Answer:
[tex]\omega = \frac{mv\frac{L}{2}\sin(29^\circ)}{\frac{1}{3}ML^2 + m(\frac{L}{2})^2} = 0.91~{\rm rad/s}[/tex]
Explanation:
The angular speed of the system can be found by conservation of angular momentum. Since the ball and the rod are stick together, the collision is completely inelastic, ergo kinetic energy is not conserved.
[tex]L_{initial} = L_{final}\\m\vec{v}\times \vec{r} = I\omega[/tex]
The moment of inertia of the combined objects is equal to the sum of moment of inertia of the separate objects.
[tex]I = \frac{1}{3}ML^2 + m(\frac{L}{2})^2[/tex]
The cross product in the left-hand side can be written as a sine of the angle.
Therefore;
[tex]mv\frac{L}{2}\sin(29^\circ) = (\frac{1}{3}ML^2 + m(\frac{L}{2})^2)\omega\\(19\times 10^{-3})(5)(\frac{47\times 10^{-2}}{2})\sin(29^\circ) = (\frac{1}{3}(146\times 10^{-3})(47\times 10^{-2})^2 + (19\times 10^{-3})(\frac{47\times 10^{-2}}{2})^2)\omega\\\omega = 0.91~{\rm rad/s}[/tex]In terms of system parameters:
[tex]\omega = \frac{mv\frac{L}{2}\sin(29^\circ)}{\frac{1}{3}ML^2 + m(\frac{L}{2})^2}[/tex]
sky divers jump out of planes at an altitude of 4000m. How much timewill passuntill they deploy their parachutes at an altitutde of 760m?
Answer:
Explanation:
Given
Altitude from which Diver steps out [tex]H=4000\ m[/tex]
altitude at which diver opens the Parachutes [tex]h=760\ m[/tex]
total height covered by diver[tex]=H-h=4000-760=3240\ m[/tex]
Initial velocity of diver is zero
using equation of motion
[tex]s=ut+\frac{1}{2}at^2[/tex]
where s=displacement
u=initial velocity
a=acceleration
t=time
[tex]3240=0+\frac{1}{2}\times 9.8\times t^2[/tex]
[tex]t^2=661.224[/tex]
[tex]t=25.71\ s[/tex]
You are going to an outdoor concert, and you'll be standing near a speaker that emits 60 W of acoustic power as a spherical wave. What minimum distance should you be from the speaker to keep the sound intensity level below 94 dB?
Answer:
[tex]r=44m[/tex]
Explanation:
β is calculated as:
[tex]\beta =(10dB)log_{10}(I/I_{o} )\\ I=I_{o}10^{\frac{\beta }{10dB} }\\ I=(1.0*10^{-12}W/m^{2} )10^{\frac{\(94dB }{10dB} }\\I=2.51mW/m^{2}[/tex]
The distance r is defined as the radius of spherical wave.solve for r
We have
[tex]I=\frac{P_{source} }{4\pi r^{2} }\\ r=\sqrt{\frac{P_{source}}{4\pi I} }\\ r=\sqrt{\frac{60W}{4\pi (2.51mW/m^{2} )} }\\r=44m[/tex]
If the balloon takes 0.19 s to cross the 1.6-m-high window, from what height above the top of the window was it dropped?
Answer:
[tex]heigth=2.86m[/tex]
Explanation:
Given data
time=0.19 s
distance=1.6 m
To find
height
Solution
First we need to find average velocity
[tex]V_{avg}=\frac{distance}{time}\\V_{avg}=\frac{1.6m}{0.19s}\\V_{avg}=8.42m/s[/tex]
Also we know that average velocity
[tex]V_{avg}=(V_{i}+V_{f})/2\\[/tex]
Where
Vi is top of window speed
Vf is bottom of window speed
Also we now that
[tex]V_{f}=V_{i}+gt\\V_{f}=V_{i}+(9.8)(0.19)\\V_{f}=V_{i}+1.862[/tex]
Substitute value of Vf in average velocity
So
[tex]V_{avg}=(V_{i}+V_{f})/2\\where\\V_{f}=V_{i}+1.862\\and\\V_{avg}=8.42m/s\\So\\8.42m/s=(V_{i}+V_{i}+1.862)/2\\2V_{i}+1.862=16.84\\V_{i}=(16.84-1.862)/2\\V_{i}=7.489m/s\\[/tex]
Vi is speed of balloon at top of the window
Now we need to find time
So
[tex]V_{i}=gt\\t=V_{i}/g\\t=7.489/9.8\\t=0.764s[/tex]
So the distance can be found as
[tex]distance=(1/2)gt^{2}\\ distance=(1/2)(9.8)(0.764)^{2}\\ distance=2.86m[/tex]
To determine the height from which the balloon was dropped, we can use the equation for vertical motion: h = 0.5*g*t^2, where h is the height, g is the acceleration due to gravity, and t is the time of flight. Given that the balloon takes 0.19 s to cross the 1.6 m high window, we can solve for the initial height to be approximately 0.01485 m or 14.85 cm above the top of the window.
Explanation:To determine the height from which the balloon was dropped, we can use the equation for vertical motion: h = 0.5*g*t^2, where h is the height, g is the acceleration due to gravity, and t is the time of flight.
Given that the balloon takes 0.19 s to cross the 1.6 m high window, we can plug in these values into the equation to solve for the initial height:
1.6 = 0.5*9.8*(0.19)^2
Simplifying the equation, we find that the balloon was dropped from a height of approximately 0.01485 m or 14.85 cm above the top of the window.
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A ball is thrown vertically upwards from the edge of the cliff and hits the ground at the base of the cliff with a speed of 30 m/s four seconds later. How high was the cliff?
To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.
[tex]v = v_0 -gt[/tex]
Here,
v = Final velocity
[tex]v_0[/tex] = Initial velocity
g = Acceleration due to gravity
t = Time
At t = 4s, v = -30m/s (Downward)
Therefore the initial velocity will be
[tex]-30 = v_0 -9.8(4)[/tex]
[tex]v_0 = 9.2m/s[/tex]
Now the position can be calculated as,
[tex]y = h +v_0t -\frac{1}{2}gt^2[/tex]
When it has the ground, y=0 and the time is t=4s,
[tex]0 = h+(9.2)(4)-\frac{1}{2} (9.8)(4)^2[/tex]
[tex]h = 41.6m[/tex]
Therefore the cliff was initially to 41.6m from the ground
The height of the cliff from which the ball was thrown is approximately 78.4 meters, as determined by the laws of Physics and specifically equations describing motion in a gravitational field.
Explanation:To calculate the height of the cliff you can use the laws of Physics, specifically the equations that deal with motion in a gravitational field. The question states that the ball is thrown upward and it hits the ground with a speed of 30 m/s. Since the ball is thrown upwards, the initial speed is 0 m/s. The only force acting on the ball is gravity, hence, the acceleration is 9.8 m/s² (downward). So let's put these values into motion equation. Our known values are final speed (v) = 30 m/s, acceleration (a) = -9.8 m/s² (we take negative as it’s downward motion) and time (t) = 4 s. We want to find out the distance (s). We can use the second equation of motion: s = ut + 0.5*at². Substituting the known values, we get 0*4 + 0.5*-9.8*4² = -78.4 m. The negative sign indicates that the ball is below the starting point, which makes sense since it fell off a cliff. So, the height of the cliff is 78.4 meters.
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A resultant vector is 8.00 units long and makes an angle of 43.0 degrees measured ������� – ��������� with respect to the positive � − ����. What are the magnitude and angle (measured ������� – ��������� with respect to the positive � − ����) of the equilibrant vector? Please show all steps in your calculations
Answer:
223 degree
Explanation:
We are given that
Magnitude of resultant vector= 8 units
Resultant vector makes an angle with positive -x in counter clockwise direction
[tex]\theta=43^{\circ}[/tex]
We have to find the magnitude and angle of the equilibrium vector.
We know that equilibrium vector is equal in magnitude and in opposite direction to the given vector.
Therefore, magnitude of equilibrium vector=8 units
x-component of a vector=[tex]v_x=vcos\theta[/tex]
Where v=Magnitude of vector
Using the formula
x-component of resultant vector=[tex]v_x=8cos43=5.85[/tex]
y-component of resultant vector=[tex]v_y=vsin\theta=8sin43=5.46[/tex]
x-component of equilibrium vector=[tex]v_x=-5.85[/tex]
y-component of equilibrium vector=[tex]-v_y=-5.46[/tex]
Because equilibrium vector lies in III quadrant
[tex]\theta=tan^{-1}(\frac{v_x}{v_y})=tan^{-1}(\frac{-5.46}{-5.85})=43^{\circ}[/tex]
The angle [tex]\theta'[/tex]lies in III quadrant
In III quadrant ,angle =[tex]\theta'+180^{\circ}[/tex]
Angle of equilibrium vector measured from positive x in counter clock wise direction=180+43=223 degree
A crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one end. What is the fundamental frequency if the tube is 0.203-m long, by taking air temperature to be 37.0ºC?
Answer:
435.467980296 Hz
Explanation:
T = Temperature = 37.0ºC
L = Length of the tube = 0.203 m
Speed of sound at a specific temperature is given by
[tex]v=331.4+0.6\times T\\\Rightarrow v=331.4+0.6\times 37\\\Rightarrow v=353.6\ m/s[/tex]
Frequency is given by (one end open other end closed)
[tex]f=\dfrac{v}{4L}\\\Rightarrow f=\dfrac{353.6}{4\times 0.203}\\\Rightarrow f=435.467980296\ Hz[/tex]
The fundamental frequency is 435.467980296 Hz
The fundamental frequency of a tube, considered as an approximation for the vocal apparatus and closed at one end, with a length of 0.203 meters and at an air temperature of 37°C, is calculated using the formula for frequency f=Vw/(4L). Substituting in the values, the fundamental frequency is about 432 Hz.
Explanation:The given question involves a concept from physics, specifically sound waves and resonance. When considering the breathing passages and mouth as a resonating tube closed at one end, the fundamental resonant frequency can be calculated using the formula Vw = fa, where Vw is the speed of sound, f represents frequency, and a is the wavelength.
For a tube closed at one end, the wavelength (λ) is four times the length of the tube. So, λ = 4L. We can rearrange the formula to solve for fundamental frequency: f = Vw / a = Vw / (4L).
If the air temperature is 37°C, the speed of sound (Vw) in air is approximately 351 m/s. Substituting these values in, you get: f = 351m/s / (4*0.203m) ≈ 432 Hz. So, the fundamental frequency of this tube (our vocal apparatus approximation) at 37°C would be about 432 Hz.
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If two such generic humans each carried 2.5 coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 600 weight? *An average human weighs about 600 r= .............. km
Answer:
They would be [tex]r=9.7\,km [/tex] apart
Explanation:
Electric force between two charged objects is:
[tex] F_{e}=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}[/tex] (1)
With q1, q2 the charges of the humans, r the distance between them and k the constant [tex] k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}[/tex], so if we want the electric force between them will be equal to their 600 N weight, we should make W=Fe=600 N on (1):
[tex]600=k\frac{\mid q_{1}q_{2}\mid}{r^{2}} [/tex]
solving for r:
[tex]r^{2}=k\frac{\mid q_{1}q_{2}\mid}{600}[/tex]
[tex] r=\sqrt{k\frac{\mid q_{1}q_{2}\mid}{600}}[/tex]
[tex]r=\sqrt{(9.0\times10^{9})\frac{\mid(-2.5)(2.5)\mid}{600}} =9682 m[/tex]
[tex]r=9.7\,km [/tex]
The space shuttle fleet was designed with two booster stages. if the first stage provides a thrust of 53 kilo-newtons and the space shuttle has an acceleration of 18,000 miles per hour squared, what is the mass of the spacecraft in units of pounds-mass?
Use Newton's second law to find the mass in kilograms, and finally convert it to pounds-mass, which is approximately 14.52 pounds-mass.
The question asks us to calculate the mass of the spacecraft given the thrust of the first booster stage and the acceleration of the space shuttle. To find the mass, we use Newton's second law of motion, which states that Force = [tex]mass imes acceleration (F = m imes a).[/tex] However, we first need to convert the given acceleration from miles per hour squared to meters per second squared, and then convert the mass obtained in kilograms to pounds-mass.
First, let's convert the acceleration: 18,000 miles/hour2 is approximately 8,046.72 m/s2 (using the conversion factor 1 mile = 1,609.34 meters and 1 hour = 3600 seconds). Next, we can calculate the mass (in kilograms) by rearranging the formula to m = F / a, which gives us 53,000 N / 8,046.72 m/s2
= approximately 6.59 kilograms. We then convert kilograms to pounds-mass (1 kilogram = 2.20462 pounds), resulting in the mass of the spacecraft being approximately 14.52 pounds-mass.
The mass of the spacecraft is approximately [tex]261,313.24\ pounds[/tex] mass.
To find the mass of the spacecraft, we can use Newton's second law of motion:
[tex]\[ F = ma \][/tex]
where:
[tex]- \( F \)[/tex] is the force (thrust) provided by the first stage ([tex]53 kN[/tex]),
[tex]- \( m \)[/tex] is the mass of the spacecraft,
[tex]- \( a \)[/tex] is the acceleration ([tex]18,000 \ mph^2[/tex]).
First, let's convert the thrust from kilo-newtons (kN) to newtons (N), since the unit of acceleration is meters per second squared (m/s²):
[tex]\[ 1 \text{ kN} = 1000 \text{ N} \][/tex]
So, [tex]53 kN[/tex] is equivalent to [tex]\(53 \times 1000 = 53000\) N.[/tex]
Now, let's convert the acceleration from miles per hour squared (mph²) to meters per second squared (m/s²).
[tex]\[ 1 \text{ mph} = \frac{1609.34}{3600} \text{ m/s} \][/tex]
[tex]1 mph^2 = \left(\frac{1609.34}{3600}\right)^2 \text{ m/s²}[/tex]
[tex]1 mph^2 = 0.447 m/s^2}[/tex]
We can rearrange Newton's second law to solve for the mass [tex]\( m \)[/tex]
[tex]\[ m = \frac{F}{a} \][/tex]
[tex]\[ m = \frac{53000 \text{ N}}{0.447 \text{ m/s²}} \][/tex]
[tex]\[ m = 118488 \text{ kg} \][/tex]
To convert kilograms to pounds-mass, we use the conversion factor:
[tex]\[ 1 \text{ kg} = 2.20462 \text{ pounds-mass} \][/tex]
So,
[tex]\[ \text{mass (pounds-mass)} = 118488 \text{ kg} \times 2.20462 \][/tex]
[tex]\[ \text{mass (pounds-mass)} = 261313.24 \text{ pounds-mass} \][/tex]
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cmcm . The explorer finds that the pendulum completes 103 full swing cycles in a time of 132 s.What is the value of the acceleration of gravity on this planet?
Answer:
13.01 m/s²
Explanation:
The period of a simple pendulum is given as
T = 2π√(L/g) .......................... Equation 1
Where T = Period of the simple pendulum, L = Length of the pendulum, g = acceleration due to gravity of the planet.
Given; T = 132/103 = 1.28 s, L = 54 cm = 0.54 m, π = 3.14
Substitute into equation 1
1.28 = (2×3.14)√(0.54/g)
1.28 = 6.28√(0.54/g)
Making g the subject of the equation,
√(0.54/g) = 1.28/6.28
√(0.54/g) = 0.2038
0.54/g = (0.2038)²
0.54/g = 0.0415
g = 0.54/0.0415
g = 13.01 m/s²
Hence the value of the acceleration due to gravity on the planet = 13.01 m/s²
An astronaut takes her bathroom scales to the moon, where g = 1.6 m/. On the moon, compared to at home on earth:
(A) Her weight is less, and her mass is less.
(B) Her weight is the same, and her mass is less.
(C) Her weight is the same, and her mass is the same.
(D) Her weight is less, and her mass is the same.
(E) Her weight is zero, and her mass is the same.
Answer: (D) Her weight is less, and her mass is the same.
Explanation:
Mass is the total amount of matter contained in the body.
Weight is the force exerted by gravity on a mass.
[tex]weight=mass\times gravity[/tex]
When the value of g is more, weight is more.
For moon g= [tex]1.6m/s^2[/tex] , whereas for earth [tex]g=9.8m/s^2[/tex]
Thus weight is more on earth as compared to moon.
On the moon, compared to at home on earth: Her weight is less, and her mass is the same.
Answer:
Explanation:
Weight is defined as the force with which a planet pulls the object towards its centre.
Weight = mass x acceleration due to gravity of the planet
W = m x g
The amount of matter contained in the substance is called mass.
the mass of the substance remains same at every planet but the weight is changed as the value of acceleration due to gravity is different for all the planets.
As the acceleration due to gravity on moon is 1.6 m/s^2 and it is less than the acceleration due to gravity on earth. So, the weight of the astronaut is less than the weight on earth and mass remains same.
Weight is less but the mass is same.
Option (D) is true.
A baseball is hit with a speed of 33.6 m/s. Calculate its height and the distance traveled if it is hit at angles of 30.0°, 45.0°, and 60.0°.
The height and distance for 30 degrees are 14.4 m and 100 m, for 45 degrees are 28.77 m and 115.11 m, and for 60 degrees are 43.17 m and 100 m, respectively.
In projectile motion, consider the motion along vertical and horizontal separately.
Given:
Initial velocity, [tex]u =33.6\ m/s\\[/tex]
For angle [tex]30^o[/tex], the initial and final velocities are calculated as:
[tex]u_x =33.6\times cos(30.0)\\=29.14m/s\\u_y =33.6\times sin30.0\\ =16.80m/s[/tex]
The height and distance traveled are computed as:
[tex]h=\frac{u_y^2}{2g}\\=\frac{16.80^2}{2\times9.8}\\=14.4\ m\\R= \frac{u^2sin^22\theta}{g}\\= \frac{2\times16.80\times29.14}{9.8}\\=100\ m[/tex]
For angle [tex]45^o[/tex], the initial and final velocities are calculated as:
[tex]u_x =33.6\times cos(45.0)\\=23.75m/s\\u_y =33.6\times sin45\\ =23.75m/s[/tex]
The height and distance traveled are computed as:
[tex]h=\frac{u_y^2}{2g}\\=\frac{23.75^2}{2\times9.8}\\=28.77\ m\\R= \frac{u^2sin^22\theta}{g}\\= \frac{2\times23.75\times23.75}{9.8}\\=115.11\ m[/tex]
For angle [tex]60^o[/tex], the initial and final velocities are calculated as:
[tex]u_x =33.6\times cos(60)\\=16.8m/s\\u_y =33.6\times sin60\\ =29.09m/s[/tex]
The height and distance traveled are computed as:
[tex]h=\frac{u_y^2}{2g}\\=\frac{29.09^2}{2\times9.8}\\=43.17\ m\\R= \frac{u^2sin^22\theta}{g}\\= \frac{2\times16.80\times29.09}{9.8}\\=100\ m[/tex]
Therefore, the distance and height for 30 degrees are 14.4 m and 100 m for 45 degrees are 28.77 m and 115.11 m, and for 60 degrees are 43.17 m and 100 m, respectively.
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To calculate the height and distance traveled by a baseball hit at different angles, we can use the equations of projectile motion. The height can be calculated using the formula h = (v²sin²θ) / (2g), where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. The distance traveled can be calculated using the formula d = v²sin(2θ) / g.
To calculate the height and distance traveled by a baseball hit at different angles, we can use the equations of projectile motion. The height can be calculated using the formula h = (v²sin²θ) / (2g), where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (approximately 9.8 m/s²). Once we have the height, we can calculate the distance traveled using the formula d = v²sin(2θ) / g.
Height (h) = (33.6²sin²30.0°) / (2 * 9.8) = 19.22 meters
Distance (d) = 33.6²sin(2 * 30.0°) / 9.8 = 152.19 meters
Height (h) = (33.6²sin²45.0°) / (2 * 9.8) = 38.45 meters
Distance (d) = 33.6²sin(2 * 45.0°) / 9.8 = 203.43 meters
Height (h) = (33.6²sin²60.0°) / (2 * 9.8) = 57.67 meters
Distance (d) = 33.6²sin(2 * 60.0°) / 9.8 = 228.05 meters
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A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is fixed in place at the bottom of the incline. If the inclined plane makes an angle θ of 35.0 ∘ with the horizontal, what is the magnitude of the acceleration of the box when it is 61.0 cm from the bottom of the incline?
The magnitude of the acceleration of the box is [tex]a=6.41\ m/s^2.[/tex]
First, we need to calculate the force between the two charges using Coulomb's law:
[tex]\[ F = \frac{qQ}{4\pi\epsilon_0r^2} \][/tex]
Given that [tex]\( q = +2.50 \, \mu C = 2.50 \times 10^{-6} \, C \), \( Q = +75.0 \, \mu C = 75.0 \times 10^{-6} \, C \)[/tex], and [tex]\( r = 61.0 \, cm = 0.610 \, m \)[/tex], we can plug these values into the equation:
[tex]\[ F = \frac{(2.50 \times 10^{-6} \, C)(75.0 \times 10^{-6} \, C)}{4\pi(8.85 \times 10^{-12} \, F/m)(0.610 \, m)^2} \][/tex]
Solving this, we get the force [tex]\( F \)[/tex] in newtons.
Next, we need to find the component of this force parallel to the incline, which is given by:
[tex]\[ F_{\parallel} = F \sin(\theta) \][/tex]
However, since the box is on an incline, we use [tex]\( \tan(\theta) \)[/tex] instead of [tex]\( \sin(\theta) \)[/tex] to find the acceleration:
[tex]\[ F_{\parallel} = F \tan(\theta) \][/tex]
Given [tex]\( \theta = 35.0^{\circ} \)[/tex], we can calculate [tex]\( F_{\parallel} \)[/tex].
Finally, the acceleration \( a \) of the box is given by Newton's second law:
[tex]\[ a = \frac{F_{\parallel}}{m} \][/tex]
where[tex]\( m = 495 \ , g = 0.495 \, kg \)[/tex].
Combining all the steps, we have:
[tex]\[ a = \frac{qQ}{4\pi\epsilon_0mr^2} \tan(\theta) \][/tex]
Plugging in the values, we get:
[tex]\[ a = \frac{(2.50 \times 10^{-6} \, C)(75.0 \times 10^{-6} \, C)}{4\pi(8.85 \times 10^{-12} \, F/m)(0.495 \, kg)(0.610 \, m)^2} \tan(35.0^{\circ}) \][/tex]
Calculating the value of [tex]\( a \)[/tex], we find the magnitude of the acceleration of the box.
[tex]a=6.41\ m/s^2[/tex]
hich one of the following statements could be an operational definition of electric current?View Available Hint(s)Which one of the following statements could be an operational definition of electric current?The magnitude of the physical quantity of electric current I in a wire equals the magnitude of the electric charge q that passes through a cross section of the wire divided by the time interval Δt needed for that charge to pass.
The statement "The magnitude of the physical quantity of electric current \( I \) in a wire equals the magnitude of the electric charge q that passes through a cross-section of the wire divided by the time interval [tex]\( \Delta t \)[/tex] needed for that charge to pass" could be an operational definition of electric current.
This definition precisely defines electric current as the rate of flow of electric charge through a conductor over a specific time period.
By quantifying the amount of charge passing through a cross-section of the wire in a given time interval, this definition provides a measurable and practical way to understand and quantify electric current.
It aligns with the fundamental concept that electric current represents the flow of charge, making it a suitable operational definition in the context of electrical systems and circuits.
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, downward. (a) How many electrons were added to or removed from the honeybee? (b) What is the ratio of the electric force on the bee to the bee’s weight (Fe/Fg)? (c) What electric field strength and direction would allow the bee to hang suspended in the air without effort?
The ratio of the electric force on the bee to the bee’s weight (Fe/Fg) is; 2.041 * 10⁻⁶
The electric field strength is; 4.9 * 10⁷ N/C
What is the electric field strength?
We are given;
Mass of bee; m = 0.12g = 0.00012 kg
Charge acquired by the bee q = +24 pC = 24 * 10⁻¹² C
Electric field; E = 100 N/C
B) The force F on a charge in electric field E is given by:
F = qE
F = 24 * 10⁻¹² * 100
F = 24 * 10⁻¹⁰ N
Weight is;
F_g = mg
F_g = 0.00012 * 9.8
F_g = 11.76 × 10⁻⁴ N
Ratio of force to weight is;
F_e/F_g = (24 * 10⁻¹⁰)/(11.76 × 10⁻⁴)
F_e/F_g = 2.041 * 10⁻⁶
C) To get the electric field strength, we will use the formula;
E' = mg/q
E' = (0.00012 * 9.8)/(24 * 10⁻¹²)
E' = 4.9 * 10⁷ N/C
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Final answer:
The honeybee gained 15,000 electrons. The ratio of the electric force on the bee to its weight is (100 N/C) x (24 x 10⁻¹² C) / [(0.12 g) x (9.8 m/s²)]. To hang suspended in the air without effort, the electric field strength should have the same magnitude as the gravitational field strength (100 N/C downward), but in the opposite direction (upward).
Explanation:
(a) How many electrons were added to or removed from the honeybee?
To determine the number of electrons added to or removed from the honeybee, we can use the formula:
q = ne
Where q is the charge in Coulombs, n is the number of electrons, and e is the elementary charge, which is the charge of a single electron.
Given that the charge acquired by the honeybee is +24 pC (pC = picocoulombs = 10⁻¹² C), we can substitute the values into the formula:
24 x 10⁻¹² C = n x (1.6 x 10⁻¹⁹ C)
Solving for n, we find:
n = 15,000 electrons
Therefore, the honeybee gained 15,000 electrons.
(b) What is the ratio of the electric force on the bee to the bee’s weight (Fe/Fg)?
To calculate the ratio of the electric force on the bee to its weight, we can use the formula:
F = Eq
Where F is the force, E is the electric field strength, and q is the charge.
Given that the electric field near the surface of the Earth is 100 N/C downward, we can substitute the values into the formula:
F = (100 N/C) x (24 x 10⁻¹² C)
Using the weight formula Fg = mg, where m is the mass and g is the acceleration due to gravity, we can find the weight of the bee.
Fg = (0.12 g) x (9.8 m/s²)
Taking the ratio of Fe to Fg, we get:
Fe/Fg = (100 N/C) x (24 x 10⁻¹² C) / [(0.12 g) x (9.8 m/s²)]
(c) What electric field strength and direction would allow the bee to hang suspended in the air without effort?
The electric field needed for the bee to hang suspended in the air without any effort would need to balance the gravitational force acting on the bee. The gravitational force on the bee can be calculated using the formula Fg = mg, where m is the mass of the bee and g is the acceleration due to gravity. We can then set this force equal to the electric force using the formula F = Eq. Rearranging the formula, we get E = F/q.
Since we want the bee to hang suspended without any effort, the electric force needs to be equal in magnitude and opposite in direction to the gravitational force. This means the electric field strength should have the same magnitude as the gravitational field strength (100 N/C downward), but in the opposite direction (upward).
A sperm whale can accelerate at about 0.120 m/s 2 0.120 m/s2 when swimming on the surface of the ocean. How far will a sperm whale travel if it starts at a speed of 1.10 m/s 1.10 m/s and accelerates to a speed of 2.05 m/s 2.05 m/s ? Assume the whale travels in a straight line.
Answer:
The distance traveled by the sperm whale is 12.46 meters.
Explanation:
Given that,
Acceleration of a sperm whale, [tex]a=0.12\ m/s^2[/tex]
Initial speed of the whale, u = 1.1 m/s
Final speed of the whale, v = 2.05 m/s
We need to find the distance traveled by the sperm whale. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2ad[/tex]
d is the distance covered
[tex]d=\dfrac{v^2-u^2}{2a}[/tex]
[tex]d=\dfrac{(2.05)^2-(1.1)^2}{2\times 0.12}[/tex]
d = 12.46 meters
So, the distance traveled by the sperm whale is 12.46 meters. Hence, this is the required solution.
A bullet is fired with a muzzle velocity of 1178 ft/sec from a gun aimed at an angle of 26° above the horizontal. Find the horizontal component of the velocity.
Answer:
1058.78 ft/sec
Explanation:
Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis
The horizontal velocity of a body can be calculated as shown below.\
Vh = Vcos∅.......................... Equation 1
Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.
Given: V = 1178 ft/sec, ∅ = 26°
Substitute into equation 1
Vh = 1178cos26
Vh = 1178(0.8988)
Vh = 1058.78 ft/sec
Hence the horizontal component of the velocity = 1058.78 ft/sec
What are the two angles between the direction of the current and the direction of a uniform 0.0400 T magnetic field for which the magnetic force on the wire has magnitude 0.0250 N?
Answer:
The two values of θ are 41.03° and 138.97°.
Explanation:
The force on a current carrying wire is given by the following equation:
[tex]\vec{F} = I\vec{L}\times \vec{B}[/tex]
The cross-product can be written with a sine term:
[tex]F = ILB\sin(\theta)\\0.025 = IL(0.04)\sin(\theta)\\\sin(\theta) = \frac{0.025}{0.04IL}\\\theta = \arcsin(\frac{0.025}{0.04IL})[/tex]
If we assume that the wire is 0.28 m long and the current is 3.40 A, then sin(θ) becomes 0.6565.
Finally, the two values of θ are 41.03° and 138.97°.
A half-full recycling bin has mass 3.0 kg and is pushed up a 40.0^\circ40.0 ∘ incline with constant speed under the action of a 26-N force acting up and parallel to the incline. The incline has friction. What magnitude force must act up and parallel to the incline for the bin to move down the incline at constant velocity?
Final answer:
A 26-N force must act down and parallel to the incline for the bin to move down at constant velocity since this force would balance out the 26-N frictional force acting up the incline.
Explanation:
The subject of this question is Physics, specifically the concepts related to mechanics and forces on inclines with friction. The student is in High School level, most likely studying the fundamentals of Newtonian mechanics.
Given that the bin is already being pushed up the incline with a 26-N force and moves at a constant speed, this means the net force on the bin is zero, therefore the force of friction must also be 26 N but acting downwards along the incline. When the bin moves down at a constant velocity, the force of friction still acts up the incline (opposite to the bin's movement) and therefore, to maintain a constant velocity, the force applied must be equal in magnitude to the frictional force but directed down the incline. Hence, a 26-N force must be applied down and parallel to the incline for the bin to move down the incline at constant velocity.
A plane flies 1.4hours at 110mph on a bearing of 40degrees.It then turns and flies 1.7hours at the same speed on a bearing of 130degrees.How far is the plane from its starting point?
Answer:
Distance from starting point= 242.249871 miles
Explanation:
Distance covered on bearing of 40 degree=a= [tex]1.4*110\ mph[/tex]
Distance covered on bearing of 40 degree=a=154 miles
Distance covered on bearing of 130 degree=b=[tex]1.7*110\ mph[/tex]
Distance covered on bearing of 40 degree=b=187 miles
Angle between bearing=[tex]130-40[/tex]
Angle between bearing=90 degree
With the angle of 90 degree, Distance from starting point can be calculated from Pythagoras theorem.
Distance from starting point=[tex]\sqrt{a^2+b^2}[/tex]
Distance from starting point=[tex]\sqrt{154^2+187^2}[/tex]
Distance from starting point= 242.249871 miles
Answer:
Distance from starting point= 242.249871 miles
Distance covered on bearing of 40 degree=a=
Distance covered on bearing of 40 degree=a=154 miles
Distance covered on bearing of 130 degree=b=
Distance covered on bearing of 40 degree=b=187 miles
Angle between bearing=
Angle between bearing=90 degree
With the angle of 90 degree, Distance from starting point can be calculated from Pythagoras theorem.
Distance from starting point=
Distance from starting point=
Distance from starting point= 242.249871 miles
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What is the resolution of an analog-to-digital converter with a word length of 12 bits and an analogue signal input range of 100V? Show work.
The resolution of an ADC with a word length of 12 bits and an input range of 100V is approximately 0.0244V. Calculating this gives us a resolution of approximately 0.0244V.
Explanation:The resolution of an analog-to-digital converter (ADC) is determined by the number of bits used to represent the digital output.
In this case, the ADC has a word length of 12 bits.
The resolution can be calculated using the formula:
Resolution = Full Scale Range / (2^Word Length)
In this case, the Full Scale Range is 100V. Plugging in the values:
Resolution = 100V / (2^12)
Calculating this gives us a resolution of approximately 0.0244V.
Problem 12.78 Determine the mass of the earth knowing that the mean radius of the moon’s orbit about the earth is 238,910 mi and that the moon requires 27.32 days to complete one full revolution about the earth.
The actual mass of the Earth (5.97 \times 10^24 kg).
To find the mass of the Earth using the given information about the moon's orbit and revolution period, we can use Kepler's third law, which states that the square of the orbital period of a planet or moon is proportional to the cube of its average distance from the planet or sun.
Let's call the mass of the Earth M and the mass of the moon m. The distance between the Earth and the moon is R = 238,910 mi.
According to Kepler's third law, we have:
(27.32 days)^2 = (238,910 mi)^3 / (M + m)
(38.79 days)^2 = (6,022,140,778 mi)^3 / M
(where 38.79 days is the sidereal period of Earth's rotation)
Dividing these two equations gives:
(38.79 days)^2 / (6,022,140,778 mi)^3 = (27.32 days)^2 / (238,910 mi)^3
Simplifying this expression:
M = (6,022,140,778 mi)^3 / [(38.79 days)^2 - (27.32 days)^2 / (1 + m/M)]^(1/3)
Since we know that m/M is very small (the mass of the moon is only about 1/81 of the mass of the Earth), we can approximate m/M as zero in this equation:
M = (6,022,140,778 mi)^3 / (38.79 days)^2 = 5.97 \times 10^24 kg
This is very close to the actual mass of the Earth (5.97 \times 10^24 kg).
A tetrahedron has an equilateral triangle base with 27.0-cm-long edges and three equilateral triangle sides. The base is parallel to the ground, and a vertical uniform electric field of strength 280 N/C passes upward through the tetrahedron.What is the electric flux through each of the three sides?
Answer:
The electric flux through each of the 3 sides is 2.95 Wb
Solution:
As per the question:
Length of the edges, l = 27.0 cm = 0.27 m
Strength of the electric field, E = 280 N/C
To calculate the electric field through each of the three sides:
Area of the equilateral triangle is given by:
[tex]A = \frac{\sqrt{3}}{4}l^{2}[/tex]
[tex]A = \frac{\sqrt{3}}{4}\times (0.27)^{2} = 0.0316\ m^{2}[/tex]
Now, the electric flux that passes through the base is given by:
[tex]\phi = - E\cdot A = - EA[/tex]
[tex]\phi = 280\times 0.0316 = - 8.85\ Wb[/tex]
Now, the overall flux that passes through the surface and the base of the tetrahedron is zero.
[tex]\phi_{total} = \phi + 3phi_{surface}[/tex]
[tex]0 = \phi + 3phi_{surface}[/tex]
[tex]phi_{surface} = -\frac{\phi }{3}[/tex]
[tex]phi_{surface} = -\frac{- 8.85}{3} = 2.95\ Wb[/tex]
A vector has an x component of -24.0 units and a y component of 40.4 units. Find the magnitude and direction of this vector.
Answer:
The magnitude is 47.0 and the direction is 120.7° above the x-axis counterclockwise.
Explanation:
Hi there!
The magnitude of any vector is calculated as follows:
[tex]|v| = \sqrt{x^{2} + y^{2} }[/tex]
Where:
v = vector.
x = x-component of v.
y = y-component of v.
Then, the magnitude of this vector will be:
[tex]|v| = \sqrt{24.0^{2} + 40.4^{2} } = 47.0[/tex]
The magnitude is 47.0
The direction can be found by trigonometry (see attached figure).
According to the trigonometry rules of right triangles (as the one in the figure):
cos θ = adjacent side to the angle θ / hypotenuse
In this case:
adjacent side = x-component of v.
hypotenuse = magnitude of v.
Then:
cos θ = 24.0 / 47.0
θ = 59.3°
Measured in counterclockwise:
180° - 59.3° = 120.7°
The magnitude is 47.0 and the direction is 120.7° above the x-axis counterclockwise.
Two cars start 200 m apart and drive toward each other at a steady 10 m/s. On the front of one of them, an energetic grasshopper jumps back and forth between the cars (he has strong legs!) with a constant horizontal velocity of 15 m/s relative to the ground. The insect jumps the instant he lands, so he spends no time resting on either car. What total distance does the grasshopper travel before the cars hit?
Answer:
Total distance does the grasshopper travel before the cars hit is 150 m
Explanation:
Each car moves x=100 m before they collide. Both the cars moving in constant velocity. time taken t by each car is
[tex]t=\frac{x}{v}[/tex]
where x is the distance traveled with velocity v
[tex]t=\frac{100}{10}\\t=10 sec[/tex]
The insect is moving through this time period with a constant velocity of 15 m/s
The distance traveled by grasshopper is
[tex]distance=V_{gh} \times t\\distance=15 \times 10\\distance=150 m[/tex]
The grasshopper travels a total distance of 2600 m before the cars collide.
Explanation:We can solve this problem by calculating the time it takes for the cars to collide. Since the cars are moving towards each other at a combined speed of 20 m/s (10 m/s + 10 m/s), and they start 200 m apart, it will take them 200 m / 20 m/s = 10 seconds to collide.
During these 10 seconds, the grasshopper keeps jumping back and forth with a velocity of 15 m/s relative to the ground. To find the total distance the grasshopper travels, we need to calculate the number of jumps the grasshopper can make in 10 seconds. Since the grasshopper jumps the instant it lands, the number of jumps is equal to the number of times the grasshopper crosses from one car to the other. Given that the grasshopper has a velocity of 15 m/s, we divide the total distance the grasshopper travels by this velocity to find the number of jumps: 200 m / (15 m/s) = 13.33 jumps.
The grasshopper is not able to make a fraction of a jump, so we take only the whole number of jumps that the grasshopper can make: 13 jumps. Therefore, the total distance the grasshopper travels before the cars hit is 13 jumps · 200 m per jump = 2600 m.
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A glow-worm of mass 5.0 g emits red light (650nm) with a power of 0.10W entirely in the backward direction. To what speed will it have accelerated after 10 y if released into free space and assumed to live
Answer:
The speed the glow-worm would have accelerated after 10 years if released into free space and assumed to live is 3.09 X 10³⁴ m/s
Explanation:
Energy associated with photon of lights, is given as
E = hc/λ
Where;
h is Planck's constant = 6.626 x 10⁻³⁴ m2 kg/s
C is the speed of light = ?
λ is the light's wavelength = 650nm = 650 X10⁻⁹ m
Also Energy = Power X time
Time given = 10 years
Time in seconds = 10 yrs X 365 days X 24 hrs X 60mins X 60 secs
Time in seconds = 315360000 seconds
P*t = hc/λ
c = (P*t*λ)/h
c = (0.1 X 315360000 X 650 X 10⁻⁹)/(6.626 x 10⁻³⁴)
c = 3.09 X 10³⁴ m/s
The speed the glow-worm would have accelerated after 10 y if released into free space and assumed to live is 3.09 X 10³⁴ m/s
A long string is pulled so that the tension in it increases by a factor of four. If the change in length is negligible, by what factor does the wave speed change?
To solve this problem we will apply the concepts related to wave velocity as a function of the tension and linear mass density. This is
[tex]v = \sqrt{\frac{T}{\mu}}[/tex]
Here
v = Wave speed
T = Tension
[tex]\mu[/tex] = Linear mass density
From this proportion we can realize that the speed of the wave is directly proportional to the square of the tension
[tex]v \propto \sqrt{T}[/tex]
Therefore, if there is an increase in tension of 4, the velocity will increase the square root of that proportion
[tex]v \propto \sqrt{4} = 2[/tex]
The factor that the wave speed change is 2.
In building a particle accelerator, you manage to produce a uniform electric field of magnitude 6.03 × 10 5 N/C in one 35.5 cm section. Calculate the magnitude of the electric potential difference across the length of the accelerator's section. How much work is required to move a proton through the section?
Answer:
V = 2.14×10⁵ V.
W = 3.424×10⁻¹⁴ J.
Explanation:
Electric Potential: This can be defined as the work done in bringing a unit positive charge from infinity to that point, against the action of a field.
The S.I unit is V.
The expression containing electric potential, distance and electric field is given as,
V = E×r .............. Equation 1
Where V = Electric potential difference across the length of the accelerator's section, E = Electric Field, r = Length of the section.
Given: E = 6.03×10⁵ N/C, r = 35.5 cm = 0.355 m.
Substitute into equation 1
V = 6.03×10⁵×0.355
V = 2.14065×10⁵ V.
V ≈ 2.14×10⁵ V.
amount of Work required to move a proton through the section is given as,
W = qV ............... Equation 2
Where W = work required to move a proton through the section, q = charge on a proton V = Electric potential.
Given: V = 2.14×10⁵ V, q = 1.60 x 10⁻¹⁹ C.
Substitute into equation 2
W = (2.14×10⁵)(1.60 x 10⁻¹⁹)
W = 3.424×10⁻¹⁴ J.
A vector has a magnitude of 46.0 m and points in a direction 20.0° below the positive x-axis. A second vector, , has a magnitude of 86.0 m and points in a direction 42.0° above the negative x-axis. a) Sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗ .
b) Using the component method of vector addition, find the magnitude of the vector C⃗ .
c) Using the component method of vector addition, find the direction of the vector C
Answer with Step-by -step explanation:
We are given that
b.[tex]\mid A\mid=46 m[/tex]
[tex]\theta=20^{\circ}[/tex] below the positive x-axis
Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis=[tex]x=360-20=340^{\circ}[/tex]
x-component of vector A=[tex]A_x=\mid A\mid cosx=46cos 340=46\times 0.94=43.24[/tex]
y-Component of vector A=[tex]A_y=\mid A\mid sinx=46sin340=46(-0.34)=-15.64[/tex]
Magnitude of vector B=86 m
The vector B makes angle with positive x- axis=[tex]x'=42^{\circ}[/tex]
x-component of vector B=[tex]B_x=86cos42=63.64[/tex]
y-Component of vector B=[tex]B_y=86sin42=57.62[/tex]
Vector A=[tex]A_xi+A_yj=43.24i-15.64j[/tex]
Vector B=[tex]B_xi+B_yj=63.64i+57.62j[/tex]
Vector C=A+B
Substitute the values
[tex]C=43.24i-15.64j+63.64i+57.62j[/tex]
[tex]C=106.88i+41.98j[/tex]
c.Direction=[tex]\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{41.98}{106.88})=21.5^{\circ}[/tex]
The direction of the vector C=21.5 degree
To sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗, start by drawing the x and y axes. Use the component method of vector addition to find the magnitude of C⃗. Use the inverse tangent function to find the direction of C⃗.
Explanation:To sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗, we start by drawing the x and y axes. Vector A⃗ has a magnitude of 46.0 m and points 20.0° below the positive x-axis, so we draw A⃗ starting from the origin and make an angle of 20.0° with the positive x-axis in a downward direction. Vector B⃗ has a magnitude of 86.0 m and points 42.0° above the negative x-axis, so we draw B⃗ starting from the origin and make an angle of 42.0° with the negative x-axis in an upward direction. To find the vector C⃗=A⃗+B⃗, we add the x-components and the y-components of A⃗ and B⃗ separately. Then we use the Pythagorean theorem to find the magnitude of C⃗ and the inverse tangent function to find the direction of C⃗ in relation to the positive x-axis.
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Learning Goal: How do 2 ordinary waves build up a "standing" wave? A very generic formula for a traveling wave is: y1(x,t)=Asin(kx−ωt). This general mathematical form can represent the displacement of a string, or the strength of an electric field, or the height of the surface of water, or a large number of other physical waves!
Part C Find ye(x) and yt(t). Remember that yt(t) must be a trig function of unit amplitude. Express your answers in terms of A, k, x, ω, and t. Separate the two functions with a comma. Use parentheses around the argument of any trig functions.
Part E At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)? Part G From
Part F we know that the string is perfectly straight at time t=π2ω. Which of the following statements does the string's being straight imply about the energy stored in the string?
a.There is no energy stored in the string: The string will remain straight for all subsequent times.
b.Energy will flow into the string, causing the standing wave to form at a later time.
c.Although the string is straight at time t=π2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string.
d.The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle.
Answer:
Explanation:
=Asin(kx−ωt). This general mathematical form can represent the displacement of a string, or the strength of an electric field, or the height of the surface of water, or a large number of other physical waves!
Part C Find ye(x) and yt(t). Remember that yt(t) must be a trig function of unit amplitude. Express your answers in terms of A, k, x, ω, and t. Separate the two functions with a comma. Use parentheses around the argument of any trig functions.
Part E At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)? Part G From
Part F we know that the string is perfectly straight at time t=π2ω. Which of the following statements does the string's being straight imply about the energy stored in the strJHJMNMMUJJHTGGHing?
a.There is no energy stored in the string: The string will remain straight for all subsequent times.
b.Energy will flow into the string, causing the standing wave to form at a later time.
c.Although the string is straight at time t=π2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string.
d.The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle.
Here, the spatial and temporal parts of the wave function are A sin(kx) and sin(ωt), respectively. The displacement of the string at x=0 is 0. Although the string is straight at time t=π/2ω, there is still energy stored in it due to parts of the string having nonzero velocity.
Explanation:Part C: The spatial part of the wave function, ye(x), can be expressed as A sin(kx), and the temporal part, yt(t), is represented by sin(ωt). These expressions are valid for a wave of unit amplitude.
Part E: At position x=0, the displacement of the string is 0. This occurs because, for the wave equation, when inputting x=0 into the spatial part, the value for y becomes 0.
Part G: The correct answer is option c. Although the string is straight at time t=π/2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string. Option d is incorrect as the total mechanical energy in a perfectly straight string does not oscillate, but remains constant.
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A cylinder with a piston holds 0.50 m3 of oxygen at an absolute pressure of 4.0 atm. The piston is pulled outward, increasing the volume of the gas until the pressure drops to 1.0 atm. If the temperature stays constant, what new volume does the gas occupy?
Answer:
2 cubic meters
Explanation:
Suppose ideal gas, when the temperature stays constant, then the product of pressure and volume is the same as before
[tex]P_1V_1 = P_2V_2[/tex]
where [tex]P_1 = 4 atm, V_1 = 0.5m^3[/tex] are the pressure and volume of the gas before. [tex]P_2 = 1 atm, V_2[/tex] are the pressure and volume of the gas after. We can solve for the volume after
[tex]V_2 = \frac{P_1V_1}{P_2} = \frac{4*0.5}{1} = 2 m^3[/tex]