a) A hockey player bounces back off the boards after running into them:
When a hockey player bounces back off the boards after running into them, the action force is the force the player exerts on the board, the boards exert an equal and opposite force on the player which causes the player to bounce back off the boards.
b) A runner can speed up: When a runner speeds up, the action force is the force that the runner exerts on the ground with their feet. The reaction force is the force that the ground exerts back on the runner.
Two students are standing on skateboards facing each other at rest.
Student A (mass of 50.0 kg) pushes on student B (mass of 60.0 kg) with a force of 150 N [W).
You may assume there is no friction.
a) FBD of each skateboarder:
(The diagram below shows the FBD of each skateboarder. )
b) Acceleration of each skateboarder:
For skateboarder A:
Acceleration =[tex]Force / mass = 150 N / 50 kg = 3.0 m/s^2 [W[/tex]]For skateboarder B:
Acceleration = [tex]Force / mass = 150 N / 60 kg = 2.5 m/s^2 [E][/tex]12.
A helicopter of mass 4500 kg is hovering at rest above the ground.
a) FBD of the hovering helicopter: (The diagram below shows the FBD of the hovering helicopter. )
b) Forces acting on the helicopter while it is hovering: The forces acting on the helicopter while it is hovering are the weight of the helicopter, which acts downwards, and the upward force from the rotors, which cancels out the weight of the helicopter to keep it hovering.
c) Cause of the upward force on the helicopter: The upward force on the helicopter is caused by the rotors pushing air downwards, according to Newton's third law, which states that every action has an equal and opposite B. the rotors cause an upward force on the helicopter by pushing air downwards.
d) Acceleration of the helicopter if the propeller blades apply a force of 49000 N [down] on the air:
The acceleration of the helicopter if the propeller blades apply a force of 49000 N [down] on the air can be found using Newton's second law:
Force = mass x acceleration Rearranging this equation to solve for acceleration:
Acceleration = [tex]Force / mass Acceleration = 49000 N / 4500 kg Acceleration = 10.9 m/s^2 [down][/tex]
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In an experiment, you find the value for the resistance of a wire to be 1.2 V/A. The manufacturer's value for this resistor is given as 1.5 V/A. What is the percent difference from the accepted value? 25% 30% 3% 20%
In an experiment, you find the value for the resistance of a wire to be 1.2 V/A. The manufacturer's value for this resistor is given as 1.5 V/A. 20% (option D) is the percent difference from the accepted value.
To calculate the percent difference from the accepted value, you can use the formula:
Percent Difference = [(Experimental Value - Accepted Value) / Accepted Value] * 100
Using the given values:
Experimental Value = 1.2 V/A
Accepted Value = 1.5 V/A
Percent Difference = [(1.2 - 1.5) / 1.5] * 100
The negative sign indicates that the experimental value is lower than the accepted value.
Percent Difference = (-0.3 / 1.5) * 100
Percent Difference = -0.2 * 100
Percent Difference = -20%
Therefore, the percent difference from the accepted value is 20%.
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A seat with an occupant during crash landing is modeled as a SDOF system undergoing vertical motion. The seat has a damper with adjustable damping. At a given damping ratio, the amplitude decays to 50% in one cycle. Determine the amplitude decay (in percentage) in one cycle if the damping ratio is now doubled.
The amplitude decay in one cycle (in percentage) when the damping ratio is doubled is 34.1%.
Given the damping ratio at which the amplitude decays to 50% in one cycle is 0.69.
Now, the damping ratio is doubled i.e, 2 x 0.69 = 1.38.
To find the percentage of the amplitude decay in one cycle when the damping ratio is doubled, we can use the formula of the damped vibration of a single-degree-of-freedom system given below:
[tex]$x(t) = x_0e^{-\zeta \omega_n t} cos(\omega_d t)[/tex]
Where,
x0 = amplitude of the vibration at time t = 0
ζ = damping ratio ω
n = natural frequency of the system
ωd = damped natural frequency of the system
At a given damping ratio, the amplitude decays to 50% in one cycle.
Since one cycle corresponds to the time taken for the argument of the cosine function to change by 2π radians, we have
[tex]$\omega_d T = 2\pi[/tex]
where T is the time period of one cycle.
Hence, we have
[tex]$x(T) = x_0e^{-\zeta \omega_n T} cos(2\pi)[/tex]
[tex]= -x_0e^{-\zeta \omega_n T}[/tex]
This means that the amplitude of vibration has decayed to 50% of its initial value.
Therefore,
[tex]$x(T) = x_0e^{-\zeta \omega_n T} cos(2\pi)[/tex]
[tex]= -x_0e^{-\zeta \omega_n T}[/tex]
Also, we know that
[tex]$\omega_d = \omega_n \sqrt{1-\zeta^2}[/tex]
Therefore,
[tex]$\frac{\omega_d}{\omega_n} = \sqrt{1-\zeta^2}[/tex]
Squaring both sides, we get
[tex]$1-\zeta^2 = \left(\frac{\omega_d}{\omega_n}\right)^2[/tex]
Substituting the value of ζωnT from equation (1), we have
[tex]$\zeta^2 = \left(ln(2)\frac{\omega_n}{\omega_d}\right)^2 - 1[/tex]
[tex]$\zeta^2 = \left(ln(2)\frac{\omega_n}{\omega_d}\right)^2 - 1[/tex]
Let
[tex]$\zeta^2 = \left(ln(2)\frac{\omega_n}{\omega_d}\right)^2 - 1[/tex]
Squaring both sides, we have
[tex]$1-\zeta^2 = p^2[/tex]
[tex]$\zeta^2 = 1-p^2[/tex]
[tex]$p = \sqrt{1-\zeta^2}[/tex]
[tex]$ln(2) = \frac{\zeta \omega_n T}{p}[/tex]
[tex]$p = \frac{\zeta \omega_n T}{ln(2)}[/tex]
Substituting the value of p in equation (2),
we have
[tex]$\zeta^2 = \left(\frac{\zeta \omega_n T}{ln(2)}\right)^2 - 1[/tex]
Solving for ζ, we get
[tex]$\zeta = 0.97[/tex]
Now, when the damping ratio is doubled, we have
[tex]$\zeta' = 2\zeta[/tex]
[tex]= 1.94[/tex]
Therefore, the percentage of the amplitude decay in one cycle when the damping ratio is doubled is given by
[tex]$\frac{x'(T)}{x'(0)} = e^{-\zeta' \omega_n T}[/tex]
[tex]$= e^{-1.94\left(\frac{ln(2)}{T}\right)}[/tex]
[tex]$$= 34.1\%$$[/tex]
Hence, the amplitude decay in one cycle (in percentage) when the damping ratio is doubled is 34.1%.
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1. A block of mass 3.14 kg lies on a frictionless horizontal surface. The block is connected by a cord passing over a pulley to another block of mass 6.75 kg which hangs in the air, as shown. Assume the cord to be light (massless and weightless) and unstretchable and the pulley to have no friction and no rotational inertia. Calculate the acceleration of the first block. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s 2 .
Calculate the tension in the cord. Answer in units of N.
2. The block of mass 3.80061 kg has an acceleration of 3.3 m/s 2 as shown. What is the magnitude of F? Assume the acceleration due to gravity is 9.8 m/s 2 and the surface is frictionless. Answer in units of N.
3. A light, inextensible cord passes over a light, frictionless pulley with a radius of 9.7 cm. It has a(n) 24 kg mass on the left and a(n) 5.3 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 4.7 m apart. The acceleration of gravity is 9.8 m/s 2 . At what rate are the two masses accelerating when they pass each other? Answer in units of m/s 2 .
1. The tension in the cord is approximately 3.11 N. 2. The magnitude of force (F) acting on the block is approximately 12.54013 N. 3. The rate at which the two masses are accelerating when they pass each other is approximately 475.26 m/s^2.
1. To calculate the acceleration of the first block and the tension in the cord, we can use Newton's second law of motion and the concept of tension in a system.
Let's denote:
m1 = mass of the first block = 3.14 kg
m2 = mass of the hanging block = 6.75 kg
a = acceleration of the system (common acceleration)
T = tension in the cord
Using Newton's second law for both blocks:
m1 * a = T (equation 1)
m2 * g - T = m2 * a (equation 2)
Solving the equations simultaneously, we can find the values of acceleration (a) and tension (T):
From equation 1, we have T = m1 * a
Substituting this value into equation 2:
m2 * g - m1 * a = m2 * a
g = (m1 + m2) * a
a = g / (m1 + m2)
Substituting the given values:
a = 9.8 m/s^2 / (3.14 kg + 6.75 kg)
a ≈ 9.8 m/s^2 / 9.89 kg
a ≈ 0.99 m/s^2
Therefore, the acceleration of the first block is approximately 0.99 m/s^2.
To calculate the tension in the cord, we can substitute the value of acceleration (a) into equation 1:
T = m1 * a
T = 3.14 kg * 0.99 m/s^2
T ≈ 3.11 N
Therefore, the tension in the cord is approximately 3.11 N.
2. To determine the magnitude of force (F) acting on the block with a known mass and acceleration, we can again use Newton's second law of motion.
m = mass of the block = 3.80061 kg
a = acceleration of the block = 3.3 m/s^2
g = acceleration due to gravity = 9.8 m/s^2
Using Newton's second law:
F = m * a
Substituting the given values:
F = 3.80061 kg * 3.3 m/s^2
F ≈ 12.54013 N
Therefore, the magnitude of force (F) acting on the block is approximately 12.54013 N.
3. To determine the rate at which the two masses are accelerating when they pass each other, we can use the concept of relative motion and the conservation of mechanical energy.
m1 = mass on the left = 24 kg
m2 = mass on the right = 5.3 kg
r = radius of the pulley = 9.7 cm = 0.097 m
d = initial vertical distance between the center of masses = 4.7 m
g = acceleration due to gravity = 9.8 m/s^2
Using the conservation of mechanical energy:
(m1 + m2) * g * d = (m1 + m2) * a * r
Simplifying the equation:
a = (g * d) / r
Substituting the given values:
a = (9.8 m/s^2 * 4.7 m) / 0.097 m
a ≈ 475.26 m/s^2
Therefore, the rate at which the two masses are accelerating when they pass each other is approximately 475.26 m/s^2.
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you compare two strings s1 and s2 using ________. compare(s1, s2) s1.compare(s2) compareto(s1, s2) s1.compareto(s2)
The correct way to compare two strings in C++ is by using the s1.compare(s2) function. The function returns an integer value that indicates whether the two strings are equal or if one is greater or less than the other.
Let's understand how the s1.compare(s2) function works. The function takes two arguments - the first is the string s1 that we want to compare, and the second is the string s2 that we want to compare s1 with. The function returns an integer value that indicates the result of the comparison.
If s1 is greater than s2, the function returns a positive integer value. If s1 is less than s2, the function returns a negative integer value. And if s1 is equal to s2, the function returns 0.The function compares the two strings lexicographically, that is, it compares the corresponding characters of the two strings starting from the first character.
If the characters at the same position in both strings are equal, it moves on to compare the next character, and so on until it finds a mismatched character. Then it returns the difference between the ASCII values of the two mismatched characters.Here is an example of using the s1.compare(s2) function:```#include
#include
using namespace std;
int main()
{
string s1 = "hello";
string s2 = "world";
int result = s1.compare(s2);
if (result == 0)
cout << "s1 and s2 are equal" << endl;
else if (result > 0)
cout << "s1 is greater than s2" << endl;
else
cout << "s1 is less than s2" << endl;
return 0;
}```In the above example, we compare two strings s1 and s2 using the s1.compare(s2) function. Since s1 is less than s2 lexicographically, the function returns a negative integer value, and we print "s1 is less than s2".I hope this helps!
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write a two-paragraph summary about an object experiencing acceleration. Describe how the object’s motion behaves due to this acceleration. Include source /Citation
Acceleration is described as the rate at which velocity changes over a given amount of time. The acceleration of an object, whether it is increasing or decreasing, results in changes in its motion.
The motion of an object changes in terms of speed, direction, or both as a result of acceleration. When an object is experiencing acceleration, its motion is changing. The rate of change of motion is the same as the acceleration. An object may accelerate in the direction of motion, in the opposite direction of motion, or even in a direction different from its original motion. In the simplest case, where an object has a positive acceleration, its velocity is increasing. An example would be a car accelerating from a stationary position, where the velocity goes from 0 to some positive value over a certain period. During this acceleration, the car's speed, as well as the distance it travels in a given period of time, both increase.
When the object has negative acceleration or deceleration, its velocity is decreasing. If the velocity is decreasing in the same direction as its motion, then the object is still moving, but its speed is decreasing. The object comes to a stop if the acceleration is large enough and its speed reaches 0. The direction of the acceleration is the direction in which the velocity is changing. If the direction of acceleration is opposite to the direction of motion, then the object comes to a stop faster. A body experiencing acceleration behaves differently according to the direction of the acceleration. The car that accelerates moves at a faster speed while the speed of a car that decelerates will decrease.
Source: State University of New York College at Buffalo, Acceleration.
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3. Will one be able to distinguish two points that are 1.5×10
−2
μm apart using a microscope fit with a filter for 575 nm and observing at 1000× Total Magnification?
575(1.25+1.25)
2.3×10
2
nm×(10
3
1 mm)
5
=2.3×10
5
1.5×10
−2
2.2×10
5
=2.3×10
2
nm
No you would nat ba. able to digtirguigh to points from ore anothat 4. A microscope is now fit with a filter for 660 nm, will the researcher be able to differentiate two cells at 8.1×10
−6
m apart while observing at 400× Total Magnification? 5. A student is using a plain microscope (no filter, assume average wavelength of 4.35×10
−3
μm ), would the student be able to see two objects that are 2.8×10
3
nm apart using 100X Total Magnification?
(3)The two points that are 1.5 × 10^−2 μm apart are smaller than the resolution of the microscope, so they would not be able to be distinguished.(4)The two cells that are 8.1 × 10^−6 m apart are smaller than the resolution of the microscope, so they would not be able to be distinguished.(5)The two objects that are 2.8 × 10^3 nm apart are larger than the resolution of the microscope, so they would be able to be distinguished.
(3)The resolution of a microscope is determined by the wavelength of light used and the magnification. The formula is
Resolution = Wavelength / Magnification
So, for the first question, the resolution of the microscope would be:
Resolution = 575 nm / 1000x = 0.0575 μm
The two points that are 1.5 × 10^−2 μm apart are smaller than the resolution of the microscope, so they would not be able to be distinguished.
(4) the resolution of the microscope would be:
Resolution = 660 nm / 400x = 0.165 μm
The two cells that are 8.1 × 10^−6 m apart are smaller than the resolution of the microscope, so they would not be able to be distinguished.
(5) the resolution of the microscope would be:
Resolution = 4.35 × 10^−3 μm / 100x = 0.0435 μm
The two objects that are 2.8 × 10^3 nm apart are larger than the resolution of the microscope, so they would be able to be distinguished.
Therefore, the answers to your questions are:
No, the two points cannot be distinguished. No, the two cells cannot be distinguished. Yes, the two objects can be distinguished.The question should be:
3. Will one be able to distinguish two points that are 1.5×10^−2μm apart using a microscope fit with a filter for 575 nm and observing at 1000× Total Magnification?
4. A microscope is now fit with a filter for 660 nm, will the researcher be able to differentiate two cells at 8.1×10^−6 m apart while observing at 400× Total Magnification?
5. A student is using a plain microscope (no filter, assume average wavelength of 4.35×10^−3μm ), would the student be able to see two objects that are 2.8×10^3 nm apart using 100X Total Magnification?
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A crate with mass 31.0 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 160 N. What acceleration is produced? Part B How far does the crate travel in 12.5 s ? What is its speed at the end of 12.5 s ?
Part (a) The acceleration produced by the net horizontal force of 160 N acting on the crate of mass 31.0 kg is 5.16 m/s².
Part (b) The distance traveled by the crate in 12.5 s is 1020.31 m and its speed at the end of 12.5 s is 64.5 m/s.
Part (a) Newton's second law of motion states that, the force acting on an object is directly proportional to its acceleration, given as, F = ma, Where, F is the force acting on the object, m is the mass of the object, a is the acceleration produced by the force. Now, substituting the given values of force and mass in the above equation, we get160 = 31.0 aa = 160/31.0a = 5.16 m/s².
Therefore, the acceleration produced by the net horizontal force of 160 N acting on the crate of mass 31.0 kg is 5.16 m/s².
Part (b) Now, to find the distance traveled by the crate and its speed at the end of 12.5 s, we need to use the following kinematic equations: v = u + ats = ut + 1/2 at²v² = u² + 2as, Where, v is the final velocity of the crate, u is the initial velocity of the crates is the distance traveled by the crate, t is the time taken by the crate, a is the acceleration produced by the force applied. Substituting the given values of u and t in the equation 1, we get
v = u + at
v = 0 + (5.16)(12.5)v = 64.5 m/s.
Now, substituting the given values of u, t, and a in equation 2, we gets = ut + 1/2 at²s = 0(12.5) + 1/2 (5.16)(12.5)²s = 1020.31 m.
Therefore, the distance traveled by the crate in 12.5 s is 1020.31 m and its speed at the end of 12.5 s is 64.5 m/s.
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Case 1: A proton E=1.5∗10 3N/C Released from rest at x=−2 cm Find the change in electric potential energy when proton reaches x=5 cm 1.6 ∗10 ∧ −19C= proton charge Change in potential energy =−qE displacement Ans =−1.68×10 −17Joules Case 2: An electron E= same as case one Electron is now fired in the same direction and position Find change in potential energy when electron reaches 12 cm3.36×10 −17 (Voltage) to accelerate a charged particle.
The change in potential energy when the electron reaches 12 cm is approximately -6.4512 × 10^-36 Joules.
In Case 1, the change in electric potential energy can be calculated using the formula:
ΔPE = -qEΔx
where ΔPE is the change in potential energy, q is the charge of the proton (1.6 × 10^-19 C), E is the electric field (1.5 × 10^3 N/C), and Δx is the displacement.
Δx = 5 cm - (-2 cm) = 7 cm = 0.07 m
Plugging in the values:
ΔPE = -(1.6 × 10^-19 C)(1.5 × 10^3 N/C)(0.07 m)
= -1.68 × 10^-17 J
Therefore, the change in electric potential energy when the proton reaches x = 5 cm is -1.68 × 10^-17 Joules.
In Case 2, the change in potential energy for the electron can be calculated in a similar way using the given electric field (3.36 × 10^-17 V/m) and displacement (12 cm = 0.12 m):
ΔPE = -(1.6 × 10^-19 C)(3.36 × 10^-17 V/m)(0.12 m)
= -6.4512 × 10^-36 J
Therefore, The change in potential energy when the electron reaches 12 cm is approximately -6.4512 × 10^-36 Joules.
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Question 3 (10 points): Compare and contrast the composition of clouds in the atmospheres of the four gas giant planets. Why are they different from each other?
The composition of clouds in the atmospheres of the four gas giant planets - Jupiter, Saturn, Uranus, and Neptune - differs significantly. These variations arise due to differences in temperature, pressure, and chemical makeup in their atmospheres.
The gas giant planets in our solar system have atmospheres predominantly composed of hydrogen and helium, with traces of other compounds. However, their cloud compositions differ due to variations in temperature, pressure, and chemical makeup.
Jupiter, the largest planet, has clouds primarily consisting of ammonia, ammonium hydrosulfide, and water vapour. These clouds form colourful bands, including the famous Great Red Spot. Saturn, known for its beautiful rings, has an atmosphere with ammonia ice clouds, as well as methane and water vapour clouds. The presence of these compounds gives Saturn its distinctive yellowish hue.
Uranus and Neptune, often referred to as ice giants, have colder atmospheres. Uranus has methane clouds that contribute to its pale blue colour, while Neptune has methane and ethane clouds, giving it a vibrant blue appearance. These clouds are located in the upper layers of the atmosphere, where temperatures and pressures are suitable for the formation of these compounds.
The differences in cloud composition among these gas giants are primarily due to variations in temperature, pressure, and the availability of specific chemical compounds in their atmospheres. Understanding these variations provides insights into the complex dynamics and atmospheric conditions of these fascinating planets.
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A ball, originally held about 2.6 feet above the ground, is thrown vertically upward with an initial velocity of 55 feet per second. The height h above the ground after t seconds is approximated by the following function:
h(t) = 2.6+55t - 16t2
At what time t will the ball strike the ground?
Considering the definition of zeros of a quadratic function, the time the ball will strike the ground is 3.484 seconds.
Definition of zeros of a functionThe points where a polynomial function crosses the axis of the independent term (x) represent the zeros of the function.
The roots or zeros of the quadratic function are those values of x for which the expression is equal to 0. Graphically, the roots correspond to the abscissa of the points where the parabola intersects the x-axis.
In a quadratic function that has the form:
f(x)= ax² + bx + c
the zeros or roots are calculated by:
[tex]x1,x2=\frac{-b+-\sqrt{b^{2} -4*a*c} }{2*a}[/tex]
Time the ball will strike the groundTo know the time the ball will strike the ground, yo need to calculated the zeros of the function h(t) = 2.6 + 55t - 16t²
Being:
a= -16b=55c= 2.6the zeros or roots are calculated as:
[tex]x1=\frac{-55+\sqrt{55^{2} -4*(-16)*2.6} }{2*(-16)}[/tex]
[tex]x1=\frac{-55+\sqrt{3191.4} }{2*(-16)}[/tex]
[tex]x1=\frac{-55+56.4925}{-32}[/tex]
x1= -0.047
and
[tex]x2=\frac{-55-\sqrt{55^{2} -4*(-16)*2.6} }{2*(-16)}[/tex]
[tex]x2=\frac{-55-\sqrt{3191.4} }{2*(-16)}[/tex]
[tex]x2=\frac{-55-56.4925}{-32}[/tex]
x2= 3.484
Finally, since time is not negative, the time the ball will strike the ground is 3.484 seconds.
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UNIT CONVERSION (round each answer to two significant digits, use conversion at the beginning of the lab. Show work for full credit) 5. 175lb kg Example work: 175/x×0 a or (175
∗
×00)− 6. 88.6 kg tb 7. 78
∘
F
∘
C 8. 39.2
∘
C 9. 187.6 m
−min
−1
mile.hr
−1
10. 6.5 mile-hr −1 m⋅m
−1
−1
175 pounds is approximately equal to 79.38 kilograms.
88.6 kilograms is approximately equal to 0.0886 tons.
78 degrees Fahrenheit is approximately equal to 13.89 degrees Celsius.
39.2 degrees Celsius is approximately equal to 102.16 degrees Fahrenheit.
187.6 meters per minute is approximately equal to 6.60 miles per hour.
6.5 miles per hour is approximately equal to 2.91 meters per second.
5. To convert 175 pounds to kilograms, we'll use the conversion factor of 1 pound = 0.4536 kilograms.
175 pounds * 0.4536 kilograms/pound = 79.38 kilograms
Therefore, 175 pounds is approximately equal to 79.38 kilograms.
6. To convert 88.6 kilograms to tons, we'll use the conversion factor of 1 ton = 1000 kilograms.
88.6 kilograms * (1 ton/1000 kilograms) = 0.0886 tons
Therefore, 88.6 kilograms is approximately equal to 0.0886 tons.
7. To convert 78 degrees Fahrenheit to degrees Celsius, we'll use the formula: Celsius = (Fahrenheit - 32) * 5/9.
Celsius = (78 - 32) * 5/9 = 25 * 5/9 = 13.89 degrees Celsius
Therefore, 78 degrees Fahrenheit is approximately equal to 13.89 degrees Celsius.
8. To convert 39.2 degrees Celsius to degrees Fahrenheit, we'll use the formula: Fahrenheit = Celsius * 9/5 + 32.
Fahrenheit = 39.2 * 9/5 + 32 = 70.16 + 32 = 102.16 degrees Fahrenheit
Therefore, 39.2 degrees Celsius is approximately equal to 102.16 degrees Fahrenheit.
9. To convert 187.6 meters per minute to miles per hour, we'll use the conversion factors: 1 mile = 1609.34 meters and 1 hour = 60 minutes.
187.6 meters/minute * (1 mile/1609.34 meters) * (60 minutes/1 hour) = 6.60 miles per hour
Therefore, 187.6 meters per minute is approximately equal to 6.60 miles per hour.
10. To convert 6.5 miles per hour to meters per second, we'll use the conversion factors: 1 mile = 1609.34 meters and 1 hour = 3600 seconds.
6.5 miles/hour * (1609.34 meters/1 mile) * (1 hour/3600 seconds) = 2.91 meters per second
Therefore, 6.5 miles per hour is approximately equal to 2.91 meters per second.
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The position of a particle is given by the expression \( x=2.00 \cos (6.00 \pi t+\pi / 2) \), where \( x \) is in meters and \( t \) is in seconds. (a) Determine the frequency. \[ \mathrm{Hz} \] (b) D
(a). The frequency of the motion is 1 Hz.
(b). The period of the motion is 1 second.
(c). The amplitude of the motion is 6.00 meters.
(d). The phase constant is π radian.
(e). The position of the particle at t = 0.350 s is 6.00 meters.
(a) The frequency of an oscillating motion is the number of complete cycles it completes per unit of time. In this case, the expression for the position of the particle is given by
x = 6.00 cos(2.00πt + π).
The general form of a cosine function is given by cos(ωt + φ), where ω is the angular frequency and φ is the phase constant. Comparing this with the given expression, we can see that the angular frequency is 2.00π.
The frequency (f) is related to the angular frequency (ω) by the equation
f = ω/2π.
Therefore, the frequency is given by:
f = (2.00π)/(2π)
f = 1 Hz
So, the frequency is 1 Hz.
(b) The period of an oscillating motion is the time it takes to complete one full cycle.
The period (T) is the inverse of the frequency,
T = 1/f.
In this case, the frequency is 1 Hz, so the period is:
T = 1/1 = 1 second
Therefore, the period is 1 second.
(c) The amplitude of an oscillating motion is the maximum displacement from the equilibrium position. In this case, the amplitude can be determined from the given expression for the position of the particle,
x = 6.00 cos(2.00πt + π).
The coefficient of the cosine function represents the amplitude, so the amplitude is:
Amplitude = 6.00 meters
Therefore, the amplitude is 6.00 meters.
(d) The phase constant (φ) represents the initial phase of the motion. In this case, the phase constant can be determined from the given expression for the position of the particle,
x = 6.00 cos(2.00πt + π).
Comparing this with the general form of a cosine function, we can see that the phase constant is π radian.
Therefore, the phase constant is π radian.
(e) To determine the position of the particle at t = 0.350 s, we substitute
t = 0.350 s into the given expression for the position of the particle,
x = 6.00 cos(2.00πt + π):
x = 6.00 cos(2.00π(0.350) + π)
x = 6.00 cos(0.700π + π)
x = 6.00 cos(1.700π)
Using the trigonometric identity cos(θ + π) = -cos(θ), we can simplify the expression:
x = 6.00 (-cos(1.700π))
Since cos(π) = -1, we have:
x = 6.00 (-(-1))
x = 6.00 (1)
x = 6.00 meters
Therefore, the position is 6.00 meters.
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Complete question is,
The position of a particle is given by the expression x = 6.00 cos (2.00πt + π), where x is in meters and t is in seconds.
(a) Determine the frequency.Hz
(b) Determine period of the motion.s
(c) Determine the amplitude of the motion.m
(d) Determine the phase constant.rad
(e) Determine the position of the particle at t = 0.350 s.m
A copper wire has a diameter of 1.422 mm. What magnitude current flows when the drift velocity is 1.54 mm/s ? Take the density of copper to be 8.92×10
3
kg/m
3
.
The magnitude current flows, when the drift velocity is 1.54 mm/s, is 20.3 A.
When the drift velocity is 1.54 mm/s.
We will make use of the formula that relates drift velocity with the current.
We have:
vd = (I / n * A * q )
Where vd is the drift velocity
I is the current
n is the density of free electrons
A is the cross-sectional area of the wire
q is the charge carried by an electron
For copper, the value of n is 8.5 × 1028 electrons/meter cube.
Given that the wire is circular, its cross-sectional area is A = πr2 = πd2/4
where d is the diameter of the wire.
Hence we have:
d = 1.422 mm = 1.422 × 10-3 mA = π/4 x (1.422 × 10-3 m)2= 1.59 x 10-6 m^2
Now we can calculate the current as follows:
I = n * A * q * vd/I = (8.5 x 10^28)(π/4)(1.422 x 10^-3)^2(1.602 x 10^-19)(1.54 x 10^-3)=20.3A
Approximately, the magnitude of the current flowing in the copper wire is 20.3A.
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x axis, and the positive charge is farther from the line of charge than the negative charge. Find the net force exerted on the dipole.
F
=−0.0588×N
The net force exerted on the dipole can be determined by the formula:
F = 2k(q1q2/d^2)
where:
F is the force
k is the electrostatic constant (k = 9 × 10^9 Nm^2/C^2)
q1 and q2 are the magnitudes of the charges
d is the distance between the charges
In this case, since the positive charge is farther from the line of charge, we consider the positive charge as q1 and the negative charge as q2.
Let's assume that the magnitudes of the charges are q1 and q2, and the distance between them is d.
The net force exerted on the dipole is given as F = -0.0588 N.
We can set up the equation:
-0.0588 = 2k(q1q2/d^2)
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What is the electric field (in N/C) at a point midway between them? N/C (b) What is the force (in N ) on a charge q
3
=23μC situated there?
(a) The electric field at a point midway between the point charges is 3.33 N/C, directed towards the positive charge.
(b) The force on the charge q3=22 μC at the same point is 73.26 N, directed towards the negative charge.
(a) To find the electric field at a point midway between the charges, we can calculate the electric field due to each charge individually and then add them together. The electric field at a point due to a point charge is given by the equation E = kq/[tex]r^2[/tex], where E is the electric field, k is the electrostatic constant (approximately 9 × [tex]10^9[/tex] [tex]Nm^2/C^2[/tex]), q is the charge, and r is the distance from the point charge.
The electric field due to q1 is E1 = ([tex]9 * 10^9 Nm^2/C^2[/tex]) * ([tex]52 * 10^{-6} C[/tex]) / [tex](1.5 m)^2[/tex], which simplifies to approximately 6.52 N/C directed towards q1.
The electric field due to q2 is E2 = ([tex]9 * 10^9 Nm^2/C^2[/tex]) * ([tex]-29 * 10^{-6} C[/tex]) / [tex](1.5 m)^2[/tex], which simplifies to approximately -3.19 N/C directed towards q2.
Adding the electric fields together, we get [tex]E_{total[/tex] = E1 + E2 = 6.52 N/C - 3.19 N/C = 3.33 N/C. The electric field is directed towards the positive charge q1.
(b) To calculate the force on q3, we can use the equation F = qE, where F is the force, q is the charge, and E is the electric field.
The force on q3 is given by F3 = (22 × 10^-6 C) * (3.33 N/C), which simplifies to approximately 73.26 N. The force is directed towards the negative charge q2.
Therefore, at the midpoint between the charges, the electric field is 3.33 N/C directed towards q1, and the force on q3=22 μC is 73.26 N directed towards q2.
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Two 2.0 kg masses are 1.5 m apart (center to center) on a frictionless table. Each has +9.1μC of charge. For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Levitation. What is the magnitude of the electric force on one of the masses? Express your answer with the appropriate units. What is the initial acceleration of the mass if it is released and allowed to move? Express your answer with the appropriate units.
The magnitude of the electric force between the two 2.0 kg masses, each with +9.1μC of charge and 1.5 m apart, is 3.369 N. When one of the masses is released and allowed to move, it experiences an initial acceleration of 1.685 m/s².
To find the magnitude of the electric force between the two masses, we can use Coulomb's law, which states that the magnitude of the electric force between two point charges is given by the equation:
F = k * (|q1| * |q2|) / r²
Where F is the force, k is the electrostatic constant (8.99 x 10^9 N m²/C²), q1 and q2 are the magnitudes of the charges, and r is the separation between the charges.
Plugging in the values, we have:
F = (8.99 x 10^9 N m²/C²) * (9.1 x 10^-6 C)² / (1.5 m)²
Calculating this, we find that the magnitude of the electric force is approximately 3.369 N.
When one of the masses is released and allowed to move, it experiences an initial acceleration due to the electric force acting on it. According to Newton's second law of motion, the force acting on an object is equal to its mass multiplied by its acceleration:
F = m * a
Rearranging the equation, we have:
a = F / m
Substituting the values, we get:
a = (3.369 N) / (2.0 kg)
Calculating this, we find that the initial acceleration of the mass is approximately 1.685 m/s².
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13. A 25Ω resistor and a 50Ω resistor are connected in series to a 7.5 V battery. Assuming that the internal resistance of the battery can be neglected, what is the intensity of the current flowing through each resistor? A. not enough data to find out B. 0.1 A C. 1 A D. 10 A
The current flowing through each resistor is equal since they are connected in series. Therefore, the correct option is B. 0.1 A.
Given that the 25Ω resistor and a 50Ω resistor are connected in series to a 7.5 V battery.
We need to find the intensity of the current flowing through each resistor.
Assuming that the internal resistance of the battery can be neglected.
The total resistance is given by, `R = R1 + R2`
Where R1 = 25Ω and R2 = 50Ω.R = 25Ω + 50Ω = 75Ω
According to Ohm's law, we know that `V = IR`
Where V = 7.5 V and R = 75Ω.
Substituting the values, we get `I = V / R`I = 7.5 / 75 = 0.1 A
The current flowing through each resistor is equal since they are connected in series.
Therefore, the correct option is B. 0.1 A.
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the tension in a string is 400N and the mass per unit length is 0.01kg/m. determine the speed of the wave on the string.
The speed of the wave on the string is approximately 200 m/s.
To determine the speed of the wave on the string, we can use the equation:
v = √(T/μ)
where:
- v is the speed of the wave
- T is the tension in the string
- μ is the mass per unit length of the string
Given that the tension in the string is 400N and the mass per unit length is 0.01kg/m, we can substitute these values into the equation to find the speed of the wave.
v = √(400N / 0.01kg/m)
Simplifying this equation, we have:
v = √(40000 N·m/kg)
Taking the square root of 40000 N·m/kg, we find that:
v ≈ 200 m/s
Therefore, the speed of the wave on the string is approximately 200 m/s.
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An archer shoots an arrow at a 79.0 m distant target; the bull's-eye of the target is at same height as the release height of the arrow. (a) At what angle in degrees must the arrow be released to hit the bull's-eye if its initial speed is 36.0 m/s ? 0 over under
Therefore, the arrow should be released at an angle of approximately 9.29 degrees to hit the bull's-eye if its initial speed is 36.0 m/s.
In projectile motion, we can analyze the horizontal and vertical components of the motion separately. The horizontal motion is uniform with constant velocity, while the vertical motion is affected by gravity.The time of flight (t) can be determined using the horizontal distance and the horizontal component of the velocity Substituting the given values into the equation, we can calculate the launch angle (θ) at which the arrow must be released to hit the bull's-eye.Since the equation involves the arcsine function, there might be multiple solutions. In this case, we assume the positive angle that corresponds to the arrow being released above the horizontal. If the result is zero, it means the arrow should be released horizontally.
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Vector
A
has x and y components of −8.70 cm and 15.0 cm, respectively. Vector
B
has x and y components of 13.2 cm and −6.60 cm, respectively. If
A
−
B
+3
C
=0, what are the components of
C
?
The components of vector C are Cx = 7.3 cm and Cy = -7.2 cm.
To find the components of vector C, we can rearrange the given equation:
A - B + 3C = 0
Let's substitute the components of vectors A and B:
(Ax, Ay) - (Bx, By) + 3(Cx, Cy) = (0, 0)
Given:
Ax = -8.70 cm
Ay = 15.0 cm
Bx = 13.2 cm
By = -6.60 cm
Substituting these values into the equation, we have:
(-8.70 cm, 15.0 cm) - (13.2 cm, -6.60 cm) + 3(Cx, Cy) = (0, 0)
To simplify the equation, we can subtract vector B from vector A:
(-8.70 cm - 13.2 cm, 15.0 cm - (-6.60 cm)) + 3(Cx, Cy) = (0, 0)
Simplifying further, we have:
(-21.9 cm, 21.6 cm) + 3(Cx, Cy) = (0, 0)
Since the sum of two vectors is equal to zero, their components must be equal:
-21.9 cm + 3Cx = 0 (equation 1)
21.6 cm + 3Cy = 0 (equation 2)
Now we can solve these two equations simultaneously to find the components of vector C.
From equation 1:
3Cx = 21.9 cm
Cx = 7.3 cm
From equation 2:
3Cy = -21.6 cm
Cy = -7.2 cm
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The actual question is:
Vector A has x and y components of −8.70 cm and 15.0 cm, respectively. Vector B has x and y components of 13.2 cm and −6.60 cm, respectively.
If A−B+3C =0,
What are the components of C?
In proton-beam therapy, a high-energy beam of protons is fired at a tumor. The protons come to rest in the tumor, depositing their Kinetic energy and breaking apart the tumor's DNA, thus killing its cells. For one patient, it is desired that (1.3×10∧−1)J of proton energy be deposited in a tumor. To create the proton beam, the protons are accelerated from rest through a (1.300×10∧7)∨ potential difference. What is the total charge of the protons that must be fired at the tumor to deposit the required energy? - No text entered - The correct answer is not displayed for Written Response type questions.
The total charge of protons that must be fired at the tumor to deposit the required energy is 6.444 x [tex]10^14[/tex] Coulombs.The energy of proton to be deposited in tumor = E = 1.3 x [tex]10^(-1)[/tex] J.
The potential difference through which protons are accelerated = V = 1.300 x [tex]10^7[/tex] volts. Charge on a single proton is given as e = 1.6 x [tex]10^(-19)[/tex] Coulombs.
We know that the potential difference is equal to the kinetic energy per unit charge.
Hence, the kinetic energy of the proton can be written as:K.E = qV Where,q = charge on proton V = potential difference.
Now, we need to calculate the number of protons required to deposit the energy of 1.3 x [tex]10^(-1)[/tex] J in the tumor.
The kinetic energy of a single proton is given by:
E = K.E. = (1/2)[tex]mv^2[/tex] Where,m = mass of proton = 1.67 x [tex]10^(-27)[/tex] kg v = velocity of proton.
Therefore,v = sqrt((2E)/m).
Hence,Charge on proton can be written asq =
K.E. / V = [(1/2)][tex]mv^2[/tex] / V = (m[tex]V^2[/tex]) / 2E = [1.67 x [tex]10^(-27)[/tex] kg x (1.300 x [tex]10^7[/tex]volts)^2] / (2 x 1.3 x [tex]10^(-1)[/tex] J) = 2.021 x [tex]10^(-16)[/tex] Coulombs.
Now, to calculate the total charge of protons required to deposit the energy of 1.3 x [tex]10^(-1)[/tex] J, we use the formula:
Charge = (Energy to be deposited) / (Charge on a single proton) = (1.3 x [tex]10^(-1)[/tex] J) / (2.021 x [tex]10^(-16)[/tex] Coulombs) = 6.444 x [tex]10^14[/tex].
Therefore, the total charge of protons that must be fired at the tumor to deposit the required energy is 6.444 x [tex]10^14[/tex] Coulombs.
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Two thin uniformly charged rods, each with length L and total charge +Q, are parallel and separated by a distance a. The first rod has one end at the origin and its other end on the positive y-axis. T - Part C Suppose that \( a \) \&s. Determine the net force \( \vec{F}_{3} \) on the second rod. Keep only the lading term. Express your answer in terms of some or all of the variables \( Q, L, a \), t
The net force on the second rod will depend on the specific values of Q, L, and a, and the distance between the small sections.
The force between two charges is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
1. Divide the second rod into small sections, each with length Δx.
2. Calculate the charge of each small section. Since the total charge of the rod is +Q, the charge per unit length (λ) is Q/L.
3. Determine the force exerted on each small section of the second rod by the corresponding section of the first rod.
This force is given by [tex]F = k * (λ1 * λ2) * Δx / r^2[/tex], where k is the Coulomb constant, λ1 is the charge density of the first rod, λ2 is the charge density of the second rod, Δx is the length of each small section, and r is the distance between the two small sections.
4. Sum up all the forces on each small section to find the net force on the second rod. Since the forces are in opposite directions, the net force will be the vector sum of all the individual forces.
Therefore, to express the answer in terms of these variables, you would need to perform the calculations using the given values and substitute them into the equation for the net force.
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The crankshaft in a race car goes from rest to Part A 3000rpm in 3.0 s. What is the angular acceleration of the crankshaft? Express your answer in radians per second squared. Part B How many revolutions does it make while reaching 3000 rpm? Express your answer in revolutions.
Part A: The angular acceleration of the crankshaft is (100π / 3) radians/second².
Part B: The crankshaft makes 50 revolutions while reaching 3000 rpm.
Part A:To find the angular acceleration of the crankshaft, we can use the formula:
angular acceleration (α) = (final angular velocity - initial angular velocity) / time
First, let's convert the final angular velocity of 3000 rpm to radians per second:
final angular velocity = 3000 rpm * (2π radians / 1 minute) * (1 minute / 60 seconds)
= 3000 * 2π / 60 radians/second
= 100π radians/second
The initial angular velocity is 0 since the crankshaft starts from rest.
Plugging in the values into the formula:
angular acceleration = (100π radians/second - 0 radians/second) / 3.0 seconds
= (100π / 3) radians/second²
So the angular acceleration of the crankshaft is (100π / 3) radians/second².
Part B:To find the number of revolutions the crankshaft makes while reaching 3000 rpm, we can use the formula:
number of revolutions = (final angular velocity - initial angular velocity) / (2π)
Plugging in the values:
number of revolutions = (100π radians/second - 0 radians/second) / (2π)
= 50 revolutions
Therefore, the crankshaft makes 50 revolutions while reaching 3000 rpm.
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A 10.0 kg metal object rests against a lovel wooden surface. What is the minimum amount of force (in N) required to begin sliding this object? (Hint use Table 5.1 for your coefficient of frction) QUESTION 2 A 10.0 kg metal object slides against a level wooden surface. How much force (in N ) is required for this object to maintain a constant speed? (Mint: use Table 5.1 for your coefficient of friction) QUESTION 3 Using values from Table 5. 2, approximate the drag force (in N ) expenionced by a Honda Civic falling front-first from the sky at a rate of 100.0 m/s The projected area of the vehicle's front is 21.80f
2
( 1 looked it up), which is approximately 2.025 m
2
. Afso assume a vehicle woight of 3,000.0 ibs, which results from approximately 1361 kg of mass. For density of ar, uso 1.210 kg/m
3
.
1: The minimum amount of force required to begin sliding the 10.0 kg metal object on the level wooden surface can be determined using the coefficient of friction from Table 5.1. 2: To maintain a constant speed while sliding, the force required is equal to the force of kinetic friction. This force can be calculated using the coefficient of kinetic friction. 3: The drag force experienced by a falling Honda Civic can be approximated using the drag coefficient, projected area, density of air, and velocity of the object.
1: The minimum amount of force required to begin sliding the 10.0 kg metal object on the level wooden surface can be determined using the coefficient of friction from Table 5.1. Let's assume the coefficient of friction between the two surfaces is μ.
The minimum force required to overcome static friction can be calculated as:
F = μN
where N is the normal force acting on the object. Since the object is at rest on a level surface, the normal force is equal to the weight of the object, which is given by:
N = mg
Plugging in the values, we have:
F = μmg
2: To maintain a constant speed while sliding, the force required is equal to the force of kinetic friction. The force of kinetic friction can be calculated as:
F = μkN
where μk is the coefficient of kinetic friction and N is the normal force.
Plugging in the values, we have:
F = μkmg
3: The drag force experienced by a falling Honda Civic can be approximated using the drag coefficient from Table 5.2, the projected area, and the density of air.
The drag force can be calculated as:
F = 0.5 * Cd * A * ρ * v^2
where Cd is the drag coefficient, A is the projected area, ρ is the density of air, and v is the velocity of the object.
Plugging in the values, we have:
F = 0.5 * Cd * A * ρ * v^2
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. A student throws a ball vertically upwards from the top of the 7 m high CPUT roof. (a) If, after 2 seconds, he catches the ball on its ways down again, with what speed was thrown? (b) What was the velocity of the ball when its was caught? [9,8 m/s] (c) If the student fails to catch the ball with what speed will it hit the ground?
The answers to the given questions are as follows:
(a) The ball was thrown upwards from the roof with an initial velocity of -6.3 m/s.
(b) When the ball was caught on its way down after 2 seconds, its velocity was 13.3 m/s in the downward direction.
(c) If the student fails to catch the ball, it will hit the ground with a speed of approximately 13.31 m/s.
To solve the problem, we can use the equations of motion for vertical motion under constant acceleration. In this case, the acceleration is due to gravity and is equal to 9.8 m/s² (assuming no air resistance).
Given:
Initial height (h) = 7 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 2 s
(a) To find the initial velocity at which the ball was thrown upwards:
Using the equation of motion:
h = ut + (1/2)gt², where u is the initial velocity.
Plugging in the known values, we have:
7 = u(2) + (1/2)(9.8)(2)²
7 = 2u + 19.6
2u = 7 - 19.6
2u = -12.6
u = -6.3 m/s
Therefore, the ball was thrown upwards with an initial velocity of -6.3 m/s (negative sign indicates the upward direction).
(b) To find the velocity of the ball when it was caught:
Since the ball is caught on its way down after 2 seconds, we can use the equation of motion:
v = u + gt, where v is the final velocity (when caught).
Plugging in the values, we have:
v = -6.3 + (9.8)(2)
v = -6.3 + 19.6
v = 13.3 m/s
Therefore, the velocity of the ball when it was caught is 13.3 m/s (positive sign indicates the downward direction).
(c) If the student fails to catch the ball, it will continue to fall freely under gravity until it hits the ground. To find the speed at which it will hit the ground, we can use the equation:
v² = u² + 2gh,
where
v is the final velocity,
u is the initial velocity,
g is the acceleration due to gravity, and
h is the initial height.
Plugging in the values, we have:
v² = (-6.3)² + 2(9.8)(7)
v² = 39.69 + 137.2
v² = 176.89
v = √176.89
v ≈ 13.31 m/s
Therefore, if the student fails to catch the ball, it will hit the ground with a speed of approximately 13.31 m/s.
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A certaln freely falling ebject, released from rest, requices 1.30 s to frover the fast 19.6 m before it hits the ground. (a) Find the velocty of the object ahen in is 35,0 m above the oround. (Indicate the difecten with the sigh of your answed, Let the positive cirection be upwart.) m
2
/s (b) fins the total alalance the object treveis suring the feni:
(a) The velocity of the object when it is 35.0 m above the ground is approximately -26.2 m/s (downward).
(b) The total distance traveled by the object during the fall is 54.6 meters.
(a) To find the velocity of the object when it is 35.0 m above the ground, we can use the equation of motion:
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and s is the displacement.
Acceleration due to gravity (a) = 9.8 m/s² (downward)
Displacement (s) = 35.0 m (upward from the ground)
Since the object is initially at rest, the initial velocity (u) is 0.
Plugging the values into the equation:
v² = 0 + 2(-9.8 m/s²)(35.0 m)
v² = -686 m²/s²
Taking the square root:
v ≈ -26.2 m/s
The velocity of the object when it is 35.0 m above the ground is approximately -26.2 m/s. The negative sign indicates that the velocity is directed downward.
(b) To find the total distance traveled by the object during the fall, we can calculate the sum of the distances covered during the upward and downward motions.
During the upward motion, the object covers a distance of 35.0 m.
During the downward motion, the object covers a distance of 19.6 m.
Therefore, the total distance traveled by the object during the fall is:
Total distance = 35.0 m + 19.6 m = 54.6 m
The object travels a total distance of 54.6 meters during the fall.
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The correct question is:
A certain freely falling object, released from rest, requires 1.30 s to frover the fast 19.6 m before it hits the ground. (a) Find the velocity of the object when it is 35,0 m above the ground. (Indicate the direction with the sign of your answer, Let the positive direction be upward.) (b) Find the total distance the object travels during the fall.
The gravitational force on a body located at distance R from the center of a uniform spherical mass is due solely to the mass lying at distance r≤R, measured from the center of the sphere. This mass exerts a force as if it were a point mass at the origin. (a) Use the above result to show that if you drill a hole through the Earth and then fall in, you will execute simple harmonic emotion about the Earth's center. Find the time it takes you to return to your point of departure and show that this is the time needed for a satellite to circle the Earth in a low orbit with r∼R
⊕
, the radius of the Earth. You may treat the Earth as a uniformly dense sphere, neglect friction and any effects due to the Earth's rotation. (10 points) (b) Show that you will also execute simple harmonic motion with the same period even if the straight hole passes far from the Earth's center.
If you drill a hole through the Earth and fall in, you will execute simple harmonic motion around the Earth's center.
If you consider a straight hole drilled through the Earth, it can be concluded that you will perform simple harmonic motion even if the straight hole passes far from the Earth's center. The motion is such that when a mass is released, it falls to the center of the Earth, overshoots, and oscillates back and forth, executing simple harmonic motion. This is possible because the gravitational force on a body located at distance R from the center of a uniform spherical mass is due solely to the mass lying at a distance r≤R, measured from the center of the sphere.
So, a simple harmonic motion can be executed about the Earth's center. The time taken by an object to complete one revolution around the Earth is given by the time taken by a satellite to circle the Earth in a low orbit with r ∼ R (the radius of the Earth). Thus, the time taken by the object to return to its point of departure is given by the time taken by the satellite to circle the Earth in a low orbit.
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f an object of 80.0 kg is hanging from a rope and the tension on the rope is
900 N in upward direction. What is the net force acting on the object? which
direction the man will move?
The answer is that the net force acting on the object is 116 N in the upward direction. The net force acting on an object can be calculated by subtracting the force of gravity from the upward force of tension on the rope. Here, the object of 80.0 kg is hanging from a rope, and the tension on the rope is 900 N in the upward direction.
The force of gravity on the object is given by the formula: F_gravity = m x g; where, m is the mass of the object, and g is the acceleration due to gravity, which is approximately 9.8 m/s². Therefore, the force of gravity on the object is: F_gravity = 80.0 kg x 9.8 m/s² = 784 N
Now, we can calculate the net force acting on the object by subtracting the force of gravity from the tension on the rope:
F_net = tension - F_gravity⇒F_net = 900 N - 784 N = 116 N
Therefore, the net force acting on the object is 116 N in the upward direction. Since the net force is acting in the upward direction, the object will not move in any direction. The man will stay still in the same position.
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A motorcyclist drives at 22 m/s in a direction 43∘ east of north relative to a car, and at 8.5 m/s due north relative to the Earth.
What is the magnitude of the car’s velocity relative to Earth, ⃗ C relative to E? ⃗ C relative to E= 3.7 m/s
What is the direction of the car’s velocity relative to Earth, measured as an angle theta counterclockwise from due east?
motorcyclist's velocity, v1 = 22 m/s and direction, θ1 = 43° east of north relative to car velocity of motorcyclist relative to Earth, v2 = 8.5 m/s Using relative velocity formula, we can find the car's velocity relative to Earth, vC/E. Therefore, the direction of the car’s velocity relative to Earth, measured as an angle θ counterclockwise from due east is 43° east of north.
So, v2² = v1² + vC/E² - 2v1vC/E cos(θ1)Putting the given values, vC/E = sqrt((22 m/s)² + (8.5 m/s)² - 2(22 m/s)(8.5 m/s)cos(43°)) = 18.7 m/s Thus, the magnitude of the car's velocity relative to Earth is 18.7 m/s. Now, we are asked to find the direction of the car's velocity relative to Earth.
Let the direction be θ measured as an angle counterclockwise from due east. To find θ, let's first calculate the angle between the car's velocity relative to the Earth and the velocity of motorcyclist relative to the Earth.
The angle between them is given by sinθ = (v1/v2)sinθ1Putting the given values, sinθ = (22 m/s/8.5 m/s)sin(43°) = 1.06Thus, there is no angle between them. Hence, the direction of the car's velocity relative to Earth is the same as the direction of the velocity of motorcyclist relative to the Earth, which is 43° east of north.
Therefore, the direction of the car’s velocity relative to Earth, measured as an angle θ counterclockwise from due east is 43° east of north.
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Find the electric potential at point P in the figure (Figure 1). Part B Suppose the three charges shown in the figure are held in place. A fourth charge, with a charge of +6.82μC and a mass of 4.23 g, is released from rest at point P. What is the speed of the fourth charge when it has moved infinitely far away from the other three charges?
The speed of the fourth charge when it has moved infinitely far away from the other three charges is 46.3 m/s.
The electric potential at point P is the sum of the electric potentials due to each of the three charges. The electric potential due to a point charge is given by:
V = kQ/r
where k is the Coulomb constant, Q is the charge of the point charge, and r is the distance from the point charge to the point where the potential is being measured.
In this case, the three charges are located at (0, 0), (3, 0), and (0, 4). The distances from these points to point P are 3, 4, and 5, respectively.
The total electric potential at point P is then:
V = k(6.82 μC) / 3 + k(6.82 μC) / 4 + k(6.82 μC) / 5
= 20.4 V
When the fourth charge is released from rest at point P, it will accelerate away from the other three charges. The potential energy of the fourth charge is given by:
U = kQq/r
where k is the Coulomb constant, Q is the charge of the fourth charge, q is the charge of one of the other three charges, and r is the distance between the fourth charge and that other charge.
The total potential energy of the fourth charge is then:
U = k(6.82 μC)(-6.82 μC) / 3 + k(6.82 μC)(-6.82 μC) / 4 + k(6.82 μC)(-6.82 μC) / 5 = -85.9 V
When the fourth charge moves infinitely far away from the other three charges, its potential energy will be zero. The change in potential energy of the fourth charge is then:
ΔU = 0 - (-85.9 V) = 85.9 V
This change in potential energy is equal to the kinetic energy of the fourth charge when it is infinitely far away from the other three charges. The kinetic energy of the fourth charge is given by:
K = 1/2 mv^2
where m is the mass of the fourth charge and v is its speed.
Solving for v, we get:
v = sqrt(2K/m) = sqrt(2(85.9 V)(1/2)) = 46.3 m/s
Therefore, the speed of the fourth charge when it has moved infinitely far away from the other three charges is 46.3 m/s.
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