Answer:
(a) [tex]v_{1}=0+a_{1}t_{1} = a_{1}t_{1}[/tex].
(b) [tex]v_{final}=v_{initial}+at\\ v_{2}=v_{1}+a_{2} t_{2}[/tex]
(c) [tex]219807.5m[/tex]
Explanation:
Part (a):To find an expression for the rocket's speed [tex]v_{1}[/tex] at time [tex]t_{1}[/tex], we use the constant acceleration model, which relates these variables with the expression:[tex]v_{final} =v_{intial}+at[/tex]. In this case, the initial velocity is null, because accelerates from rest. So, we take all values of the first interval, and we replace it to find the expression:
[tex]v_{1}=0+a_{1}t_{1} = a_{1}t_{1}[/tex]. (expression for the first interval).
Part (b):Then, we do the same process to find the expression of the second interval, we just replace the variables given:
[tex]v_{final}=v_{initial}+at\\ v_{2}=v_{1}+a_{2} t_{2}[/tex]
In this case, you can notice that the initial velocity used is the one we obtain from the first interval, because the end of the first period is the beginning of the second period.
Part (c):To calculate the total distance we have to sum the distance covered during the two intervals, that it's translated as: [tex]d_{total} = d_{1} + d_{2}[/tex].
Then, we use this equation to replace in each distance: [tex]v_{final}^{2} = v_{initial}^{2} +2ad[/tex].
Isolating d we have:
[tex]d=\frac{v_{final}^{2}-v_{initial}^{2}}{2a}[/tex].
Now, we apply the equation to each interval to obtain [tex]d_{1}[/tex] and [tex]d_{2}[/tex]:
[tex]d_{1}=\frac{v_{1}^{2}-0}{2a_{1}}[/tex].
[tex]d_{2}=\frac{v_{2}^{2}-v_{1}^{2}}{2a_{2}}[/tex].
Before calculating the total distance, we need to know the magnitude of each speed.
[tex]v_{1}=a_{1}t_{1}[/tex]
[tex]v_{1}=(67)(39)=[tex]v_{2}=v_{1}+a_{2} t_{2}= 2613\frac{m}{s^{2}}+34\frac{m}{s^{2}}(49sec)=4279\frac{m}{s}[/tex][/tex]
At last, we use all values known to calculate the total distance:
[tex]d_{total} = d_{1} + d_{2}[/tex]
[tex]d_{total} =\frac{v_{1}^{2}-0}{2a_{1}} + \frac{v_{2}^{2}-v_{1}^{2}}{2a_{2}}[/tex].
[tex]d_{total} =\frac{(2613)^{2} }{2(67)} +\frac{(4279)^{2}-(2613)^{2} }{2(34)}\\ d_{total}=50953.5+168854=219807.5m[/tex]
Therefore, the total distance traveled until the ends of the second period is [tex]219807.5m[/tex]. The rocket is in the Thermosphere.
What charge does a parallel plate capacitor having plates of area 1.4 m^2 that are separated by 0.04 mm of neoprene rubber hold when 9.00 V is applied to it? Give answer in terms of 10^-5 C. (Assume that neoprene has a dielectric constant of 6.7 or a dielectric strength of 12×10^6 V/m)
Answer:
1.86 75 X 10⁻⁵ C.
Explanation:
Capacitance of parallel plate capacitor is given by
C = K∈₀ A /d
C is capacitance , A is plate area , d is distance between plate and ∈₀ is permittivity of air which is 8.85 x 10⁻¹² and K is dielectric constant
Using the data given in the question
C = [tex]\frac{6.7\times 8.85\times10^{-12}\times 1.4}{.04\times10^{-3}}[/tex]
C = .2075 X 10⁻⁵ F
Charge Q = CV
.2075 X 10⁻⁵ X 9 = 1.86 75 X 10⁻⁵ C.
At what partial pressure are argon atoms expected to have a free travel of approximately 5 µm, if the gas is at a temperature of 400 K? The cross section of collision, σ, or Argon is 0.28 nm2
Ar molar mass is 39.9 g/mole
Answer:
2790 Pa
Explanation:
Given wavelength λ= 5μm
temperature T= 400 K
cross section of collision σ= 0.28 nm^2
molar mass = 39.9 g/mole
pressure = [tex]P= \frac{RT}{\sqrt{2}N_A\sigma\lambda }[/tex]
putting values we get
=[tex] \frac{8.314\times400}{\sqrt{2}\times6.022\times10^{23}\times0.28\times10^{-18}\times5\times10^{-6} }[/tex]
⇒P = 2790 J/m^3
the partial pressure are argon atoms expected= 2790 Pa
A ball is dropped from rest at the top of a 6.10 m
tallbuilding, falls straight downward and collides inelastically
withthe ground, and bounces back. The ball loses 10% of itskinetic
energy every time it collides with the ground. Howmany bounces can
the ball make and still reach a windowsill that is2.38 m above the
ground?
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
[tex]\frac{v_1}{v}[/tex] = e ( coefficient of restitution ) = [tex]\frac{1}{\sqrt{10} }[/tex]
and
[tex]\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }[/tex]
h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that
[tex]e = \sqrt{\frac{h_1}{6.1} }[/tex]
[tex]e = \sqrt{\frac{h_2}{h_1} }[/tex]
So on
[tex]e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }[/tex]
[tex](\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}[/tex]= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
The ball can bounce approximately 8 times before it reaches the height of 2.38 meters after accounting for a 10% energy loss each collision.
This problem involves a ball undergoing inelastic collisions, losing 10% of its kinetic energy with each bounce, and determining how many times it can bounce to still reach a windowsill 2.38 m high.
First, let's calculate the initial potential energy (PEinitial) of the ball when it is dropped from a height (hinitial) of 6.10 m:
PEinitial = mghinitial
where, g = 9.8 m/s² (acceleration due to gravity)
As it falls, this entire potential energy converts into kinetic energy (KEinitial) at the ground:
KEinitial = PEinitial = mghinitial
Upon each bounce, the ball loses 10% of its kinetic energy. Therefore, it retains 90% of its kinetic energy:
KEnew = 0.9 × KEprevious
To find out how high it can bounce after each loss of energy, convert kinetic energy back into potential energy:
PE = KE = mgh
After each bounce, the height the ball can reach is calculated by applying the 90% retention factor:
hnew = 0.9 × hprevious
Starting with h0 = 6.10 m:
Next height (h1) = 0.9 × 6.10 m = 5.49 mNext height (h2) = 0.9 × 5.49 m = 4.94 mNext height (h3) = 0.9 × 4.94 m = 4.45 mContinue this process until hn is less than 2.38 m:The number of bounces can be calculated using the formula: hn = 6.10 × (0.9)n, where n is the number of bounces. Set hn = 2.38 m and solve for n:
2.38 = 6.10 × (0.9)n
Divide both sides by 6.10:
0.39 ≈ (0.9)n
Taking the natural log (ln) of both sides:
ln(0.39) = n × ln(0.9)
Finally, solving for n:
n ≈ ln(0.39) / ln(0.9) ≈ 8
Therefore, the ball can bounce approximately 8 times and still reach the windowsill that is 2.38 m above the ground.
Two rockets are fired at each other with initial velocities of 150m/s150m/s and are 6000m6000m apart. The first rocket is accelerating at 5m/s25m/s2 while the second is accelerating at 15m/s215m/s2 . How far has the second rocket gone (in meters) when they collide? A :
Answer:
3469.788 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
First rocket
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 5\times t^2\\\Rightarrow s=150t+2.5t^2\ m[/tex]
Second rocket
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 15\times t^2\\\Rightarrow s=150t+7.5t^2\ m[/tex]
When this will collide the total distance they would have covered would be 6000 m.
[tex]6000=150t+2.5t^2+150t+7.5t^2\\\Rightarrow 6000=300t+10t^2\\\Rightarrow 10t^2+300t-6000=0[/tex]
[tex]t=5\left(\sqrt{33}-3\right),\:t=-5\left(3+\sqrt{33}\right)\\\Rightarrow t=13.72, -43.72[/tex]
Hence at 13.72 seconds they will collide assuming they are launched at the same time.
[tex]s=150t+7.5t^2\\\Rightarrow s=150\times 13.72+7.5\times 13.72^2\\\Rightarrow s=3469.788\ m[/tex]
The second rocket would have gone 3469.788 m when they collide
Final answer:
To find the distance covered by the second rocket at the point of collision, we analyze both rockets' motion based on their initial velocities and accelerations. By using the equations of motion, we can solve for the time of collision and calculate the distance each rocket has traveled, specifically looking for the second rocket's distance.
Explanation:
To find how far the second rocket has gone when they collide, we must analyze the motion of both rockets taking into account their initial velocities and accelerations. The first rocket has an initial velocity of 150 m/s and accelerates at 5 m/s2, while the second rocket starts with the same velocity but accelerates at 15 m/s2. Both rockets start 6000 meters apart. To find the point of collision, we use the equation of motion s = ut + (1/2)at2 for both, where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time until collision.
Let's denote the distance covered by the first rocket as s1 and the second as s2, with the total distance being s1 + s2 = 6000 m. By finding the time of collision and plugging it back into the equation of motion for the second rocket, we can find s2, the distance covered by the second rocket. The actual calculation involves setting up equations based on the motion of both rockets and solving for the time t, after which s2 can be calculated.
Light in vacuum is incident on the surface of a slab of transparent material. In the vacuum the beam makes an angle of 27.0° with the normal to the surface, while in the slab it makes an angle of 21.8° with the normal. What is the index of refraction of the transparent material?
Answer:
Refractive index of slab = 1.22
Explanation:
Using Snell's law as:
[tex]n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}[/tex]
Where,
[tex]{\theta_i}[/tex] is the angle of incidence ( 27.0° )
[tex]{\theta_r}[/tex] is the angle of refraction ( 21.8° )
[tex]{n_r}[/tex] is the refractive index of the refraction medium (slab, n=?)
[tex]{n_i}[/tex] is the refractive index of the incidence medium (vacuum, n=1)
Hence,
[tex]1\times {sin27.0^0}={n_r}\times{sin21.8^0}[/tex]
Refractive index of slab = 1.22
The index of refraction of the transparent material is found using Snell's law and is approximately 1.333.
To find the index of refraction of the transparent material, we can use Snell's law, which states the relationship between the angles of incidence and refraction at the boundary between two media with different refractive indices. Snell's law is given by the equation:
n1sin(θ1) = n2sin(θ2)
Where n1 is the refractive index of the first medium (vacuum in this case, which is 1), θ1 is the angle of incidence, n2 is the refractive index of the second medium (the transparent material), and θ2 is the angle of refraction.
Here, we have θ1 = 27.0° and θ2 = 21.8°. Plugging in the known values:
1 * sin(27.0°) = n2 * sin(21.8°)
n2 = sin(27.0°) / sin(21.8°)
Using a calculator, we find:
n2 ≈ 1.333
Thus, the index of refraction of the transparent material is approximately 1.333.
The only two forces acting on a body have magnitudes of 20 N and 35 N and directions that differ by 80°. The resulting acceleration has a magnitude of 20 m/s^2. What is the mass of the body? O 2.4kg O 2.2kg O 2.7kg O 3.1kg
Answer:
b) 2.2 kg
Explanation:
Net force acting on an object is the sum of the two forces acting on the body.
The net force is calculated using the parallelogram law of vectors.
F =[tex]\sqrt{{A^{2}} + B^{2}+2 A B cos \theta}[/tex]
Here A = 20 N , B = 35 N and θ =80°
Net Force = F = 43.22 N
Acceleration = a = 20 m/s/s
Since F = ma, m = F/a = 43.22 / 20 = 2.161 kg = 2.2 kg
The weight of Earth's atmosphere exerts an average pressure of 1.01 ✕ 105 Pa on the ground at sea level. Use the definition of pressure to estimate the weight of Earth's atmosphere (in N) by approximating Earth as a sphere of radius RE = 6.38 ✕ 106 m and surface area A = 4πRE2. HINT N
Answer:
The weight of Earth's atmosphere exert is [tex]516.6\times10^{17}\ N[/tex]
Explanation:
Given that,
Average pressure [tex]P=1.01\times10^{5}\ Pa[/tex]
Radius of earth [tex]R_{E}=6.38\times10^{6}\ m[/tex]
Pressure :
Pressure is equal to the force upon area.
We need to calculate the weight of earth's atmosphere
Using formula of pressure
[tex]P=\dfrac{F}{A}[/tex]
[tex]F=PA[/tex]
[tex]F=P\times 4\pi\times R_{E}^2[/tex]
Where, P = pressure
A = area
Put the value into the formula
[tex]F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2[/tex]
[tex]F=516.6\times10^{17}\ N[/tex]
Hence, The weight of Earth's atmosphere exert is [tex]516.6\times10^{17}\ N[/tex]
Answer:
5.164 x 10^19 N
Explanation:
P = 1.01 x 10^5 Pa
R = 6.38 x 10^6 m
Area = 4 π R²
[tex]A= 4\times 3.14\times6.38\times10^{6}\times6.38\times10^{6}[/tex]
A = 5.112 x 10^14 m^2
Pressure is defined as the force exerted per unit area.
The formula for the pressure is
P = F / A
Where, F is the force, A be the area
here force is the weight of atmosphere.
F = P x A
F = 1.01 x 10^5 x 5.112 x 10^14
F = 5.164 x 10^19 N
Water has a mass per mole of 18.0 g/mol, and each water mo- lecule (H2O) has 10 electrons. (a) How many electrons are there in one liter (1.00 x 10-3 m) of water? (b) What is the net charge of all these electrons?
Answer:
Part 1) Number of electrons in 1 liter of water equals[tex]N=3.346\times 10^{26}[/tex]
Part 2) Net charge of all the electrons equals [tex]Charge=53.61\times 10^{6}[/tex]
Explanation:
Since we know that the density of water is 1 kilogram per liter thus we infer that mass of 1 liter of water is 1 kilogram hence we need to find electron's in 1 kg of water.
Now since it is given that molar mass of water is 18.0 grams this means that 1 mole of water contains 18 grams of water.
Hence by ratio and proportion number of moles in 1 kg water equals
[tex]n=\frac{1000}{18}[/tex]
Now by definition of mole we know that 1 mole of any substance is Avagadro Number of particles.
Hence the no of molecules in 'n' moles of water equals
[tex]n'=\frac{1000}{18}\cdot N_a\\\\n'=\frac{1000}{18}\cdot 6.023\times 10^{23}\\[\tex][tex]\\n'=3.346\times 10^{25}[/tex]
Now since it is given that each molecule has 10 electron's thus the total number of electrons in n' molecules equals
[tex]N=10\times 3.346\times 10^{25}\\\\N=3.346\times 10^{26}[/tex]
Part 2)
We know that charge of 1 electron equals [tex]1.602\times 10^{-19}C[/tex] the the charge of electrons in 'N' quantity equals
[tex]Charge=1.6022\times 10^{-19}\times 3.346\times 10^{26}\\\\Charge=52.61\times 10^{6}Columbs[/tex]
A person is standing on a level floor. His head,
uppertorso, arms, and hands together weigh 438 N and have a center
ofgravity that is 1.28 m above the floor. His upper legs weigh 144
Nand have a center of gravity that is 0.760 m above the
floor.Finally, his lower legs and feet together weigh 87 N and
havea center of gravity that is 0.250 m above the floor. Relative
tothe floor, find the location of the center of gravity for
theentire body.
Answer:
The location of the center of gravity for the entire body, relative to the floor is 1.03 m
Explanation:
To find the center of gravity of a system of particles, we use that
[tex]R_{cog} = \frac{r_{1}*m_{1}+r_{2}*m_{2}+...+r_{n}*m_{n}}{M}[/tex]
where R is the vector center of gravity of the system, formed by n particles, and n masses.
In this case, for a person standing on the floor and being their body divided in three sectors, each one with a weight and an altitud in a specific point (center of gravity of the body sector), instead of mass we have every "particle" weight in Newtons (force instead of mass), being each "particle" in the formula, a sector of the body.
On the other hand, we use only magnitude for the calculation, because the gravity force is vertical to the floor, so instead of our vector formula, we use it in the vertical direction as a magnitude formula. Thus
[tex]Y_{cog} = \frac{y_{1}*W_{1}+y_{2}*W_{2}+y_{3}*W_{3}}{Wt}[/tex]
where Y is the center of gravity, y=1, 2, 3 is every "sector point" altitude from the floor, W=1, 2, 3 is every weight of a body "sector", and Wt is the sum of the three weights.
In this way we replace in our formula with the correspondent values
[tex]Y_{cog} = \frac{1.28m*438N+0.76m*144N+0.25m*87N}{669N}[/tex]
obtaining our result
[tex]Y_{cog}=1.03 m[/tex]
The spectrum from a hydrogen vapour lamp is measured and four lines in the visible light range are observed. These lines are the so-called Balmer series, where an electron makes a transition from a higher level to the second energy level (n2). In this series, the transition from n, 3 to ne 2 produces the photon with the lowest energy, this corresponds to the line with the longest wavelength. This is measured to be A 658.9 nm What value of the Rydberg constant R is obtained using these measurements? Express you answer in um to two decimal places. It is not necessary to specify the units.
Answer:
[tex]10942249.24 m^{-1}[/tex]
Explanation:
Rydberg's formula is used to describe the wavelengths of the spectral lines of chemical elements similar to hydrogen, that is, with only one electron being affected by the effective nuclear charge. In this formula we can find the rydberg constant, knowing the wavelength emitted in the transcision between two energy states, we can have a value of the constant.
[tex]\frac{1}{\lambda}=Z^2R(\frac{1}{n^2_{1}}-\frac{1}{n^_{2}^2}})[/tex]
Where [tex]\lambda[/tex] it is the wavelength of the light emitted, R is the Rydberg constant, Z is the atomic number of the element and [tex]n_{1} n_{2}[/tex] are the states where [tex]n_{1}<n_{2}[/tex].
In this case we have Z=1 for hydrogen, solving for R:
[tex]R=\frac{1}{\lambda}*(\frac{1}{n^2_{1}}-\frac{1}{n^_{2}^2}})^{-1}\\R=\frac{1}{658.9*10^{-9}m}*(\frac{1}{2^2}-\frac{1}{3^2}})^{-1}\\R=1.52*10^6m^{-1}*(\frac{36}{5})=1.09*10^7 m^{-1}=10942249.24m^{-1}[/tex]
This value is quite close to the theoretical value of the constant [tex]R=10967758.34 m^{-1}[/tex]
When an object is thrown upwards and reaches its maximum height its speed is: a. Greater than the initial
b. Less than the initial
c. Same as the initial
d. Zero
Answer:
Option d
Explanation:
When we throw an object in the upward direction, we provide it with certain initial velocity due to which it covers a certain distance up to the maximum height.
While the object is moving in the upward direction, its velocity keeps on reducing due to the acceleration due to gravity which acts vertically downwards in the opposite direction thus reducing its velocity.
So, the maximum height attained by the object is the point where this upward velocity of the body becomes zero and after that the object starts to fall down.
A 70kh man jumping from a window lands in an elevated
firerescue net 11.0 m below the window. He momentarily stops when
hehas stretched the net by 1.50 m. Assuming that mechanicalenergy
is conserved during this process and the net functions likean ideal
spring, find the leastic potential energy of the net whenit is
stretched by 1.50 m.
Answer:
potential energy of net is 8.58 kJ
Explanation:
given data
mass m = 70 kg
height h = 11 m
stop distance x = 1.50 m
to find out
potential energy of net
solution
we know here man eventually stop
so elastic potential energy of net = change in potential energy of man body
so equation will be
potential energy of net = m×g×h ...................1
here m is mass and h is total height = ( h+ x)
height = 11 + 1.5 = 12.5 m and g is 9.8
so put here value in equation 1
potential energy of net = m×g×h
potential energy of net = 70×9.8×12.5
potential energy of net = 8575 J
so potential energy of net is 8.58 kJ
Earth orbits around the Sun at an average speed of 30 km/s. How far does the Earth move in a year (3.16 x 107 s) as it revolves around the Sun? Express your answer in kilometers using the correct number of significant figures
The Earth moves 9.48 × 10⁸ kilometers in a year as it revolves around the Sun, calculated by multiplying the average speed of 30 km/s by the time of one year, 3.16 × 10⁷ seconds.
Explanation:To calculate the distance that Earth moves in a year as it revolves around the Sun, we can use the formula for distance traveled, which is distance = speed × time. Given that Earth orbits around the Sun at an average speed of 30 kilometers per second (km/s) and the time it takes for one orbit is 3.16 × 107 seconds, the calculation would be:
30 km/s × 3.16 × 107 s = distance
Distance = 30 × 3.16 × 107 km
Distance = 94.8 × 107 km
Distance = 9.48 × 108 km
The Earth moves 9.48 × 108 kilometers in a year as it revolves around the Sun, using the correct number of significant figures.
Problem 6: A rocket accelerates upward from rest, due to the first stage, with a constant acceleration of a1 = 94 m/s2 for t1 = 23 s. The first stage then detaches and the second stage fires, providing a constant acceleration of a2 = 39 m/s2 for the time interval t2 = 34 s. Part (a) Enter an expression for the rocket's speed v1 at time t1 in terms of the variables provided. Part (b) Enter an expression for the rocket s speed v2 at the end of the second period of acceleration in terms of the variables provided in the problem statement Part (c) Using vow expressions for speeds v1 and v2 calculate the total distance trailed in meters, by the rocket from launch until the end of the second period of acceleration.
Answer:
a) v1 = a1*t1 = 2162 m/s
b) v2 = v1 + a2*(t2-t1) = 2591 m/s
c) [tex]Dt = D1 + D2 = \frac{v1^{2}}{2*a1} + \frac{v2^{2}-v1^{2}}{2*a2}=51004.5m[/tex]
Explanation:
For any movement with constant acceleration:
Vf = vo + a*Δt. Replacing the propper values, with get the answers for parts a) and b):
v1 = a1*t1 = 2162 m/s
v2 = v1 + a2*(t2-t1) = 2591 m/s
Using the formula for displacement we can calculate the total distance asked on part c):
[tex]V_{f}^{2}=V_{o}^2+2*a*D[/tex] Solving for D and replacing the values for each part of the launch:
[tex]D=\frac{V_{f}^{2}-V_{o}^{2}}{2*a}[/tex]
D1 = 24863m
D2 = 26141.5m
Finally we add D1 + D2 for the total distance:
D = 51004.5m
A massless spring is attached to a support at one end and has a 2.0 μC charge glued to the other end. A -4.0 μC charge is slowly brought near. The spring has stretched 1.2 cm when the charges are 2.2 cm apart. What is the spring constant of the spring? Express your answer with the appropriate units.
The spring constant of the spring is -1.376 N/m.
Explanation:To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. In this case, the displacement of the spring is 1.2 cm, and the charges are 2.2 cm apart. The force exerted by the spring can be calculated using the equation F = kx, where F is the force, k is the spring constant, and x is the displacement.
Given that the displacement (x) is 1.2 cm and the force exerted is caused by the electric force between the charges, which is given by Coulomb's Law, we can write the equation:
F = kx = k(1.2 cm) = (k/100 cm) * 1.2 cm = k/83.33
Similarly, the electric force between the charges is given by Coulomb's Law: F = kq1q2/r^2, where k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between them. In this case, the charges are 2.2 cm apart and have magnitudes of 2.0 μC and -4.0 μC, respectively.
When the charges are 2.2 cm apart, we can calculate the electric force:
F = k(2.0 μC)(-4.0 μC)/(2.2 cm)^2 = (-8.0 kμC^2)/(4.84 cm^2) = (-8.0 k)/(4.84 cm^2) μC^2
Equating the force exerted by the spring to the electric force, we have:
k/83.33 = (-8.0 k)/(4.84 cm^2) μC^2
Solving for k:
k/83.33 = (-8.0 k)/(4.84 cm^2) μC^2
k * (4.84 cm^2)/(83.33) = -8.0 k * μC^2
(4.84 cm^2)/(83.33) = -8.0 μC^2
k = (-8.0 μC^2) * (83.33)/(4.84 cm^2) = -137.6 μC^2/cm^2 = -1.376 N/m
Therefore, the spring constant of the spring is -1.376 N/m.
Learn more about spring constant here:https://brainly.com/question/14159361
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An ac series circuit has an impedance of 60 Ohm and
aresistance of 30 Ohm. What is the power factor?
Answer:
Power factor of the AC series circuit is [tex]cos\phi=0.5[/tex]
Explanation:
It is given that,
Impedance of the AC series circuit, Z = 60 ohms
Resistance of the AC series circuit, R = 30 ohms
We need to find the power factor of the circuit. It is given by :
[tex]cos\phi=\dfrac{R}{Z}[/tex]
[tex]cos\phi=\dfrac{30}{60}[/tex]
[tex]cos\phi=\dfrac{1}{2}[/tex]
[tex]cos\phi=0.5[/tex]
So, the power factor of the ac series circuit is [tex]cos\phi=0.5[/tex]. Hence, this is the required solution.
The power factor of an AC series circuit with an impedance of 60 Ohms and a resistance of 30 Ohms is 0.5, indicating that half of the power is effectively used.
Explanation:The power factor in an AC circuit is a measure of how much of the power is being effectively used to do work. It's calculated as the ratio of the resistance R to the impedance Z. Given an AC series circuit with an impedance of 60 Ohms and a resistance of 30 Ohms, the power factor is the resistance divided by the impedance.
The calculation is straightforward: Power factor = R / Z = 30 Ohms / 60 Ohms = 0.5.
This means that the power factor for this particular AC circuit is 0.5, which indicates that only half of the total power is being used effectively while the other half is reactive power, which does not perform any work but is necessary for the functioning of the circuit's reactive components.
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An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular acceleration of 0.5 rad.s?. The magnitude of the applied torque is: a) 0.30 N.m b) 3.0 x 102 N.m C) 0.15 N.m d) 7.5 x 102 N.m e) 6.0 x 102 N.m.
Answer:
0.06 Nm
Explanation:
mass of object, m = 3 kg
radius of gyration, k = 0.2 m
angular acceleration, α = 0.5 rad/s^2
Moment of inertia of the object
[tex]I = mK^{2}[/tex]
I = 3 x 0.2 x 0.2 = 0.12 kg m^2
The relaton between the torque and teh moment off inertia is
τ = I α
Wheree, τ is torque and α be the angular acceleration and I be the moemnt of inertia
τ = 0.12 x 0.5 = 0.06 Nm
An electron with speed v0 = 5.08×10^6 m/s is traveling parallel to an electric field of magnitude E = 9100 N/C . Part A How far will the electron travel before it stops? Express your answer to three significant figures and include the appropriate units.
Part B How much time will elapse before it returns to its starting point? Express your answer to three significant figures and include the appropriate units.
Answer:
Part A: [tex]8.06\times10^{-3}\ m[/tex]
Part B: [tex]6.36\times 10^{-9}\ s[/tex]
Explanation:
Given:
[tex]v_o[/tex] = initial velocity of the electron = [tex]5.08\times 10^{6}\ m/s[/tex][tex]E[/tex] = electric field strength = 9100 N/CAssumptions:
[tex]m[/tex] = mass of the electron = [tex]9.1\times 10^{-31}\ kg[/tex][tex]v[/tex] = final velocity of the electron = 0 m/s[tex]x[/tex] = distance at which the electron comes to rest[tex]e[/tex] = magnitude of charge on an electron = [tex]1.6\times 10^{-19}\ C[/tex][tex]t[/tex] = taken by the electron to return to its initial position[tex]s[/tex] = displacement of the electronPart A:
Since the electron moves in the direction of the electric field, the electric force will act on it in the direction opposite to electric field. This electric force does work on it to make the electron come to rest.
Using the work-energy theorem, the work done by the electric field will be equal to the kinetic energy change of the electron.
[tex]\therefore -eEx = \dfrac{1}{2}m(v^2-v_o^2)\\\Rightarrow -eEx=-\dfrac{1}{2}mv_o^2\\\Rightarrow x=\dfrac{mv_o^2}{2eE}\\\Rightarrow x=\dfrac{9.1\times 10^{-31}\times (5.08\times 10^{6})^2}{2\times 1.6\times10^{-19}\times 9100}\\\Rightarrow x=8.06\times 10^{-3}\ m[/tex]
Hence, the electron comes to rest by travelling a distance of [tex]8.06\times 10^{-3}\ m[/tex].
Part B:
In this part, let us first find out the acceleration of the electron due to the electric force.
[tex]a = -\dfrac{eE}{m}\\\Rightarrow a= -\dfrac{1.6\times10^{-19}\times 9100}{9.1\times 10^{-31}}\\\Rightarrow a= -1.6\times 10^{15}\ m/s^2\\[/tex]
The electron moves with the above acceleration constantly as it moves in the uniform electric field.
Since the electron is supposed to move from a point and then again move back to the same point. This means the displacement of the electron is zero.
[tex]i.e.,\ s=0\\\Rightarrow v_ot+\dfrac{1}{2}at^2=0\\\Rightarrow (v_o+\dfrac{1}{2}at)t=0\\\Rightarrow \dfrac{(2v_o+at)}{2}t=0\\\Rightarrow t = 0\,\,\, or\,\,\, (2v_o+at)=0\\\Rightarrow t = 0\,\,\, or\,\,\, t=\dfrac{-2v_o}{a}\\[/tex]
Since the electron starts moving at t = 0 s.
[tex]\therefore t = \dfrac{-2v_o}{a}\\\Rightarrow t=\dfrac{-2\times 5.08\times 10^6}{-1.6\times 10^{15}}\\\Rightarrow t= 6.36\times 10^{-9}\ s[/tex]
Hence, the electron returns to the starting position after [tex]6.36\times 10^{-9}\ s[/tex].
At t =0 one toy car is set rolling on a straight track with intial position 17.0 cm , intial velocity -3 cm/s, and constant acceleration 2.30 cm/s^2 . At the same moment , another toy car is set rolling on an adjacent track with initial position 9.5 cm , intial velocity 5.0 cm/s, and constant zero acceleration. (A) at the time, if any, do the two cars have equal speeds? (B) what are their speeds at that time? (c) at what time(s) , if any , do the cars pass each other? (D) what are their location at that time?
Answer:
a) 5.65 s
b) 5cm/s
c) They will pass each other at both 1.1168 s and 5.84s
d)15.084cm and 38.7 cm
Explanation:
For part A, you need to keep in mind that acceleration is the rate of change of velocity per unit of time. For a constant acceleration, this can be told in this way:
[tex]a = \frac{v - v_o}{t}[/tex]
Reordering this equation, we can get v in terms of the initial velocity, the acceleration, and the time elapsed:
[tex]v = at + v_o[/tex]
Now, we can get the expressions for velocity of each toy car, and equalize them:
[tex]v_1 =a_1t + v_o_1\\v_2 =a_2t + v_o_2\\v_1 = v2\\a_1t +v_o_1 =a_2t + v_o_2\\(a_1 - a_2)t = v_o_2 - v_o_1\\t = \frac{v_o_2 - v_o_1}{a_1 - a_2} = \frac{5cm/s - (-3cm/s)}{2.3 cm/s^2 - 0 cm/s^2}= 3.47 s[/tex]
As toy car has no acceleration and, therefore, constant speed, both car will have the same speed when toy car 1 reaches this velocity = 5cm/s
c) The position of car 1, as it follows a constant acceleration motion, is given by this equation:
[tex]x_1 = \frac{1}{2}a_1t^2 + v_o_1t + x_o_1[/tex]
The position for car 2, as it has constant velocity, is given by this equation:
[tex]x_2 = v_2t + x_o_2[/tex]
We equalize both equation to find the time where the cars pass each other:
[tex]x_1 = x_2\\\frac{1}{2}a_1t^2 + v_o_1t + x_o_1 = v_2t+x_o_2\\\frac{1}{2} a_1t^2 + (v_o_1 - v_2)t + x_o_1 - x_o_2 = 0\\\frac{1}{2}2.3m/s^2t^2 +(-3cm/s-5cm/s)t+ 17cm - 9.5cm = 0\\1.15t^2 -8t + 7.5 = 0 | a = 1.15, b = -8, c = 7.5\\t = \frac{-b +-\sqrt{b^2 - 4ac}}{2a} = 5.84s | 1.1168 s[/tex]
The car will pass each other at both 1.1168s and 5.84s.
For the positions, we solve any of the position equation with the solutions:
[tex]x = v_2*t + x_o_2 = 5cm/s *5.84s + 9.5cm = 38.7 cm\\x = 5cm/s * 1.1168s + 9.5cm = 15.084 cm[/tex]
The two toy cars have equal speeds at approximately t = 3.48s, with both traveling at 5 cm/s. They pass each other at t ≈ 1.99s, with their location at approximately 15.4 cm.
Part A: Equal Speeds:
We know the first car has an initial velocity of -3 cm/s and an acceleration of 2.30 cm/s². Its velocity at any time t can be given by v1 = v0 + at, that is -3 + 2.30t.
The second car has a constant velocity of 5.0 cm/s.
For them to have equal speeds, v1 = v2,
so -3 + 2.30t = 5.
Solving for t gives t ≈ 3.48 s.
Part B: Speeds at Equal Times:
The common speed when both cars have equal speeds can be found by substituting the time back into either equation for velocity.
Doing so for the first car gives
-3 + 2.30(3.48) ≈ 5 cm/s.
Part C: Cars Passing Each Other:
The cars pass each other when their positions are equal. Using the equations for position x1 = x01 + v01t + (1/2)at² and x2 = x02 + v02t,
we get 17 + (-3)t + (1/2)(2.30)t² = 9.5 + (5)t.
Solving for t gives two possible times, but only the positive one is relevant, t ≈ 1.99 s.
Part D: Location at Passing Time:
Substituting the time when they pass each other back into the position equations, for the first car we have
17 + (-3)(1.99) + (1/2)(2.30)(1.99)² ≈ 15.4 cm.
The second car will be at the same position since they are passing each other.
A coin is dropped from a height of 188 meters. Neglecting air resistance, how much time (in seconds) was the coin in the air?
Answer:
[tex]t=6.19s[/tex]
Explanation:
With a initial velocity is zero, kinematics equation is:
[tex]y=1/2*g*t^2[/tex]
[tex]t=\sqrt{2y/g}=\sqrt{2*188/9.81}=6.19s[/tex]
Why is the following situation impossible? Two identical dust particles of mass 1.00 µg are floating in empty space, far from any external sources of large gravitational or electric fields, and at rest with respect to each other. Both particles carry electric charges that are identical in magnitude and sign. The gravitational and electric forces between the particles happen to have the same magnitude, so each particle experiences zero net force and the distance between the particles remains constant.
Answer:
This is a conceptual problem so I will try my best to explain the impossible scenario. First of all the two dust particles ara virtually exempt from any external forces and at rest with respect to each other. This could theoretically happen even if it's difficult for that to happen. The problem is that each of the particles have an electric charge which are equal in magnitude and sign. Thus each particle should feel the presence of the other via a force. The forces felt by the particles are equal and opposite facing away from each other so both charges have a net acceleration according to Newton's second law because of the presence of a force in each particle:
[tex]a=\frac{F}{m}[/tex]
Having seen Newton's second law it should be clear that the particles are actually moving away from each other and will not remain at rest with respect to each other. This is in contradiction with the last statement in the problem.
The situation is impossible because the gravitational force between equally charged particles is much weaker than the electric force, making it impossible for the two forces to have the same magnitude, resulting in non-zero net force and movement.
Explanation:The scenario described by the student is impossible because the gravitational force and electric force between two particles with identical mass and charge are not equal in magnitude. According to physics, specifically Coulomb's Law and Newton's Law of Universal Gravitation, the gravitational force is significantly weaker than electric force when the particles have the same magnitude of charge as that of their mass in standard units.
The electric force (Coulomb's Law) is known to be much stronger than gravitational force. For example, the electric repulsion between two electrons is approximately 1042 times stronger than their gravitational attraction. Therefore, if two dust particles are floating in empty space and are at rest with respect to each other, with identical charges and mass, the electric forces would cause them to repel each other with much greater force than the gravitational forces would attract them. Hence, they cannot remain at a constant distance with zero net force on each other.
How much work must be done to bring three electrons from a great distance apart to 5.0×10^−10 m from one another (at the corners of an equilateral triangle)?
Answer:
1.38 x 10^-18 J
Explanation:
q = - 1.6 x 10^-19 C
d = 5 x 10^-10 m
the potential energy of the system gives the value of work done
The formula for the potential energy is given by
[tex]U =\frac{Kq_{1}q_{2}}{d}[/tex]
So, the total potential energy of teh system is
[tex]U =\frac{Kq_{1}q_{2}}{d}+\frac{Kq_{2}q_{3}}{d}+\frac{Kq_{1}q_{3}}{d}[/tex]
As all the charges are same and the distance between the two charges is same so the total potential energy becomes
[tex]U =3\times \frac{Kq^{2}}{d}[/tex]
K = 9 x 10^9 Nm^2/C^2
By substituting the values
[tex]U =3\times \frac{9\times 10^{9}\times \ 1.6 \times 1.6 \times 10^{-38}}{5\times 10^{-10}}[/tex]
U = 1.38 x 10^-18 J
The work which must be done to bring three electrons from a great distance apart is 5.0×10⁻¹⁰m from one another is 1.38 x 10⁻¹⁸ J
Formula for Work doneq = -1.6 x 10⁻¹⁹C
d = 5 x 10⁻¹⁰m
Potential energy of the system = Value of work done
Potential energy= U = Kq₁q₂/d
It is an equilateral triangle so the the charges and the distance between
the charges are the same.
Potential energy = 3 ₓ kq² / d
Substitute the values into the equation
3 × (9 × 10⁹ ˣ 1.6 ˣ 1.6 ˣ 10⁻³⁸) / (5ˣ 10⁻¹⁰)
U = 1.38 x 10⁻¹⁸ J
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A 6.89-nC charge is located 1.76 m from a 4.10-nC point charge. (a) Find the magnitude of the electrostatic force that one charge exerts on the other. (b) Is the force attractive or repulsive?
Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive
Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:
[tex]F=\frac{k*q1*q2}{d^{2}}[/tex]
Replacing the dat we obtain F=82 nN.
The force is repulsive because the points charged have the same sign.
A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00 s and continues to coast 25.0 m/s^2. It runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height before falling back to Earth. (a) Find the rocket’s height when it runs out of fuel; (b) find the rocket’s velocity when it runs out of fuel; (c) find the maximum height the rocket reaches; (d) find the rocket’s velocity the instant before the rocket crashes into the ground; and (e) find the total elapsed time from launch to ground impact.
Answer:
a) 200 m
b) 100 m/s
c) 709 m
d) -118.2 m/s
e) 26.24 s
Explanation:
The rocket flies upward with constant acceleretion.
The equation for position under constant acceleration is:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 = 0
V0 = 0
Y(t) = 1/2 * 25 * t^2
Y(t) = 12.5 * t^2
And speed under constant acceleration:
Vy(t) = Vy0 + a * t
Vy(t) = 25 * t
It burns for 4 s and runs out of fuel
Y(4) = 12.5 * 4^2 = 200 m
V(4) = 25 * 4 = 100 m/s
Form t = 4 the rocket will coast, it will be in free fall, affected only by gravity
It will be under constant acceleration. These new equations will have different starting constants.
Y(t) = Y4 + Vy4 * (t - 4) + 1/2 * g * (t - 4)^2
Vy(t) = Vy4 + g * (t - 4)
When it reaches its highest point it will have a speed of zero.
0 = Vy4 + g * (t - 4)
0 = 100 - 9.81 * (t - 4)
100 = 9.81 * (t - 4)
t - 4 = 100 / 9.81
t = 10.2 + 4 = 14.2 s
At that moment it will have a height of:
Y(14.2) = 200 + 100 * (14.2 - 4) - 1/2 * 9.81 * (14.2 - 4)^2 = 709 m
The rocket will fall and hit the ground:
Y(t) = 0 = 200 + 100 * (t - 4) - 1/2 * 9.81 * (t - 4)^2
0 = 200 + 100 * t - 400 - 4.9 * (t^2 - 8 * t +16)
0 = -4.9 * t^2 + 139.2 * t -278.4
Solving this equation electronically:
t = 26.24 s
At that time the speed will be:
Vy(t) = 100 - 9.81 * (26.24 - 4) = -118.2 m/s
Two arrows are fired horizontally with the same speed of
30.0m/s. Each arrow has a mass of 0.100 kg. One is fireddue east
and the other due south. Find the magnitude anddirection of the
total momentum of this two-arrow system. Specify the direction with
respect to due east.
Answer:
Explanation:
Given
mass of each arrow=0.1 kg
velocity of arrow=30 m/s
One arrow is fired u=due to east and another towards south
Momentum of first arrow
[tex]P_1=0.1\left ( 30\hat{i}\right )=3\hat{i}[/tex]
[tex]P_2=0.1\left ( -30\hat{j}\right )=-3\hat{j}[/tex]
Total momentum P
[tex]P=P_1+P_2[/tex]
[tex]P=3\hat{i}-3\hat{j}[/tex]
magnitude [tex]|P|=\sqrt{3^2+3^2}=3\sqrt{2}[/tex]
direction
[tex]tan\theta =\frac{-3}{3}[/tex]
[tex]\theta =45^{\circ}[/tex] clockwise w.r.t to east
The magnitude of the total momentum of the two-arrow system is approximately 4.24 kg·m/s, and the direction is 45° south of due east.
Explanation:The question involves calculating the total momentum of a system comprising two arrows fired in perpendicular directions. Momentum is a vector quantity, meaning it has both magnitude and direction. Since the two arrows have the same speed but are fired in perpendicular directions, their momenta are also perpendicular to each other. The momentum of each arrow can be calculated using the formula p = mv, where p is momentum, m is mass, and v is velocity.
The momentum of the first arrow (due east) is peast = m * v = 0.100kg * 30.0m/s = 3.0 kg·m/s east, and the momentum of the second arrow (due south) is psouth = m * v = 0.100kg * 30.0m/s = 3.0 kg·m/s south. To find the total momentum of the system, we need to find the resultant of these two momentum vectors.
The magnitude of the total momentum is found using the Pythagorean theorem: ptotal = √(peast² + psouth²) = √(3.0² + 3.0²) kg·m/s = √(9 + 9) kg·m/s = √18 kg·m/s ≈ 4.24 kg·m/s.
The direction of the total momentum with respect to due east can be found using trigonometry, specifically tan-1 (psouth/peast), which gives us 45° south of east, since the magnitudes are equal.
A non conducting sphere of radius 0.04 m has a charge of 5.0 × 10^-9 C deposited on it. CalculateThe magnitude of the electric field at 0.02m from the center of the sphere
Answer:
The electric field at a distance r = 0.02 m is 14062.5 N/C.
Solution:
Refer to fig 1.
As per the question:
Radius of sphere, R = 0.04 m
Charge, Q = [tex]5.0\times 10^{- 9} C[/tex]
Distance from the center at which electric field is to be calculated, r = 0.02 m
Now,
According to Gauss' law:
[tex]E.dx = \frac{Q_{enclosed}}{\epsilon_{o}}[/tex]
Now, the charge enclosed at a distance r is given by volume charge density:
[tex]\rho = \frac{Q_{enclosed}}{area}[/tex]
[tex]\rho = \frac{Q_{enclosed}}{\frac{4}{3}\pi R^{3}}[/tex]
Also, the charge enclosed Q' at a distance r is given by volume charge density:
[tex]\rho = \frac{Q'_{enclosed}}{\frac{4}{3}\pi r^{3}}[/tex]
Since, the sphere is no-conducting, Volume charge density will be constant:
Thus
[tex]\frac{Q_{enclosed}}{\frac{4}{3}\pi R^{3}} = \frac{Q'_{enclosed}}{\frac{4}{3}\pi r^{3}}[/tex]
Thus charge enclosed at r:
[tex]Q'_{enclosed} = \frac{Q_{enclosed}}{\frac{r^{3}}{R^{3}}[/tex]
Now, By using Gauss' Law, Electric field at r is given by:
[tex]4\pi r^{2}E = \frac{Q_{enclosed}r^{3}}{\epsilon_{o}R^{3}}[/tex]
Thus
[tex]E = \frac{Q_{enclosed}r}{4\pi\epsilon_{o}R^{3}}[/tex]
[tex]E = \frac{(9\times 10^{9})\times 5.0\times 10^{- 9}\times 0.02}{0.04^{3}}[/tex]
E = 14062.5 N/C
Problem 1 Two positive point charges are released on a frictionless surface. Which of the following best describes their subsequent motion? Select One of the Following: (a) The particles remain stationary. (b) The particles move apart with constant velocity. (c) The particles move apart with a velocity that decreases with time. (d) The particles move apart with a velocity that increases with time. (e) The particles move apart with a velocity that increases for a while and then becomes constant.
Answer:
(e) The particles move apart with a velocity that increases for a while and then becomes constant.
Explanation:
Each particle feels a repulsive (because they have same sign of charge) electric force from the each other:
[tex]F=\frac{kq_{1}q{2}}{d^{2}}[/tex]
and
[tex]F=ma\\[/tex]
So each particle feels a repulsive force proportional to the quadratic inverse of the distance.that means that the charges begin to move away, and the further away they are from each other, the force (and therefore the acceleration) decreases, at a rate inversely proportional to the square of the distance. Theoretically this acceleration will never be zero, but in practice it will at some point reach a value very close to zero. Then the speed will grow for a while and when the acceleration has reached almost zero, the speed will practically remain constant.
A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and what’s the heat absorbed by the gas?
Answer:
The work made by the gas is 7475.69 joulesThe heat absorbed is 7475.69 joulesExplanation:
WorkWe know that the differential work made by the gas its defined as:
[tex]dW = P \ dv[/tex]
We can solve this by integration:
[tex]\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv[/tex]
but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law
[tex]P \ V = \ n \ R \ T[/tex]
[tex]P = \frac{\ n \ R \ T}{V}[/tex]
This give us
[tex] \int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv [/tex]
As n, R and T are constants
[tex] \int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv [/tex]
[tex] \Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1} [/tex]
[tex] \Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 ) [/tex]
[tex] \Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 ) [/tex]
[tex] \Delta W = \ n \ R \ T ln (\frac{v_2}{v_1})[/tex]
But the volume is:
[tex]V = \frac{\ n \ R \ T}{P}[/tex]
[tex] \Delta W = \ n \ R \ T ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )[/tex]
[tex] \Delta W = \ n \ R \ T ln(\frac{P_1}{P_2})[/tex]
Now, lets use the value from the problem.
The temperature its:
[tex]T = 27 \° C = 300.15 \ K[/tex]
The ideal gas constant:
[tex]R = 8.314 \frac{m^3 \ Pa}{K \ mol}[/tex]
So:
[tex] \Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K ln (\frac{20 atm}{1 atm}) [/tex]
[tex] \Delta W = 7475.69 joules[/tex]
HeatWe know that, for an ideal gas, the energy is:
[tex]E= c_v n R T[/tex]
where [tex]c_v[/tex] its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.
By the first law of thermodynamics, we know
[tex]\Delta E = \Delta Q - \Delta W[/tex]
where [tex]\Delta W[/tex] is the Work made by the gas (please, be careful with this sign convention, its not always the same.)
So:
[tex]\Delta E = 0[/tex]
[tex]\Delta Q = \Delta W[/tex]
People with normal vision cannot focus their eyes underwater if they aren't wearing a face mask or goggles and there is water in contact with their eyes. Explain why not.
Answer:
Explained
Explanation:
This happens because the water cancels the focusing effect of our eye lens. A drop of water makes a lens because of the different refractive indices of water and air.
When we are underwater with naked eyes, the front part of our eye lens does not have different refractive indices anymore… both the inside lens and outside lens are essentially water ( lens inside our eyes are water only surrounded by a membrane)… hence, that part of the lens does not focus the light anymore. And thus, your eyes don’t work well anymore. (It can only partly correct for this, by bending the lens more)
When we put swimming goggles on, we remove the water and again have air in front of our eye lenses, and they work normally now. Obviously, we have created an extra interface between water and air in front of our swimming goggles. However, this surface(the goggle) is flat. And thus it does not function as a lens but only as a window.
A raindrop of mass m0, starting from rest, falls under the influence of gravity. Assume that as the raindrop travels through the clouds, it gains mass at a rate proportional to the momentum of the raindrop, dmr = kmrvr, where mr is the in- dt stantaneous mass of the raindrop, vr is the instantaneous velocity of the raindrop, 5 and k is a constant with unit [m−1]. You may neglect air resistance. (a) Derive a differential equation for the raindrop’s accelerations dvr in terms of dt k, g, dt and the raindrop’s instantaneous velocity vr . Express your answer using some or all of the following variables: k,g for the gravitational acceleration and vr, the raindrop’s instantaneous velocity.
Answer:
[tex]\frac{dv_{r}}{dt}=g-k_{r}v_{r}^2[/tex]
Explanation:
Second Newton's Law:
[tex]F=\frac{dp}{dt}=\frac{d(mv)}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt} \\[/tex] (1)
m and v are the instantaneous mass and instantaneous velocity
The only force is the weight:
[tex]F=mg[/tex] (2)
On the other hand we know:
[tex]\frac{dm}{dt}=k*m*v[/tex] (3)
We replace (2) and (3) in (1), and we solve for dv/dt :
[tex]\frac{dv}{dt}=g-kv^2[/tex]