Answer:
D) The potential difference between the plates increases.
Explanation:
The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.
Where ϵ0 is the permittivity of free space.
A capacitor with increased distance, will have a new capacitance C1=ϵ0kA/d1
Where d1 = nd
since d1 > d
therefore n >1
n is a factor derived as a result of the increased distance
Therefore the new capacitance becomes:
C1=ϵ0A/d1
C1= ϵ0A/nd
C1= C/n -------1
Where C1 is the capacitance with increased distance.
This implies that the charge storing capacity of the capacitor with increased plate separation decreases by a factor of (1/n) compared to that of the capacitor with original distance.
Given points
The charge stored in the original capacitor Q=CV
The charge stored in the original capacitor after inserting dielectric Q1=C1V1
The law of conservation of energy states that the energy stored is constant:
i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.
Charge before plate separation increase same as after plate separation increase
Q = Q1
CV = C1V1
CV = C1V1 -------2
We derived C1=C/n in equation 1. Inserting this into equation 2
CV = (CV1)/n
V1 = n(CV)/C
= n V
Since n > 1 as a result of the derived new distance, the new voltage will increase
A train leaves the train station at noon and travels at a constant speed of vt = 50 mi/hr on a straight track. 2 hr later, a car leaves the same train station and travels in the same direction at vc = 70 mi/hr on a road next to the train track. How far from the station is the place where the car catches up with the train? x =
Final answer:
The distance from the train station where the car catches up with the train is 250 miles.
Explanation:
To find the distance from the train station where the car catches up with the train, we need to determine the time it takes for the car to catch up.
Let's assume the car catches up with the train after t hours. Since the train leaves 2 hours before the car, the train has already been traveling for t + 2 hours.
Distance traveled by the train = speed of the train × time = 50t + 100 miles
Distance traveled by the car = speed of the car × time = 70t miles
Since the car catches up with the train, the distances traveled by both must be equal. Equating the distances, we get:
50t + 100 = 70t
20t = 100
t = 5
Now, substitute the value of t back into one of the distance formulas to find the distance from the train station:
Distance from the train station = speed of the train × time = 50 × 5 = 250 miles
Final answer:
The car catches up with the train in 5 hours, and they are 350 miles from the station when the car catches up.
Explanation:
The problem you presented is related to finding when two objects, moving at different constant speeds, meet. To solve this, we use the concept of relative speed.
The train travels for 2 hours before the car starts, so by that time, the train would have covered a distance of (50 mi/hr × 2 hr) = 100 miles. When the car starts, it has to cover not only the distance to the train but also the additional distance the train covers as the car approaches.
Let's consider the time it takes for the car to catch up with the train to be 't' hours. In that time, the car would travel 70t miles, and the train would travel an additional 50t miles.
Since the car needs to cover the initial 100 miles plus whatever distance the train covers in 't' hours, we can set up the equation:
70t = 100 + 50t
Subtracting 50t from both sides gives us:
20t = 100
Dividing both sides by 20:
t = 5
So, the car catches up with the train in 5 hours. To find out how far from the station they are when the car catches up with the train:
x = 70 mi/hr × 5 hr = 350 miles
A 2.0 m × 4.0 m horizontal plastic sheet has a charge of −10 μC , uniformly distributed. A tiny 4.0 μg plastic sphere is suspended motionless just above the center of the sheet. What is the charge on the sphere?
Answer:
Explanation:
Given
Dimension of Plastic sheet is [tex]2\times 4\ m^2[/tex]
Charge on sheet [tex]Q=-10\ \mu C[/tex]
Charge density [tex]\sigma =\frac{q}{A}[/tex]
[tex]\sigma =\frac{-10\times 10^{-6}}{8}=1.25\times 10^{-6}\ C/m^2[/tex]
Sphere has mass of [tex]m=4\ \mug[/tex]
If sphere is suspended motionless then its weight is balanced by repulsion force
Repulsive force [tex]F_r=qE[/tex]
where E=Electric field due to sheet
[tex]E=\frac{q}{2\epsilon _0}[/tex]
[tex]E=\frac{-10\times 10^{-6}}{2\times 8.85\times 10^{-12}}[/tex]
[tex]E=5.647\times 10^{5}\ N/C[/tex]
[tex]F_r=q\times 5.647\times 10^{5}[/tex]
[tex]F_r=mg[/tex]
[tex]q=\frac{mg}{E}[/tex]
[tex]q=\frac{4\times 10^{-6}\times 9.8}{5.647\times 10^{5}}[/tex]
[tex]q=6.9417\times 10^{-11}\ C[/tex]
Suppose the average mass of each of 20,000 asteroids in the solar system is 1017 kg. Compare the total mass of these asteroids to the mass of Earth. Assuming a spherical shape and a density of 3000 kg/m3, estimate the diameter of an asteroid having this average mass.
Answer:
The mass of the asteroids is 0.000334896182184 times the mass of the Earth.
39929.4542466 m
Explanation:
Total mass of the asteroids
[tex]m_a20000\times 10^{17}=2\times 10^{21}\ kg[/tex]
[tex]m_e[/tex] = Mass of Earth = [tex]5.972\times 10^{24}\ kg[/tex]
The ratio is
[tex]\dfrac{m_a}{m_e}=\dfrac{2\times 10^{21}}{5.972\times 10^{24}}\\\Rightarrow \dfrac{m_a}{m_e}=0.000334896182184[/tex]
The mass of the asteroids is 0.000334896182184 times the mass of the Earth.
Volume is given by
[tex]V=\dfrac{m}{\rho}\\\Rightarrow \dfrac{4\pi}{3\times 8} d^3=\dfrac{m}{\rho}\\\Rightarrow d^3=\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{10^{17}}{3000})^{\dfrac{1}{3}}\\\Rightarrow d=39929.4542466\ m[/tex]
The diameter is 39929.4542466 m
Suppose that at room temperature, a certain aluminum bar is 1.0000 m long. The bar gets longer when its temperature is raised. The length l of the bar obeys the following relation: l=1.0000+2.4×10−5T, where T is the number of degrees Celsius above room temperature. What is the change of the bar's length if the temperature is raised to 16.1 ∘C above room temperature?
Answer:
The change of the bar's length is [tex] 3.9\times10^{-4} m[/tex]
Explanation:
The bar length is a function of temperature T above room temperature:
[tex] L(T)=1.0000+2.4\times10^{-5} T[/tex]
So, if we evaluate at T= 16.1 C above room temperature
[tex]L(16.1)=1.0000+2.4\times10^{-5} (16.1)[/tex]
[tex]L=1.00039 m [/tex]
Now we can find the change of the bar length with the difference of L and Lo (the length at room temperature)
[tex]L- L_0=1.00039-1.0000 = 3.9\times10^{-4} m[/tex]
A nonconducting sphere has radius R = 2.36 cm and uniformly distributed charge q = +2.50 fC. Take the electric potential at the sphere's center to be V0 = 0. What is V at radial distance r = 1.45 cm?
Explanation:
It is known that inside a sphere with uniform volume charge density the field will be radial and has a magnitude E that can be expressed as follows.
E = [tex]\frac{q}{4 \p \epsilon_{o}R^{3}}r[/tex]
V = [tex]-\int_{0}^{r}E. dr[/tex]
= [tex]-\int_{0}^{r}\frac{q}{4 \p \epsilon_{o}R^{3}}r dr[/tex]
= [tex](\frac{q}{4 \p \epsilon_{o}R^{3}})(\frac{r^{2}}{2})[/tex]
= [tex]\frac{-qr^{2}}{8 \pi \epsilon_{o}R^{3}}[/tex]
At r = 1.45 cm = [tex]1.45 \times 10^{-2}[/tex] (as 1 m = 100 cm)
V = [tex]\frac{2.50 \times 10^{-15} \times 1.45 \times 10^{-2}}{8 \times 3.1416 \times 8.85 \times 10^{-12} \times 2.36 \times 10^{-2}}[/tex]
= [tex]6.905 \times 10^{-6}[/tex] mV
Thus, we can conclude that value of V at radial distance r = 1.45 cm is [tex]6.905 \times 10^{-6}[/tex] mV.
If the volume of one drop is 0.031 mL according to Stu Dent’s measurement, approximately what volume would 22 drops be? Answer with two significant digits and units of mL.
Answer:
Volume of 22 drop will be 0.68 ml
Explanation:
We have given volume of one drop = 0.031 ml
We know that 1 liter = 1000 ml
So [tex]1ml=10^{-3}L[/tex]
So 0.031 ml will be equal to [tex]0.031\times 10^{-3}L[/tex]
We have to find the volume of 22 drop
For finding volume of 22 drop we have to multiply volume of one drop by 22
So volume of 22 drop will be [tex]=22\times 0.031\times 10^{-3}=0.682\times 10^{-3}L=0.68mL[/tex]
So volume of 22 drop will be 0.68 ml
A 3.0 L cylinder is heated from an initial temperature of 273 K at a pressure of 105 kPa to a final temperature of 367 K. Assuming the amount of gas and the volume remain the same, what is the pressure (in kilopascals) of the cylinder after being heated?
Answer:
P₂= 141.15 kPa
Explanation:
Given that
Volume ,V= 3 L
V= 0.003 m³
Initial temperature ,T₁ = 273 K
Initial pressure ,P₁ = 105 kPa
Final temperature ,T₂ = 367 K
Given that volume of the cylinder is constant .
Lets take final pressure = P₂
We know that for constant volume process
[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}\\P_2=P_1\times \dfrac{T_2}{T_1}\\P_2=105\times \dfrac{367}{273}\ kPa\\\\P_2=141.15\ kPa[/tex]
Therefore the final pressure = 141.15 kPa
P₂= 141.15 kPa
A 200-m-wide river has a uniform flow speed of 0.99 m/s through a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of 4.4 m/s with respect to the water. There is a clearing on the north bank 35 m upstream from a point directly opposite the clearing on the south bank.
a. At what angle, measured relative to the direction of flow of the river, must the boat be pointed in order to travel in a straight line and land in the clearing on the north bank?
b. How long will the boat take to cross the river and land in the clearing?
Answer:
a. 1.174 rad[/tex] or 67.3 degree
b. t = 49.28 s
Explanation:
Let [tex]v_v[/tex] be the vertical component of the boat velocity with respect the the river, pointing North. Let [tex]v_h[/tex] be the horizontal component of the boat velocity with respect to the river, pointing West, aka upstream. Since the total velocity of the boat is 4.4m/s
[tex]v_v^2 + v_h^2 = 4.4^2 = 19.36[/tex]
The time it takes for the boat to cross 200m-wide river at [tex]v_v[/tex] rate is
[tex]t = 200 / v_v[/tex] or [tex]v_v = 200 / t[/tex]
This is also the time it takes for the boat to travel 35m upstream, horizontally, at the rate of [tex] v_h - 0.99[/tex] m/s
[tex]t = \frac{35}{v_h - 0.99}[/tex]
[tex]v_h - 0.99 = 35/t[/tex]
[tex]v_h = 35/t + 0.99[/tex]
We can substitute [tex]v_v,v_h[/tex] into the total velocity equation to solve for t
[tex]\frac{200^2}{t^2} + (\frac{35}{t} + 0.99)^2 = 19.36[/tex]
[tex]\frac{40000}{t^2} + \frac{35^2}{t^2} + 2*0.99*\frac{35}{t} + 0.99^2 = 19.36[/tex]
From here we can multiply both sides by [tex]t^2[/tex]
[tex]40000 + 1225 + 69.3t + 0.9801t^2 = 19.36t^2[/tex]
[tex]18.38 t^2 - 69.3t - 41225 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{69.3\pm \sqrt{(-69.3)^2 - 4*(18.3799)*(-41225)}}{2*(18.38)}[/tex]
[tex]t= \frac{69.3\pm1742.31}{36.7598}[/tex]
t = 49.28 or t = -45.51
Since t can only be positive we will pick t = 49.28
[tex]v_h = 35 / t + 0.99 = 35 / 49.38 + 0.99 = 1.7 m/s[/tex]
The angle, relative to the flow of river direction is
[tex]cos(\alpha) = \frac{v_h}{v} = \frac{1.7}{4.4} = 0.3864[/tex]
[tex]\alpha = cos^{-1}(0.3864) = 1.174 rad[/tex] or 67.3 degree
A gas has a pressure of 48atm in a 15.5L container. It was found that at 25∘C the gas occupied a volume of 25L and had a pressure of 22atm. What was the initial temperature in degrees Celsius?
Answer:
130.165636364°C
Explanation:
P = Pressure
V = Volume
n = Number of moles
R = Gas constant = 0.082 L atm/mol K
From ideal gas law we have
[tex]PV=nRT\\\Rightarrow n=\dfrac{PV}{RT}\\\Rightarrow n=\dfrac{22\times 25}{0.082\times (25+273.15)}\\\Rightarrow n=22.496451696\ moles[/tex]
[tex]PV=nRT\\\Rightarrow T=\dfrac{PV}{nR}\\\Rightarrow T=\dfrac{48\times 15.5}{22.496451696\times 0.082}\\\Rightarrow T=403.315636364\ K[/tex]
The initial temperature is [tex]403.315636364-273.15=130.165636364\ ^{\circ}C[/tex]
Answer: 130 degrees Celsius
Explanation:
P1V1 / T1 = P2V2 / T2
Let the subscript 1 represent the initial 15.5L of gas and the subscript 2 represent the gas at the final volume of 25L. Rewrite the temperature in Kelvin by adding 273.
P1=48atm, V1=15.5L, P2=22atm, V2=25L, T2=298K, and T1 is unknown.
Substitute the known values into the combined gas law equation.
P1V1 / T1=P2V2 / T2 → (48atm)(15.5L) / T1 = (22atm)(25L) / 298K
Solve the equation for T1, simplify, and round to the nearest degree.
T1 = P1V1T2 / P2V2
T1 = (48atm)(15.5L)(298K)(22atm)(25L)
T1 = 403K
To get the temperature in Celsius, subtract 273.
T1 = 130∘C
A square steel bar has a length of 9.8 ftft and a 2.6 inin by 2.6 inin cross section and is subjected to axial tension. The final length is 9.80554 ftnt . The final side length is 2.59952 in in . What is Poisson's ratio for the material? Express your answer to three significant figures.
To solve this problem we will apply the concept related to the Poisson ratio for which the longitudinal strains are related, versus the transversal strains. First we need to calculate the longitudinal strain as follows
[tex]\epsilon_x = \frac{l_f-l_i}{l_i}[/tex]
[tex]\epsilon_x = \frac{(9.80554)-(9.8)}{9.8}[/tex]
[tex]\epsilon_x = 0.0005653[/tex]
Second we will calculate the lateral strain as follows
[tex]\epsilon_y = \frac{a_f-a_i}{a_i}[/tex]
[tex]\epsilon_y = \frac{2.59952-2.6}{2.6}[/tex]
[tex]\epsilon_y = -0.0001846153[/tex]
The Poisson's ratio is the relation between the two previous strain, then,
[tex]\upsilon = -\frac{\epsilon_y}{\epsilon_x}[/tex]
[tex]\upsilon = -\frac{(-0.0001846153)}{0.0005653}[/tex]
[tex]\upsilon = 0.3265[/tex]
Therefore the Poisson's ratio for the material is 0.3265
A 15-turn circular wire loop with a radius of 3.0 cm is initially in a uniform magnetic field with a strength of 0.5 T. The field decreases to zero over a time of 0.10 sec. What is the induced emf?
Answer:
0.212V
Explanation:
The induced emf in circular loop is [tex]\epsilon=-d\frac{\phi_B}{dt}=-NA\frac{dB}{dt}[/tex]
N = Number of loops = 15
A = Cross sectional Area = [tex]\pi[/tex][tex]0.03^{2}[/tex] = 0.0009[tex]\pi[/tex]
dB = change in magnetic Field = 0-0.5 = -0.5
dt = time taken = 0.1sec
[tex]\epsilon=-15*0.0009\pi *\frac{-0.5}{0.1}[/tex] = 0.212V
Which of the following does not involve work? 1. A child is pushed on a swing. 2. A golf ball is struck. 3. A weight lifter does military presses (lifting weights over his head.) 4. A professor picks up a piece of chalk from the floor. 5. A runner stretches by pushing against a wall.
From the work theorem, this is defined as the amount of force applied on an object displaced on a longitudinal unit. Mathematically this is
[tex]W = \vec{F} \times \vec{d}[/tex]
Here,
F = Force vector
d = Displacement vector
Of all the options presented, only in the last one there is no change in distance, so the work done there is zero.
The correct option is 5.
Work in Physics is when a force causes displacement. Of the choices, the example of a runner stretching by pushing against a wall does not involve work, as there is no displacement.
Explanation:In the context of Physics, 'work' is defined as a force causing displacement on an object. Essentially, work is done when a force acts upon an object to cause or prevent motion. Among the options provided, 5. A runner stretches by pushing against a wall does not involve 'work'. This is because, despite the runner exerting a force against the wall, there is no displacement of the wall in response to this force.
Work (W) is calculated by multiplying the force (F) that is applied to an object and the distance (d) that the object is moved, i.e., W = F * d. In this case, since the displacement (d) is zero, the work done is also zero.
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The temperature of a sample of silver increased by 24.0 °C when 269 J of heat was applied. What is the mass of the sample?
Answer:
Mass of the silver will be equal to 46.70 gram
Explanation:
We have given heat required to raise the temperature of silver by 24°C is 269 J , so [tex]\Delta T=24^{\circ}C[/tex]
Specific heat of silver = 0.240 J/gram°C
We have to find the mass of silver
We know that heat required is given by
[tex]Q=mc\Delta T[/tex], here m is mass, c is specific heat of silver and [tex]\Delta T[/tex] is rise in temperature
So [tex]269=m\times 0.240\times 24[/tex]
m = 46.70 gram
So mass of the silver will be equal to 46.70 gram
What are the three longest wavelengths for standing sound waves in a 120-cm-long tube that is (a) open at both ends and (b) open at one end, closed at the other?
Answer
given,
length of tube = 120 cm
a) Open at both ends.
distance between two successive nodes or anti nodes = λ/2
for first possibility = λ/2 = 120
λ =2 x 120 = 240 cm
for second Possibility, distance between two node or antinode acting symmetrically
λ = 120 cm
for three nodes in the tube, distance between them is equal to
[tex]\dfrac{3}{2}\lambda = 120[/tex]
[tex]\lambda = 80 cm[/tex]
b) Open at one end
First possibility, The distance between one node or anti node
[tex]\dfrac{\lambda}{4} = 120[/tex]
λ = 480 cm
Second Possibility, distance between two node or anti node is equal to
[tex]\dfrac{3}{4}\lambda = 120[/tex]
λ = 160 cm
Third possibility, distance between three nodes and three anti nodes is equal to
[tex]\dfrac{5}{4}\lambda = 120[/tex]
λ = 100 cm
To practice Problem-Solving Strategy 2.1 Motion with constant acceleration You are driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. Your reaction time before stepping on the brakes is 0.50 s , and the maximum deceleration of your car is 10 m/s2 . How much distance is between you and the deer when you come to a stop
a. How much distance is between you and the deer when you come to a stop?
b. What is the maximum speed you could have and still not hit the deer?
Answer:
a) [tex]\Delta s=5\ m[/tex] is the distance between deer and the vehicle
b) [tex]u'=22.36\ m.s^{-1}[/tex] is the maximum speed the driver can be at and still not hit the deer.
Explanation:
Given:
initial speed of driving, [tex]u=20\ m.s^{-1}[/tex]distance of deer from the vehicle, [tex]x=35\ m[/tex]reaction time taken to step onto the brakes, [tex]t'=0.5\ s[/tex]maximum deceleration of the car, [tex]a_m=-10\ m.s^{-2}[/tex]a)
Now the distance travelled after application of the brakes till the vehicle stops:
[tex]v^2=u^2+2a_m.s[/tex]
(assuming that the brakes are applied with maximum acceleration)
where:
[tex]s=[/tex] displacement of the vehicle after braking till it stops
[tex]v=[/tex] final velocity of the vehicle = 0 (stops)
putting the values:
[tex]0^2=20^2-2\times 10\times s[/tex]
[tex]s=20\ m[/tex]
Now before the application of the brakes 0.5 second is taken to react and the vehicle travels during this time as well.
So, distance covered before applying the brakes:
[tex]s'=u.t'[/tex]
[tex]s'=20\times 0.5[/tex]
[tex]s'=10\ m[/tex]
The distance between the deer and the vehicle:
[tex]\Delta s=x-(s+s')[/tex]
[tex]\Delta s=35-(20+10)[/tex]
[tex]\Delta s=5\ m[/tex]
b)
The maximum speed the driver can have with the vehicle and still not hit the deer is given as:
[tex]v^2=u'^2+2. a_m.(x-s')[/tex]
because s' is the distance covered before braking during the reaction time.
[tex]0^2=u'^2-2\times 10\times (35-10)[/tex]
[tex]u'=22.36\ m.s^{-1}[/tex] is the maximum speed the driver can be at and still not hit the deer.
Using the equations of motion under constant acceleration, a) the distance between the driver and the deer when the car comes to a stop is 5 m and b) the maximum speed the driver could have and still not hit the deer is approximately 23.45 m/s.
Explanation:The subject of the question, Problem-Solving Strategy 2.1 Motion with constant acceleration, involves using the equations of motion under constant acceleration. Let's break down the problem into two parts:
How much distance is between you and the deer when you come to a stop? In this scenario, you first drive at 20 m/s for 0.50 s before stepping on the brakes. So, the distance travelled during this time is v*t = 20 m/s * 0.50 s = 10 m. Then, you decelerate at 10 m/s². As you finally come to stop, the additional distance travelled can be found by the formula (v² - u²) / 2a, which gives (0 - (20²)) / 2*(-10) = 20 m. So, the total distance covered is 10 m + 20 m = 30 m. Therefore, you come to a stop 5 m away from the deer because the deer was initially 35 m away.What is the maximum speed you could have and still not hit the deer? For this, you need to calculate the stopping distance for the car in relation to the deer at 35 m and find the initial speed where the stopping distance equals the distance to the deer. If the car travels distance D in the driver's reaction time, then it travels (35 - D) while braking. The braking distance = (v² - u²) / 2a => v² = 2aD, where D = (35 - v*0.50). Solving for v in the quadratic equation gives the maximum speed, approx 23.45 m/s to not hit the deer.Learn more about Motion with Constant Acceleration here:https://brainly.com/question/25307462
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An airplane in a holding pattern flies at constant altitude along a circular path of radius 3.50 km. If the airplane rounds half the circle in 1.50 3 102 s, determine the magnitude of its (a) displacement and (b) average velocity during that time. (c) What is the airplane’s average speed during the same time interval?
Explanation:
A.
Displacement,S; Θ = S/r
= 180/360 * 2π * 3500
= 10995.6 m
Θ = 10995.6/3500
= 3.142
B.
Angular speed,w = Θ/t
= 3.142/1.503102
= 2.09 rad/s
Velocity = w * r
= 2.09 × 3500
= 7315 m/s.
C.
The same as B.
The airplane's Motion in a Circle displacement is 7.0km, its average velocity is 0.0467 km/s, and its average speed is 0.0737 km/s.
The airplane is flying half a circle, which is essentially semi-circular motion. Under this condition, we can calculate the values asked in the question as follows:
(a) Displacement: Displacement is a vector quantity and represents the shortest distance between the initial and the final points of an object's path. But since the airplane is rounding half the circle, the displacement is essentially the diameter of the circular path. The diameter will be two times the radius, therefore 2*3.50km = 7.00 km.
(b) Average velocity: Average velocity is the total displacement divided by the total time. So, the average velocity would be 7.00km / (1.50*10^2) s = 0.0467 km/s.
(c) Average speed: Average speed is defined as the total distance traveled by the object divided by the total time taken. Here, the airplane travels half the circumference of the circle in the given time. The formula for the circumference of a circle is 2*pi*r, so half the circle's circumference will be pi*3.50 km. Then, Average speed = (pi*3.50 km) / (1.50*10^2) s = 0.0737 km/s.
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In the infinitesimal neighborhood surrounding a point in an inviscid flow, the small change in pressure, dP, that corresponds to a small change in velocity, dV, is given by the differential relation: dP=−rhoVdV. (a) Using this relation, derive a differential relation for the fractional change in density, drho/rho, as a function of the fractional change in velocity, dV/V, with the compressibility τ as a coefficient. (b) The velocity at a point in an isentropic flow of air is 10 m/s, and the density and pressure are 1.23 kg/m3 and 1.01 x 105 N/m2, respectively. The fractional change in velocity at the point is 0.01. Calculate the fractional change in density. (c) Repeat part (b), except for a local velocity at the point of 1000 m/s. Compare this result with that from part (b), and comment on the differences.
Answer:
(a). differential relation becomes dρ/ρ = -τρV2 dV/V
(b). fraction change in density; dρ/ρ = -8.7 ˣ 10⁻⁶
(c). dρ/ρ = -8.7 ˣ 10⁻²
Explanation:
Let us begin,
(a). given from the question we have that dp = -ρVdV
where dρ = ρ τ dp, i.e.
dp = dρ/ρτ ...............(1)
replacing value of dp we have,
-ρVdV = dρ/ρτ
so that dρ = -τp2 VdV
finally, dρ/ρ = -τp V2 dV/V
(b). from the question here, we were given Velocity to be = 10 m/s
density (ρ) = 1.23 kg/m3
pressure (p) = 1.01 x 10⁵ N/m2
from formula,
dρ/ρ = τs ρ V2 dV/V .............(2)
but τs = 1/γp = 1/(1.4× 1.01×10⁵) = 7.07 ˣ 10⁻⁶ m²/N
substituting value of τs into equation (2) we have
dρ/ρ = τs ρ V2 dV/V = (7.07 ˣ 10⁻⁶) ˣ (1.23) ˣ (10ˣ2) (0.01) = -8.7 ˣ 10⁻⁶
dρ/ρ = -8.7 ˣ 10⁻⁶
(c). from we question we know that dρ/ρ has a large ratio of (1000/10)²
so dρ/ρ = -8.7 ˣ 10⁻⁶ × (1000/10)² = -8.7 ˣ 10⁻²
dρ/ρ = -8.7 ˣ 10⁻².
comparing this result with part (b). we can see that when we increase the velocity of a factor 100, there is an increased factorial change in the density by a factor 104.
A hydraulic turbine is used to generate power by using the water in a dam. The elevation difference between the free surfaces upstream and downstream of the dam is 131 m. The water is supplied to the turbine at a rate of 201 kg/s. If the shaft power output from the turbine is 234 kW, the efficiency of the turbine is_________.
Answer:
0.906
Explanation:
Let g = 9.81 m/s2. We can calculate the rate of change in potential energy when m = 201kg of water is falling down a distance of h = 131m per second
[tex]\dot{E_p} = \dot{m}gh = 201*9.81*131 = 258307 J/s (W) = 258.307 kW[/tex]
So the efficiency of the water turbine is the ratio of output power over input power:
[tex]\frac{234}{258.307} = 0.906[/tex]
This question involves the concepts of efficiency and potential energy.
The efficiency of the turbine is "90.6%".
The efficiency of the turbine can be given by the following formula:
[tex]Efficiency = \frac{Turbine\ shaft\ power}{Potential\ Energy\ power\ of\ Water}\\\\[/tex]
where,
Turbine shaft power = 234 KW
Potential energy power of water = mgh/t = (201 kg/s)(9.81 m/s²)(131 m)
Potential energy power of water = 258307.11 W = 258.31 KW
Therefore,
[tex]Efficiency=\frac{234\ KW}{258.31\ KW}[/tex]
Efficiency = 0.906 = 90.6%
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A cannon ball is shot straight upward with a velocity of 72.50 m/s. How high is the cannon ball above the ground 3.30 seconds after it is fired? (Neglect air resistance.)
Answer:
Explanation:
Given
Cannon is fired with a velocity of [tex]u=72.50\ m/s[/tex]
Using Equation of motion
[tex]y=ut+\frac{1}{2}at^2[/tex]
where
[tex]y=displacement[/tex]
[tex]u=initial\ velocity[/tex]
[tex]a=acceleration[/tex]
[tex]t=time[/tex]
after time [tex]t=3.3 s[/tex]
[tex]y=72.50\times 3.3-\frac{1}{2}\times 9.8\times (3.3)^2[/tex]
[tex]y=239.25-53.36[/tex]
[tex]y=185.89\ m[/tex]
So after 3.3 s cannon ball is at a height of 185.89 m
A man wandering in the desert walks 2.7 miles in the direction S 35°W. He then turns 90° and walks 3.5 miles in the direction N 55° W. At that time, how far is he from his starting point, and what is his bearing from his starting point?
To solve this problem we will make a diagram in the Cartesian plane that will allow us to find and understand more accurately the displacement and the angle of rotation.
According to Pythagoras, the distance traveled would be equivalent to
[tex]d = \sqrt{(2.7)^2+(3.5)^2}[/tex]
[tex]d = 4.4 miles[/tex]
The individual had a displacement of 4.4 thousand from the starting point.
Now the angle [tex]\theta[/tex] plus the previously given angle will allow us to find the direction of travel.
[tex]tan\theta = \frac{\text{Opposite side}}{\text{Adjacent side}}[/tex]
[tex]tan\theta = \frac{3.5}{2.7}[/tex]
[tex]\theta = tan^{-1} (\frac{3.5}{2.7})[/tex]
[tex]\theta = 52.35\°[/tex]
[tex]\angle =[/tex] [tex]\theta + 35 = 52.35+35 = 87.35\°[/tex]
Therefore the net direction of the man is S 87.35° W
A spherical conductor has a radius of 14.0 cm and a charge of 42.0 µC. Calculate the electric field and the electric potential at the following distances from the center. (a) r = 8.0 cm magnitude direction electric field MN/C electric potential MV (b) r = 30.0 cm magnitude direction electric field MN/C electric potential MV (c) r = 14.0 cm magnitude direction electric field MN/C electric potential MV
Answer:
0 MN/C, 2.697 MV
4.1953 MN/C, 1.2586 MV
19.2642857143 MN/C, 2.697 MV
Explanation:
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
Electric field at r = 8 cm
E = 0 (inside)
Electric potential is given by
[tex]V=\dfrac{kq}{R}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV[/tex]
Electric potential is 2.697 MV
r = 30 cm
[tex]E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3^2}\\\Rightarrow E=4195333.33\ MN/C=4.1953\ MN/C[/tex]
Electric field is 4.1953 MN/C
[tex]V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3}\\\Rightarrow V=1258600\ V=1.2586\ MV[/tex]
Electric potential is 1.2586 MV
r = R = 14 cm
[tex]E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14^2}\\\Rightarrow E=19264285.7143\ N/C=19.2642857143\ MN/C[/tex]
The electric field is 19.2642857143 MN/C
[tex]V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV[/tex]
The potential is 2.697 MV
To calculate the electric field and electric potential at different distances from the center of a spherical conductor, use the equations for electric field and electric potential due to a point charge. At a distance of 8.0 cm, the electric field is 7.87 * 10^3 N/C and the electric potential is 5.25 kV. At a distance of 30.0 cm, the electric field is 1.26 * 10^3 N/C, but the electric potential cannot be calculated without a reference point. At 14.0 cm, the electric field and electric potential are zero.
Explanation:To calculate the electric field and electric potential at different distances from the center of a spherical conductor with a radius of 14.0 cm and a charge of 42.0 µC, we can use the equations for electric field and electric potential due to a point charge.
(a) At a distance of 8.0 cm from the center, the magnitude of the electric field can be calculated using the equation:
E = k * (Q/r²)
Substituting the given values, we get:
E = (9 * 10^9 Nm²/C²) * (42 * 10^-6 C) / (0.08 m)² = 7.87 * 10^3 N/C
To calculate the electric potential at this distance, we can use the equation:
V = k * (Q/r)
Substituting the given values, we get:
V = (9 * 10^9 Nm²/C²) * (42 * 10^-6 C) / (0.08 m) = 5.25 kV
(b) At a distance of 30.0 cm from the center, the magnitude of the electric field can be calculated in the same way:
E = (9 * 10^9 Nm²/C²) * (42 * 10^-6 C) / (0.3 m)² = 1.26 * 10^3 N/C
But the electric potential at this distance cannot be calculated using the given information, as we need the reference point for potential measurement.
(c) At a distance of 14.0 cm (the radius of the conductor) from the center, the electric field and electric potential are zero, as the conductor is in electrostatic equilibrium.
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A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.90 s. You may ignore air resistance, so the brick is in free fall. (a) How tall, in meters, is the building? (b) What is the magnitude of the brick’s velocity just before it reaches the ground? (c) Sketch ay-t, vy-t, and y-t graphs for the motion of the brick.
Answer: Height of building = 17.69m, velocity of brick = 18.6m/s
Explanation: From the question, the body has a zero initial speed, thus initial velocity (u) all through the motion is zero.
By ignoring air resistance makes it a free fall motion thus making it to accelerate constantly with a value of [tex]a = 9.8m/s^{2}[/tex].
Time taken to fall = 1.90s
a)
thus the height of the building is calculated using the formulae below
[tex]H = ut + \frac{1}{2} gt^{2}[/tex]
but u = 0 , hence
[tex]H = \frac{1}{2} gt^{2}[/tex]
[tex]H = \frac{1}{2} *9.8* 1.9^{2} \\\\H = 17.69m[/tex]
b)
to get the value of velocity (v) as the brick hits the ground, we use the formulae below
[tex]v^{2} = u^{2} + 2aH[/tex]
but u= 0, hence
[tex]v^{2} = 2gH\\[/tex]
[tex]v^{2} = 2 * 9.8 * 17.69\\v = \sqrt{2 *9.8* 17.69} \\v = 18.62m/s[/tex]
find the attachment in this answer for the accleration- time graph, velocity- time graph and distance time graph
a. The height of building, in meters, is equal to 17.69 meters.
b. The magnitude of the brick’s velocity just before it reaches the ground is equal to 18.72 m/s.
Given the following data:
Time = 1.90 secondsInitial velocity = 0 m/sWe know that acceleration due to gravity (a) for an object in free fall is equal to 9.8 meter per seconds square.
a. To determine the height of building, in meters, we would use the second equation of motion:
Mathematically, the second equation of motion is given by the formula;
[tex]S = ut + \frac{1}{2} at^2[/tex]
Where:
S is the distance covered.u is the initial velocity.a is the acceleration.t is the time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]S = 0(1.90) + \frac{1}{2} \times 9.8 \times 1.90^2\\\\S = 0 + 4.9 \times 3.61[/tex]
Distance, S = 17.69 meters.
b. To determine the magnitude of the brick’s velocity just before it reaches the ground, we would use the third equation of motion;
[tex]V^2 = U^2 + 2aS[/tex]
Where:
V is the final speed.U is the initial speed.a is the acceleration.S is the distance covered.
Substituting the given parameters into the formula, we have;
[tex]V^2 = 0^2 + 2 \times 9.8 \times 17.69\\\\V^2 = 350.26\\\\V = \sqrt{350.26}[/tex]
V = 18.72 m/s
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A tuner first tunes the A string very precisely by matching it to a 440 Hz tuning fork. She then strikes the A and E strings simultaneously and listens for beats between the harmonics. What beat frequency between higher harmonics indicates that the E string is properly tuned
Answer:
The beat frequency is 2 Hz.
Explanation:
Given that,
Frequency of A = 440 Hz
Frequency of E = 659 Hz
Suppose, piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note A should have a frequency of 440 Hz and the note E should be at 659 Hz.
We need to calculate the third harmonic of A
Using formula of harmonic
[tex]f_{3}=n\times f[/tex]
Put the value into the formula
[tex]f_{3} =3\times440[/tex]
[tex]f_{3}=1320\ Hz[/tex]
We need to calculate the second harmonic of E
Using formula of harmonic
[tex]f_{2}=n\times f[/tex]
Put the value into the formula
[tex]f_{2} =2\times659[/tex]
[tex]f_{2}=1318\ Hz[/tex]
We need to calculate the beat frequency
Using formula of beat frequency
[tex]f_{b}=f_{3}-f_{2}[/tex]
Put the value into the formula
[tex]f_{b}=1320-1318[/tex]
[tex]f_{b}=2\ Hz[/tex]
Hence, The beat frequency is 2 Hz.
The beat frequency indicating proper tuning of the E string relative to the A string tuned to 440 Hz should be zero. This corresponds to a situation where the harmonics of both strings match exactly, with the second harmonic of the E string (659.26 Hz) aligning with the third harmonic of the A string (660 Hz), reducing the beat frequency to zero.
Explanation:To determine the beat frequency that indicates proper tuning for the E string of a guitar, we must consider the fundamental frequencies of the A and E strings. The A string is tuned to 440 Hz. The E string above it should be tuned to a fundamental frequency of 329.63 Hz.
When the A and E strings are played together, harmonics of these frequencies are heard. The second harmonic of the E string would be 2 * 329.63 Hz = 659.26 Hz. This is very close to the 660 Hz, which is the third harmonic of the A string (3 * 440 Hz = 1320 Hz). For the strings to resonate without beats, the harmonics should match as closely as possible. Any difference in these harmonic frequencies would cause beats corresponding to the frequency difference.
Therefore, if the second harmonic of the E string (659.26 Hz) matches the third harmonic of the A string (660 Hz) closely, there will be no beats. In this ideal case, the beat frequency would be 660 Hz - 659.26 Hz = 0.74 Hz. However, for the E string to be perfectly in tune, we would want the beat frequency to be zero, indicating no difference between the expected harmonic frequency and the actual harmonic frequency.
An Atwood machine is constructed using a disk of mass 2 kg and radius 24.8 cm. What is the acceleration of the system? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s 2 .
Answer:
a = (m₂-m₁) / (m₂ + m₁ + ½ m)
a = (m₂-m₁) / (m₂ + m₁ + 1)
Explanation:
An Atwood machine consists of two masses of different m1 and m2 value that pass through a pulley, in this case with mass. Let's use Newton's second law for this problem.
Assume that m₂> m₁, so the direction of descent of m₂ is positive, this implies that the direction of ascent of m₁ is positive
Equation of the side of m₁
T₁ - W₁ = m₁ a
Equation of the side of m₂
W₂ - T₁ = m₂ a
Mass pulley equation m; by convention the counterclockwise rotation is positive
τ = I α
T₁ R - T₂ R = I (-α)
The moment of inertia of a disk is
I = ½ m R²
Angular and linear acceleration are related
a = α R
α = a / R
The rotation is clockwise, so it is negative
We replace
(T₁ –T₂) R = ½ m R² (-a / R)
T₁ -T₂ = - ½ m a
Let's write our three equations together
T₁ - m₁ g = m₁ a
m₂ g - T₂ = m₂ a
T₁ –T₂ = -½ m a
Let's multiply the last equation by (-1) and add
m₂ g - m₁ g = m₂ a + m₁ a + ½ m a
a = (m₂-m₁) / (m₂ + m₁ + ½ m)
calculate
a = (m₂ - m₁)/ (m₁ +m₂ + 1)
Based on the calculations, the acceleration of this system is equal to 0.58 [tex]m/s^2[/tex].
Given the following data:
Mass of disk = 2 kg.Radius of disk = 24.8 cm.Acceleration of gravity = 9.8 [tex]m/s^2[/tex]How to calculate the acceleration of the system.First of all, we would determine the moment of inertia of this disk by using this formula:
[tex]I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 2 \times (0.248)^2\\\\I=0.0615\;Kgm^2[/tex]
Next, we would use a free body diagram to determine the tensional forces acting on the disk by applying Newton's Second Law of Motion as follows:
For the first force:
[tex]F_1g-F_{T1}=m_1a\\\\m_1g-F_{T1}=m_1a\\\\1.61(9.8)-F_{T1}=1.61a\\\\15.8-F_{T1}=1.61a\\\\F_{T1}=15.8-1.61a[/tex]
For the second force:
[tex]F_{T2}-F_2g=m_2a\\\\F_{T2}-m_2g=m_2a\\\\F_{T2}-1.38(9.8)=1.38a\\\\F_{T2}-13.5=1.38a\\\\F_{T2}=13.5+1.38a[/tex]
For the torque, we have:
[tex]\sum T =I\alpha \\\\F_{T1}r-F_{T2}r=I\alpha\\\\(F_{T1}-F_{T2})r=I\alpha\\\\(15.8-1.61a-[13.5+1.38a])0.248=0.0615 \times \frac{a}{0.248} \\\\(15.8-1.61a-13.5-1.38a)0.248=0.0615 \times \frac{a}{0.248} \\\\(2.3-2.99a)0.248=0.0615 \times \frac{a}{0.248}\\\\0.5704-0.7415a=\frac{0.0615a}{0.248}\\\\0.1415-0.1839a=0.0615a\\\\0.1839a+0.0615a=0.1415\\\\0.2454a=0.1415\\\\a=\frac{0.1415}{0.2454}[/tex]
Acceleration, a = 0.58 [tex]m/s^2[/tex]
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the separation between the plates is now increased to 4.50 mmmm . How much energy is stored in the capacitor now?
Complete question:
A parallel-plate capacitor has plates with an area of 405 cm² and an air-filled gap between the plates that is 2.25 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery. (a) How much energy is stored in the capacitor? (b) The separation between the plates is now increased to 4.50 mm. How much energy is stored in the capacitor now?
Answer:
The energy stored in the capacitor when the plates is increased to 4.50 mm is 1.32 X 10⁻⁵ J
Explanation:
Given:
Area of the plates = 405 cm² = 405 X (10⁻²)² m²= 405 X 10⁻⁴m² = 0.0405m²
Energy stored in a capacitor = CV²/2
Where;
V is the voltage across the plates = 575 V
C is the capacitor =?
C = Kε(A/d)
K is constant = 1.0
ε is permittivity of free space = 8.885 X 10⁻¹²
d is the diameter of the two plates = 2.25 mm = 0.00225m
C = 1.0 x 8.885 X 10⁻¹² x (0.0405/0.00225)
C = 1.5993 X 10⁻¹⁰ F
(a) Energy stored in a capacitor = 0.5 X 1.5993 X 10⁻⁹ X 575²
= 2.64 X 10⁻⁵ J
(b) The separation between the plates is now increased to 4.50 mm.
C = Kε(A/d)
New diameter, d = 4.5 mm = 0.0045 m
C = 1.0 x 8.885 X 10⁻¹² x (0.0405/0.0045)
C = 7.9965 X 10 ⁻¹¹ F
Energy stored in a capacitor = 0.5 X 7.9965 X 10 ⁻¹¹ X 575²
= 1.32 X 10⁻⁵ J
Therefore, the energy stored in the capacitor when the plates is increased to 4.50 mm is 1.32 X 10⁻⁵ J
A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a height H, directly above the point where the first ball was thrown upward. There is no air resistance. (a) Find the time at which the two balls collide. (b) Find the value of H in terms of v0 and g such that at the instant when the balls collide, the first ball is at the highest point of its motion.
Answer:
(a) [tex]t=\frac{H}{v_0}[/tex]
(b) [tex]H=\frac{v_0^2}{g}[/tex]
Explanation:
Let the two balls collide at a height x from the ground. Therefore, ball 2 travels a distance of (H-x) before colliding with ball 1.
Using the following Newton's law of motion,
[tex]S=ut+\frac{1}{2}at^2[/tex]
where,
[tex]S[/tex] = displacement
[tex]u[/tex] = initial velocity
[tex]a[/tex] = acceleration
[tex]t[/tex] = time
we can write the equations of motion of the two balls(ball 1 and ball 2 respectively):
[tex]x=v_0t-\frac{1}{2}gt^2[/tex] ......(1) ([tex]a=-g[/tex], ball is moving against gravity)
[tex]H-x=\frac{1}{2} gt^2[/tex] .......(2) (initial velocity is zero; [tex]a=+g[/tex])
Substituting [tex]x[/tex] from equation (1) in (2),
[tex]H-v_0t+\frac{1}{2}gt^2=\frac{1}{2}gt^2[/tex]
or, [tex]t=\frac{H}{v_0}[/tex] ......(a)
(b) Now, it is said that the collision will occur when ball 1 is at it's highest point. That is, it's final velocity must be zero.
This time we shall have to use another equation of motion given by,
[tex]v^2=u^2+2aS[/tex]
where, [tex]v[/tex] = final velocity
therefore, we get for ball 1,
[tex]0=v_0^2-2gx[/tex] ([tex]u=v_0,v=0,a=-g[/tex])
or, [tex]x=\frac{v_0^2}{2g}[/tex]
Putting the value of [tex]x[/tex] in equation (2) and rearranging, we get,
[tex]\frac{g}{2v_0^2}H^2-H+\frac{v_0^2}{2g}=0[/tex]
which is a quadratic equation, whose solution is given by,
[tex]H=\frac{+1\pm\sqrt{(-1)^2-(4\times\frac{g}{2v_0^2} \times\frac{v_0^2}{2g}) } }{2\times\frac{g}{2v_0^2} }[/tex]
[tex]=\frac{v_0^2}{g}[/tex]
(a) The time at which the balls collide is H/[tex]v_{0}[/tex]
(b) The height H is equal to [tex]\frac{v_{0} ^{2} }{g}[/tex]
Let the balls collide at a height x above the ground.
Then the distance traveled by the ball thrown above is x.
And the distance traveled by the ball dropped from height H is (H-x).
(i) Both the balls will take the same time to travel respective distances in order to collide.
[tex]H-x=\frac{1}{2}gt^{2}[/tex]
[tex]x = v_{0}t - \frac{1}{2}gt^{2}[/tex]
We get:
[tex]x=v_{0}t-(H-x)[/tex]
[tex]t=\frac{H}{v_{0}}[/tex] , is the time after which the balls collide.
(ii) Let the ball thrown up attains its maximum height x at the time of thecollision
[tex]v^{2} = u^{2}-2gx[/tex] here v is the final velocity which is 0 when the ball attains maximum height
[tex]0=v_{0} ^{2}-2gx[/tex]
[tex]x=\frac{v_{0} ^{2} }{2g}[/tex] is the maximum height attained.
Now, the ball thrown downward travels distance (H-x) just before collision:
[tex]H-x=\frac{1}{2}gt^{2}[/tex]
[tex]H-\frac{v_{0} ^{2} }{2g}=\frac{1}{2}g\frac{H^{2} }{v_{0} ^{2} }[/tex]
Solving the quadratic equation we get:
[tex]H=\frac{v_{0} ^{2} }{g}[/tex]
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You are on your balcony and notice some bad squirrels digging in your garden directly below. You start tossing your pistachios at them to get them out of your yard. If the height of your guard rail is 43 inches above the deck, which is at a standard height of 10 feet above the ground, and you toss the pistachios straight down with a speed of 7.4 m/s, then how long does it take for the pistachio to reach the ground?
Answer:
0.43 s
Explanation:
We have the following parameters:
Initial velocity, u = 7.4 m/s
Acceleration of gravity, g = 9.8 [tex]m/s^2[/tex]
Distance, s = 43 in + 10 ft = 1.092 m + 3.048 m = 4.14 m
Time, t = ?
Using the equation of motion [tex]s=ut +\frac{1}{2}gt^2[/tex], we have
[tex]4.14 = 7.4t + 0.5\times9.8t^2[/tex]
[tex]4.9t^2 + 7.4t - 4.14 =0[/tex]
Using the quadratic formula [tex]\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex] where a = 4.9, b = 7.4 and c = - 4.14, and solving for the positive value of t only, we have
[tex]t = 0.43[/tex] s
A vector that is 7.1 units long and a vector that is 5.0 units long are added. Their sum is a vector 7.6 units long. (a) Show graphically at least one way that the vectors can be added. (Do this on paper. Your instructor may ask you to turn in this work.) (b) Using your sketch in Part (a), determine the angle between the original two vectors. °
Answer:
[tex]\theta=119.8138^{\circ}[/tex]
Explanation:
Given:
magnitude of first vector, [tex]A=7.1\ units[/tex]magnitude of second vector, [tex]B=5.0\ units[/tex]resultant of magnitude of the two vectors, [tex]R=7.6\ units[/tex]a)
From the vector addition rule we know:
[tex]R^2={A^2+B^2+2A.B\cos\theta}[/tex]
where
[tex]\theta=[/tex] angle between the two vectors with their tail at common point.
b)
Putting respective values:
[tex]7.6^2=7.1^2+5^2+7.1\times 5\times \cos \theta[/tex]
[tex]\theta=119.8138^{\circ}[/tex]
To add the two vectors, draw them graphically and connect the tip of one vector to the tail of the other. The angle between the original two vectors can be determined using the adjacent and hypotenuse sides of a right triangle.
Explanation:Part (a)To add the two vectors, draw the first vector as a line segment with a length of 7.1 units and draw the second vector as a line segment with a length of 5.0 units. Place the tail of the second vector at the tip of the first vector. The sum of the two vectors is the line segment connecting the tail of the first vector to the tip of the second vector. This sum vector should have a length of 7.6 units, as mentioned in the question.
Part (b)Using the sketch from Part (a), draw a horizontal line from the tail of the first vector to the tip of the second vector. Next, draw a vertical line from the tip of the first vector to the tip of the second vector. Connect the tail of the first vector to the tip of the second vector to form a right triangle. The angle between the original two vectors is the angle opposite to the horizontal line. To determine this angle, we can use the trigonometric inverse function. Let's call the angle 'theta'. We have the adjacent side as 5.0 units and the hypotenuse as 7.6 units. Using the arccosine function (inverse cosine), we can calculate theta using the equation: theta = arccos(5.0/7.6).
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In what regions of the electromagnetic spectrum is the atmosphere transparent enough to allow observations from the ground?
Answer:
Visible Light and Radio waves
Explanation:
The earth's atmosphere is transparent to a few windows in the electromagnetic spectrum. it is completely transparent to allow observation from the ground in visible light rang 380 to 740 nano meters. Also in the range of radio wave as communication are done from space to ground in the form of radio waves.
it is Partially transparent to Microwave and infrared range.
The atmosphere is transparent to certain regions of the electromagnetic spectrum, such as visible light, radio waves, and parts of the infrared and microwave spectra.
The Earth's atmosphere is transparent to certain regions of the electromagnetic spectrum, allowing observations to be made from the ground. These regions include the visible light spectrum, the radio waves spectrum, and parts of the infrared and microwave spectra.
In these regions, electromagnetic waves can pass through the Earth's atmosphere relatively easily, allowing ground-based observations to be conducted.
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What is the boiling point of an aqueous solution that freezes at -2.05 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your answer using 2 decimal places!!!!
To solve this problem we will apply the concepts of Boiling Point elevation and Freezing Point Depression. The mathematical expression that allows us to find the temperature range in which these phenomena occur is given by
Boiling Point Elevation
[tex]\Delta T_b = K_b m[/tex]
Here,
[tex]K_b[/tex] = Constant ( Different for each solvent)
m = Molality
Freezing Point Depression
[tex]\Delta T_f = K_f m[/tex]
Here,
[tex]K_f =[/tex] Constant ( Different for each solvent)
m = Molality
According to the statement we have that
[tex]\Delta T_f = T_0 -T_f = 0-(-2.05)[/tex]
[tex]\Delta T_f = 2.05\°C[/tex]
From the two previous relation we can find the ratio between them, therefore
[tex]\frac{\Delta T_b}{\Delta T_f} = \frac{K_b}{K_f}[/tex]
[tex]\frac{\Delta T_b}{\Delta T_f} = \frac{0.512}{1.86}[/tex]
[tex]\frac{\Delta T_b}{\Delta T_f} = 0.275[/tex]
We already know the change in the freezing point, then
[tex]\Delta T_b = 0.275 (\Delta T_f)[/tex]
[tex]\Delta T_b = 0.275 (2.05)[/tex]
[tex]\Delta T_b = 0.5643\°C[/tex]
The temperature difference in the boiling point is 100°C (Aqueous solution), therefore
[tex]T_b -100 = 0.5643[/tex]
[tex]T_b = 100.56\°C[/tex]
Therefore the boiling point of an aqueous solution is [tex]100.56\°C[/tex]
Final answer:
The boiling point of the aqueous solution is 99.44 °C.
Explanation:
The boiling point of an aqueous solution can be determined using the formula:
Tbp = Tbpsolvent + (Kbp * m)
where:
Tbp is the boiling point of the solution
Tbpsolvent is the boiling point of the pure solvent
Kbp is the ebullioscopic constant for the solvent
m is the molality of the solution
In this case, we are given that the freezing point of the solution is -2.05 degrees C. We can use the formula for freezing point depression to find the molality:
AT = Kf * m
Rearranging the formula, we can solve for m:
m = AT / Kf
Substituting the given values:
m = (-2.05 °C) / (1.86 °C kg.mol-¹)
Rounding to 2 decimal places:
m = -1.10 mol/kg
Now, we can use the formula for boiling point elevation to find the boiling point:
Tbp = Tbpsolvent + (Kbp * m)
Substituting the given values:
Tbp = 100.00 °C + (0.512 °C/m * -1.10 mol/kg)
Calculating:
Tbp = 100.00 °C - 0.56 °C = 99.44 °C
Therefore, the boiling point of the aqueous solution is 99.44 °C.