Explanation:
A nucleus that absorbs 11.45δ is less shielded than a nucleus that absorbs at 4.13δ.
the nucleus that absorbs at 11.45δ requires weaker applied field strength to come into resonance than the nucleus that absorbs at 4.13δ.
A nucleus absorbing at 4.13 δ is more shielded than one absorbing at 11.45 δ because it is in a weaker local magnetic field and resonates at a lower frequency.
Explanation:In the context of nuclear magnetic resonance (NMR) spectroscopy, the chemical shift is given in units of delta (δ) which represents the resonance frequency of a nucleus relative to a standard reference compound. When a nucleus is surrounded by a dense cloud of electrons, it is considered to be ‘shielded’. A shielded nucleus is influenced by a smaller local magnetic field because the electrons repel some of the external magnetic field. As a result, shielded nuclei resonate at a lower frequency (higher δ values) when compared to de-shielded nuclei. Therefore, a nucleus that absorbs at 4.13 δ is more shielded than a nucleus that absorbs at 11.45 δ because the latter is in a stronger local magnetic field and is, hence, de-shielded.
A load consists of a 70-Ω resistor in parallel with a 90-μF capacitor. If the load is connected to a voltage source vs(t) = 160cos 2000t, find the average power delivered to the load.
Answer:
Power delivered by the source will be 182.912 watt
Explanation:
We have given a load is consist of a resistor of 70 ohm in parallel with [tex]90\mu F[/tex] capacitance
Voltage source is given [tex]v_s(t)=160cos(2000t)[/tex]
So maximum value of voltage source is 160 volt
So rms value [tex]v_{r}=\frac{v_m}{\sqrt{2}}=\frac{160}{1.414}=113.154volt[/tex]
We know that average power delivered by the source will be equal to average power absorbed by the resistor
So power absorbed by the resistor [tex]P=\frac{v_r^2}{R}=\frac{113.154^2}{70}=182.912watt[/tex]
So power delivered by the source will be 182.912 watt
A 20 μF capacitor has previously been charged up to contain a total charge of Q=100 μC on it. The capacitor is then discharged by connecting it directly across a 100kΩ resistor. Calculate the charge remaining on the capacitor exactly 3.00 seconds after being connected to the resistor.
Answer:
Q= 22.3 μC
Explanation:
Given that
C= 20 μF
Qo= 100 μC
R= 100 kΩ
t= 3 s
T= R C
T= 100 x 1000 x 20 x 10⁻⁶ s
T=2 s
We know that charge on the capacitor is given as
[tex]Q=Q_0e^{\dfrac{-t}{T}}[/tex]
[tex]Q=100\times 10^{-6}\times e^{\dfrac{-3}{2}}[/tex]
Q= 0.0000223 C
Q= 22.3 μC
Final answer:
The charge remaining on the capacitor after 3 seconds of discharging is approximately 22.31 µC, calculated using the formula for exponential decay of charge in an RC circuit.
Explanation:
To calculate the charge remaining on a 20 µF capacitor after 3 seconds of discharging through a 100 kΩ resistor, we'll use the formula for exponential decay of charge in an RC circuit, which is Q(t) = Q0 e-(t/RC), where Q0 is the initial charge, t is the time, and RC is the time constant of the circuit. The time constant (RC) for this circuit can be calculated as (100 × 103 Ω)(20 × 10-6 F) = 2 seconds. Plugging in the given values, we find that Q(3s) = 100 µC × e-(3/2), which after calculation gives a remaining charge of approximately 22.31 µC.
The information on a one-gallon paint can is that the coverage, when properly applied, is 440 ft2. One gallon is 231 in3. What is the average thickness of the paint in such an application
Answer:
t= 0.00364 in
Explanation:
Given that
The volume of the paint ,V= 231 in³
The surface area ,A = 440 ft²
We know that
1 ft = 12 in
That is why
A= 440 x 12 x 12 in²
A= 63360 in²
Lets take the thickness of the paint = t in
We know that
V= A t
[tex]t=\dfrac{V}{A}[/tex]
[tex]t=\dfrac{231}{63360}\ in[/tex]
t= 0.00364 in
Therefore the thickness ,t= 0.00364 in
To find the average thickness of the paint, divide the volume of paint by the area covered. This results in an average thickness of approximately 0.003645 inches, or about 0.000304 feet when the paint is applied as per the can's instructions.
Explanation:To find the average thickness of the paint applied, we need to calculate the volume of paint used per unit of area covered. The volume of one gallon of paint is 231 cubic inches, and the coverage is 440 square feet. To convert the coverage to square inches, we multiply by the number of square inches in a square foot, which is 144 (12 inches × 12 inches).
The total coverage in square inches is 440 ft2 × 144 in2/ft2 = 63,360 in2. We can then find the average thickness by dividing the volume of paint by the area covered: Thickness (in inches) = Volume (in cubic inches) / Area (in square inches).
This gives us an average thickness of 231 in3 / 63,360 in2, which simplifies to approximately 0.003645 inches. This can also be converted to feet by knowing that 1 inch is equal to 1/12 of a foot, so the average paint thickness is roughly 0.003645/12 feet when applied as instructed on the paint can.
The falling object in Example 2 satisfies the initial value problem dv/dt =9.8−(v/5), v(0) =0. (a) Find the time that must elapse for the object to reach 98% of its limiting velocity. (b) How far does the object fall in the time found in part (a)?
Answer:
a. [tex]t=19.56 s[/tex]
b.[tex]d=718.34[/tex]
Explanation:
The solution to the differential equation
[tex]\dfrac{dv}{dt}=9.8-\dfrac{v}{5}[/tex]
is the exponential function
[tex]v(t)=ce^{-0.2t}+49[/tex]
and we find [tex]c[/tex] from the initial condition [tex]v(0)=0:[/tex]
[tex]0=ce^{-0.2*0}+49\\\\0=c+49\\\\c=-49[/tex]
Therefore, we have
[tex]v(t)=-49e^{-0.2t}+49[/tex]
[tex]\boxed{ v(t)=49(1-e^{-0.2t})}[/tex]
Part A:
The maximum velocity that the object can reach is 49 (which the maximum value [tex]v(t)[/tex] can have).
Now, 98% of 49 is 48.02; therefore,
[tex]48.02=49(1-e^{-0.2t})[/tex]
[tex]0.98=1-e^{-0.2t}[/tex]
[tex]e^{-0.2t}=0.02[/tex]
[tex]\boxed{t=19.56 s}[/tex]
Part B:
The distance traveled is the integral of the speed:
[tex]d=\int_0^{19.56}v(t)*dt[/tex]
[tex]d=\int^{19.56}_0 {49(1-e^{-0.2t})} \, dt[/tex]
[tex]d=49[t+5e^{-0.2t}]_0^{19.56}[/tex]
[tex]\boxed{d=718.34}[/tex]
To find the time that must elapse for the object to reach 98% of its limiting velocity, we need to solve the differential equation. We can then find the distance the object falls by integrating the velocity function with respect to time.
Explanation:(a) Finding the time to reach 98% of the limiting velocityTo find the time it takes for the object to reach 98% of its limiting velocity, we need to solve the differential equation. First, we separate the variables by writing it as:
dv / (9.8 - (v/5)) = dt
Next, we integrate both sides:
∫ (1 / (9.8 - (v/5))) dv = ∫ dt
After evaluating the integrals, we can solve for v:
v = 49 - 49e^(-t/5)
Substituting v with 0.98 times the limiting velocity (which is 49), we can solve for t:
49 - 49e^(-t/5) = 0.98 * 49
Solving this equation will give us the time it takes for the object to reach 98% of its limiting velocity.
(b) Finding the distance the object fallsTo find the distance the object falls, we need to integrate the velocity function, v, with respect to time:
∫ v dt
By evaluating the integral, we can calculate the distance the object falls in the time found in part (a).
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A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.9 km , the jet is moving with a speed of 400 m/s.
a. What is the jet's acceleration, assuming it to be a constant acceleration?
b-Is your answer reasonable ? Explain.
The jet's acceleration is calculated as 7.14 m/s² using a kinematic equation. This value is reasonable for a jet plane given its powerful engines. High-speed aircraft often have high accelerations.
To determine the jet's acceleration, we can apply the kinematic equation:
vf² = vi² + 2aΔx
where:
vf is the final velocity (400 m/s)vi is the initial velocity (300 m/s)a is the accelerationΔx is the displacement (4.9 km or 4900 m)Solving for acceleration a:
400² = 300² + 2a(4900)
160000 = 90000 + 9800a
70000 = 9800a
a = 7.14 m/s²
Reasonableness of the Answer
Yes, the answer is reasonable. Accelerations for jet planes are typically high due to the powerful engines they possess. An acceleration of 7.14 m/s² aligns well with the capabilities of high-speed aircraft.
As we learned in class if a material’s crystal structure is known a theoretical density, ????, can be computed from a tiny fundamental unit using the formula ???? = ???????? ????c???????? Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its theoretical density with its experimentally measured density of 7.87 g/cm^3.
Answer : Yes, theoretical density can be computed from a tiny fundamental unit using the formula [tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex].
Explanation :
Nearest neighbor distance, r = [tex]0.124nm=1.24\times 10^{-8}cm[/tex] [tex](1nm=10^{-7}cm)[/tex]
Atomic mass (M) = 55.85 g/mol
Avogadro's number [tex](N_{A})=6.022\times 10^{23} mol^{-1}[/tex]
For BCC = Z = 2
Given density = [tex]7.87g/cm^3[/tex]
First we have to calculate the cubing of edge length of unit cell for BCC crystal lattice.
For BCC lattice : [tex]a^3=(\frac{4r}{\sqrt{3}})^3=(\frac{4\times 1.24\times 10^{-8}cm}{\sqrt{3}})^3=2.35\times 10^{-23}cm^3[/tex]
Now we have to calculate the density of unit cell for BCC crystal lattice.
Formula used :
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex] .............(1)
where,
[tex]\rho[/tex] = density
Z = number of atom in unit cell (for BCC = 2)
M = atomic mass
[tex](N_{A})[/tex] = Avogadro's number
a = edge length of unit cell
Now put all the values in above formula (1), we get
[tex]\rho=\frac{2\times (55.85g/mol)}{(6.022\times 10^{23}mol^{-1}) \times (2.35\times 10^{-23}Cm^3)}=7.89g/Cm^{3}[/tex]
From this information we conclude that, the given density is approximately equal to the given density.
Yes, theoretical density can be computed from a tiny fundamental unit using the formula [tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex].
Car 1 goes around a level curve at a constant speed of 65 km/h . The curve is a circular arc with a radius of 95 m . Car 2 goes around a different level curve at twice the speed of Car 1. How much larger will the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration
Answer:
The radius of the curve that Car 2 travels on is 380 meters.
Explanation:
Speed of car 1, [tex]v_1=65\ km/h[/tex]
Radius of the circular arc, [tex]r_1=95\ m[/tex]
Car 2 has twice the speed of Car 1, [tex]v_2=130\ km/h[/tex]
We need to find the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration. We know that the centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{r}[/tex]
According to given condition,
[tex]\dfrac{v_1^2}{r_1}=\dfrac{v_2^2}{r_2}[/tex]
[tex]\dfrac{65^2}{95}=\dfrac{130^2}{r_2}[/tex]
On solving we get :
[tex]r_2=380\ m[/tex]
So, the radius of the curve that Car 2 travels on is 380 meters. Hence, this is the required solution.
Final answer:
For Car 2 to have the same centripetal acceleration while traveling at twice the speed of Car 1, it must travel on a curve with a radius that is four times larger, which would be 380 meters.
Explanation:
The centripetal acceleration (ac) of a car going around a level curve at a constant speed is given by the formula ac = v2 / r, where v is the speed of the car and r is the radius of the circular path. If Car 1 is traveling at 65 km/h and has a centripetal acceleration on a curve with a radius of 95 m, and Car 2 is traveling at twice the speed of Car 1, we are asked to find the required radius of the curve for Car 2 to maintain the same centripetal acceleration.
To keep the same centripetal acceleration for Car 2, which is moving at twice the speed, the radius r2 must be increased proportionally to the square of the speed ratio. Since Car 2 is traveling at twice the speed, the radius must be increased by a factor of 22, or 4 times larger than that for Car 1. Therefore, if Car 1's radius is 95 m, Car 2's radius needs to be 95 m * 4, which is 380 m.
A heavy neutral atom, such as iron, produces many spectral lines compared to light elements like hydrogen and helium. Why?
Answer:
Due to a larger number of electrons in the heavy atoms.
Explanation:
Spectral lines are caused by the emission of light by electrons when they transit from a higher energy state(excited state) to a lower energy state.
Hence, the more electrons an atom has, the more the emission and spectral lines. The less the electrons an atom has, the less the emission and spectral lines.
Therefore, heavy nuclei (which contain more electrons) such as Iron will emit more light and so will have more spectral lines than light atoms like Hydrogen and Helium.
Answer:
more electrons in heavy atoms
Explanation:
A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field zero?
Final answer:
To find where the electric field is zero between two point charges (5.0 μC and -4.0 μC), set up an equation based on Coulomb's Law and solve it considering that the point lies closer to the smaller magnitude charge.
Explanation:
The question requires the application of concepts from electrostatics, specifically the properties of the electric field generated by point charges. To determine the point at which the electric field is zero, one must consider the magnitudes of the charges and their distances from the point of interest. Since electric fields generated by individual charges superpose, the point where the electric field is zero is where the electric field due to one charge balances out the electric field due to the other charge.
For two charges Q1 (+5.0 μC) and Q2 (-4.0 μC) separated by 50 cm (0.5 meters), the electric field is zero at a point that is closer to the smaller magnitude charge. Let's call the distance from Q1 to the point where the field is zero 'd'. Using Coulomb's Law and the concept of superposition, we can set up the equation:
E1 = E2
|k * Q1 / d^2| = |k * Q2 / (0.5 - d)^2|
|Q1/d^2| = |Q2/(0.5 - d)^2|
Substituting Q1 and Q2 with their respective values, we can solve for 'd' using algebraic methods to find the point on the meter stick where the electric field equals zero. However, since the solution requires calculations, it is important to apply the proper mathematical steps to reach the correct conclusion.
Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mass of 6.90 10-2 kg and a charge of +5.35 10-5 C. The particle has a speed of 2.00 m/s on surface A. A nonconservative outside force is applied to the particle, and it moves to surface B, arriving there with a speed of 3 m/s. How much work is done by the outside force in moving the particle from A to B?
Explanation:
Formula for the change in potential energy from point A to B is as follows.
P.E = [tex](V_{A} - V_{B}) \times q[/tex]
Putting the given values into the above formula as follows.
P.E = [tex](V_{A} - V_{B}) \times q[/tex]
= [tex](5650 - 7850) \times 5 \times 10^{-5}[/tex]
= -0.11 J
Now, we will calculate the change in kinetic energy as follows.
K.E = [tex]0.5 \times m \times (v^{2}_{B} - v^{2}_{A})[/tex]
= [tex]0.5 \times 6.90 \times 10^{-2} \times (2^{2} - 1^{2})[/tex]
= 0.1035 J
Therefore, supplied difference by the outside force is calculated as follows.
0.1035 J - (-0.11) J
= 0.2135 J
Thus, we can conclude that work is done by the outside force in moving the particle from A to B is 0.2135 J.
The total work done by the external force in moving the charged particle from equipotential surface A to B is 1.194 J.
Explanation:The work done by an external non-conservative force in moving a charged particle from one equipotential surface to another can be calculated in two parts. Firstly, we calculate the difference in electrical potential energy between the two points. This can be calculated using the formula ΔU = qΔV, where ΔV = Vb - Va. Following the given question, q = +5.35 x 10^-5 C, ΔV = 7850V - 5650V. Therefore, ΔU = 5.35 x 10^-5(7850 - 5650) = 1.1765 J.
Secondly, we calculate the change in kinetic energy which is given by ΔK = ½m(vb² - va²), where m = 6.90 x10^-2 kg, va = 2 m/s, and vb = 3 m/s. Therefore, ΔK = 0.5 * 6.90 x10^-2 * (3^2 - 2^2) = 0.0175 J.
Summing both gives the total work done by the external force on the particle: W = ΔU + ΔK = 1.1765 J + 0.0175 J = 1.194 J.
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In an experiment to measure the acceleration due to gravity g, two independent equally reliable measurements gave 9.67 m/s2 and 9.88 m/s2. Determine (i) the percent difference of the measurements (ii) the percent error of their mean. [Take the theoretical value of g to be 9.81 m/s
Answer:
i. +/- 1.43% and +/- 0.71% ii. +/- 0.33%
Explanation:
[tex]% Error = \frac{Error}{Measurement}* 100%\\[/tex]
An orange loses 1.2 kJ of heat as it cools per °C drop in its temperature. What is the amount of heat loss from the orange per °F drop in its temperature?
To solve this problem we will apply the conversion rate between Celcius and Fahrenheit degrees. We will use the direct relationship clearly and not the added degrees of scale conversion. We know from the statement that the orange loses heat at the rate of
[tex]Q = 1.2kJ/\°C[/tex]
We have the conversion to °F is given as
[tex]T (\°F) = 1.8T+32[/tex]
Calculate the amount of heat loss from orange per °F
[tex]Q = \frac{1.2}{1.8}[/tex]
[tex]Q = 0.667kJ/\°F[/tex]
Therefore the amount of heat loss from the orange per °F drop in its temperature is 0.667kJ/°F
The heat loss from an orange per °F drop is 0.67 kJ, calculated by taking 1.2 kJ per °C drop and dividing it by 1.8 to convert it to Fahrenheit,
Explanation:The heat loss from the orange per °F drop in its temperature can be found by converting 1.2 kJ lost per 1 °C drop in temperature to kJ lost per 1 °F drop. This can be achieved using the formula that 1 °C equals 1.8 °F.
Therefore, the heat loss per degree Fahrenheit will be less than the heat loss per degree Celsius. We calculate this as follows:
(1.2 kJ / °C) / 1.8 = 0.67 kJ per °F.
So for every degree Fahrenheit that the orange cools, it will lose 0.67 kilojoules of heat.
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A fox locates its prey, usually a mouse, under the snow by slight sounds the rodents make. The fox then leaps straight into the air and burrows its nose into the snow to catch its next meal. In your calculations ignore the effects of air resistance. 1) If a fox jumps to a height of 81.0 cm. Calculate the speed at which the fox leaves the snow. (Express your answer to three significant figures.)
Answer:
The fox leaves the snow at 3.99 m/s
Explanation:
Hi there!
The equation of height and velocity of the fox are the following:
h = h0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
h = height of the fox at a time t.
h0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s considering the upward direction as positive)
v = velocity of the fox at a time t.
We know that at the maximum height of the fox, its velocity is zero, so using the equation of velocity we can obtain an expression of v0 in function of t:
v = v0 + g · t
At the maximum height, v = 0
0 = v0 + g · t
Solving for v0:
-g · t = v0
We know the maximum height of the fox, 0.810 m. So, using the equation of height and replacing v0 by (-g · t), we can obtain the time at which the fox is at the maximum height and then calculate the initial velocity:
h = h0 + v0 · t + 1/2 · g · t²
When t is the time at which the fox is at the maximum height, h = 0.810 m and v0 = (-g · t). Let´s consider the ground as the origin of the frame of reference so that h0 = 0.
0.810 m = (-g · t) · t + 1/2 · g · t²
0.810 m = -g · t² + 1/2 · g · t²
0.810 m = - 1/2 · g · t²
t² = -2 · 0.810 m / -9.81 m/s²
t = 0.406 s
And the initial velocity will be:
v0 = -g · t
v0 = -(-9.81 m/s²) · 0.406 s
v0 = 3.99 m/s
The fox leaves the snow at 3.99 m/s
The speed at which the fox leaves the snow is approximately 4.52 m/s.
Explanation:To calculate the speed at which the fox leaves the snow, we can use the principle of conservation of energy. Since the fox jumps straight up, its initial vertical velocity is zero. The final velocity can be calculated using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. In this case, the displacement is the height jumped, which is 81.0 cm or 0.81 m. The acceleration is equal to g, the acceleration due to gravity (approximately 9.8 m/s^2). Solving for v, we find that the speed at which the fox leaves the snow is approximately 4.52 m/s.
Suppose that one sphere is held in place; the other sphere, with mass 1.40 g, is shot away from it. What minimum initial speed would the moving sphere need to escape completely from the attraction of the fixed sphere?
Answer:
The minimum initial speed is 20.0 m/s.
Explanation:
Given that,
Mass of sphere = 1.40 g
Suppose a system of two small spheres, one carrying a charge of 1.70 μC and the other a charge of -4.40 μC , with their centers separated by a distance of 0.240 m .
We need to calculate the potential energy
Using formula of potential energy
[tex]P.E=\dfrac{kq_{1}q_{2}}{d}[/tex]
Put the value into the formula
[tex]P.E=\dfrac{9\times10^{9}\times1.70\times10^{-6}\times4.40\times10^{-6}}{0.240}[/tex]
[tex]P.E=-280.5\times10^{-3}\ J[/tex]
We need to calculate the minimum initial speed
Using formula of energy
[tex]K.E=-P.E[/tex]
[tex]\dfrac{1}{2}mv^2=-280.5\times10^{-3}[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times1.40\times10^{-3}\times v^2=280.5\times10^{-3}[/tex]
[tex]v=\sqrt{\dfrac{2\times280.5\times10^{-3}}{1.40\times10^{-3}}}[/tex]
[tex]v=20.0\ m/s[/tex]
Hence, The minimum initial speed is 20.0 m/s.
What is a constellation as astronomers define it today? What does it mean when an astronomer says, "I saw a comet in Capricorn last night?"
A constellation, in astronomy, is a conventional grouping of stars, whose position in the night sky is apparently invariable. The peoples, generally of ancient civilizations, decided to link them through imaginary strokes, thus creating virtual silhouettes on the celestial sphere. From 1928, the International Astronomical Union (UAI) decided to officially regroup the celestial sphere into 88 constellations with precise limits, such that every point in the sky would be within the limits of a figure. When an astronomer says he saw a comet in Capricorn last night, it means that he saw a comet in the direction of the constellation of Capricorn.
A runner in a relay race runs 20 m north, turns around and runs south for 30 m, then turns north again and runs 40 m. The entire run took 30 seconds. Draw a sketch. What was the average speed of the runner?
Answer:
3.33m/s
Explanation:
The total distance that he runs is
30m north + 30m south + 40m north = 100 m
If the entire run takes 30 seconds then the average speed is distance over a unit of time
100 / 30 = 3.33 m/s
You can draw a sketch of a line going north 2 spaces, then going south by 3 spaces, finally going north again by 4 spaces
What are the largest optical telescopes in use today? Why do astronomers want their telescopes to be as large as possible?
The largest telescope currently used is the Gran Telescopio Canarias, (also known as GTC or GRANTECAN). It is 10.4m in diameter, slightly larger than the Keck telescopes in Hawaii.
The telescope observes the visible and infrared light coming from space and has a primary mirror of 10.4 meters, segmented into 36 hexagonal glass-ceramic pieces, 1.9 m between vertices, 8 cm thick, and 470 kg of mass each. The optical system is completed with two mirrors (secondary and tertiary) that form an image in seven focal stations.
Optically the diameter directly influences the magnification of the image. This added to the fact that astronomical objects are quite far away, a telescope of this magnitude allows to obtain more precise images of what is observed in space
The distance between Pluto and the Sun is 39.1 times more than the distance between the Sun and Earth. Calculate the time taken by Pluto to orbit the Sun in Earth days.
Answer:
Explanation:
Given
Distance between Pluto and sun is 39.1 times more than the distance between earth and sun
According to Kepler's Law
[tex]T^2=kR^3[/tex]
where k=constant
T=time period
R=Radius of orbit
Suppose [tex]R_1[/tex] is the radius of orbit of earth and sun
so Distance between Pluto and sun is [tex]R_2=39.1\cdot R_1[/tex]
[tex]T_1[/tex] and [tex]T_2[/tex] is the time period corresponding to [tex]R_1[/tex] and R_2[/tex]
[tex](T_1)^2=k(R_1)^3---1[/tex]
[tex](T_2)^2=k(R_2)^3---2[/tex]
divide 1 and 2
[tex](\frac{365}{T_2})^2=(\frac{R_1}{39.1})^3[/tex]
[tex]T_2^2=365^2\times 39.1^3[/tex]
[tex]T_2=89239.67\ Earth\ days[/tex]
Final answer:
Using Kepler's Third Law of Planetary Motion, we can calculate that it takes Pluto approximately 92,446 Earth days to complete one orbit around the Sun.
Explanation:
To calculate the time taken by Pluto to orbit the Sun, we use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Given that the distance between Pluto and the Sun is approximately 39.1 times more than the distance between the Earth and the Sun, we could simplify and use the approximation that Pluto's distance in astronomical units (AU) is roughly 40, as given in the reference information.
Pluto's orbital period (P) can be calculated using the average distance from the Sun (a) with the formula P² = a³. Here, we assume Earth's orbital period to be 1 Earth year and its distance as 1 AU by definition. Therefore, for Pluto:
a = 40 AU (Pluto's distance from the Sun) P² = 40³ P² = 64,000 P = √64,000 P ≈ 253 (Pluto's orbital period in Earth years)
To convert this period into Earth days, we multiply by the number of days in one Earth year (365.25 days, accounting for the leap year cycle):
P (in days) = 253 years × 365.25 days/year P (in days) ≈ 92,446.25 days
Hence, it takes Pluto approximately 92,446 Earth days to complete one orbit around the Sun.
If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision?
Using the physics equation of motion and the given initial velocity, reaction time, and deceleration, one can determine whether a truck can stop in time to avoid a collision.
Explanation:The question focuses on stopping distance and acceleration required to avoid a collision, indicating its base in Physics. If we have a truck moving at a constant velocity and it brakes at a certain distance from an obstacle, the minimum acceleration needed to avoid a collision can be calculated using the equation of motion v^2 = u^2 + 2as. Here, 'v' is the final velocity (0 m/s as the truck needs to stop), 'u' is the initial velocity, 'a' is the acceleration, and 's' is the distance over which the truck needs to stop.
To determine if the truck will hit the child, we must account for the driver's reaction time as well. During this reaction time, the truck continues to travel at its initial speed. After the reaction time, the truck will begin decelerating until it comes to a stop. The total stopping distance is the distance covered during the reaction time plus the distance covered during deceleration. The latter can be found using the deceleration rate and the formula mentioned above.
For the given scenario of the truck with an initial velocity of 10 m/s, a braking distance of 50 m, reaction time of 0.5 seconds, and deceleration of -1.25 m/s^2, we can calculate whether or not the truck will be able to stop in time to avoid hitting the child.
An astronaut is in space with a baseball and a bowling balL The astronaut gives both objects an equal push in the samedirection. Does the baseball have the same inertia as the bowling ball? Why? Does the baseball have the sameacceleration as the bowling ball from the push? Why? If both balls are traveling at the same speed, does the baseball havethe same momentum as the bowling ball?
For all solutions the answer is NO. And this is easily intuited because for the three conditions there is the dependence of the mass against some physical property of movement. Both bodies do not have the same mass.
In the case of Inertia, it is understood that it is the tendency of an object to resist change and is mass dependent. The object with greater mass will tend to resist change. Since the mass of the bowling ball is greater than the base ball, the bowling ball has greater inertia compared to the base ball.
For the second part, remember that force, according to Newton's second law, is defined as the product between mass and acceleration, so the bowling ball will accelerate less by having a greater mass.
Finally, momentum is defined as the product between mass and velocity. The mass is greater than one of its objects even though the speeds are the same. Therefore, the momentum of the bowling ball is greater than the momentum of baseball.
In space, the baseball has less inertia, greater acceleration, and less momentum compared to the bowling ball when both receive an equal push, due to their differences in mass.
The question relates to the concept of inertia, acceleration, and momentum in physics. When the astronaut in space gives an equal push to a baseball and a bowling ball, the baseball does not have the same inertia as the bowling ball, because inertia is a measure of an object's resistance to changes in its state of motion and is directly proportional to the object's mass. Since the bowling ball has a greater mass than the baseball, it has more inertia.
As a result, the baseball will have a greater acceleration than the bowling ball from the same force, due to Newton's second law of motion (F=ma), which states that force equals mass times acceleration. If we consider both balls traveling at the same speed, the baseball will not have the same momentum as the bowling ball, because momentum is the product of mass and velocity (p=mv), and the bowling ball has a greater mass.
A metal ball with diameter of a half a centimeter and hanging from an insulating thread is charged up with 1010 excess electrons. An initially uncharged identical metal ball hanging from an insulating thread is brought in contact with the first ball, then moved away, and they hang so that the distance between their centers is 20 cm. (a) Calculate the electric force one ball exerts on the other, and state whether it is attractive or repulsive. (Enter the magnitude of the force.) Entry field with correct answer 1.44e-7 N This force is Entry field with correct answer attractive repulsive (b) Now the balls are moved so that as they hang, the distance between their centers is only 5 cm. Naively one would expect the force that one ball exerts on the other to increase by a factor of 42
Answer:
a) F = 1.44 10⁻⁷ N, the charges are of the same sign the force is repulsive
b) F₂ / F₁ = 16
Explanation:
a) When the balls are touched the load is distributed evenly between the two balls, therefore when separating each ball has a load of
q₁ = q₂ = ½ 10¹⁰ 1.6 10⁻¹⁹ C
q₁ = q₂ = 0.8 10⁻⁹ C
As the charges are of the same sign the force is repulsive
To calculate the force let's use Coulomb's law
F = k q₁ q₂ / r²
F = 8.99 10⁹ 0.8 10⁻⁹ 0.8 10⁻⁹ / 0.20²
F = 1.44 10⁻⁷ N
b) Let us seek strength for the distress of
r₂ = 0.05 m
F₂ = 8.99 10⁹ 0.8 0.8 10⁻¹⁸ / 0.05²
F₂ = 2,301 10⁻⁶ N
b) The relationship between these two forces is
F₂ / F₁ = 23.01 10⁻⁷ / 1.44 10⁻⁷
F₂ / F₁ = 16
By standard convention, both the electric potential and the the electric potential energy between two charges is taken to be zero in what configuration?
Answer: at when distance r = infinity.
Explanation: The formulae for the electric potential of an electric charge to an arbitrary point is given by the formulae below
V = q/4πεr
V = electric potential (volts)
q = magnitude of electric charge
ε = permittivity of free space
r = distance between arbitrary point and charge.
In the equation above, it can be seen that only electric potential (v) and distance (r) is a variable, and there is an inverse relationship between them (an increase in one leads to a decrease in the other)
Thus to have zero value of electric potential (v= 0) we have to have the largest value of r ( r = infinity).
Same goes for electric potential energy between two charges, the formulae is given below as
W = q1 *q2/4πεr
W= electric potential energy
q1 = magnitude of first charge.
q2 = magnitude of second charge
ε = permittivity of free space
r = distance between arbitrary point and charge.
Also, all values are constant aside from electric potential energy (w) and distance (r) which have an inverse relationship.
Thus to have zero value of electric potential energy (w =0), we have to get an infinite value of distance ( r =infinity)
A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the transverse speed and acceleration of an element of the string at t = 0.160 s for the point on the string located at x = 1.40 m.Your response differs from the correct answer by more than 10%. Double check your calculations. m/sm/s2(b) What are the wavelength, period, and speed of propagation of this wave?msm/s
Explanation:
(a) It is known that equation for transverse wave is given as follows.
y = [tex](0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)[/tex]
Now, we will compare above equation with the standard form of transeverse wave equation,
y = [tex]A sin(kx + \omega t)[/tex]
where, A is the amplitude = 0.09 m
k is the wave vector = [tex]\frac{\pi}{11}[/tex]
[tex]\omega[/tex] is the angular frequency = [tex]4\pi[/tex]
x is displacement = 1.40 m
t is the time = 0.16 s
Now, we will differentiate the equation with respect to t as follows.
The speed of the wave will be:
v(t) = [tex]\frac{dy}{dt}[/tex]
v(t) = [tex]A \omega cos(kx + \omega t)[/tex]
v(t) = [tex](0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)[/tex]
v(t) = -0.84 m/s
The acceleration of the particle in the location is
a(t) = [tex]\frac{dv}{dt}[/tex]
a(t) = [tex]-A \omega 2sin(kx + \omega t)[/tex]
a(t) = [tex]-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)[/tex]
a(t) = -9.49 [tex]m/s^{2}[/tex]
Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 [tex]m/s^{2}[/tex] .
(b) Wavelength of the wave is given as follows.
[tex]\lambda = \frac{2\pi}{k}[/tex]
[tex]\lambda = (frac{2\pi}{\frac{\pi}{11}) [/tex]
[tex]\lambda[/tex] = 22 m
The period of the wave is
T = [tex]\frac{2 \pi}{\omega}[/tex]
T = [tex]\frac{2 \pi}{4 \pi}[/tex]
= 0.5 sec
Now, we will calculate the speed of propagation of wave as follows.
v = [tex]\frac{\lambda}{T}[/tex]
= [tex]\frac{22 m}{0.5 s}[/tex]
= 44 m/s
therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.
To find the transverse speed and acceleration of an element on a string given its wave function and specific values for time and position, we differentiate the wave function with respect to time. The first derivative gives us speed, and the second derivative provides acceleration. To determine the wave's wavelength, period, and propagation speed, we compare the wave function to its standard form and use the wave number and angular frequency.
Explanation:To determine the transverse speed and acceleration of a string element for the given wave function y = (0.090 m) sin (px/11 + 4pt) at t = 0.160 s and x = 1.40 m, we need to take the first and second derivatives of the wave function with respect to time. We also need to convert the given wave function into SI units for consistency.
First, let's tackle the transverse speed, which is given by the first derivative of the displacement y with respect to time t. Acceleration is given by the second derivative with respect to time.
(a) At t = 0.160 s and x = 1.40 m:
Vy = ∂y/∂t = 4p(0.090m)cos(px/11 + 4pt)
Acceleration, ay = ∂^2 y/∂t^2 = -16π^2(0.090m)sin(px/11 + 4pt)
Now plug in the values of x and t to get the numerical measure of speed and acceleration.
(b) To find the wavelength, period, and wave propagation speed, compare the wave function's format to the standard form y(x, t) = A sin (kx - ωt), where k is the wave number related to wavelength (λ) by k = 2π/λ and ω is the angular frequency related to period (T) by ω = 2π/T. The wave speed (v) is given by v = λ/T.
Identify k and ω from the wave function and calculate the wavelength, period, and wave speed.
A rock is thrown straight up into the air with an initial speed of 55 m/s at time t = 0. Ignore air resistance in this problem. At what times does it move with a speed of 36 m/s? Note: There are two answers to this problem.
Answer:
After 1.938 sec velocity of rock will be 36 m/sec
Explanation:
We have given initial velocity at which rock is thrown u = 55 m/sec
Final velocity v = 36 m/sec
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
According to first equation of motion we know that [tex]v=u+gt[/tex], here v is final velocity, u is initial velocity, g is acceleration due to gravity and t is time
So [tex]36=55-9.8t[/tex] ( Negative sign is due to rock is thrown upward )
So [tex]9.8t=19[/tex]
t = 1.938 sec
So after 1.938 sec velocity of rock will be 36 m/sec
A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistance is negligible. Your answer should contain the gravitational constant ????.)
a. The time at which the maximum height is achieved is functionsequation editor s.
b. The maximum height achieved by the projectile is functionsequation editor m.
c. The time when the projectile hits the ground is functionsequation editor s.
d. The range of the projectile is functionsequation editor m.
e. The speed of the projectile on impact with the ground is functionsequation editor m/s.
Answer:
a. 2.0secs
b. 20.4m
c. 4.0secs
d. 141.2m
e. 40m/s, ∅= -30°
Explanation:
The following Data are giving
Initial speed U=40m/s
angle of elevation,∅=30°
a. the expression for the time to attain the maximum height is expressed as
[tex]t=\frac{usin\alpha }{g}[/tex]
where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at
[tex]t=40sin30/9.81\\t=2.0secs[/tex]
b. the expression for the maximum height is expressed as
[tex]H=\frac{u^{2}sin^{2}\alpha }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m[/tex]
c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,
Hence T=2t
T=2*2.0
T=4.0secs
d. The range of the projectile is expressed as
[tex]R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m[/tex]
e. The landing speed is the same as the initial projected speed but in opposite direction
Hence the landing speed is 40m/s at angle of -30°
A phonograph record has an initial angular speed of 37 rev/min. The record slows to 14 rev/min in 1.6 s. What is the record’s average angular acceleration during this time interval? Answer in units of rad/s 2
Answer:
Acceleration will be [tex]\alpha =-1.50rad/sec^2[/tex]
Explanation:
We have given initial angular velocity [tex]\omega _i=37rpm[/tex]
In radian/sec initial angular velocity will be [tex]\omega _i=\frac{2\times \pi 37}{60}=3.873rad/sec[/tex]
Angular velocity after 1.6 sec is 14 rpm
So final angular velocity [tex]\omega _f=\frac{2\times \pi\times 14}{60}=1.465rad/sec[/tex]
Time t = 1.6 sec
We have to find the angular angular acceleration
From first equation of motion we know that
[tex]\omega _f=\omega _+\alpha t[/tex]
[tex]1.465=3.873+\alpha \times 1.6[/tex]
[tex]\alpha =-1.50rad/sec^2[/tex] here negative sign indicates that motion is deaccelerative in nature
At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravity") on test pilots and astronauts. In this device, an arm 8.84 m long rotates about one end in a horizontal plane, and an astronaut is strapped in at the other end. Suppose that he is aligned along the centrifuge’s arm with his head at the outermost end. The maximum sustained acceleration to which humans are subjected in this device is typically 12.5 g. (a) How fast must the astronaut’s head be moving to experience this maximum acceleration? (b) What is the difference between the acceleration of his head and feet if the astronaut is 2.00 m tall? (c) How fast in rpm (rev/min) is the arm turning to produce the maximum sustained acceleration?
Answer:
32.9242311983 m/s
1.40506567727g
35.3405770759 rpm
Explanation:
v = Linear Velocity of the capsule
[tex]a_c[/tex] = Centripetal acceleration = [tex]12.5g=12.5\times 9.81[/tex]
r = Radius of the centrifuge = 8.84 m
l = Person's height = 2 m
Centripetal acceleration is given by
[tex]a_c=\frac{v^2}{r}\\\Rightarrow v=a_cr\\\Rightarrow v=\sqrt{12.5\times 9.81\times 8.84}[/tex]
The linear speed of the capsule is 32.9242311983 m/s
The radius would be
[tex]r=\sqrt{r^2+\dfrac{l^2}{4}}\\\Rightarrow r=\sqrt{8.84^2+\dfrac{2^2}{4}}\\\Rightarrow r=8.89638128679\ m[/tex]
The centripetal acceleration
[tex]a_{c2}=\dfrac{32.9242311983^2}{8.89638128679\times 9.81}g\\\Rightarrow a_{c2}=12.4207805891g[/tex]
Change in acceleration from Pythagoras law
[tex]a=\sqrt{a_{ch}^2-a_{c2}^2}\\\Rightarrow a=\sqrt{12.5^2g^2-12.4207805891^2g^2}\\\Rightarrow a=1.40506567727g[/tex]
The difference is 1.40506567727g
Velocity
[tex]v=\omega r\\\Rightarrow v=2\pi Nr\\\Rightarrow N=\dfrac{v}{2\pi r}\\\Rightarrow N=\dfrac{32.9242311983}{2\pi 8.89638128679}\\\Rightarrow N=0.589009617932\ rev/s[/tex]
[tex]N=0.589009617932\times 60=35.3405770759\ rpm[/tex]
The speed is 35.3405770759 rpm
The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\times 10^{-5}~\text{N}\cdot\text{m}^2/\text{C}3×10 −5 N⋅m 2 /C when the area is parallel to the sheet of charge. Find the charge density on the sheet.
Answer:
Explanation:
Given
side of square shape [tex]a=5\ cm[/tex]
Electric flux [tex]\phi =3\times 10^{-5}\ N.m^2/C[/tex]
Permittivity of free space [tex]\epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}[/tex]
Flux is given by
[tex]\phi =EA\cos \theta [/tex]
where E=electric field strength
A=area
[tex]\theta [/tex]=Angle between Electric field and area vector
[tex]E=\frac{\phi }{A\cos (0)}[/tex]
[tex]E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}[/tex]
[tex]E=0.012\ N/C[/tex]
and Electric field by a uniformly charged sheet is given by
[tex]E=\frac{\sigma }{2\epsilon_0}[/tex]
where [tex]\sigma[/tex]=charge density
[tex]=\frac{\sigma }{\epsilon_0}[/tex]
[tex]\sigma =0.012\times 8.85\times 10^{-12}[/tex]
[tex]\sigma =2.12\times 10^{-13}\ C/m^2[/tex]
A uniform line charge extends from x = - 2.6 cm to x = + 2.6 cm and has a linear charge density of = 5.5 nC/m.(a) Find the total charge.Find the electric field on the y axis at the following distances.(b) y = 4 cm(c) y = 12 cm
Answer with Explanation:
We are given that
x=-2.6 cm to x=2.6 cm
Linear charge density=[tex]\lambda=5.5nC/m=5.5\times 10^{-9} C/m[/tex]
[tex]1nC=10^{-9} C[/tex]
Length of wire=[tex]2.6-(-2.6)=2.6+2.6=5.2cm[/tex]
Length of wire=[tex]\frac{5.2}{100}=0.052m[/tex]
1 m=100 cm
We know that
a.Linear charge density=[tex]\frac{Q}{L}[/tex]
Where Total charge =Q
Length=L
Total charge,Q=[tex]\lambda L[/tex]
Using the formula
Total charge,Q=[tex]5.5\times 10^{-9}\times 0.052=2.86\times 10^{-10}C[/tex]
b.y=4 cm=[tex]\frac{4}{100}=0.04m[/tex]
1 m=100 cm
Electric field on the y-axis is given by
[tex]E=\frac{2\lambda}{4\pi\epsilon_0 y}(\frac{L}{\sqrt{4y^2+L^2}}[/tex]
[tex]\frac{1}{4\pi\epsilon_0}=9\times 10^9 Nm^2/C^2[/tex]
Using the formula
[tex]E=\frac{2\times 5.5\times 10^{-9}\times 9\times 10^9\times 0.052}{0.04\times \sqrt{4(0.04)^2+(0.052)^2}}[/tex]
[tex]E=1348.8N/C[/tex]
c.y=12 cm=[tex]\frac{12}{100}=0.12m[/tex]
Using the formula
[tex]E=\frac{2\times 5.5\times 10^{-9}\times 9\times 10^9\times 0.052}{0.12\times \sqrt{4(0.12)^2+(0.052)^2}}[/tex]
[tex]E=174.7N/C[/tex]
To find the total charge of the uniform line charge, multiply the linear charge density by the length of the line. To find the electric field at a certain distance on the y axis, use the formula for electric field due to a line charge.
Explanation:(a) To find the total charge of the line, we need to multiply the linear charge density by the length of the line:
Charge = (Linear charge density)*(Length)
Convert the length to meters:
Length = (2.6 cm + 2.6 cm) = 0.052 m
Total Charge = (5.5 nC/m)*(0.052 m) = 0.286 nC
(b) To find the electric field at y = 4 cm, we can use the formula:
Electric Field = (Linear charge density)/(2πε₀y)
Convert the linear charge density to C/m:
Charge density = 5.5 nC/m = 5.5 x 10-9 C/m
Electric Field = (5.5 x 10-9 C/m)/(2πε₀(0.04 m))
(c) To find the electric field at y = 12 cm, we can use the same formula:
Electric Field = (5.5 x 10-9 C/m)/(2πε₀(0.12 m))
A woman exerts a horizontal force of 4 pounds on a box as she pushes it up a ramp that is 10 feet long and inclined at an angle of 30 degrees above the horizontal.Find the work done on the box.
Answer:
W = 34.64 ft-lbs
Explanation:
given,
Horizontal force = 4 lb
distance of push, d = 10 ft
angle of ramp, θ = 30°
Work done on the box = ?
We know,
W = F.d cos θ
W = 4 x 10 x cos 30°
W = 40 x 0.8660
W = 34.64 ft-lbs
Hence, work done on the box is equal to W = 34.64 ft-lbs
Final answer:
The work done on the box by the woman as she pushes it up the ramp with a horizontal force of 4 pounds is 34.64 foot-pounds, using the work calculation with the cosine of the ramp's angle. So, the final answer is 34.64 foot-pounds.
Explanation:
To calculate the work done on a box by a woman pushing it up a ramp, we need to use the formula Work = Force * Distance *cos(Ф), where Ф is the angle of the applied force relative to the direction of motion. Since the woman is exerting a horizontal force and the ramp is inclined at a 30-degree angle, the work done is the horizontal component of the force times the distance moved up the ramp.
In this scenario, the force is 4 pounds and the distance is 10 feet. The angle Ф the force makes with the displacement is 30 degrees as the ramp is inclined at this angle to the horizontal, and the force is horizontal. Therefore, the work done is calculated as:
Work = 4 lbs*10 ft *cos(30 degrees)
Using the cosine of 30 degrees (approximately 0.866), the calculation simplifies to:
Work = 4 lbs*10 ft*0.866
Work = 34.64 foot-pounds