The question requires determining angles for bright and dark fringes in a double-slit experiment using the formula d sin(\theta) = m\lambda. The student should apply this formula, adjusting it for dark fringes, and solve for the angles for each order m, considering the wavelength and slit separation given.
Explanation:The question relates to Young's double-slit experiment, which demonstrates the wave nature of light through the interference pattern produced when light passes through two closely spaced slits. The angles locating the fringes can be calculated using the formula
d sin(\theta) = m\lambda, where d is the separation between the slits, \theta is the angle of the fringe from the central maximum, m is the order of the fringe (which can be an integer or half-integer value depending on whether it's a bright or dark fringe), and \lambda is the wavelength of the light.
(a) For the dark fringe with m=0, we expect no fringe to appear as m=0 corresponds to the central maximum which is bright, not dark.
(b) For the bright fringe with m=1, we rearrange the formula to \theta = arcsin(m\lambda / d). Substituting the given values, the angle can be calculated.
(c) For the dark fringe with m=1, the condition is changed to d sin(\theta) = (m + 0.5)\lambda, as dark fringes occur at half-wavelength shifts from the bright fringes.
(d) The calculation for the bright fringe with m=2 follows the same procedure as for m=1, using the appropriate value for m.
Find the coefficient of x3 y4 in the expansion of ( x+2y )^7.
Answer:
The coefficient of x³y⁴ in the expansion of ( x+2y)⁷ is 560.
Explanation:
The given expression is
[tex](x+2y)^7[/tex]
According to binomial expansion,
[tex](a+b)^n=^nC_0a^nb^0+^nC_1a^{n-1}b^1+...+^nC_{n-1}a^1b^{n-1}+^nC_0a^0b^n[/tex]
The r+1th term of the expansion is
[tex]^nC_rx^{n-r}(2y)^r=^nC_r(2^r)x^{n-r}(y)^r[/tex] ... (1)
In the term x³y⁴ the power of x is 3 and the power of y is 4. It means the value of r is 4 and the value n-r is 3.
[tex]n-r=3[/tex]
[tex]n-4=3\Rightarrow n=7[/tex]
Put n=7 and r=4 in equation (1)
[tex]^7C_4(2^4)x^{7-4}(y)^4[/tex]
[tex]\frac{7!}{4!(7-4)!}(16)x^3y^4[/tex]
[tex]\frac{7\times 6\times 5\times 4!}{4!(3)!}(16)x^3y^4[/tex]
[tex]560x^3y^4[/tex]
Therefore the coefficient of x³y⁴ in the expansion of ( x+2y)⁷ is 560.
The coefficient of x^3 y^4 in the expansion of (x + 2y)^7 is calculated using the binomial theorem, resulting in 560.
To find the coefficient of x^3 y^4 in the expansion of (x + 2y)^7, we use the binomial theorem. The general term in the expansion of (a + b)^n is given by T(r+1) = C(n, r) * a^(n-r) * b^r, where C(n, r) is the binomial coefficient n choose r. In this question, we need to find the term where x is raised to the power of 3 and y to the power of 4.
Plug in the values: a = x, b = 2y, n = 7, and r = 4 (since y^4). Thus, T(5) = C(7, 4) * x^(7-4) * (2y)^4 = C(7, 4) * x^3 * 16y^4. The binomial coefficient C(7, 4) is 35.
Now multiply 35 by 16 to get the coefficient: 35 * 16 = 560. Therefore, the coefficient of x^3 y^4 is 560.
A meteorite has a speed of 95.0 m/s when 750 km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 3.35 m . Part A What is its speed just before striking the sand?
The meteorite's speed just before striking the sand is approximately 3836.82 m/s, calculated using conservation of energy principles from its initial potential energy at 750 km above Earth.
Let's break down the solution step by step:
Step 1: Calculate the initial potential energy of the meteorite when it is 750 km above the Earth's surface.
[tex]\[PE_{initial} = mgh\][/tex]
Where:
- [tex]\(m\)[/tex] is the mass of the meteorite,
- [tex]\(g\)[/tex] is the acceleration due to gravity, and
- [tex]\(h\)[/tex] is the height above the Earth's surface.
Given:
- [tex]\(h = 750 \, \text{km} = 750 \times 10^3 \, \text{m}\)[/tex] (converted to meters)
- [tex]\(g = 9.8 \, \text{m/s}^2\)[/tex] (acceleration due to gravity)
[tex]\[PE_{initial} = mg \times h\][/tex]
[tex]\[PE_{initial} = (m)(9.8 \, \text{m/s}^2)(750 \times 10^3 \, \text{m})\][/tex]
Step 2: Calculate the final kinetic energy of the meteorite just before it strikes the sand.
[tex]\[KE_{final} = \frac{1}{2}mv^2\][/tex]
Where:
- [tex]\(v\)[/tex] is the final velocity of the meteorite just before it strikes the sand.
Given:
- [tex]\(d = 3.35 \, \text{m}\)[/tex] (distance traveled while coming to rest)
Step 3: Apply the principle of conservation of energy.
Since energy is conserved, the initial potential energy of the meteorite is converted into kinetic energy just before it strikes the sand.
[tex]\[PE_{initial} = KE_{final}\][/tex]
[tex]\[mg \times h = \frac{1}{2}mv^2\][/tex]
[tex]\[9.8 \times 750 \times 10^3 = \frac{1}{2} \times v^2\][/tex]
[tex]\[v^2 = \frac{2 \times 9.8 \times 750 \times 10^3}{1}\][/tex]
[tex]\[v^2 = 2 \times 9.8 \times 750 \times 10^3\][/tex]
[tex]\[v^2 = 2 \times 7.35 \times 10^6\][/tex]
[tex]\[v^2 = 14.7 \times 10^6\][/tex]
[tex]\[v^2 = 1.47 \times 10^7\][/tex]
[tex]\[v = \sqrt{1.47 \times 10^7}\][/tex]
[tex]\[v \approx 3836.82 \, \text{m/s}\][/tex]
So, the speed of the meteorite just before striking the sand is approximately [tex]\(3836.82 \, \text{m/s}\)[/tex].
A 4.0 × 102-nm thick film of kerosene (n = 1.2) is floating on water. White light is normally incident on the film. What is the visible wavelength in air that has a maximum intensity after the light is reflected? Note: the visible wavelength range is 380 nm to 750 nm.
Answer:
the visible wavelength is 480 nm
Explanation:
Given data
thick film = 4.0 × 10² nm
n = 1.2
wavelength range = 380 nm to 750 nm
to find out
the visible wavelength in air
solution
we know that index of water is 1 and kerosene is 1.2
we can say that when light travel reflected path difference is = 2 n t
and for maximum intensity it will be k × wavelength
so it will be 2 n t = k × wavelength
2 × 1.2 × 4.0 × 10² = k × wavelength
wavelength = 2 × 1.2 × 4.0 × 10² / k
here k is 2 for visible
so wavelength = 2 × 1.2 × 4.0 × 10² / 2
wavelength = 480 nm
the visible wavelength is 480 nm
It takes 0.16 g of helium (He) to fill a balloon. How many grams of nitrogen (N2) would be required to fill the balloon to the same pressure, volume, and temperature?
Answer: The mass of nitrogen gas required to fill the balloon is 1.12 grams.
Explanation:
We are given:
Mass of helium gas = 0.16 g
We need to calculate the mass of nitrogen gas that can fill the balloon at same pressure, volume and temperature. This means that the moles of both the gases filling up balloon will be same.
So, to calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of Helium = 0.16 g
Molar mass of helium = 4.00 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of helium}=\frac{0.16g}{4g/mol}=0.04mol[/tex]
Now, calculating the mass of nitrogen gas using same above equation, we get:
Moles of nitrogen gas = 0.04 moles
Molar mass of nitrogen gas = 28.01 g/mol
Putting values in above equation, we get:
[tex]0.04mol=\frac{\text{Mass of }N_2}{28.01g/mol}\\\\\text{Mass of }N_2=1.12g[/tex]
Hence, the mass of nitrogen gas required to fill the balloon is 1.12 grams.
To fill the balloon with nitrogen (N2) under the same conditions as helium (He), you would need 1.12 grams, based on the molar masses of each gas and Avogadro's law.
Explanation:The subject matter of your question lies in the field of chemistry, specifically the concepts of Avogadro's law and the ideal gas law. Avogadro's law states that equal volumes of gases, at the same temperature and pressure, contain the same number of molecules. Therefore, given that N2 and He are both gases, the same volume of each gas will contain the same number of molecules.
The ideal gas law can also be applied here. The molar mass of nitrogen (N₂) is 28.01 g/mol, and the molar mass of helium (He) is 4 g/mol. Given that we know the mass of helium required to fill the balloon (0.16 g), we can use the relationship between molar mass and the actual mass to determine the equivalent mass for nitrogen.
This relationship is represented by the ratio of the molar mass of nitrogen to the molar mass of helium, which is 28.01 / 4 = 7. Thus, to fill the balloon with nitrogen under the same conditions of pressure, volume, and temperature, the required mass would be 7 times that of the mass of the helium, or 0.16 * 7 = 1.12 g.
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A diffraction grating with 140 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles in the third-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm? (angles in radians) The 434 nm line:
Answer:
0.003181 radians
0.003005 radians
Explanation:
Number of slits = 140 /cm
λ = Wavelength = 434 nm = 434×10⁻⁹ m
m = 3 Third order spectrum
Space between slits
[tex]d=\frac{0.01}{140} =7.14\times 10^{-5}\ m[/tex]
Now,
[tex]dsin\theta = m\lambda\\\Rightarrow \theta=sin^{-1}\left(\frac{m\lambda}{d}\right)\\\Rightarrow \theta=sin^{-1}\left(\frac{3\times 434\times 10^{-9}}{7.14\times 10^{-5}}\right)\\\Rightarrow \theta=sin^{-1}0.018228\\\Rightarrow \theta=0.01823^{\circ}=0.01823\times \frac{\pi}{180}=0.003181 radians[/tex]
0.003181 radians
When λ = 410 nm = 410×10⁻⁹ m
[tex]dsin\theta = m\lambda\\\Rightarrow \theta=sin^{-1}\left(\frac{m\lambda}{d}\right)\\\Rightarrow \theta=sin^{-1}\left(\frac{3\times 410\times 10^{-9}}{7.14\times 10^{-5}}\right)\\\Rightarrow \theta=sin^{-1}0.01722\\\Rightarrow \theta=0.01722^{\circ}=0.01722\times \frac{\pi}{180}=0.003005 radians[/tex]
0.003005 radians
What is the gravitational force between two 10 kg Iron balls separated by a distance of 0.1 m?
Answer:
Gravitational force, [tex]F=6.67\times 10^{-7}\ N[/tex]
Explanation:
Masses of two iron balls, m₁ = m₂ = 10 kg
Distance between balls, d = 0.1 m
We need to find the gravitational force between two balls. It is given by :
[tex]F=G\dfrac{m_1m_2}{d^2}[/tex]
[tex]F=6.67\times 10^{-11}\times \dfrac{(10\ kg)^2}{(0.1\ m)^2}[/tex]
[tex]F=6.67\times 10^{-7}\ N[/tex]
Hence, this is the required solution.
Final answer:
The gravitational force between two 10 kg Iron balls separated by a distance of 0.1 m is 6.67 x 10^-11 N, calculated using Newton's law of universal gravitation.
Explanation:
The gravitational force between two 10 kg Iron balls separated by a distance of 0.1 m can be calculated using Newton's law of universal gravitation.
The formula for gravitational force is F = G(m1 * m2) / d^2, where G is the gravitational constant, m1 and m2 are the masses of the objects, and d is the distance between their centers.
Plugging in the values, the force would be 6.67 x 10^-11 N.
When 0.511 g of sodium metal is added to an excess of hydrochloric acid, 5310 J of heat are produced. What is the enthalpy of the reaction as written?
Answer:
The the enthalpy of the reaction is 241.36 KJ.
Explanation:
Given that,
Weight of sodium = 0.511 g
We need to calculate the number of moles
Using formula of moles
[tex]moles=\dfrac{given\ mass}{molar\ mass}[/tex]
[tex]moles=\dfrac{0.511}{23}[/tex]
[tex]moles= 0.022[/tex]
We need to calculate the energy in 1 mole
In 0.022 moles of sodium metal = 5310 J
In 1 moles of sodium metal =[tex]\dfrac{5310}{0.022}[/tex]
The Energy in 1 moles of sodium is 241363.63 J.
Hence, The the enthalpy of the reaction is 241.36 KJ.
Final answer:
To find the reaction enthalpy per mole of sodium reacting with hydrochloric acid, divide the heat produced (5310 J) by the number of moles of sodium (0.0222 mol). This calculation results in an enthalpy change of -239 kJ/mol for sodium.
Explanation:
To calculate the enthalpy of the reaction per mole of sodium metal reacting with hydrochloric acid, we can use the information provided about the heat produced during the reaction. Given that 0.511 g of sodium metal yields 5310 J of heat, we first need to determine the amount in moles of sodium. We know that the molar mass of sodium (Na) is approximately 23.0 g/mol. Thus, 0.511 g Na amounts to 0.511 g / 23.0 g/mol = 0.0222 mol Na.
Given the heat released, we can now calculate the enthalpy change per mole of sodium reacted. The enthalpy change (H) for the reaction can be calculated by dividing the heat by the number of moles of sodium:
H = Heat Produced / Moles of Sodium
H = 5310 J / 0.0222 mol Na
H = -239,189 J/mol Na (note the negative sign because the reaction is exothermic)
Converting this result to kilojoules, the enthalpy change per mole of sodium is approximately -239.189 kJ/mol Na, which we often express as -239 kJ/mol to match conventional significant figures for such values.
If the torque required to loosen a nut holding a wheel on a car is 48 N · m, what force must be exerted at the end of a 0.23 m lug wrench to loosen the nut when the angle between the force and the wrench is 41◦ ? Answer in units of N.
Answer:
F = 318.1 N
Explanation:
As we know that torque to open the nut is given by formula
[tex]\tau = \vec r \times \vec F[/tex]
so we can write it as
[tex]\tau = rFsin\theta[/tex]
now we know that
[tex]\tau = 48 Nm[/tex]
r = 0.23 m
angle between force and the wrench is 41 degree
so we have
[tex]48 = (0.23)Fsin41[/tex]
[tex]F = \frac{48}{(0.23)sin41}[/tex]
[tex]F = 318.1 N[/tex]
Problem One: A beam of red light (656 nm) enters from air into the side of a glass and then into water. wavelength, c. and speed in both the glass and the water. Find the a. the frequencies, b. the
Answer:
Part a)
[tex]f_w = f_g = 4.57 \times 10^{14} Hz[/tex]
Part b)
[tex]\lambda_w = 492 nm[/tex]
[tex]\lambda_g = 437.3 nm[/tex]
Part c)
[tex]v_w = 2.25 \times 10^8 m/s[/tex]
[tex]v_g = 2.0 \times 10^8 m/s[/tex]
Explanation:
Part a)
frequency of light will not change with change in medium but it will depend on the source only
so here frequency of light will remain same in both water and glass and it will be same as that in air
[tex]f = \frac{v}{\lambda}[/tex]
[tex]f = \frac{3 \times 10^8}{656 \times 10^{-9}}[/tex]
[tex]f = 4.57 \times 10^{14} Hz[/tex]
Part b)
As we know that the refractive index of water is given as
[tex]\mu_w = 4/3[/tex]
so the wavelength in the water medium is given as
[tex]\lambda_w = \frac{\lambda}{\mu_w}[/tex]
[tex]\lambda_w = \frac{656 nm}{4/3}[/tex]
[tex]\lambda_w = 492 nm[/tex]
Similarly the refractive index of glass is given as
[tex]\mu_w = 3/2[/tex]
so the wavelength in the glass medium is given as
[tex]\lambda_g = \frac{\lambda}{\mu_g}[/tex]
[tex]\lambda_g = \frac{656 nm}{3/2}[/tex]
[tex]\lambda_g = 437.3 nm[/tex]
Part c)
Speed of the wave in water is given as
[tex]v_w = \frac{c}{\mu_w}[/tex]
[tex]v_w = \frac{3 \times 10^8}{4/3}[/tex]
[tex]v_w = 2.25 \times 10^8 m/s[/tex]
Speed of the wave in glass is given as
[tex]v_g = \frac{c}{\mu_g}[/tex]
[tex]v_g = \frac{3 \times 10^8}{3/2}[/tex]
[tex]v_g = 2 \times 10^8 m/s[/tex]
The weight of a product is measured in pounds. A sample of 50 units is taken from a recent production. The sample yielded X⎯⎯⎯ = 75 lb, and we know that σ2 = 100 lb. Calculate a 99 percent confidence interval for μ.
Answer:
The 99% confidence interval for the weights = [71.36lb, 78.64lb]
Explanation:
Mean weight
[tex]\bar{x} =75 lb[/tex]
Variance of weights
[tex]\sigma^2 =100lb[/tex]
Standard deviation,
[tex]\sigma =\sqrt{100}=10lb[/tex]
Confidence interval is given by
[tex]\bar{x}-Z\times \frac{\sigma}{\sqrt{n}}\leq \mu\leq \bar{x}+Z\times \frac{\sigma}{\sqrt{n}}[/tex]
For 99% confidence interval Z = 2.576,
Number of weights, n = 50
Substituting
[tex]75-2.576\times \frac{10}{\sqrt{50}}\leq \mu\leq 75+2.576\times \frac{10}{\sqrt{50}}\\\\71.36\leq \mu\leq 78.64[/tex]
The 99% confidence interval for the weights = [71.36lb, 78.64lb]
The 99 percent confidence interval for the population mean (μ) can be calculated using the formula X ± Z*(σ/√n) where X is the sample mean, Z is the Z-score for a 99% confidence interval (approximately 2.57), σ is the standard deviation of the population and n is the sample size. Substituting the given values results in 75 ± 2.57*(10/√50).
Explanation:To calculate a 99 percent confidence interval for the population mean (μ), you can use the formula:
X± Z*(σ/√n)
Where:
X is the sample mean, which is given as 75 pounds.Z is the Z-score associated with the desired confidence level. For a 99% confidence interval, the Z-score is approximately 2.57. This can be found using the standard normal distribution table or online Z-score calculator.σ is the standard deviation of the population, the square root of given variance 100 lb, which is 10 lb.n is the sample size, which is given as 50.Substituting these values into the formula gives the following 99 percent confidence interval:
75 ± 2.57*(10/√50).
Calculate the range and you will find your answer for the 99 percent confidence interval for μ.
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Two identical loudspeakers are some distance apart. A person stands 5.80 m from one speaker and 3.90 m from the other. What is the fourth lowest frequency at which destructive interference will occur at this point? The speed of sound in air is 343 m/s.
Answer:
f = 632 Hz
Explanation:
As we know that for destructive interference the path difference from two loud speakers must be equal to the odd multiple of half of the wavelength
here we know that
[tex]\Delta x = (2n + 1)\frac{\lambda}{2}[/tex]
given that path difference from two loud speakers is given as
[tex]\Delta x = 5.80 m - 3.90 m[/tex]
[tex]\Delta x = 1.90 m[/tex]
now we know that it will have fourth lowest frequency at which destructive interference will occurs
so here we have
[tex]\Delta x = 1.90 = \frac{7\lambda}{2}[/tex]
[tex]\lambda = \frac{2 \times 1.90}{7}[/tex]
[tex]\lambda = 0.54 m[/tex]
now for frequency we know that
[tex]f = \frac{v}{\lambda}[/tex]
[tex]f = \frac{343}{0.54} = 632 Hz[/tex]
An engineering firm is designing a ski lift. The wire rope needs to travel with a linear velocity of 2.0 meters per second, and the angular velocity of the bullwheel will be 10 revolutions per minute. What diameter bullwheel should be used to drive the wire rope?
Answer:
The diameter of the bull-wheel is 3.82
Explanation:
Given that,
Velocity = 2.0 m/s
Angular velocity = 10 rev/m
[tex]\omega=10\times\dfrac{2\pi}{60}[/tex]
[tex]\omega=1.0472\ rad/s[/tex]
We need to calculate the diameter of bull-wheel
Using formula of angular velocity
[tex]v= r\omega[/tex]
[tex]r=\dfrac{v}{\omega}[/tex]
Put the value into the formula
[tex]r=\dfrac{2.0}{1.0472}[/tex]
[tex]r=1.91\ m[/tex]
The diameter of the bull-wheel
[tex]D=2r[/tex]
[tex]D=2\times1.91[/tex]
[tex]D=3.82\ m[/tex]
Hence, The diameter of the bull-wheel is 3.82 m.
Radio receivers are usually tuned by adjusting the capacitor of an LC circuit. If C = C1 for a frequency of 600 kHz, then for a frequency of 1200 kHz C must be adjusted to what value?
Answer:
[tex]C_2=\frac{C_1}{4}[/tex]
Explanation:
Given:
Initial capacitance, C = C₁
Initial Frequency, F₁ = 600kHz
Final Frequency, F₂ = 1200kHz
let the adjusted capacitance be, C₂
Now
the frequency (f) is given as:
[tex]f={\frac{1}{2\pi \sqrt{LC}}}[/tex]
where, L = inductance (same for the same material)
thus substituting the values we have
[tex]600={\frac{1}{2\pi \sqrt{LC_1}}}[/tex] ...............(1)
and
[tex]1200={\frac{1}{2\pi \sqrt{LC_2}}}[/tex] ..............(2)
dividing the equation 1 with 2, we get
[tex]\frac{600}{1200}=\frac{{\frac{1}{2\pi \sqrt{LC_1}}}}{{\frac{1}{2\pi \sqrt{LC_2}}}}[/tex]
or
[tex]\frac{1}{2}=\sqrt{\frac{C_2}{C_1}}[/tex]
or
[tex]\frac{C_2}{C_1}=\frac{1}{4}[/tex]
or
[tex]C_2=\frac{C_1}{4}[/tex]
hence, the new capacitance C, must be one-fourth times the initial capacitance
The capacitance for the frequency of 1200 kHz must be one-fourth times the capacitance for the frequency of 800 kHz.
What is capacitance?Capacitance is the ability of a circuit to store energy in the form of an electrical charge.it is an energy-storing device. generally, it is defined by the ratio of electric charge stored to the potential difference.
Given:
Initial capacitance is C
Capacitance for the frequency of 1200 kHz is [tex]\rm{C=C_1}[/tex]
The capacitance for the frequency of 800 kHz is [tex]\rm{C=C_2}[/tex]
The frequency (f) is given by
[tex]f=\frac{1}{2\pi \sqrt{LC} }[/tex]
The capacitance for the frequency of 1200 kHz finds by
[tex]f_1=\frac{1}{2\pi \sqrt{LC_1} }[/tex]
[tex]1200=\frac{1}{2\pi \sqrt{LC_1} }[/tex] ------------------1
The capacitance for the frequency of 800 kHz
[tex]f_2=\frac{1}{2\pi \sqrt{LC_2} }[/tex]
[tex]600=\frac{1}{2\pi \sqrt{LC_2} }[/tex] ------------------2
On dividing the equation 2 by 1
[tex]\frac{1}{2} =\sqrt{\frac{C_2}{C_1} }[/tex]
[tex]{\frac{C_2}{C_1} }=\frac{1}{4}[/tex]
[tex]C_2=\frac{C_1}{4}[/tex]
Hence the capacitance for the frequency of 1200 kHz must be one-fourth times the capacitance for the frequency of 800 kHz.
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A boat sails south with the help of a wind blowing in the direction S36°E with magnitude 300 lb. Find the work done by the wind as the boat moves 130 ft. (Round your answer to the nearest whole number.) ft-lb
To find the work done by the wind on the boat, we first find the southward component of the wind's force, since the boat is sailing south. We find this to be about 243 lb. Multiplying this force by the distance the boat travels, we find that the wind does approximately 31590 ft-lb of work on the boat.
Explanation:In order to find the work done by the wind on the boat, we need to find the component of the wind force that acts in the same direction as the displacement of the boat. The wind is blowing in the direction of S36°E, meaning it has a southward component and an eastward component. However, since the boat is moving south, only the southward component of the wind's force will do work on the boat.
We use the equation F_s = F * cos(θ) to find the southward component of the force, where F is the magnitude of the total wind force and θ is the angle between the force of the wind and the direction of displacement. Plugging in the given values, we get F_s = 300 lb * cos(36°) = 243 lb.
To find work, we use the equation W = F * d * cos(θ), where F is the force, d is the distance traveled, and θ is the angle between the force and the displacement. Since the force and the displacement are in the same direction, the angle θ is 0, so cos(θ) is 1. Plugging in the appropriate values, we get W = 243 lb * 130 ft * 1 = 31590 ft-lb.
Rounded to the nearest whole number, the wind does approximately 31590 ft-lb of work on the boat.
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A battery charger can produce 3A at 12 Volt and charges a battery fer 2 hr. Calculate work in KJ.
Answer: 259.2 KJ
Explanation:
The formula calculate work don in a circuit is given by :-
[tex]W=QV[/tex], where Q is charge and V is the potential difference.
The formula to calculate charge in circuit :-
[tex]Q=It[/tex], where I is current and t is time.
Given : Current : [tex]I=3A[/tex]
Potential difference : [tex]V=12\ V[/tex]
Time : [tex]t=2\ hr=2(3600)\text{ seconds}=7200\text{ seconds}[/tex]
Now, [tex]Q=3(7200)=21,600\ C[/tex]
Then, [tex]W=(21600)(12)=259,200\text{ Joules}=259.2\text{ KJ}[/tex]
Hence, the work done = 259.2 KJ
The work done by the battery charger is 259.2 kilojoules.
Explanation:To calculate the work done by the battery charger, we can use the formula:
Work (W) = Power (P) x Time (t)
In this case, the power is given as 3A (current) and 12 Volts (voltage), and the time is given as 2 hours. We need to convert the time to seconds:
2 hours = 2 x 60 x 60 = 7200 seconds
Now we can substitute the values into the formula:
W = 3A x 12V x 7200s = 259,200 Joules
To convert the work into kilojoules, we divide by 1000:
W = 259,200 J = 259.2 kJ
Therefore, the work done by the battery charger is 259.2 kilojoules.
A positive charge (q = +6.0 µC) starts from point A in a constant electric field and accelerates to point B. The work done by the electric force is WAB = +2.2 × 10-3 J. Determine the potential difference VB - VA between the two points. Be sure to include the proper algebraic sign.
Vb - Va = -366.7 V.
Vab = Va - Vb, the potential of a with respect to b, is equal to the work done by the electric force when a unit of charge moves from a to b, it is given by:
Vab = Va - Vb = Wab/q,
So, in order to determinate the potential difference Vb - Va we have to multiply by -1 both side of the equation above:
- (Va - Vb) = - (Wab/q)
Resulting
Vb - Va = -(Wab/q)
Given a positive charge q = 6.0μC = 6.0x10⁻⁶C, Wab = 2.2x10⁻³J. Determine Vb - Va.
Vb - Va = - (2.2x10⁻³J/6.0x10⁻⁶C)
Vb - Va = -366.7 J/C = -366.7 V
The potential difference between point A and point B is 366.67 V; point B is at a lower potential than point A as derived from the work done WAB and the charge q.
Explanation:The potential difference (also known as voltage) between two points in an electric field, such as point A to point B (VB - VA), can be calculated using the work done by or against the electric field to move a charge q from point A to point B. The formula for work done by the electric force is W = q(VB - VA), and given that work WAB = +2.2 × 10-3 J and the charge q = +6.0 µC (or +6.0 × 10-6 C), we can rearrange the formula to solve for the potential difference: VB - VA = WAB / q.
The calculation yields VB - VA = +2.2 × 10-3 J / (+6.0 × 10-6 C) which equals +366.67 V. Therefore, the electric potential difference between point A and point B is 366.67 V, with point B being at a lower potential than point A since the charge is positive and the work done is positive, indicating that it has moved in the direction of the electric field, from higher to lower potential.
A swimming duck paddles the water with its feet once every 1.6 s, producing surface waves with this period. The duck is moving at constant speed in a pond where the speed of surface waves is 0.32 m/s, and the crests of the waves ahead of the duck are spaced 0.12 m apart. (a) What is the duck's speed? (b) How far apart are the crests behind the duck?
Answer:
a)
0.245 m/s
b)
0.904 m
Explanation:
a)
[tex]v_{d}[/tex] = speed of duck ahead of wave
[tex]v_{s}[/tex] = speed of surface wave = 0.32 m/s
T = time for paddling = 1.6 s
d = spacing between the waves = 0.12 m
speed of duck ahead of wave is given as
[tex]v_{d}[/tex] = [tex]v_{s}[/tex] - [tex]\frac{d}{T}[/tex]
[tex]v_{d}[/tex] = 0.32 - [tex]\frac{0.12}{1.6}[/tex]
[tex]v_{d}[/tex] = 0.245 m/s
b)
[tex]v_{w}[/tex] = speed of wave behind the duck
speed of wave behind the duck is given as
[tex]v_{w}[/tex] = [tex]v_{s}[/tex] + [tex]v_{d}[/tex]
[tex]v_{w}[/tex] = 0.32 + 0.245
[tex]v_{w}[/tex] = 0.565 m/s
D = spacing between the crests
spacing between the crests is given as
D = [tex]v_{w}[/tex] T
D = (0.565) (1.6)
D = 0.904 m
A GPS tracking device is placed in a police dog to monitor its whereabouts relative to the police station. At time t1 = 23 min, the dog's displacement from the station is 1.2 km, 33° north of east. At time t2 = 57 min, the dog's displacement from the station is 2.0 km, 75° north of east. Find the magnitude and direction of the dog's average velocity between these two times.
Answer:
Explanation:
Let east be towards X-axis ant north be Y-axis. Let initial position of Dog be at
A . O be the police station (centre). Vector OA can be written as follows
OA = 1.2 Cos 33 + 1.2 Sin 33
O B =2 Cos 75 + 2 Sin 75.
Displacement A → B = OB - OA =2Cos75i +2 Sin 75j -1.2 Cos 33i -1.2 Sin 33j
= 2 x .2588 i+ 2x .966j - 1.2 x .8387i - 1.2 x .5446j = .5176i + 1.932 j- 1.0064i - .65356j
= -.4888 i + 1.27844j
Magnitude of displacement =√( .4888)² + ( 1.27844)²
= 1.405 km
Average velocity =1.405 / 57-23 km / min = 1.405 /34 x 60 =2.48 km/h
angle with x-axis ( east towards north ) ∅
Tan∅ =- 1.27844/.4888 = - 2.615
∅ = -69° or 111° towards north from east or 21° towards west from north.
Suppose 8.50 ✕ 10^5 J of energy are transferred to 1.63 kg of ice at 0°C. The latent heat of fusion and specific heat of water are Lf = 3.33 ✕ 105 J/kg and c = 4186 J (kg · °C) . HINT (a) Calculate the energy (in J) required to melt all the ice into liquid water. (Enter your answer to at least three significant figures.) J (b) How much energy (in J) remains to raise the temperature of the liquid water? (Enter your answer to at least three significant figures.) J (c) Determine the final temperature of the liquid water in Celsius. °C
Answer:
(a) 5.43 x 10⁵ J
(b) 3.07 x 10⁵ J
(c) 45 °C
Explanation:
(a)
[tex]L_{f}[/tex] = Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg
m = mass of ice = 1.63 kg
[tex]Q_{f}[/tex] = Energy required to melt the ice
Energy required to melt the ice is given as
[tex]Q_{f}[/tex] = m [tex]L_{f}[/tex]
[tex]Q_{f}[/tex] = (1.63) (3.33 x 10⁵)
[tex]Q_{f}[/tex] = 5.43 x 10⁵ J
(b)
E = Total energy transferred = 8.50 x 10⁵ J
Q = Amount of energy remaining to raise the temperature
Using conservation of energy
E = [tex]Q_{f}[/tex] + Q
8.50 x 10⁵ = 5.43 x 10⁵ + Q
Q = 3.07 x 10⁵ J
(c)
T₀ = initial temperature = 0°C
T = Final temperature
m = mass of water = 1.63 kg
c = specific heat of water = 4186 J/(kg °C)
Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J
Using the equation
Q = m c (T - T₀)
3.07 x 10⁵ = (1.63) (4186) (T - 0)
T = 45 °C
(a) The energy to melt the ice is 5.43 × 10^5 J.
(b) The remaining energy is 3.07 × 10^5 J used to raise the temperature of the water.
(c) The final temperature of the water is approximately 45 °C.
(a) Energy to Melt the Ice:
To find the energy required to melt the ice, we use the formula Q_melt = mass * latent heat of fusion.
Given:
Mass of the ice = 1.63 kg
Latent heat of fusion (Lf) = 3.33 × 10^5 J/kg
Q_melt = 1.63 kg * (3.33 × 10^5 J/kg) = 5.43 × 10^5 J
(b) Energy to Raise the Temperature of Water:
The remaining energy after melting is used to raise the temperature of the water. Subtracting Q_melt from the total energy transferred gives Q_raise.
Q_raise = 8.50 × 10^5 J - 5.43 × 10^5 J = 3.07 × 10^5 J
(c) Final Temperature:
To find the temperature increase, we use the formula ΔT = Q_raise / (mass * specific heat).
Given:
Mass of the water = 1.63 kg
Specific heat of water (c) = 4186 J/(kg · °C)
ΔT = (3.07 × 10^5 J) / (1.63 kg * 4186 J/(kg · °C)) ≈ 45 °C
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The question probable may be:
Suppose 8.50 ✕ 10^5 J of energy are transferred to 1.63 kg of ice at 0°C. The latent heat of fusion and specific heat of water are Lf = 3.33 ✕ 1065 J/kg and c = 4186 J (kg · °C) . HINT
(a) Calculate the energy (in J) required to melt all the ice into liquid water. (Enter your answer to at least three significant figures.)
(b) How much energy (in J) remains to raise the temperature of the liquid water? (Enter your answer to at least three significant figures.)
c) Determine the final temperature of the liquid water in Celsius.
Assume the amplitude of the electric field in a plane electromagnetic wave is E1 and the amplitude of the magnetic field is B1. The source of the wave is then adjusted so that the amplitude of the electric field doubles to become 2E1. (i) What happens to the amplitude of the magnetic field in this process? It becomes four times larger. It becomes two times larger. It can stay constant. It becomes one-half as large. It becomes one-fourth as large. (ii) What happens to the intensity of the wave? It becomes four times larger. It becomes two times larger. It can stay constant. It becomes one-half as large. It becomes one-fourth as large. Need Help?
Answer:
I) It becomes two times larger
II) It becomes four times larger
Explanation:
I) Electric field is directly proportional to Magnetic field and as such, if one is increased, the other is also increased by the same proportion.
The formula is given as
[tex]E_{1} = cB_{1}[/tex]
c = speed of light, therefore
[tex]2E_{1} = c2B_{1}[/tex]
II) Intensity of is proportional to the square of amplitude, as such
If amplitude is doubled ([tex]X2^{2}[/tex]), intensity is [tex]X4^{}[/tex]
This means the intensity becomes four times bigger.
A student drops a ball from the top of a 10-meter tall building. The ball leaves the thrower's hand with a zero speed. What is the speed of the ball at the moment just before it hits the ground?
Answer:
14 m/s
Explanation:
u = 0, h = 10 m, g = 9.8 m/s^2
Use third equation of motion
v^2 = u^2 + 2 g h
Here, v be the velocity of ball as it just strikes with the ground
v^2 = 0 + 2 x 9.8 x 10
v^2 = 196
v = 14 m/s
Which of the following is a property of an object not a result of something happening to the object? Select one: weight a. b. inertia C. momentum d. force
Answer:
option (b)
Explanation:
Th weight of the body is equal to the product of mass of the body and acceleration due to gravity. So, it depends on the acceleration due to gravity.
Inertia is the inherent property of the body which always resists any change in the body. It does not depend on any factor. mass is the measure of inertia.
Momentum is the product of mass of body and the velocity of the body.
Force is the product of mass and the acceleration of the body.
A 200 N trash can is pulled across the sidewalk by a person at constant speed by a force of 75 N. What is the coefficient of friction between the trash can and the sidewalk in the problem above.
Answer:
μ = 0.375
Explanation:
F = Applied force on the trash can = 75 N
W = weight of the trash can = 200 N
f = frictional force acting on trash can
Since the trash can moves at constant speed, force equation for the motion of can is given as
F - f = 0
75 - f = 0
f = 75 N
μ = Coefficient of friction
frictional force is given as
f = μ W
75 = μ (200)
μ = 0.375
A certain gasoline engine has an efficiency of 35.3%. What would the hot reservoir temperature be for a Carnot engine having that efficiency, if it operates with a cold reservoir temperature of 160°C? (°C)
Answer:
The temperature of hot reservoir is 669.24 K.
Explanation:
It is given that,
Efficiency of gasoline engine, [tex]\eta=35.3\%=0.353[/tex]
Temperature of cold reservoir, [tex]T_C=160^{\circ}C=433\ K[/tex]
We need to find the temperature of hot reservoir. The efficiency of Carnot engine is given by :
[tex]\eta=1-\dfrac{T_C}{T_H}[/tex]
[tex]T_H=\dfrac{T_C}{1-\eta}[/tex]
[tex]T_H=\dfrac{433}{1-0.353}[/tex]
[tex]T_H=669.24\ K[/tex]
So, the temperature of hot reservoir is 669.24 K. Hence, this is the required solution.
A parallel-plate capacitor is formed from two 6.0-cm-diameter electrodes spaced 2.0 mm apart. The electric field strength inside the capacitor is 1.0×106 N/C1.0×106 N/C. What is the charge (in nC) on each electrode?
Answer:
2.5 x 10^-8 C
Explanation:
Diameter = 6 cm, radius = 3 cm = 0.03 m, d = 2 mm = 2 x 10^-3 m
E = 1 x 10^6 N/C, q = ?
q = C V
As we know that, V = E x d and C = ∈0 A / d
q = ∈0 x A x E x d / d
q = ∈0 x A x E
q = 8.854 x 10^-12 x 3.14 x 0.03 x 0.03 x 1 x 10^6
q = 2.5 x 10^-8 C
Electric field exerts a force on all charged particles. The charge on the electrode is 2.505 x 10⁻⁸ C.
What is an electric field?An electric field can be thought to be a physical field that surrounds all the charged particles and exerts a force on all of them.
Given to us
Plate dimensions diameter, d = 6 cm
Area of the plate, A = πr² = π(0.03)² = 0.00283 m²
Distance between the two plates, d = 2 mm = 0.002 m
Electric field strength, E = 1.0 x 10⁶ N/C
We know that electric field inside a parallel plate capacitor is given as,
[tex]E = \dfrac{Q}{A\epsilon_0}[/tex]
Substitute the value,
[tex]1 \times 10^6 = \dfrac{Q}{0.00283 \times 8.854 \times 10^{-12}}[/tex]
Q = 2.505 x 10⁻⁸ C
Hence, the charge on the electrode is 2.505 x 10⁻⁸ C.
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Which of the following is NOT true for a spherical concave mirror? a. it cannot produce virtual images
b. its focal length is half its radius
c. its focal length is positive
d. it can produce upright images
Answer:
Its focal length is positive
Explanation:
A concave mirror is shown in attached figure. The distance from the pole to the focus of the mirror is called its focal length. Spherical mirrors are a part of a sphere.
As per conventions, we know that the axis opposite to x axis is taken as negative.
So, it is clear that the focal length of spherical concave mirror is negative.
Hence, the incorrect option is (c) " its focal length is positive".
A rope pulls a 82.5 kg skier at a constant speed up a 18.7° slope with μk = 0.150. How much force does the rope exert?
Answer:
374 N
Explanation:
N = normal force acting on the skier
m = mass of the skier = 82.5
From the force diagram, force equation perpendicular to the slope is given as
N = mg Cos18.7
μ = Coefficient of friction = 0.150
frictional force is given as
f = μN
f = μmg Cos18.7
F = force applied by the rope
Force equation parallel to the slope is given as
F - f - mg Sin18.7 = 0
F - μmg Cos18.7 - mg Sin18.7 = 0
F = μmg Cos18.7 + mg Sin18.7
F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7
F = 374 N
A bat can detect small objects, such as an insect, whose size is approximately equal to one wavelength of the sound the bat makes. If bats emit a chirp at a frequency of 7.84 104 Hz, and if the speed of sound in air is 343 m/s, what is the smallest insect a bat can detect?
Answer:
0.4375 cm
Explanation:
f = frequency of the chirp emitted by the bats = 7.84 x 10⁴ Hz
v = speed of sound in air = 343 m/s
λ = smallest wavelength = size of the smallest insect a bat can detect
Using the equation
v = f λ
inserting the values
343 = (7.84 x 10⁴) λ
λ = [tex]\frac{343}{7.84\times 10^{4}}[/tex]
λ = 43.75 x 10⁻⁴ m
λ = 0.4375 cm
In a circus performance, a large 3.0 kg hoop with a radius of 1.3 m rolls without slipping. If the hoop is given an angular speed of 6.8 rad/s while rolling on the horizontal and is allowed to roll up a ramp inclined at 24◦ with the horizontal, how far (measured along the incline) does the hoop roll? The acceleration of gravity is 9.81 m/s2 .
Answer:
The distance is 19.58 m.
Explanation:
Given that,
Mass = 3.0 kg
Radius = 1.3 m
Angular speed = 6.8 rad/s
Angle = 24°
Acceleration of gravity = 9.81 m/s²
We need to calculate the distance
Using formula of kinetic energy
Total Initial kinetic energy,
[tex]K.E = \dfrac{1}{2}(mv^2+I\omega^2)[/tex]
[tex]K.E = \dfrac{1}{2}(mv^2+mr^2\omega^2)[/tex]
[tex]K.E =\dfrac{(mr^2+mr^2)\omega^2}{2}[/tex]
[tex]K.E =mr^2\omega^2[/tex]....(I)
Now, Total potential energy
[tex]P.E=mgs\sin\theta[/tex]
[tex]P.E=mgs\sin24^{\circ}[/tex]....(II)
Equating equation (I) and (II)
[tex]mr^2\omega^2==mgs\sin24^{\circ}[/tex]
[tex](1.3)^2\times6.8^2=9.81\times s\times\sin24^{\circ}[/tex]
[tex]s = \dfrac{1.3^2\times6.8^2}{9.81\times\sin24^{\circ}}[/tex]
[tex]s=19.58\ m[/tex]
Hence, The distance is 19.58 m.
Which of the following represents a fully plastic collision? 1. Two billiard balls hit each other then move in opposite directions. 2. Two train cars hit then detach. 3. A ball hits a wall with no loss of energy. 4. A bullet hits a block and becomes embedded. 5. None of these.
Answer:
Option (4)
Explanation:
There are two types of collision.
Perfectly elastic collision: the collision in which the momentum and kinetic energy is conserved. There is no loss of energy in other forms of energy.
Perfectly plastic collision: The collision in which the momentum is conserved and kinetic energy is not conserved. The two bodies stick after the collision.
Here, the bullet hits the block and then embedded in the block, it is the example of plastic collision.