Answer:
a) [tex]x=26m[/tex]
b) [tex]v_{y}=-21.5m/s[/tex]
Explanation:
From the exercise we know initial velocity, initial height
[tex]y_{o}=20m[/tex]
[tex]v_{o} =12m/s[/tex] [tex]\beta =45[/tex]º
a) The range of the stone is defined by how far does it goes. From the theory of free falling objects, we have:
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]
The stone strike the water at y=0
[tex]0=20+12sin(45)t-\frac{1}{2}(9.8)t^{2}[/tex]
Solving for t, using the quadratic formula
[tex]t=-1.33s[/tex] or [tex]t=3.06s[/tex]
Since time can't be negative, the answer is t=3.06s
Now, we can calculate the range of the stone
[tex]x=v_{o}t=(12cos(45))m/s(3.06s)=26m[/tex]
b) We can calculate the velocity were the stone strike the water using the following formula
[tex]v_{y}=v_{oy}+gt=12sin(45)m/s-(9.8m/s^{2})(3.06s)=-21.5m/s[/tex]
The negative sign indicates that the stone is going down
Answer:
(a) x= 26m
(b) vf= 23.15m/s
Explanation:
Given data
h=20m
Ɵ=45°
to find
(a) range of stone=x=?
(b) velocity=vf=?
Solution
For part (a)
You need to solve for time first using
yf = yi + visinƟt + 1/2gt^2
0 = 20m + 12sin45t + 1/2(-9.8)t^2
and use the quadratic equation to solve for t
t = 3.064 sec
To solve for the distance traveled use
x = xi + vicosƟt + 1/2at^2 there is no acceleration in the x direction so that cancels
x = 12cos(45)(3.064)
x= 26m
For part(b)
For b I'm not sure if you what direction you want the final velocity in the x, y, or the direction its traveling so I'll just give all 3.
Theres no change in the velocity in the x direction so its just vfx = vixcosƟ = 12cos45 = 8.49m/s
For the y direction its vfy^2 = viy^2 + 2g(Δy)
vfy = sqrt((12sin(45))^2 + 2(-9.8)(0-20m)) = 21.54m/s
The velocity the direction the stone is traveling is vf = sqrt(vx^2 + vy^2) = sqrt(8.49^2 + 21.54^2)
vf= 23.15m/s
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.
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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Point charge A with a charge of +4.00 μC is located at the origin. Point charge B with a charge of +7.00 μC is located on the x axis at x = 7.00 cm. And point charge C with a charge of +3.00 μC is located on the y axis at y = 6.00 cm. What is the direction of the net force exerted on charge A by the others?
Answer:
210.3 degrees
Explanation:
The net force exerted on charge A = 59.5 N
Use the x and y coordinates of net force to get the direction
arctan (y/x)
Answer:
F = -51.357i -29.958j
abs(F) = 59.45 N
Explanation:
To solve the problem we use coulomb's law with vectorial notation, F = q1*q2/(4*pi*eo*r^2) where q1 and q2 are the charges and r is the distance between them:
Point B exerts a force on A in '-i' direction
Point C exerts a force on A in '-j' direction
Fba = 4*7/(4*pi*eo*0.07^2) = 51.357N
Fca = 4*3/(4*pi*eo*0.06^2) = 29.958N
F = -51.357i -29.958j
abd(F) = 59.45 N
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]
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.
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.
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.1 m/s . Two seconds later the bicyclist hops on his bike and accelerates at 2.4 m/s^2 until he catches his friend. a) How much time does it take until he catches his friend (after his friend passes him)?
Express your answer using two significant figures.
Answer:
5.91 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Distance = Speed × Time
⇒Distance = 3.1t
Distance traveled by bicycle that passes through = 3.1t
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 2.4\times (t-2)^2\\\Rightarrow s=1.2(t-2)^2[/tex]
They both travel the same distance
[tex]3.1t=1.2(t-2)^2\\\Rightarrow 31t=12\left(t-2\right)^2\\\Rightarrow 12t^2-79t+48=0[/tex]
[tex]t=\frac{79+\sqrt{3937}}{24},\:t=\frac{79-\sqrt{3937}}{24}\\\Rightarrow t=5.91, 0.67[/tex]
Hence, time taken by the bicyclist to catch the other bicyclist is 5.91 seconds
Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.4 m/s2 while Kathy maintains an acceleration of 4.49 m/s2. (a) Find the time at which Kathy overtakes Stan. 6.7 Correct: Your answer is correct. seenKey 6.7 s from the time Kathy started driving
Answer:
time at which Kathy overtakes Stan is 6.70 sec
Explanation:
given data
time = 1 sec
acceleration= 3.4 m/s²
acceleration = 4.49 m/s²
to find out
the time at which Kathy overtakes Stan
solution
we consider here travel time for kathy = t1
and travel time for stan is = t2
and we know initial velocity = 0
so
t1 = 1 + t2
and distance travel equation by kinematic is
d1 = ut + [tex]\frac{1}{2}[/tex] at²
d1 = 0+ [tex]\frac{1}{2}[/tex] 4.49 (t2)² ................1
and
d2 = 0 + [tex]\frac{1}{2}[/tex] 3.4 (1+t2)² ..................2
and when overtake distance same so from equation 1 and 2
[tex]\frac{1}{2}[/tex] 4.49 (t2)² = [tex]\frac{1}{2}[/tex] 3.4 (1+t2)²
t2 = 6.703828 sec
so time at which Kathy overtakes Stan is 6.70 sec
Two boats start together and race across a 48-km-wide lake and back. Boat A goes across at 48 km/h and returns at 48 km/h. Boat B goes across at 24 km/h, and its crew, realizing how far behind it is getting, returns at 72 km/h. Turnaround times are negligible, and the boat that completes the round trip first wins.
(1) Which boat wins? (Or is it a tie?)
(A) boat A
(B) boat B
(C) it's a tie
.
(2) By how much? answer in km
(3) What is the average velocity of the winning boat? answer needs to be km/h
Answer:
Part 1)
Boat A will win the race
Part 2)
Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line
Part 3)
average velocity must be zero
Explanation:
As we know that the distance moved by the boat is given as
[tex]d = 48 km[/tex]
now the time taken by the boat to move to and fro is given as
[tex]t = \frac{d}{v}[/tex]
[tex]t = \frac{48 + 48}{48}[/tex]
[tex]t = 2 hrs[/tex]
Time taken by Boat B to cover the distance
[tex]t = \frac{48}{24} + \frac{48}{72}[/tex]
[tex]t = 2.66 h[/tex]
Part 1)
Boat A will win the race
Part 2)
Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line
Part 3)
Since the displacement of Boat A is zero
so average velocity must be zero
Final answer:
Boat A wins the race since it takes a total of 2 hours for the round trip, which is faster than Boat B's 2.6667 hours. Boat A wins by approximately 48 km when Boat B finishes. The average velocity of the winning boat (Boat A) is 48 km/h.
Explanation:
To determine which boat wins and by how much, we need to calculate the time each boat takes to make the round trip across the 48-km-wide lake. For Boat A, since it travels at 48 km/h both ways, the time to go across and back is simply the total distance (48 km there and 48 km back, for a total of 96 km) divided by the speed (48 km/h):
Time for Boat A = Total Distance / Speed = 96 km / 48 km/h = 2 hours.
For Boat B, the time to go across at 24 km/h and come back at 72 km/h must be calculated separately and then summed:
Time to go across for Boat B = Distance / Speed = 48 km / 24 km/h = 2 hours.
Time to come back for Boat B = Distance / Speed = 48 km / 72 km/h = 0.6667 hours.
Total time for Boat B = 2 hours + 0.6667 hours = 2.6667 hours.
Boat A wins since it takes less time to complete the round trip.
To find out by how much Boat A wins, we need to convert the time difference into distance. Since Boat A has already finished when Boat B has 0.6667 hours to go, we calculate how far Boat B would travel in that time at their average speed on the way back (72 km/h):
Difference in distance = Time difference x Speed of Boat B on return = 0.6667 hours x 72 km/h = 48 km.
For the average velocity of the winning boat (Boat A), considering the round trip without any rest or waiting time, it remains constant at 48 km/h because it travels at that speed in both the outward and return journeys.
The acceleration due to gravity on the Moon's surface is
one-sixth that on Earth. An astronaut's life support backpack
weighs300lb on Earth. What does it weigh on the moon?
The weight of the astronaut's life support backpack on the moon would be one-sixth of its weight on Earth, which equates to 50lb.
Explanation:The acceleration due to gravity on the moon is one-sixth that on Earth. Therefore, the weight of any object on the moon will be one-sixth of its weight on Earth. Since weight is a product of mass and gravity, if gravity is reduced, weight is also reduced accordingly.
To calculate the weight of an astronaut's life support backpack on the moon, you need to divide its Earth weight (300lb) by 6 (the relative gravity of the moon compared to Earth). So, on the moon, the backpack would weigh 300/6 = 50lb.
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Use dimensional analysis to determine how the linear acceleration a in m/s2 of a particle traveling in a circle depends on some, or all, of the following properties: the radius r of the circle, the angular frequency ω in s−1 with which the particle orbits about the circle, and the mass m of the particle. There is no dimensionless constant involved in the relation.
I have to say, i love this kind of problems.
So, we got the linear acceleration, as we all know, linear the acceleration its, in dimensional units:
[tex][a]=[\frac{distance}{time^2}][/tex].
Now, we got the radius
[tex][r] = [distance][/tex]
the angular frequency
[tex][\omega] = \frac{1}{s}[/tex]
and the mass
[tex][m]=[mass][/tex].
Now, the acceleration doesn't have units of mass, so it can't depend on the mass of the particle.
The distance in the acceleration has exponent 1, and so does in the radius. As the radius its the only parameter that has units of distance, this means that the radius must appear with exponent 1. Lets write
[tex]a \propto r[/tex].
The time in the acceleration has exponent -2 As the angular frequency its the only parameter that has units of time, this means that the angular frequency must appear, but, the angular frequency has an exponent of -1, this means it must be squared
[tex]a \propto r \omega^2[/tex].
We are almost there. If this were any other problem, we would write:
[tex]a = A r \omega^2[/tex]
where A its an dimensionless constant. Its common for this constants to appears if we need an conversion factor. If we wanted the acceleration in cm/s^2, for example. Luckily for us, the problem states that there is no dimensionless constant involved, so:
[tex]a = r \omega^2[/tex]
Final answer:
The linear acceleration a of a particle in circular motion depends on the radius r of the circle and the square of the angular frequency ω, which is expressed by the formula a = rω².
Explanation:
To determine how the linear acceleration a in m/s² of a particle traveling in a circle depends on the properties such as the radius r of the circle, the angular frequency ω in s⁻¹, and the mass m of the particle, we can use dimensional analysis and known relationships from physics.
Firstly, we recognize that mass m does not directly affect a, because a depends on the force per unit mass. By considering the relationship between linear and angular variables, we know that linear velocity v is related to ω by v = rω, and the centripetal acceleration of an object moving in a circle is ac = v²/r. Substituting v = rω into the expression for centripetal acceleration, we get ac = (rω)²/r = rω².
Therefore, the linear acceleration a of a particle undergoing circular motion is directly proportional to the radius r and the square of the angular frequency ω. This relationship can be expressed as a = rω², where a is the linear acceleration, r is the radius of the circle, and ω is the angular frequency.
A car starts out when the air temperature is 288 K. The air pressure in the tires is 460 kPa. After driving a while, the temperature of the air in the tires increases to 298 K. What is the pressure in the tires at that point, assuming the volume remains constant?
(A) 476 kPa
(B) 498 kPa
(C) 488 kPa
(D) 563 kPa
(E) 512 kPa
Answer:
(A) 476 kPa
Explanation:
If the volume remains constant, the ideal gas law says:
P/T=constant
so: P1/T1=P2/T2
P2=P1*T2/T1=460*298/288=476KPa
Answer:
A. The final pressure at the tire would be 476 kPa
Explanation:
Since the volume is constant the ideal gas equation would be used to obtain the final pressure at the tire.
Given
the initial temperature [tex]T_{1}[/tex] = 288 K
the initial pressure [tex]P_{1}[/tex] = 460 k Pa
the final temperature [tex]T_{2}[/tex] = 298 K
the final pressure [tex]P_{2}[/tex] = ?
Using the ideal gas equation;
PV = nRT
[tex]P_{1}[/tex] / [tex]T_{1}[/tex] = [tex]P_{2}[/tex] / [tex]T_{2}[/tex]
Making [tex]P_{2}[/tex] the subject formula
[tex]P_{2}[/tex] = ([tex]P_{1}[/tex] x [tex]T_{2}[/tex] ) / [tex]T_{1}[/tex]
[tex]P_{2}[/tex] = (460 x 298) / 288
[tex]P_{2}[/tex] = 475.972 kPa
[tex]P_{2}[/tex] ≈ 476 kPa
Therefore the final pressure at the tire would be 476 kPa
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.
An ideal, monotomic gas initially at a temperature of 450K, a pressure of 4.00 atm and a volume of 10.0L, undergoes an adiabatic compression to 1/3 its original volume. Find the final temperature of the gas. A. 72 K B. 150 K C. 216 K D. 936 K E. 1350 K
Answer: D) 936 K
Explanation:
Given:
Initial temperature of the gas, [tex]T = 450\ K[/tex]
Initial Pressure of the gas, [tex]P=4\ atm[/tex]
initial volume of the gas, [tex]V=10\ L[/tex]
It it given that the process is adiabatic, so for a adiabatic process we have
Let [tex]T_f \ \ and\ \ V_f [/tex] be the final temperature and volume of the gas.
[tex]T_iV_i^{\gamma -1}=T_fV_f^{\gamma -1}[/tex]
For monotomic gas [tex]\gamma=1.67[/tex]
[tex]450\times V_i^{1.67 -1} =T_f\left (\dfrac{V_i}{3} \right )^{1.67-1}\\T_f=936 K[/tex]
Hence the final temperature of the gas is 936 K. So option D is correct
An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accelerated by the presence of the electric field. 50% Part (a) What is the change in the electron’s kinetic energy, in kiloelectron volts, if it travels over a distance of 0.25 m in this field? ΔK = - 4.63 * 105|
Answer:
[tex]W = 462.5 keV[/tex]
Explanation:
As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron
So here we have to find the work done by electric field on moving electron
So we have
[tex]F = qE[/tex]
[tex]F = (1.6 \times 10^{-19})(1.85 \times 10^6)[/tex]
[tex]F = 2.96 \times 10^{-13} N[/tex]
now the distance moved by the electron is given as
[tex]d = 0.25 m[/tex]
so we have
[tex]W = F.d[/tex]
[tex]W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)[/tex]
[tex]W = 7.4 \times 10^{-14} J[/tex]
now we have to convert it into keV units
so we have
[tex]1 keV = 1.6 \times 10^{-16} J[/tex]
[tex]W = 462.5 keV[/tex]
If your car accelerates from rest at a steady rate of 4.2 m/s^2, how soon will it reach 72 km/h ( 44.7 mph or 20 m/s)?
Answer:
The car will reach 20 m/s in 4.76 seconds
Explanation:
The car moves following an uniformly accelerated motion, therefore we know that [tex]v=v_0+ a(t-t_0)[/tex]
Where [tex]v_0 = 0[/tex] are [tex]t_0 =0[/tex]
Then [tex]t= \frac{v}{a}=\frac{20 m/s}{4.2 m/s^{2}}=4.76s[/tex]
Suppose a person pushes thumbtack that is 1/5 centimeter long into a bulletin board, and the force (in dynes) exerted when the depth of the thumbtack in the bulletin board is x centimeters is given by F(x) = 1000 (1 + 2x)2 for 0 ≤ x ≤ 1 5 . Find the work W done by pushing the thumbtack all the way into the board.
Answer:
W = 290.7 dynes*cm
Explanation:
d = 1/5 cm = 0.2 cm
The force is in function of the depth x:
F(x) = 1000 * (1 + 2*x)^2
We can expand that as:
F(x) = 1000 * (1 + 4*x + 4x^2)
F(x) = 1000 + 4000*x + 4000*x^2
Work is defined as
W = F * d
Since we have non constant force we integrate
[tex]W = \int\limits^{0.2}_{0} {(1000 + 4000*x + 4000*x^2)} \, dx[/tex]
W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2
W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3
W = 200 + 80 + 10.7 = 290.7 dynes*cm
In this exercise, we have to use work knowledge to calculate the total, in this way:
[tex]W = 290.7[/tex]
Given some information about the exercise we find that:
d = 1/5 cm = 0.2 cm[tex]F(x) = 1000 * (1 + 2*x)^2[/tex]Using the working definition we find that:
[tex]W = F * d\\W = [1000*x + 2000*x^2 + 1333*X^3]\\W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3\\W = 200 + 80 + 10.7 = 290.7[/tex]
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What is the weight of a 2.5-kg pumpkin? Remember to always show all work.
Final answer:
The weight of a 2.5-kg pumpkin is calculated by multiplying the mass by the acceleration due to gravity, resulting in 24.5 newtons (N) when rounded to one decimal place.
Explanation:
To find out the weight of a 2.5-kg pumpkin, you would need to multiply its mass by the acceleration due to gravity. In most physics problems, this is approximately 9.8 meters per second squared (m/s²). Therefore, the weight W can be calculated using the formula W = m×g, where m is the mass in kilograms and g is the acceleration due to gravity.
So, for a 2.5-kg pumpkin, W = 2.5 kg × 9.8 m/s² = 24.5 kg·m/s². Since weight is a force, it is measured in newtons (N) in the International System of Units (SI). Therefore, the pumpkin weighs 24.5 N.
Final answer:
The weight of a 2.5-kg pumpkin is calculated by multiplying its mass by the gravitational acceleration, which results in a weight of 24.5 Newtons (N) on Earth.
Explanation:
The weight of a pumpkin is the force with which it is pulled towards the Earth due to gravity. To calculate the weight, you can use the formula:
Weight (W) = mass (m) × gravitational acceleration (g)
Where the gravitational acceleration on Earth is typically 9.8 meters per second squared (9.8 m/s²). For a 2.5-kg pumpkin, the calculation would be:
Weight (W) = 2.5 kg × 9.8 m/s²
Weight (W) = 24.5 kg·m/s²
Since weight is measured in Newtons (N), and 1 kg·m/s² is equivalent to 1 Newton, the weight of the pumpkin is 24.5 Newtons (N). When converting measurements and rounding, it's important to consider significant figures and the least precise measurement. In this calculation, both 2.5 kg and 9.8 m/s² are exact enough, so our final answer in Newtons does not need additional rounding.
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].
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.
A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 27 ft/s. Its height in feet after t seconds is given by y = 27 t − 20 t 2 y=27t-20t2. A. Find the average velocity for the time period beginning when t=3 and lasting .01 s: Preview .005 s: Preview .002 s: Preview .001 s: Preview NOTE: For the above answers, you may have to enter 6 or 7 significant digits if you are using a calculator. Estimate the instanteneous velocity when t=3. Preview Get help: Video
Answer:
[tex]V_{3.01}=-93.2m/s[/tex]
[tex]V_{3.005}=-93.1m/s[/tex]
[tex]V_{3.002}=-93.04m/s[/tex]
[tex]V_{3.001}=-93.02m/s[/tex]
[tex]V_{3}=-93m/s[/tex]
Explanation:
To calculate average velocity we need the position for both instants t0 and t1.
Now we will proceed to calculate all the positions we need:
[tex]Y_{3}=-99m/s[/tex]
[tex]Y_{3.01}=-99.932m/s[/tex]
[tex]Y_{3.005}=-99.4655m/s[/tex]
[tex]Y_{3.002}=-99.18608m/s[/tex]
[tex]Y_{3.001}=-99.09302m/s[/tex]
Replacing these values into the formula for average velocity:
[tex]V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s[/tex]
[tex]V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s[/tex]
[tex]V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s[/tex]
[tex]V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s[/tex]
To know the actual velocity, we derive the position and we get:
[tex]V=27-40t = -93m/s[/tex]
A stunt man drives a car at a speed of 20 m/s off a 28-m-high cliff. The road leading to the cliff is inclined upward at an angle of 20°. How far from the base of the cliff does the car land? What is the car's impact speed?
The car lands approximately 45.67 meters away from the base of the cliff, and its impact speed is 36.1 m/s.
Explanation:The situation described involves typing a projectile motion. Ignore any effect of air resistance for simplicity's sake. The stunt man's car exits the ramp at an angle, and continues its journey downwards under the force of gravity.
Firstly, we need to break down the initial velocity of the car into its horizontal and vertical components. Horizontal component (Ux) is governed by the formula Ux = U cos θ which results to 20 cos 20° = 18.79 m/s. Vertical component (Uy) is calculated by Uy = U sin θ which gives us 20 sin 20° = 6.84 m/s.
Following this, we need to calculate the total time the car is in the air (t), governed by the equation h = Uy * t - (1/2) * g * t², where g represents gravity (9.8 m/s²) and h is the height of the cliff (28m). Solving for t gives us approximately 2.43 seconds.
Finally, we find the horizontal distance (X) the car covers during the time it's in the air: X = Ux * t = 18.79 m/s * 2.43 s ≈ 45.67 m.
To calculate the impact speed, we take the square root of the sum of the squares of the final horizontal and vertical velocities. The final vertical velocity (Vy) = Uy + g * t = 6.84 m/s + 9.8 m/s² * 2.43 s = 30.95 m/s. The impact speed then equals √((18.79 m/s)² + (30.95 m/s)²) = 36.1 m/s.
Learn more about Projectile Motion here:https://brainly.com/question/29545516
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A ceiling fan has 5-inch blades (so the radius of the circular fan is inches). Suppose the fan turns at a rate of 2 revolutions per minute. (a) Find the angular speed of the fan in radians per minute. (b) Find the linear speed of the tip of a blade in inches per minute.
Answer:
a) 12.56 rad per minute
b)62.8 inches per minute
Explanation:
Given:
Radius of the blades, R=5 inchesFrequency of the blades, f=2 revolutions per minutea)
Let [tex]\omega[/tex] be the angular velocity of the blade which is given by
[tex]\omega =2\pi f\\\omega=2\times 3.14 \times 2\\\omega=12.56\ \rm rad/min[/tex]
b)
Let v be the linear speed of the tip of the blade which is given by
[tex]v=\omega \times R\\v=12.56\times 5\\v=62.8\ \rm inches/min[/tex]
Hence we have calculated the angular velocity and linear speed.
Two ice skaters, Frank and Mary, are initially facing each other and separated by 2000 m on a smooth frozen lake. Mary begins skating towards Frank at t = 0 s with an acceleration of 0.75 m/s^2. After she reaches a speed of 12 m/s, she continues skating at a constant 12 m/s. Three seconds after Mary begins to skate, Frank starts skating. He accelerates at 1.1 m/s^2 for 9 s and then skates at a constant speed. a) When do the skaters pass each other? b) How far from Mary’s starting position is the point at which the skaters pass each other?
Answer:
a) 99.1 s
b) 1093.2 m
Explanation:
The equation for speed with constant acceleration is
V(t) = V0 + a * t
Mary starts accelerating with a speed of zero, so
V0 = 0
V(t) = a * t
To reach a speed of 12 m/s
t = V(t) / a
t = 12 / 0.75 = 16 s
The equation for position under constant acceleration is
X(t) = X0 + V0 * t + 1/2 * a * t^2
Mary's starting position is zero
X0 = 0
X(t) = 1/2 * a * t^2
X(16) = 1/2 * 0.75 * 16^2 = 96 m
Frank starts skating 3 seconds after Mary, and he accelerates at 1.1 m/s^2 for 9 s. He will stop accelerating at second 12 (9 + 3).
His position after accelerating will be:
X(12) = X0 + V0 * (t - 3) + 1/2 * a * (t - 3)^2
His initial position is 2000, and his initial speed is zero
x(12) = 2000 - 1/2 1.1 * (12 - 3)^2 = 1955.5 m
Shi speed will be
V(12) = -1.1 * (12 - 3) = -9.9 m/s
From there they will move at constant speed from these positions. We can consider them as moving at constant speed starting at t0 = 16 and t0 = 12 respectively.
For Mary:
X(t) = X0 + V0 * (t - t0)
X(t) = 96 + 12 * (t - 16)
For Frank:
X(t) = 1955.5 - 9.9 * (t - 12)
Equating these two we can find the time when they meet:
96 + 12 * (t - 16) = 1955.5 - 9.9 * (t - 12)
96 + 12*t - 192 = 1955.5 - 9.9*t + 118.8
21.9*t = 2170.3
t = 2170.3 / 21.9 = 99.1 s
Replacing this time value on either equation we get the position:
X(99.1) = 96 + 12 * (99.1 - 16) = 1093.2 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
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
In physics, why would an equation like y=mx+b be plotted as a straight line rather than as a parabola, as is done in math?
Answer:
Explanation:
In Both Physics and Math
y=mx+b is plotted as straight line where
m=slope of line
b=intercept on Y-axis
whereas Equation of parabola is something like this
[tex]y^2=4ax[/tex]
or
[tex]x^2=4ay[/tex]
Math is a tool to solve Physics problems so equations are same in math and physics
Carl is eating lunch at his favorite cafe when his friend Isaac calls and says he wants to meet him. Isaac is calling from a city 155 miles155 miles away, and wants to meet Carl somewhere between the two locations. Isaac says he will start driving right away, but Carl needs 31.0 min31.0 min to finish his lunch before he can begin driving. Isaac plans to drive at 70.0 mph70.0 mph , and Carl plans to drive at 50.0 mph50.0 mph . Ignore acceleration and assume the highway forms a straight line. How long will Isaac be driving before he meets Carl?
Answer:
1.507 h
Explanation:
We can think that they move at constant speed, so their positions will follow
X(t) = Xo + v * t
However Carl will start moving 31 minutes later, so we can adjust the equation like this
X(t) = Xo + v * (t - t1)
Where t1 is the time at which each of them will start moving, 0 for Isaac since he starts driving right away, and 31 minutes (0.5167 h) for Carl
Also, we can considre the initial position of Carl to be 0 and his speed to be 50, while Isaac will have an initial position of 155 and a speed of -70 (negative because he is driving towards the origin of coordinates).
Then we have these two equations:
x(t) = 0 + 50 * (t - 0.5167)
x(t) = 155 - 70 * t
These are equations of lines, the point where they intersect determines the place and time they meet.
50 * (t -0.5167) = 155 - 70 * t
50 * t - 25.83 = 155 - 70 * t
120 * t = 180.83
t = 180.83 / 120 = 1.507 h
This is the time Isaac will be driving
Issac will drive for [tex]\boxed{1.507\text{ hr}}[/tex] before he meets carl.
Further Explanation:
The acceleration of the driving is ignored. It means that the motion of the cars is purely linear on the highways.
Given:
The distance between Carl and Issac is [tex]155\text{ miles}[/tex].
The speed of Carl is [tex]50\text{ mph}[/tex].
The speed of Issac is [tex]70\text{ mph}[/tex].
The time taken by Carl to finish lunch before starting to drive is [tex]31\text{ min}[/tex].
Concept:
Since Carl is eating lunch and he will start after 31 minutes, Issac will cover some distance within this time.
The distance covered by Issac before Carl starts to drive is,
[tex]\begin{aligned}d_1&=\text{speed}\times\text{time}\\&=50\text{ mph}\times\left(\dfrac{31}{60}\right)\text{hr}\\&=36.16\text{ mile}\end{aligned}[/tex]
Now after Issac has covered [tex]36.16\text{ miles}[/tex], Carl starts to drive. So, the distance left between Carl and Issac to be covered after carl starts to drive is,
[tex]\begin{aligned}d&=155\text{ miles}-36.16\text{ miles}\\&=118.84\text{ miles}\end{aligned}[/tex]
So, now the sum of the distances covered by Issac and Carl in time [tex]t[/tex] should be equal to the [tex]118.84\text{ miles}[/tex].
[tex]\begin{aligned}D_{\text{Issac }}+D_{\text{Carl}}&=118.84\text{ miles}\\(V_{Issac}\times t)+(V_{Carl}\times t)&=118.84\text{ miles}\end{aligned}[/tex]
Substituting the values of speed of Issac and Carl,
[tex]\begin{aligned}(70\times t)+(50\times t)&=118.84\text{ miles}\\120t&=118.84\text{ miles}\\t&=\frac{118.84}{120}\text{ hr}\\t&=0.990\text{ hr}\end{aligned}[/tex]
So, Issac and Carl meet after [tex]0.990\text{ hr}[/tex] after Carl starts driving.
So, the time for which Issac has been driving is,
[tex]\begin{aligned}T_{\text{ Issac}}&=0.990\text{ hr}+\dfrac{31}{60}\text{ hr}\\&=(0.990+0.516)\text{ hr}\\&=1.507\text{ hr}\end{aligned}[/tex]
Thus, Issac will drive for [tex]\boxed{1.507\text{ hr}}[/tex] before he meets carl.
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Answer Details:
Grade: College
Subject: Physics
Chapter: Linear motion in one dimension
Keywords:
Carl, Issac, Speed, meet, 155 miles, eating lunch, 70 mph, 50 mph, 31 min, between two locations, highway forms a straight line.
Joe and Max shake hands and say goodbye. Joe walks east 0.40 km to a coffee shop, and Max flags a cab and rides north 3.65 km to a bookstore. How fare apart is there destination?
Answer:
3.67 km
Explanation:
Joe distance towards coffee shop is,
[tex]OB=0.40 km[/tex]
And the Max distance towards bookstore is,
[tex]OA=3.65 km[/tex]
Now the distance between the Joy and Max will be,
By applying pythagorus theorem,
[tex]AB=\sqrt{OB^{2}+OA^{2}}[/tex]
Substitute 0.40 km for OB and 3.65 km for OA in the above equation.
[tex]AB=\sqrt{0.40^{2}+3.65^{2}}\\AB=\sqrt{13.4825} \\AB=3.67 km[/tex]
Therefore the distance between there destination is 3.67 km.
Two narrow slits are separated by a distance of 0.10 mm. When illuminated with 550 nm light, a diffraction pattern is observed on a screen 2.2 m from the slits. What is the lateral separation of the first and second order maxima seen on the screen (in cm)?
Answer:
1.21 cm.
Explanation:
Given that the wavelength of light is 550 nm.
And the distance between the slits is 0.10 mm.
And the distance to the screen is 2.2 m.
As we know that the lateral separation can be defined by the formula.
[tex]LS=\dfrac{m\lambda D}{d}[/tex]
Here, m is the order of maxima, [tex]\lambda[/tex] is the wavelength , d is the distance between the slits and D is the distance to the screen.
For first maxima m=1, and for second maxima m=2.
Difference in the lateral maxima will be,
[tex]\Delta LS=\dfrac{2\lambda D}{d}-\dfrac{1\lambda D}{d}\\\Delta LS=\dfrac{\lambda D}{d}[/tex]
Substitute all the variables in the above equation.
[tex]\Delta LS=\dfrac{550 nm\times 2.2m}{0.10mm}\\\Delta LS=\dfrac{550\times 10^{-9} nm\times 2.2m}{0.10\times 10^{-3} }\\ \Delta LS=121\times 10^{-4}m\\ \Delta LS=1.21 cm[/tex]
Therefore, the lateral separation difference between between first order and second order maxima is 1.21 cm.