect absorber and emitter of radiation.
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An electric motor exerts a constant torque of 10 Nm on a grindstone mounted on its shaft. The moment of inertia of the grindstone is I = 2.0 kgm2 . The system starts from rest. A. Determine the angular acceleration of the shaft when this torque is applied. B. Determine the kinetic energy of the shaft after 8 seconds of operation. C. Determine the work done by the motor in this time. D. Determine the average power delivered by the motor over this time interval
Answer:
Angular acceleration = 5 rad /s ^2
Kinetic energy = 0.391 J
Work done = 0.391 J
P =6.25 W
Explanation:
The torque is given as moment of inertia × angular acceleration
angular acceleration = torque/ moment of inertia
= 10/2= 5 rad/ s^2
The kinetic energy is = 1/2 Iw^2
w = angular acceleration/time
=5/8= 0.625 rad /s
1/2 × 2× 0.625^2
=0.391 J
The work done is equal to the kinetic energy of the motor at this time
W= 0.391 J
The average power is = torque × angular speed
= 10× 0.625
P = 6.25 W
A. The angular acceleration is 5 rad/s². B. The kinetic energy of the shaft after 8 seconds is 16000 J. C. The work done by the motor in this time is 1600 J. D. The average power delivered by the motor over this time interval is 200 W.
Explanation:A. Angular acceleration can be found using the rotational analog to Newton's second law a = net τ/I. The moment of inertia I is given and the torque τ can be found from the given torque value. So, τ = 10 Nm and I = 2.0 kgm2. Substituting these values, we get a = 10 Nm / 2.0 kgm2 = 5 rad/s².
B. Using the equation K = (1/2) I ω², where I is the moment of inertia and ω is the angular velocity, we can calculate the initial angular velocity ω₁ as 0 rad/s. Then, the final angular velocity ω can be found using the relationship ω = ω₁ + αt. So, ω = 0 rad/s + (5 rad/s²)(8 s) = 40 rad/s. Finally, substituting these values into the kinetic energy equation, we have K = (1/2)(2.0 kgm2)(40 rad/s)² = 16000 J.
C. The work done by the motor can be found using the equation W = τθ, where τ is the torque and θ is the angular displacement. In this case, θ = ω₁t + (1/2)αt² = 0 rad/s(8 s) + (1/2)(5 rad/s²)(8 s)² = 160 rad. Therefore, W = (10 Nm)(160 rad) = 1600 J.
D. The average power can be calculated using the equation P = W/t, where W is the work done and t is the time interval. Substituting the values we found in the previous calculations, we have P = 1600 J / 8 s = 200 W.
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The energy of a photon is inversely proportional to the wavelength of the radiation. (T/F)
Answer:
true
Explanation:
The statement "the energy of a photon is inversely proportional to the wavelength of the radiation" is definitely true.
What is Wavelength?Wavelength may be characterized as the space or distance between two successive crests of a wave that especially includes the points in a sound wave or electromagnetic wave. The wavelength may be represented by a letter known as lambda (λ).
When the energy of the radiation increases, the waves travel faster which directs the small gap between two successive crests. This means that the wavelength decreases.
On contrary, when the energy of the radiation decreases, the waves travel slower which leads to a huge gap between two successive crests. This means that the wavelength increases.
Therefore, the statement "the energy of a photon is inversely proportional to the wavelength of the radiation" is definitely true.
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A model of a helicopter rotor has four blades, each 3.40 m long from the central shaft to the blade tip. The model is rotated in a wind tunnel at 550 rev/min. (a) What is the linear speed of the blade tip, in m/s? (b) What is the radial acceleration of the blade tip expressed as a multiple of g?
the radial acceleration of the blade tip is approximately 1210.50 times g .
To find the linear speed of the blade tip and the radial acceleration of the blade tip, we can use the following formulas:
(a) Linear speed of the blade tip:
[tex]\[ \text{Linear speed} = \text{Angular speed} \times \text{Radius} \]\[ v = \omega \times r \][/tex]
(b) Radial acceleration of the blade tip:
[tex]\[ \text{Radial acceleration} = \omega^2 \times r \]\[ a_r = \omega^2 \times r \][/tex]
Given:
- Angular speed [tex](\( \omega \))[/tex] = 550 rev/min
- Radius [tex](\( r \))[/tex] = 3.40 m
- Acceleration due to gravity[tex](\( g \))[/tex] = 9.81 m/s²
First, let's convert the angular speed from rev/min to rad/s:
[tex]\[ \omega = \frac{550 \text{ rev/min} \times 2\pi \text{ rad/rev}}{60 \text{ s/min}} \]\[ \omega = \frac{550 \times 2\pi}{60} \text{ rad/s} \]\[ \omega \approx 57.91 \text{ rad/s} \][/tex]
(a) Linear speed of the blade tip:
[tex]\[ v = \omega \times r \]\[ v = 57.91 \times 3.40 \]\[ v \approx 197.04 \text{ m/s} \][/tex]
So, the linear speed of the blade tip is approximately 197.04 m/s.
(b) Radial acceleration of the blade tip:
[tex]\[ a_r = \omega^2 \times r \]\[ a_r = (57.91)^2 \times 3.40 \]\[ a_r \approx 11854.76 \text{ m/s}^2 \][/tex]
Now, express the radial acceleration as a multiple of ( g ):
[tex]\[ \text{Multiple of } g = \frac{a_r}{g} \]\[ \text{Multiple of } g = \frac{11854.76}{9.81} \]\[ \text{Multiple of } g \approx 1210.50 \][/tex]
So, the radial acceleration of the blade tip is approximately 1210.50 times ( g ).
Two balanced three-phase loads that are connected in parallel are fed by a three-phase line having a series impedance of (0.4+????2.7) Ω per phase. One of the loads absorbs 560 kVA at 0.707 power factor lagging, and the other 132 kW at unity power factor. The line-to-line voltage at the load end of the line is 2200√3 V. Compute (a) the line-to-line voltage magnitude (in RMS value) at the source end of the line, (b) the total real and reactive power losses in the three-phase line, and (c) the total real and reactive power supplied at the sending end of the line.
Answer:
find answers below
Explanation:
a)
S1 560 e^(j acos 0.707 ) ⋅kVA S1 = ( ) 395.92 396.04j kW+ ⋅
S2 := 132 kW⋅ S2 = 132 kW⋅
Sd S1 +S2 :=
Sd = 527.92 396.04j kW
Sd = 660 kVA ⋅
arg Sd = 36.877 deg ⋅
Vd 2.2 e^(− j⋅0⋅deg)⋅kV
current in the line =(Sd/3 Vd) ⋅
⎯
:=
I Line = 79.988 60.006j A
Line = 99.994 A
arg I( ) Line = −36.877⋅deg
r Line := 0.4⋅Ω resistance in the line xLine := 2.7⋅Ω
Vsan Vd+ (r Line+ j xLine) ILine
Vsan = ( ) kV 2.394 0.192j + ⋅ Vsan = 2.402 kV⋅ arg V( ) san = 4.584 deg ⋅
Vsab 3 e
j 30 ⋅ ⋅deg ⋅ Vsan := ⋅
Vsab = ( ) kV 3.425 2.361j + ⋅ Vsab = 4.16 kV⋅ arg V( ) sab = 34.584 deg
b)
PLine 3 I ( ) Line
2 ⋅ r
Line := ⋅
PLine = 12 kW⋅
QLine 3 I ( ) Line
2 ⋅ xLine := ⋅ QLine = 80.99 kVAR ⋅
c)
Ss 3 Vsan ⋅ I
Line := ⋅
⎯ Ss = ( ) kVA 539.919 477.03j + ⋅
Ss = 720.5 kVA ⋅ arg S( )s = 41.461 deg ⋅
Ps Re S( )s := Ps = 540 kW⋅
Qs Im S( )s := Qs = 477 kVAR ⋅
S1 S2 + PLine + j QLine + ⋅ = ( ) kVA 539.919 477.03j + ⋅ Check
Normal human body temperature is about 37°C. What is this temperature in kelvins? What is the peak wavelength emitted by a person with this temperature? In what part of the spectrum does this lie?
Normal human body temperature is about 37°C, which is equivalent to 310.15 K. The peak wavelength emitted by a person with this temperature is 9.35 μm, which lies in the infrared part of the spectrum.
Explanation:Normal human body temperature is about 37°C. To convert this temperature to kelvin, we can use the formula K = °C + 273.15. So, the temperature in kelvins would be:
K = 37 + 273.15 = 310.15 K.
The wavelength emitted by a person with this temperature can be determined using Wien's displacement law: λ = b/T, where b is Wien's constant and T is the absolute temperature. The peak wavelength (λmax) is the wavelength at which the emitted radiation is the greatest. For a temperature of 310.15 K, the peak wavelength can be calculated as: λmax = b/T = 2897.8/T = 9.35 μm.
The peak wavelength of 9.35 μm corresponds to the infrared part of the electromagnetic spectrum.
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For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.100 kg of turkey. The slices of turkey are weighed on a plate of mass 0.400 kg placed atop a vertical spring of negligible mass and force constant of 200 N/m . The slices of turkey are dropped on the plate all at the same time from a height of 0.250 m . They make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (SHM). You may assume that the collision time is extremely small.a. What is the amplitude of oscillation A of the scale after the slices of ham land on the plate?b. What is the period of oscillation T of the scale?
Answer:
a) A = 0.0221 m, b) T = 0.314 s
Explanation:
For this exercise we must separate the process into two parts, one when the body falls and another when it hits the tray
Let's look for the speed with which it reaches the tray, using energy conservation
Initial. Highest point
Em₀ = U = m g h
Final. Tray Point
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v = √2gh
Now let's apply moment conservation
Initial. Just before the crash
p₀ = m v
Final. After the crash
p_{f} = (m + M) v_{f}
p₀ = p_{f}
m v = (m + M) v_{f}
v_{f} = m / (m + M) √2gh
This is the turkey + plate system speed, which is the one that will oscillate
b) The angular velocity of the oscillation is
w = √ k / m
The angular velocity is related to the frequency and period
w = 2πf = 2π / T
T = 2π √m / k
T = 2π √ ((0.100 + 0.400) / 200)
T = 0.314 s
a) To find the amplitude let's use the equation that describes the oscillatory motion
y = A cos (wt + fi)
Speed is
v = dy / dt = - A w sin (wt + fi)
At the initial point the turkey + plate system has a maximum speed, in the previous equation the speed is maximum when cos (wt + fi) = ±1
v = v_{f} = A w
m / (m + M) √ 2gh = A w
A = m / (m + M) √2gh 1 / w
w = 2π / T
Let's calculate
A = 0.100 / (0.100 + 0.400) √(2 9.8 0.250) 0.314/2π
A = 0.2 2.2136 0.049975
A = 0.0221 m
Earth's atmosphere is constantly bombarded by cosmic ray protons that originate somewhere in space. If the protons all passed through the atmosphere, assume that each square meter of Earth's surface would intercept protons at the average rate of 1110 protons per second. What would be the electric current intercepted by the total surface area of the planet?
Answer:
Explanation:
Each square meter of earth surface intercepts 1110 protons per second.
Total surface of the earth will intercept no of proton equal to
= 1110 x area of surface
= 1110 x 5.1 x 10¹⁴ m²
= 5661 x 10¹⁴
Total charge
= 1.6 x 10⁻¹⁹ x 5661 x 10¹⁴ coulomb per second
= .09 A .
electric current intercepted by the total surface area of the planet = .09 A.
An X-ray telescope located in Antarctica would not work well because of (a) the extreme cold; (b) the ozone hole; (c) continuous daylight; (d) Earth’s atmosphere.
The entrance of certain frequencies through the atmosphere varies depending on the sector. This concept of permittivity by which the rays can enter the earth is called the Radio Window and has a spectrum that ranges from 5MHz to 30GHz. Of the radiation produced by astronomical objects, one part is 'filtered' and the other has the ability to be absorbed by the earth due to its opacity effect.
Ultraviolet rays, X-rays and even gamma rays are completely affected by the ozone layer in Earth's atmosphere.
This makes it difficult to use X-ray telescopes located in Antarctica.
The correct option is D.
Earth's atmosphere limits X-ray observations, making Antarctica unsuitable due to extreme cold.
Earth's atmosphere acts as a barrier for X-ray observations, as it blocks most radiation at wavelengths shorter than visible light. Telescopes located in Antarctica would not work well due to the extreme cold which can affect their performance.
What is the Doppler effect, and how does it alter the way in which we perceive radiation?
Answer:
Explanation:
When the distance is decreasing, the frequency of the received wave form will be higher than the source wave form. Besides sound and radio waves, the Doppler effect also affects the light emitted by other bodies in space. If a body in space is "blue shifted," its light waves are compacted and it is coming towards us.
The Doppler effect is a phenomenon that describes how the perceived frequency and wavelength of waves change due to relative motion. It can result in a blue shift or a red shift, depending on the direction of motion. The Doppler effect has applications in astronomy and medicine, among other fields.
Explanation:The Doppler effect is a phenomenon in physics that describes how the perceived frequency and wavelength of waves change when there is relative motion between the source of the waves and the observer. This effect can alter the way in which we perceive radiation, such as light or sound.
When an object emitting waves moves towards an observer, the waves are compressed, resulting in a higher frequency and shorter wavelength. This is known as a blue shift. On the other hand, when an object moves away from an observer, the waves are stretched, resulting in a lower frequency and longer wavelength, which is called a red shift.
The Doppler effect has several applications in various fields. For example, it is used in astronomy to determine the speed and direction of celestial objects based on their red or blue shifts. In medicine, it is used in ultrasound technology to measure blood flow velocity. Overall, the Doppler effect allows us to understand and interpret the motion of objects emitting waves.
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Two small plastic spheres each have a mass of 1.2 g and a charge of -56.0 nC . They are placed 3.0 cm apart (center to center). Part A What is the magnitude of the electric force on each sphere? Express your answer to two significant figures and include the appropriate units.
Answer:
F_e = +/- 0.013133 N
Explanation:
Given:
- charge on each sphere q = -56 nC
- separation of spheres r = 3.0 cm
- charge constant k = 8.99*10^9
Find:
- Magnitude of Electric Force F_e on each sphere.
Solution:
The magnitude of electric Force F_e of one sphere on other is:
F_e = k*q^2 / r^2
- Plugging the given values :
F_e = (8.99*10^9) * (56*10^-19)^2 / (0.03)^2
F_e = 0.013133 N
- An equal and opposite force is experienced on the other sphere. Hence, F_e = + / - 0.013133 N
Water on Earth was (a) transported here by comets; (b) accreted from the solar nebula; (c) produced by volcanoes in the form of steam; (d) created by chemical reactions involving hydrogen and oxygen shortly after Earth formed.
Answer: Water on Earth was transported here by comets. The correct option is A.
Explanation:
Comets are made up of water with ice, rock and minerals.
Alot of research and hypotheses has been made to prove the origin of water on planet earth. Extraplanetary source such as comets, trans-Neptunian objects, and water-rich meteoroids (protoplanets) are believed to have delivered water to Earth.
Final answer:
Water on Earth primarily originated from the (b) solar nebula during planetary accretion, with minor contributions from comets and volcanic activity. Hence, (b) is the correct option.
Explanation:
As Earth coalesced from the protoplanetary disk around the young Sun, volatile compounds, including water, were incorporated into the developing planet. Comets may have contributed some water, but the predominant source was the solar nebula.
While volcanic activity released water vapor, it played a minor role compared to the water acquired during accretion. Chemical reactions involving hydrogen and oxygen likely occurred after Earth's formation, but the bulk of the water was established during the early stages of planetary accretion from the solar nebula.
A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of the same mass on the same spring be on the Moon, where the acceleration due to gravity is one sixth that of Earth? Show your work.
To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.
The extension of the spring due to the weight of the object on Earth is 0.3m, then
[tex]F_k = F_{W,E}[/tex]
[tex]kx_1 = mg[/tex]
The extension of the spring due to the weight of the object on Moon is a value of [tex]x_2[/tex], then
[tex]kx_2 = mg_m[/tex]
Recall that gravity on the moon is a sixth of Earth's gravity.
[tex]kx_2 = m\frac{g}{6}[/tex]
[tex]kx_2 = \frac{1}{6} mg[/tex]
[tex]kx_2 = \frac{1}{6} kx_1[/tex]
[tex]x_2 = \frac{1}{6} x_1[/tex]
We have that the displacement at the earth was [tex]x_1 = 0.3m[/tex], then
[tex]x_2 = \frac{1}{6} 0.3[/tex]
[tex]x_2 = 0.05m[/tex]
Therefore the displacement of the mass on the spring on Moon is 0.05m
The same spring and mass system will have a displacement of 0.05 m on the Moon compared to 0.3 m on Earth due to the Moon's lower gravitational acceleration.
Explanation:The subject of the question relates to how a spring and mass system will behave on Earth versus the Moon. The solution requires understanding of Hooke's Law and gravitation. Hooke's Law states that the displacement of a spring is directly proportional to the force applied. On Earth, when a 0.500 kg mass is suspended and displaced by 0.3 m, this sets up a certain relationship of force (F = mg, where g is Earth's gravity, 9.8 m/s²): F = 0.500 kg * 9.8 m/s² = 4.9 Newtons. Displacement, d, is proportional to the force:F = kd, so d = F/k where k is the spring constant.
On the Moon, g is not equal to Earth's, it's 1.67 m/s². The same hanging mass on the Moon would exert a force of Fm=0.500 kg * 1.67 m/s² = 0.835 Newtons. Since the spring constant doesn't change, and F = kd still holds, the new displacement on the Moon will be greater because the force is smaller. Displacement on the Moon (dM) will be dM = Fm/k, which is Fm divided by the same k we had on Earth. Not knowing k, we do know dM = (Fm/F) * d, and (Fm/F) is equal to the ratio of the Moon's gravity to Earth's gravity, 1/6, so dM = (1/6)*0.3 m = 0.05 m.
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Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at an angle of ? = 65.0 ? above the negative xaxis in the second quadrant. F? 2 has a magnitudeof 5.80 N and is directed at an angle of ? = 53.9 ? below the negative x axis in the third quadrant.A.What is the x component Fx of the resultant force?Express your answer in newtons.B. What is the y component Fy of the resultant force?Express your answer in newtons.C. What is the magnitude F of the resultant force?Express your answer in newtons.D. What is the angle ? that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis.Express your answer in degrees.
A) 7.297 N is the x component Fx of the resultant force.
Part B) - 3.654 N is the y component Fy of the resultant force.
Part C) 8.16N is the magnitude F of the resultant force.
D) 26.55° from the negative x-axis in a clockwise direction.
A) x-component of forces is 7.297 N
For x-component FX;
FX = F1 cos 65 + F2 cos 53.9
FX = 9.20 cos 65 + 5.8 cos 53.9
NB both forces are positive if resolved along the x-axis.
FX = 3.88 + 3.417 = 7.297 N
B) 6-component of forces is -3.654 N
For y-component FY
FY = -F1 sin 65 + F2 sin 53. 9
FY = - 9.2 sin 65 + 5.8 sin 53.9
NB F1 is negative along y-axis
FY = - 8.34 + 4.686 = - 3.654 N
C) The magnitude of the resultant force is 8.16 N
Magnitude :
[tex](FX^2 + FY^2)^{0.5}\\(7.297^2 + (-3.654)^2)^{0.5} \\= 8.16 \text N[/tex]
D) Angle made with negative x-axis is 26.56°
[tex]\text{Tan}^{ -1[/tex] of FY/FX gives the angle of the resultant force.
= [tex]\text{tan}^{ -1[/tex] of -3.654/7.297
=[tex]\text{tan}^{ -1[/tex] of -0.5
= -26.56°
i.e. 26.55° from the negative x-axis in a clockwise direction.
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(A) The x-component of the resultant force is approximately 7.039 N.
(B) The y-component of the resultant force is approximately 3.654 N.
(C) The magnitude of the resultant force is approximately 7.95 N.
(D) The angle θ with the negative x-axis is approximately 28.6 degrees (measured clockwise).
To find the x and y components of the resultant force, as well as its magnitude and angle with the negative x-axis, we can use vector addition. Given the forces F₁ and F₂:
F₁ = 9.20 N (65.0° above the negative x-axis)
F₂ = 5.80 N (53.9° below the negative x-axis)
(A) Finding the x-component (Fₓ) of the resultant force:
We'll first find the x-component of each force:
For F₁:
F₁ₓ = F₁ * cos(65.0°)
For F₂:
F₂ₓ = F₂ * cos(53.9°)
Now, calculate the net x-component:
Fₓ = F₁ₓ + F₂ₓ
Calculate Fₓ:
F₁ₓ = 9.20 N * cos(65.0°) ≈ 4.016 N
F₂ₓ = 5.80 N * cos(53.9°) ≈ 3.023 N
Fₓ = 4.016 N + 3.023 N ≈ 7.039 N
(B) Finding the y-component (Fᵧ) of the resultant force:
Similarly, we'll find the y-component of each force:
For F₁:
F₁ᵧ = F₁ * sin(65.0°)
For F₂:
F₂ᵧ = -F₂ * sin(53.9°) [negative because it's below the x-axis]
Now, calculate the net y-component:
Fᵧ = F₁ᵧ + F₂ᵧ
Calculate Fᵧ:
F₁ᵧ = 9.20 N * sin(65.0°) ≈ 8.111 N
F₂ᵧ = -5.80 N * sin(53.9°) ≈ -4.457 N
Fᵧ = 8.111 N - 4.457 N ≈ 3.654 N
(C) Finding the magnitude (F) of the resultant force:
The magnitude of the resultant force is given by the Pythagorean theorem:
F = √(Fₓ² + Fᵧ²)
Calculate F:
F = √(Fₓ² + Fᵧ²) ≈ √(7.039 N)² + (3.654 N)² ≈ √(49.53 N² + 13.35 N²) ≈ √62.88 N² ≈ 7.95 N
(D) Finding the angle (θ) with the negative x-axis:
The angle θ with the negative x-axis can be found using the inverse tangent function:
θ = arctan(Fᵧ / Fₓ)
Calculate θ:
θ = arctan(Fᵧ / Fₓ) ≈ arctan(3.654 N / 7.039 N) ≈ arctan(0.519) ≈ 28.6°
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A car is initially traveling west at 25 mph, and makes a turn north. As the car finishes the turn, its velocity is 20 mph north. It takes 3.5 seconds to complete the turn. What is the average acceleration of the car while it is turning, in mph/s?a. 0 -7.11+5.7.b. 7.1-5.73.c. 571-7.13.d. -5.7i+71.
Answer:
a. -7.11+5.7
Explanation:
Given:
initial velocity of the car while travelling west, [tex]v_w=25\hat i\ mph[/tex]velocity of the car after turning towards north, [tex]v_n=20\hat j\ mph[/tex]time taken to turn, [tex]t=3.5\ s[/tex]Now the acceleration:
[tex]a=\frac{v_w}{t} +\frac{v_n}{t}[/tex]
[tex]a=\frac{-25\hat i}{3.5} +\frac{20\hat j}{3.5}[/tex]
[tex]a=\frac{-50}{7} \hat i+\frac{40}{7} \hat j[/tex]
[tex]a=-7.1428\hat i+5.7142\hat j[/tex]
[tex]a=\sqrt{(\frac{50}{7} )^2+(\frac{40}{7} )^2}[/tex]
[tex]a=9.1473\ m.s^{-2}[/tex]
Whose view was that the Cosmos was based on the belief that every occurrence in the physical universe had logos behind it and that is where life originated?
Answer:
Aristotle
Explanation:
Aristotelian theory of the Universe
For two millennia, the philosophical tradition considered that the universe was eternal and did not change. The wise Aristotle said so, with total clarity and his ideas dominated Western thought for more than two thousand years.
This distinguished philosopher believed that the stars are made of an imperishable matter and that the landscapes of the sky are immutable.
From the time of Aristotle until the beginning of the twentieth century, the idea that the universe was static, that the cosmos had been eternally equal, was admitted.
In those years, the origin of the universe was not really considered in a scientific way, since it was based on the basis that the gods had created everything that exists, at the time they wanted it, according to their omnimous power.
So that all the efforts of the wise men of the time focused on discovering the existing organization in the universe created by the gods.
According to Aristotle and the thinkers of the fourth century B.C. what is below the Moon is a changing world, what is beyond the Moon is an immutable world.
How much excess charge must be placed on a copper sphere 25.0 cm in diameter so that the potential of its center, relative to infinity, is 1.50 kV?
Answer:
[tex]q=2.08*10^{-8}C\\ q=20.8nC[/tex]
Explanation:
Given data
Electric potential V=1.50 kV
Diameter d=25.0 cm
radius=diameter/2=25/2=12.5 cm=0.125 m
to find
We are asked to find excess charge
Solution
As inside the sphere the electric field is zero everywhere and potential is same at every point inside the sphere and on its surface
So we can use the value of the potential to get the charge q on the sphere
[tex]V=\frac{1}{4\pi E}\frac{q}{R}\\ q=4\pi ERV\\Where\\4\pi E=\frac{1}{9.0*10^{9}N.m^{2}/C^{2} }\\ q=(\frac{1}{9.0*10^{9}N.m^{2}/C^{2} })(0.125m)(1.50*10^{3}V )\\q=2.08*10^{-8}C\\ q=20.8nC[/tex]
To make the potential of a copper sphere's center 1.50 kV relative to infinity, an excess charge of approximately 2.09 x 10^-8 C must be placed on it.
Explanation:The potential V of a charged sphere is given by the equation V = kQ/R, where k is Coulomb's constant (8.99 x 109 Nm2/C2), Q is the charge on the sphere, and R is the radius. We can rearrange this to calculate the sphere's charge: Q = VR/k. The sphere's radius is half its diameter, hence 25.0 cm / 2 = 12.5 cm = 0.125 m. Substituting the given values in, we have Q = 1.50 x 103 V x 0.125 m / 8.99 x 109 Nm2/C2, recalling that 1 kV = 103 V. Doing the calculation, we find the excess charge Q to be approximately 2.09 x 10-8 C.
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Perform the following unit conversions:
a. 1 L to in.3
b. 650 J to Btu
c. 0.135 kW to ft · lbf/s
d. 378 g/s to lb/min
e. 304 kPa to lbf/in.2
f. 55 m3/h to ft3/s
g. 50 km/h to ft/s
h. 8896 N to ton (=2000 lbf)
Explanation:
a. 1 liter (L) is equal to 61.0237 cubic inches ([tex]in^3[/tex]), So:
[tex]1L*\frac{61.0237in^3}{1L}=61.0237in^3[/tex]
b. 1 J is equal to [tex]9.47817*10^{-4}[/tex] BTU. Thus:
[tex]650J*\frac{9.47817*10^{-4}BTU}{1J}=6.16081*10^{-1}BTU[/tex]
c. 1 kW is equal to 737.56 [tex]\frac{ft\cdot lbf}{s}[/tex]. So:
[tex]0.135kW*\frac{737.56\frac{ft\cdot lbf}{s}}{1kW}=99.57\frac{ft\cdot lbf}{s}[/tex]
d. 1 [tex]\frac{g}{s}[/tex] is equal to 0.1323 [tex]\frac{lb}{min}[/tex]. Therefore:
[tex]378\frac{g}{s}*\frac{0.1323\frac{lb}{min}}{1\frac{g}{s}}=50.01\frac{lb}{min}[/tex]
e. 1 kPa is equal to 0.145[tex]\frac{lbf}{in^2}[/tex]. Thus:
[tex]304kPa*\frac{0.145\frac{lbf}{in^2}}{1kPa}=44.08\frac{lbf}{in^2}[/tex]
f. 1 [tex]\frac{m^3}{h}[/tex] is equal to [tex]9.81*10^{-3}\frac{ft^3}{s}[/tex]. So:
[tex]55\frac{m^3}{h}*\frac{9.81*10^{-3}\frac{ft^3}{s}}{1\frac{m^3}{h}}=5.40*10^{-1}\frac{ft^3}{s}[/tex]
g. 1 [tex]\frac{km}{h}[/tex] is equal to 0.911 [tex]\frac{ft}{s}[/tex]. Therefore:
[tex]50\frac{km}{h}*\frac{0.911\frac{ft}{s}}{1\frac{km}{h}}=45.55\frac{ft}{s}[/tex]
h. 1 N is equal to [tex]1.1*10^{-4}[/tex] ton. Thus:
[tex]8896N*\frac{1.1*10^{-4}ton}{1N}=0.98ton[/tex]
The nucleus of a plutonium-239 atom contains 94 protons. Assume that the nucleus is a sphere with radius 6.64 fm and with the charge of the protons uniformly spread through the sphere. At the surface of the nucleus, what are the (a) magnitude and (b) direction (radially inward or outward) of the electric field produced by the protons
Answer:
(a) E = 3.065 * 10^21 N/C
(b) The direction is radially outward.
Explanation:
Parameters given:
Number of protons = 94
Charge of proton = 1.6023 * 10^-19 C
Radius of nucleus = 6.65 fm = 6.65 * 10^-15
At the surface of the nucleus, Electric field, E is given as:
E = kq/r^2
Where k = Coulombs constant
r = radius of nucleus
=> E = (9 * 10^9 * 94* 1.6023 * 10^-19)/(6.65 * 10^21)
E = 3.065 * 10^21 N/C
The direction of the field is radially outward.
The magnitude of the electric field at the surface of a plutonium-239 nucleus is roughly 7.94 x 10^21 NC^-1. The direction of the electric field, created by positively charged protons, is radially outward.
Explanation:The question pertains to the electric field created by the protons of a plutonium-239 atom. The electric field resulting from a spherical charge distribution can be determined by using Coulomb's law transformed to a volumetric charge distribution. The formula is E = kQ/r2, where k is the Coulomb constant (8.99 x 109 Nm2/C2), Q is the total charge, and r is the distance from the center of the field (which in this case is the radius of the nucleus).
(a) Magnitude of the electric field: Here, the total charge Q is the charge of a single proton (1.60 x 10-19 C) multiplied by the total number of protons which is 94. Therefore, upon inserting these values into the formula, the calculated electric field at the surface of the nucleus is around 7.94 x 1021 NC-1.
(b) Direction of the electric field: The direction of the electric field is always from positive to negative. Since protons are positively charged and they are creating the field, the electric field direction will be radially outward.
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2)It is known that the connecting rodS exerts on the crankBCa 2.5-kN force directed down andto the left along the centerline ofAB. Determine the moment of this force about for the two casesshown at below.
Answer:
M_c = 100.8 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two cases shown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *144
M_c = 2.5*( 42 / 150 ) *144
M_c = 100.8 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
A light beam in glass (n = 1.5) reaches an air-glass interface, at an angle of 60 degrees from the surface. What is the angle of the refracted light beam from the normal in air?Note: sin(30) = 0.5, sin(45) = 0.71, sin(60) = 0.87 (all angles are in degrees)
Answer:
θ₂ = 35.26°
Explanation:
given,
refractive index of air, n₁ = 1
refractive index of glass, n₂ = 1.5
angle of incidence, θ₁ = 60°
angle of refracted light, θ₂ = ?
using Snell's Law
n₁ sin θ₁ = n₂ sin θ₂
1 x sin 60° = 1.5 sin θ₂
sin θ₂ = 0.577
θ₂ = sin⁻¹(0.577)
θ₂ = 35.26°
Hence, the refracted light is equal to θ₂ = 35.26°
When a light beam passes from one medium to another, its angle of incidence and angle of refraction are related by Snell's law. The angle of the refracted light beam from the normal in air is 90 degrees.
Explanation:When a light beam passes from one medium to another, its angle of incidence and angle of refraction are related by Snell's law. The formula for Snell's law is:
n1×sin(θ1) = n2×sin(θ2)
In this case, the light beam is passing from glass (with a refractive index of 1.5) to air. Given that the angle of incidence is 60 degrees, we can use Snell's law to find the angle of refraction from the normal in air.
Using Snell's law, we have:
1.5×sin(60) = 1×sin(θ2)
Simplifying the equation, we find:
sin(θ2) = 1.5×sin(60) = 1.5×0.87 = 1.31
However, the maximum value for the sine function is 1, so the angle of refraction cannot exceed 90 degrees (the maximum angle for a light beam in air). Therefore, the angle of the refracted light beam from the normal in air is 90 degrees.
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A dock worker applies a constant horizontal force of 80.5 N to a block of ice on a smooth horizontal floor. The frictional force is negligible.The block starts from rest and moves a distance 13.0 m in a time 4.50 s.(a)What is the mass of the block of ice?----- I found this to be 56.9kg and got it right.(b)If the worker stops pushing at the end of 4.50 s, how far does the block move in the next 4.20s ? -
Answer:
(a). The mass of the block of ice is 62.8 kg.
(b). The distance is 24.192 m.
Explanation:
Given that,
Horizontal force = 80.5 N
Distance = 13.0 m
Time = 4.50 s
(a). We need to calculate the acceleration of the block of ice
Using equation of motion
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Put the value into the formula
[tex]13=0+\dfrac{1}{2}\times a\times(4.50)^2[/tex]
[tex]a=\dfrac{13\times2}{(4.50)^2}[/tex]
[tex]a=1.28\ m/s^2[/tex]
We need to calculate the mass of the block of ice
Using formula of force
[tex]F = ma[/tex]
[tex]m=\dfrac{F}{a}[/tex]
Put the value into the formula
[tex]m=\dfrac{80.5}{1.28}[/tex]
[tex]m=62.8\ kg[/tex]
(b). If the worker stops pushing at the end of 4.50 s,
We need to calculate the velocity
Using equation of motion
[tex]v =u+at[/tex]
Put the value into the formula
[tex]v=0+1.28\times4.50[/tex]
[tex]v=5.76\ m/s[/tex]
We need to calculate the distance
Using formula of distance
[tex]v = \dfrac{d}{t}[/tex]
[tex]d=v\times t[/tex]
Put the value into the formula
[tex]d=5.76\times4.20[/tex]
[tex]d=24.192\ m[/tex]
Hence, (a). The mass of the block of ice is 62.8 kg.
(b). The distance is 24.192 m.
What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2? (Use any variable or symbol stated above as necessary.)
Answer:
[tex]\frac{4}{3}\rho\pi(r_2^3 - r_1^3)[/tex]
Explanation:
The volume of the shell is the difference between the volume of the outer sphere and the volume of the inner sphere
[tex]V = V_o - V_i = \frac{4}{3}\pi r_2^3 - \frac{4}{3}\pi r_1^3[/tex]
[tex]V = \frac{4}{3}\pi(r_2^3 - r_1^3)[/tex]
The mass of the shell is the product of the volume and its density
[tex]m = V\rho = \frac{4}{3}\rho\pi(r_2^3 - r_1^3)[/tex]
The same couple moment of 160 lb⋅ftlb⋅ft is required to open the hatch door, independent of where the couple's forces are applied. If the forces are applied at points C and D, what force magnitude FFF is required to open the hatch door?
Answer:
F = 88.88[lb-f]
Explanation:
To solve this problem, we must first look at the attached image.
Another important data that is needed to solve the problem is the distance from the Centre O to points C and D. The distance is 0.9 [ft].
Then we do a sum of moments around Point O and it equals zero.
[tex]SM_{O} =0\\F*0.9+F*0.9-160=0\\2*F*0.9=160\\F=88.88[lb-f][/tex]
Therefrore the couple of forces needed are 88.88[lb-f]
A car of mass m push = 1200 kg is capable of a maximum acceleration of 6.00 m / s 2 . If this car is required to push a stalled car of mass m stall = 1750 kg, what is the maximum magnitude of the acceleration a of the two‑car system? a = 0.62 m / s 2
1)
first you find the maxium force that the car can produce.
f=ma
Fmax=(1100kg)(6m/s^2)
then use f = ma again to find the accel with the passengers
Fmax=(1100kg +1650kg)(a)
=> a = (1100kg)(6m/s^2)/( 1100kg +1650kg)
= 2.4 m/s^2
The maximum magnitude of the acceleration of the two‑car system is [tex]2.44 \;\rm m/s^{2}[/tex].
Given data:
The mass of car is, m = 1200 kg.
The maximum acceleration is, [tex]a =6.00 \;\rm m/s^{2}[/tex].
The mass of stall is, m' = 1750 kg.
As per the Newton's Second law of motion, the applied force on the car is equal to the product of mass and acceleration of car. So, the maximum force to push the car is given as,
F = ma
[tex]F = 1200 \times 6.00\\\\F= 7200\;\rm N[/tex]
Now for two-car system, the total mass is,
M = m + m'
M = 1200 +1750 = 2950 kg.
So, the acceleration for the two-car system is,
[tex]F = M \times a'\\\\7200 =2950 \times a'\\\\a'=2.44 \;\rm m/s^{2}[/tex]
Thus, the maximum magnitude of the acceleration a of the two‑car system is [tex]2.44 \;\rm m/s^{2}[/tex].
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The vertical deflecting plates of a typical classroom are a pair of parallel square metal plates carrying equal but opposite charges. Potential difference between the plates is 25.0V; typical dimensions are about 3.8cm on a side with a separation of about 4.6mm. Under these conditions, how much charge is on each plate? How stron is the electric field between the plates? I fan electron is ejected from the negative plates, how fast is it moving when it reaches the postive plate?
Answer:
[tex]6.945326087\times 10^{-11}\ C[/tex]
5434.78260873 N/C
2963369.48874 m/s
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
V = Voltage = 25 V
Side = 3.8 cm
d = Separation = 4.6 mm
m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
Charge is given by
[tex]Q=\dfrac{VA\epsilon_0}{d}\\\Rightarrow Q=\dfrac{25\times (3.8\times 10^{-2})^2\times 8.85\times 10^{-12}}{4.6\times 10^{-3}}\\\Rightarrow Q=6.945326087\times 10^{-11}\ C[/tex]
Charge on each plate is [tex]6.945326087\times 10^{-11}\ C[/tex]
Electric field is given by
[tex]E=\dfrac{Q}{A\epsilon_0}\\\Rightarrow E=\dfrac{6.945326087\times 10^{-11}}{(3.8\times 10^{-2})^2\times 8.85\times 10^{-12}}\\\Rightarrow E=5434.78260873\ N/C[/tex]
The electric field strength is 5434.78260873 N/C
Acceleration is given by
[tex]a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times 5434.78260873}{9.11\times 10^{-31}}\\\Rightarrow a=9.5451725291\times 10^{14}\ m/s^2[/tex]
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.5451725291\times 10^{14}\times 4.6\times 10^{-3}+0^2}\\\Rightarrow v=2963369.48874\ m/s[/tex]
The velocity is 2963369.48874 m/s