Answer:
a) During the reaction time, the car travels 21 m
b) After applying the brake, the car travels 48 m before coming to stop
Explanation:
The equation for the position of a straight movement with variable speed is as follows:
x = x0 + v0 t + 1/2 a t²
where
x: position at time t
v0: initial speed
a: acceleration
t: time
When the speed is constant (as before applying the brake), the equation would be:
x = x0 + v t
a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:
x = 0m + 26 m/s * 0.80 s = 21 m
b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):
v = v0 + a* t
0 = 26 m/s + (-7.0 m/s²) * t
-26 m/s / - 7.0 m/s² = t
t = 3.7 s
With this time, we can calculate how far the car traveled during the deacceleration.
x = x0 +v0 t + 1/2 a t²
x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = 48 m
a. The car travels 20.8 meters during the reaction time and 48.29 meters while decelerating, b. resulting in a total distance of 69.09 meters.
Part a: Distance Traveled During Reaction Time
The initial speed of the car is 26 m/s, and the driver has a reaction time of 0.80 seconds. During this reaction time, the car continues to move at the initial speed because the brakes haven't been applied yet.
Using the formula:
distance = speed × time
We get:
distance = 26 m/s × 0.80 s = 20.8 meters
Part b: Distance Traveled After Brakes Are Applied
The car decelerates at a rate of -7.0 m/s² until it comes to a stop. We need to find the distance traveled during this deceleration.
Using the kinematic equation:
v² = u² + 2as
where:
v = final velocity = 0 m/s (since the car stops)u = initial velocity = 26 m/sa = acceleration = -7.0 m/s²s = distance during decelerationRearranging the equation to solve for s:
0 = (26 m/s)² + 2(-7.0 m/s²) * s
Solving for s:
0 = 676 - 14s
14s = 676
s = 48.29 meters
The total distance the car travels includes both the distance during the reaction time and the distance while braking:
Total Distance = 20.8 meters + 48.29 meters = 69.09 meters
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 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.
To know more about the distance:
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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 mYour 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
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
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.
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
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.
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.
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
Monochromatic light passes through a double slit, producing interference, the distance between the slit centres is 1.2 mm and the distance between constructive fringes on a screen 5 m away is 0.3 cm. What is the wavelength?
Answer:
The wavelength of the light is [tex]7200\ \AA[/tex].
Explanation:
Given that,
Distance between the slit centers d= 1.2 mm
Distance between constructive fringes [tex]\beta= 0.3\ cm[/tex]
Distance between fringe and screen D= 5 m
We need to calculate the wavelength
Using formula of width
[tex]\beta=\dfrac{D\lambda}{d}[/tex]
Put the value into the formula
[tex]0.3\times10^{-2}=\dfrac{5\times\lambda}{1.2\times10^{-3}}[/tex]
[tex]\lambda=\dfrac{0.3\times10^{-2}\times1.2\times10^{-3}}{5}[/tex]
[tex]\lambda=7.2\times10^{-7}\ m[/tex]
[tex]\lambda=7200\ \AA[/tex]
Hence, The wavelength of the light is [tex]7200\ \AA[/tex].
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.
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]
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.
A 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
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
A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 52.0 ∘ above the horizontal. The firefighters want to direct the water at a blaze that is 12.0 m above ground level. How far from the building should they position their cannon? There are two possibilities (d1
Final answer:
To calculate the distance the firefighters should position their water cannon from the building, analyze the vertical motion of the water stream. Therefore, the firefighters should position their water cannon approximately 11.40 meters from the building.
Explanation:
To find the horizontal distance from the building where the firefighters should position their water cannon, we can use the projectile motion equations.
Given:
[tex]- Initial \ velocity \ of \ the \water \ cannon, \( v_0 = 25.0 \, \text{m/s} \)\\- Launch angle, \( \theta = 52.0^\circ \)\\- Vertical displacement, \( y = 12.0 \, \text{m} \)\\- Vertical acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)\\[/tex]
First, we'll find the time it takes for the water to reach a height of 12.0 m. We'll use the kinematic equation for vertical motion:
[tex]\[ y = v_{0y}t + \frac{1}{2}gt^2 \][/tex]
Where:
[tex]- \( v_{0y} \) is the initial vertical component of the velocity\\- \( t \) is the time[/tex]
Since the initial velocity is at an angle, we need to find its vertical component:
[tex]\[ v_{0y} = v_0 \sin(\theta) \][/tex]
Substitute the given values:
[tex]\[ v_{0y} = 25.0 \, \text{m/s} \times \sin(52.0^\circ) \]\[ v_{0y} \approx 25.0 \, \text{m/s} \times 0.788 \]\[ v_{0y} \approx 19.7 \, \text{m/s} \][/tex]
Now, let's use the equation for vertical motion to solve for \( t \):
[tex]\[ 12.0 \, \text{m} = (19.7 \, \text{m/s})t - \frac{1}{2}(9.81 \, \text{m/s}^2)t^2 \][/tex]
This is a quadratic equation, we can solve it to find \( t \). Let's denote [tex]\( a = -4.905 \, \text{m/s}^2 \) and \( b = 19.7 \, \text{m/s} \):[/tex]
[tex]\[ -4.905t^2 + 19.7t - 12.0 = 0 \][/tex]
Now, we can use the quadratic formula to solve for \( t \):[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Let's calculate \( t \).
We have the quadratic equation:
[tex]\[ -4.905t^2 + 19.7t - 12.0 = 0 \][/tex]
Using the quadratic formula:
[tex]\[ t = \frac{-19.7 \pm \sqrt{(19.7)^2 - 4(-4.905)(-12.0)}}{2(-4.905)} \]\[ t = \frac{-19.7 \pm \sqrt{389.21 - 235.2}}{-9.81} \]\[ t = \frac{-19.7 \pm \sqrt{154.01}}{-9.81} \]\[ t = \frac{-19.7 \pm 12.41}{-9.81} \][/tex]
Now, we have two possible values for \( t \):
[tex]1. \( t_1 = \frac{-19.7 + 12.41}{-9.81} \)\\2. \( t_2 = \frac{-19.7 - 12.41}{-9.81} \)[/tex]
Calculating each value:
[tex]1. \( t_1 = \frac{-7.29}{-9.81} \) \( t_1 \approx 0.742 \) seconds\\2. \( t_2 = \frac{-32.11}{-9.81} \) \( t_2 \approx 3.27 \) seconds[/tex]
Since the time \( t \) cannot be negative, we'll take [tex]\( t = t_1 \approx 0.742 \) seconds[/tex].
Now, to find the horizontal distance \( x \) from the building, we can use the equation for horizontal motion:
[tex]\[ x = v_{0x} \cdot t \][/tex]
where \( v_{0x} \) is the initial horizontal component of the velocity, given by:
[tex]\[ v_{0x} = v_0 \cdot \cos(\theta) \][/tex]
Substituting the given values:
[tex]\[ v_{0x} = 25.0 \, \text{m/s} \cdot \cos(52.0^\circ) \]\[ v_{0x} \approx 25.0 \, \text{m/s} \cdot 0.615 \]\[ v_{0x} \approx 15.375 \, \text{m/s} \][/tex]
Now, we can find \( x \):
[tex]\[ x = 15.375 \, \text{m/s} \cdot 0.742 \, \text{s} \]\[ x \approx 11.40 \, \text{m} \][/tex]
Therefore, the firefighters should position their water cannon approximately 11.40 meters from the building.
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]
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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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]
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.
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]
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 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]
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
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
Pulsed lasers used in science and medicine produce very short bursts of electromagnetic energy. If the laser light wavelength is 1062 nm (this corresponds to a Neodymium-YAG laser), and the pulse lasts for 34 picoseconds, how many wavelengths are found within the laser pulse?
How short would the pulse need to be to fit only one wavelength?
Explanation:
Given that,
Wavelength of the laser light, [tex]\lambda=1062\ nm=1062\times 10^{-9}\ m[/tex]
The laser pulse lasts for, [tex]t=34\ ps=34\times 10^{-12}\ s[/tex]
(a) Let d is the distance covered by laser in the given by, [tex]d=c\times t[/tex]
[tex]d=3\times 10^8\times 34\times 10^{-12}[/tex]
d = 0.0102 meters
Let n is the number of wavelengths found within the laser pulse. So,
[tex]n=\dfrac{d}{\lambda}[/tex]
[tex]n=\dfrac{0.0102}{1062\times 10^{-9}}[/tex]
n = 9604.51
(b) Let t is the time need to be fit only in one wavelength. So,
[tex]t=\dfrac{\lambda}{c}[/tex]
[tex]t=\dfrac{1062\times 10^{-9}}{3\times 10^8}[/tex]
[tex]t=3.54\times 10^{-15}\ s[/tex]
Hence, this is the required solution.
Calculate the volume of a sphere of radius R. Write out each step.
Answer:
[tex]V=\frac{4}{3}\pi R^{3}[/tex]
Explanation:
The volume of a body is defined as the capacity of the object. The amount of matter that object contains is called its volume.
All the three dimensional objects have volume.
The SI unit of volume is m^3. The volume of liquids is measured by teh unit litre or milli litre.
The volume of the sphere is given by
[tex]V=\frac{4}{3}\pi R^{3}[/tex]
where, R is the radius of the sphere.
Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. (Membranes are discussed in some detail in Nerve Conduction—Electrocardiograms.) What is the voltage (in mV) across a 7.95 nm thick membrane if the electric field strength across it is 6.60 MV/m? You may assume a uniform electric field.
Answer:
Voltage, V = 0.0524 volts
Explanation:
Thickness of the membrane, [tex]d=7.95\ nm=7.95\times 10^{-9}\ m[/tex]
Electric field strength, [tex]E=6.6\ MV/m=6.6\times 10^6\ V/m[/tex]
We need to find the voltage across it. The relationship between the voltage, electric field and the distance is given by :
[tex]V=E\times d[/tex]
[tex]V=6.6\times 10^6\ V/m\times 7.95\times 10^{-9}[/tex]
V = 0.0524 volts
So, the voltage across the thick membrane is 0.0524 volts. Hence, this is the required solution.
The voltage across a 7.95 nm thick membrane with an electric field strength of 6.60 MV/m, is found to be 52.47 mV.
To find the voltage across a membrane, we can use the relationship between electric field strength (E) and voltage (V) given by the formula:
V = E * d
where:
V is the voltage
E is the electric field strength
d is the thickness of the membrane
In this case, the electric field strength (E) is given as 6.60 MV/m and the thickness of the membrane (d) is 7.95 nm.
First, convert the thickness from nanometers to meters:
d = 7.95 nm = 7.95 * 10⁻⁹ m
Now use the formula to calculate the voltage:
V = 6.60 * 10⁶ V/m * 7.95 * 10⁻⁹ m
V = 5.247 * 10⁻² V = 52.47 mV
Therefore, the voltage across the 7.95 nm thick membrane is 52.47 mV.
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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