Answer:
The momentum is not conserved.
Explanation:
When the superhero throws the piano, he exerts a force on the piano. Newton's third law says that each action has an opposite and equal reaction. In other words, this means that when the superhero throws the piano, the piano also exerts a force on the superhero, therefore, the superhero should recoil. You can visualize a gun being shot to visualize this phenomenon, when the gun is shot, it will tend to go backwards, trying to conserve momentum.
The scenario in the movie is not physically possible and violates the principles of Newton's laws of motion and conservation of momentum.
Explanation:The scenario in the movie where the superhero hovers in the air and throws a piano while remaining stationary is not physically possible. It violates the principles of Newton's laws of motion and the conservation of momentum. In order for the superhero to hover and throw a piano, they would need some external force or propulsion, like jetpack or levitation powers, which are not commonly associated with superheroes.
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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]
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
Your favorite professor decided to throw in the air a marker above his head (vertically). If the marker leaves his hand at a height of 1.2 m from the ground with an initial velocity of 20 m/s, when would the marker reach the highest point in the air (1.6 m from the ground)?
Answer:
[tex]t=4.06s[/tex]
Explanation:
From the exercise we know
[tex]y_{o}=1.2m\\v_{o}=20m/s\\y=1.6m\\g=-9.8m/s^2[/tex]
To find how long does it takes the marker to reach the highest point we need to use the equation of position:
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]
[tex]1.6m=1.2m+(20m/s)t-\frac{1}{2}(9.8m/s^2)t^2[/tex]
[tex]0=-0.4+20t-4.9t^2[/tex]
Now, we need to use the quadratic formula:
[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]a=-4.9\\b=20\\c=-0.4[/tex]
Solving for t
[tex]t=0.020s[/tex] or [tex]t=4.06s[/tex]
So, the answer is t=4.06s because the other option is almost 0 and doesn't make any sense for the motion of the marker
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.
An engineer has a 37.0 µH inductor, a 37.0 µF capacitor, and a variable frequency AC source. Determine the source frequency (in kHz) at which the inductor and capacitor have the same reactance.
Answer:
Frequency will be 4.30 kHz
Explanation:
We have inductance [tex]L=37\mu H=37\times 10^{-6}H[/tex]
And capacitance [tex]C=37\mu H=37\times 10^{-6}F[/tex]
Inductive reactance is given by [tex]X_L=\omega L[/tex]
And capacitive reactance [tex]X_C=\frac{1}{\omega C}[/tex]
As in question it is given that inductive reactance and capacitive reactance is same
So [tex]X_L=X_C[/tex]
[tex]\omega L=\frac{1}{\omega C}[/tex]
[tex]\omega ^2=\sqrt{\frac{1}{LC}}=\frac{1}{\sqrt{37\times 10^{-6}\times 37\times 10^{-6}}}=27027.027rad/sec[/tex]
We know that angular frequency [tex]\omega =2\pi f[/tex]
[tex]2\times 3.14\times f=27027.027[/tex]
f = 4303.66 Hz =4.30 kHz
An automobile approaches a barrier at a speed of 20 m/s along a level road. The driver locks the brakes at a distance of 50 m from the barrier. What minimum coefficient of kinetic friction is required to stop the automobile before it hits the barrier?
The minimum coefficient of kinetic friction required to stop the automobile before it hits the barrier is approximately 0.2041.
Explanation:To stop the automobile before it hits the barrier, the minimum coefficient of kinetic friction needed can be found using the equation F = μk Mcg. The force of friction is equal to the product of the coefficient of kinetic friction (μk), the mass of the car (Mc), and the acceleration due to gravity (g). Since the car is stopping, its acceleration is negative, equal in magnitude to the square of its velocity divided by twice the stopping distance.
Using the given information, we can solve for the coefficient of kinetic friction:
F = μk Mcg
μk Mcg = (Mc)((-v^2)/(2d))
Substituting the values, we get:
μk = ((-v^2)/(2d))/g
where v = 20 m/s and d = 50 m.
Substituting the values in the equation, we get:
μk = ((-20^2)/(2*50))/9.8 ≈ -0.2041 ≈ 0.2041
Therefore, the minimum coefficient of kinetic friction required to stop the automobile before it hits the barrier is approximately 0.2041.
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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 distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm. What eyepiece focal length will give the microscope an overall angular magnification of 200?
Final answer:
To achieve an overall magnification of 200 with the given measurements, the eyepiece lens would need to have a focal length of approximately 5.68 cm.
Explanation:
To calculate the eyepiece focal length that will give an overall angular magnification of 200 in a microscope, we can use the formula of magnification for a compound microscope. Given that the distance between the objective and eyepiece lenses is 19 cm and the objective lens has a focal length of 5.5 mm, we can set up the equation as follows:
M = (D/fo) × (25/fe)
Where:
M is the total magnificationD is the distance between objective and eyepiece lenses (which is often approximate to the near point distance of a standard human eye, usually taken to be 25 cm for convenience)fo is the focal length of the objective lensfe is the focal length of the eyepiece lensIn the given problem, M = 200 and fo = 5.5 mm (which we should convert to centimeters to keep consistent units, thus fo = 0.55 cm). We are solving for fe.
Using the formula, we substitute the values:
200 = (25/0.55) × (25/fe)
Solving for fe, we find:
fe = (25/0.55) × (25/200)
fe = (45.45) × (0.125)
fe = 5.68 cm
Therefore, the eyepiece lens would need to have a focal length of approximately 5.68 cm to achieve an overall magnification of 200.
If a boulder is dropped straight down off the edge of a vertical cliff, how long does it take it to reach 90 Km/h? Give in seconds
Answer:
2.551 m/s
Explanation:
Given:
Boulder is dropped from the cliff
thus,
Initial speed of the boulder, u = 0 m/s
Final speed to be obtained, v = 90 km/h =[tex]90\times\frac{5}{18}[/tex] =25 m/s
also, in the case of free fall the acceleration of the boulder will be equal to the acceleration due to the gravity i.e g = 9.8 m/s²
Now, from the Newton's equation of motion
v = u + at
where, a is the acceleration = g = 9.8 m/s²
t is the time
on substituting the respective values, we get
25 = 0 + 9.8 × t
or
t = [tex]\frac{\textup{25}}{\textup{9.8}}[/tex]
or
t = 2.551 m/s
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.
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]
A child slides down a hill on a toboggan with an acceleration of 1.8 m/s^2. If she starts at rest, how far has she traveled in : (a) 1.0 s, (b) 2.0 s, and (c) 3.0 s?
The distances traveled by the child on the toboggan are [tex]0.9, 3.6, and\ 8.1[/tex] respectively.
The distance traveled by the child on the toboggan can be calculated using the equations of motion for uniformly accelerated motion. The equation that relates distance (d), initial velocity (u), acceleration (a), and time (t) is given by:
[tex]d = ut + (1/2)at^2[/tex]
In this case, the child starts at rest [tex](u = 0)[/tex] and has an acceleration of a[tex]= 1.8 m/s^2[/tex].
Calculate the distances for the given times:
(a) For t = 1.0 s:
[tex]d = (0) \times (1.0 ) + (1/2) \times (1.8 ) \times (1.0 )^2\\d = 0 + 0.9 \\d = 0.9\ m[/tex]
(b) For t = 2.0 s:
[tex]d = (0) \times (2.0 ) + (1/2) \times (1.8 ) \times (2.0 )^2\\d = 0 + 3.6\\d = 3.6\ m[/tex]
(c) For t = 3.0 s:
[tex]d = (0) \times (3.0 ) + (1/2) \times (1.8 ) \times (3.0)^2\\d = 0 + 8.1\\d = 8.1\ m[/tex]
So, the distances traveled by the child on the toboggan are [tex]0.9, 3.6, and\ 8.1[/tex] respectively.
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Final answer:
Using the kinematic equation for motion with constant acceleration, the distance traveled by the child on the toboggan is 0.9 m after 1.0 s, 3.6 m after 2.0 s, and 8.1 m after 3.0 s.
Explanation:
To solve for the distance traveled by the child on the toboggan in each case, we'll use the kinematic equation for motion with constant acceleration: d = v_i * t + \frac{1}{2} * a * t^2, where d is the distance traveled, v_i is the initial velocity, t is the time, and a is the acceleration.
Since the child starts at rest, the initial velocity v_i is 0 m/s. The acceleration a is given as 1.8 m/s^2.
(a) For t = 1.0 s: d = 1/2 * 1.8 m/s^2 * (1.0 s)^2 = 0.9 m(b) For t = 2.0 s: d = 1/2 * 1.8 m/s^2 * (2.0 s)^2 = 3.6 m(c) For t = 3.0 s: d = 1/2 * 1.8 m/s^2 * (3.0 s)^2 = 8.1 mA rigid tank contains water at 200kPa and an unknown temperature. When the tank is 100 °C the vapour starts condensing. Calculate the initial temperature of the water. Show the process on T-v diagram.
Answer:
T = 450 C
Explanation:
given,
pressure of rigid tank containing water = 200 kPa
water start condensing at 100 °C
vapor saturation pressure at 100 °C = 101421 Pa (from saturation table)
specific volume = 1.6718 m³/kg
From super heated steam table
steam table is two set of table which is of energy transfer properties of water and steam.
at v = 1.6718 m³/kg and pressure = 200 kPa
temperature is equal to T = 450 C
Two spheres are cut from a certain uniform rock, One has radius 4.85 cm. The mass of the other is three times greater. Find its radius.
Answer:7 cm
Explanation:
Given
one sphere has a radius(r) of 4.85 cm
Also another sphere has a mass 3 times the previous one
since the density remains same therefore
second sphere has a volume 3 times of former.
Let R be the radius of bigger Sphere
[tex]\frac{4\pi R^3}{3}=3\times \frac{4\pi r^3}{3}[/tex]
[tex]R=\left ( 3\right )^\frac{1}{3}r=1.44\times 4.85=6.99 \approx 7 cm[/tex]
You measure the velocity of a drag racer that accelerates with constant acceleration. You want to plot the data and determine the acceleration of the dragster. Would you use a. a) Linear equation
b) Quadratic equation
c) cubic equation
d) a higher order equation
Answer:
a) Linear equation
Explanation:
Definition of acceleration
[tex]a=\frac{dv}{dt}\\[/tex]
if a=constant and we integrate the last equation
[tex]v(t)=v_{o}+a*t[/tex]
So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.
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.
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 parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 61 m/s^2. If he reaches the ground with a speed of 17 m/s, how long was he in the air (in seconds)?
Answer:
55.66 m
Explanation:
While falling by 50 m , initial velocity u = 0
final velocity = v , height h = 50 , acceleration g = 9.8
v² = u² + 2gh
= 0 + 2 x 9.8 x 50
v = 31.3 m /s
After that deceleration comes into effect
In this case final velocity v = 17 m/s
initial velocity u = 31.3 m/s
acceleration a = - 61 m/s²
distance traveled h = ?
v² = u² + 2gh
(17)² = (31.3)² - 2x 61xh
h = 690.69 / 2 x 61
= 5.66 m
Total height during which he was in air
= 50 + 5.66
= 55.66 m
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.
Calculate the time (in seconds) needed for a car to accelerate from 0 m/s to 25 m/s at 5 m/s^2?
Answer:
Car will take 5 sec to reach from 0 m/sec to 25 m/sec
Explanation:
We have given initial velocity u = 0 m/sec
And final velocity v = 25 m/sec
Acceleration [tex]a=5m/Sec^2[/tex]
From first equation of motion we know that [tex]v=u+at[/tex], here v is final velocity, u is initial velocity, a is acceleration and t is time
So [tex]25=0+5\times t[/tex]
[tex]t=4 sec[/tex]
So car will take 5 sec to reach from 0 m/sec to 25 m/sec
Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following accelerations. (a) a = -4.0 m/s^2 (b) a = -8.0 m/s^2
Explanation:
Given that,
Initial speed of the car, u = 88 km/h = 24.44 m/s
Reaction time, t = 2 s
Distance covered during this time, [tex]d=24.44\times 2=48.88\ m[/tex]
(a) Acceleration, [tex]a=-4\ m/s^2[/tex]
We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{-(24.44)^2}{2\times -4}[/tex]
s = 74.66 meters
s = 74.66 + 48.88 = 123.54 meters
(b) Acceleration, [tex]a=-8\ m/s^2[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{-(24.44)^2}{2\times -8}[/tex]
s = 37.33 meters
s = 37.33 + 48.88 = 86.21 meters
Hence, this is the required solution.
Summarize one or more impacts of the physics of matter on aviation operations.
Answer:
Answered
Explanation:
impacts of the physics of matter on aviation operations.
Thrust, drag and lift.
Thrust:
It is the force developed by airplane engines that cause it to pull forward. With the help of huge propellers of course attached to the wings.
Drag:
It is the resistive force on the plane caused by the friction between air and plane. Its magnitude depends upon surface area, speed and viscosity of the air.
Lift:
The drag produced is utilized such that one of its component acts opposite to the weight. This causes the plane to take flight and stay in air. Lift can be deduced using Bernoulli's principle.
Bernoulli's principle is equivalent to law of conservation of energy. Meaning it tries to keep the energy of a system constant. In doing so, it produces low pressure zone above the wing. Which causes a net upward force, lift.
Calculate the change in velocity for a model rocket having zero velocity to start with to a final acceleration of 330 feet in 3.1 seconds
Answer:
213 ft/s
Explanation:
Given:
Δx = 330 ft
v₀ = 0 ft/s
t = 3.1 s
Find: Δv
x = x₀ + ½ (v + v₀)t
330 ft = ½ (v + 0 ft/s) (3.1 s)
v = 213 ft/s
The change in velocity is:
Δv = 213 ft/s − 0 ft/s
Δv = 213 ft/s
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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If you throw a rock straight up and there is no air drag, what is the acceleration of the rock on the way up? What is its acceleration on the way down? What is the velocity of the rock at the instant it reaches its highest point in its trajectory? What is the acceleration of the rock at its highest point?
Answer:
Explanation:
The rock will decelerate at the rate of g or 9.8 m s⁻²while on its way up because of gravitational attraction towards the earth. . In other words , it will accelerate with the magnitude - 9.8 m s⁻² while on its way up.
On its way down , it will accelerate due to gravitational attraction. In this case acceleration will be positive and equal to 9.8 m s⁻².
At the highest point in its trajectory , rock becomes stationary or comes to rest momentarily. So at the peak position , velocity is zero.
At the highest point , as the gravitational force continues to act on the body undiminished, its acceleration will remain the same ie 9.8 m s⁻² at the highest point as well. It will act in the downward direction.
A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is the car's average acceleration in ft/s2? (b) In 3.0 more seconds, the car is going 76 ft/s (52 mi/h). What is the car's average acceleration for this time period?
Answer:
[tex]a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}[/tex]
Explanation:
A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.
a)
[tex]a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\[/tex]
b)
[tex]a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}[/tex]
Final answer:
The car's average acceleration for part (a) is 9.5 [tex]ft/s^2[/tex], and for part (b) it is 12.67 [tex]ft/s^2[/tex], calculated using the change in velocity over the time period.
Explanation:
Car's Average Acceleration
To find the car's average acceleration in part (a), we use the formula for average acceleration, which is the change in velocity (deltaV) divided by the time (t).
Average Acceleration (a) = (deltaV) / (t)For part (a), the initial velocity (V0) is 0 ft/s, the final velocity (V) is 38 ft/s, and the time (t) is 4.0 s.
Therefore, a = (38 ft/s - 0 ft/s) / 4.0 s = 9.5 [tex]ft/s^2[/tex]
For part (b), the final velocity (V) is now 76 ft/s and the time (t) is 3.0 s (the additional time).
Average Acceleration (a) = (76 ft/s - 38 ft/s) / 3.0 s = 12.67 [tex]ft/s^2[/tex]
Which of the following statements regarding heuristics and empiricism is true? A. Heuristics allows a priori theoretical concepts B. Empiricism makes no use of logic C. Empiricism relies solely on theoretical concepts D. Heuristics makes no use of logic E. Heuristics can't be based on a trial and error approach
The statement that heuristics allow a priori theoretical concepts is true. Heuristics are mental shortcuts that can definitely be informed by a priori theoretical concepts, while empiricism emphasizes the role of experience in the formation of ideas.
Explanation:Regarding the question on the statements about heuristics and empiricism, the best choice would be 'A'. Heuristics allow a priori theoretical concepts'. Heuristics are mental shortcuts or rules of thumb that simplify decisions, particularly under conditions of uncertainty. They are not necessarily based on theoretical concepts but rather on practical, personal experience or common sense. However, they can certainly be informed by a priori theoretical concepts in the sense that our theoretical understanding can shape the rules of thumb we use.
On the other hand, empiricism is a philosophical system that emphasizes the role of experience, especially sensory perception, in the formation of ideas, while discounting a priori reasoning, intuitiveness, and innate ideas. Hence, options B, C, and D are incorrect. Option E is also incorrect as heuristics can definitely be based on a trial and error approach.
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Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the temperature and pressure changes to 800 K and 8.74 * 10 N/m². Determine the change in following across the shockwave: a. Internal energy b. Enthalpy c. Entropy
Answer:
change in internal energy 3.62*10^5 J kg^{-1}
change in enthalapy 5.07*10^5 J kg^{-1}
change in entropy 382.79 J kg^{-1} K^{-1}
Explanation:
adiabatic constant [tex]\gamma =1.4[/tex]
specific heat is given as [tex]=\frac{\gamma R}{\gamma -1}[/tex]
gas constant =287 J⋅kg−1⋅K−1
[tex]Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}[/tex]
specific heat at constant volume
[tex]Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}[/tex]
change in internal energy [tex]= Cv(T_2 -T_1)[/tex]
[tex] \Delta U = 717.5 (800-295) = 3.62*10^5 J kg^{-1}[/tex]
change in enthalapy [tex]\Delta H = Cp(T_2 -T_1)[/tex]
[tex] \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}[/tex]
change in entropy
[tex]\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})[/tex]
[tex]\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})[/tex]
[tex]\Delta S = 382.79 J kg^{-1} K^{-1}[/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]
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