A collision refers to an event where two objects come in contact and their motion gets altered. There are two types of collisions- elastic and inelastic. In an elastic collision, the kinetic energy of the colliding bodies is conserved while in an inelastic collision, the kinetic energy is not conserved. Therefore the speed of the red disk after the collision is 7.437 m/s and the speed of the green disk after the collision is 10.26 m/s.
Therefore, we can consider this to be an inelastic collision. Hence, the total kinetic energy after the collision will not be equal to the total kinetic energy before the collision. We can apply the principle of conservation of momentum for this collision to determine the speeds of the disks.
According to the principle of conservation of momentum, the total momentum of the system before and after the collision remains constant. Since there are two disks, the equation can be written as,m1u1 + m2u2 = m1v1 + m2v2where, m1 = mass of the red disk = 5.0 kgm2 = mass of the green disk = 2.0 kgu1 = initial speed of the red disk = 0 m/su2 = initial speed of the green disk = 10 m/s (as it is given that the green disk is moving to the right at a speed of 10 m/s)v1 = final speed of the red diskv2 = final speed of the green disk By applying the conservation of momentum principle, we get,5.0 x 0 + 2.0 x 10 = 5.0v1 + 2.0v2v1 + v2 = 10 --------(1)
Next, we can use the principle of conservation of kinetic energy to form another equation for the given collision. According to the principle of conservation of kinetic energy, the total kinetic energy of the system remains constant before and after the collision. Since this is an inelastic collision, the kinetic energy of the system after the collision will be less than the kinetic energy of the system before the collision. The kinetic energy of the system can be given by the equation, K.E. = 0.5m1u1^2 + 0.5m2u2^2 By substituting the values, we get, K.E. = 0.5 x 5.0 x 0^2 + 0.5 x 2.0 x 10^2 = 100 Joules The kinetic energy of the system after the collision can be given by, K.E. = 0.5m1v1^2 + 0.5m2v2^2By substituting the values, we get, K.E. = 0.5 x 5.0 x v1^2 + 0.5 x 2.0 x v2^2 ----(2)
Since this is an inelastic collision, the total kinetic energy after the collision will be less than 100 Joules. By applying the given angle, we can determine the direction of motion of the green disk after the collision. The final direction of the green disk will be at an angle of 5.0° to the horizontal axis. Hence, we can resolve the velocity of the green disk along the horizontal axis as well as the vertical axis.
The velocity along the horizontal axis will be equal to the final velocity of the red disk (v1). Therefore, we can write,v1 = v cos 5.0°where, v = final velocity of the green disk After substituting the value of v1 from equation (1), we get, v cos 5.0° + v sin 5.0° = 10v (cos 5.0° + sin 5.0°) = 10v = v = 10 / (cos 5.0° + sin 5.0°) = 10.26 m/s The velocity of the green disk after the collision is 10.26 m/s. We can substitute this value in equation (1) to get the value of v1,v1 + v2 = 10v2 = 10.26 m/sv1 = 10 - v2v1 = 10 - 2.563 = 7.437 m/s
Therefore, the speed of the red disk after the collision is 7.437 m/s and the speed of the green disk after the collision is 10.26 m/s.
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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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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 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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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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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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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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A woman is driving her van with speed 50.0mi/h on a horizontal stretch of road. (a) When the road is wet, the coefficient of static friction between the road and the tires is 0.102. Find the minimum stopping distance (in m). m (b) When the road is dry, μs=0.595. Find the minimum stopping distance (in m ). m
The minimum stopping distance on a wet road at a speed of 50.0 mi/h is calculated to be 2035.56 m, while on a dry road it is calculated to be 1359.56 m.
(a) Wet RoadA woman is driving her van with speed 50.0 mi/h on a horizontal stretch of wet road. The coefficient of static friction between the road and the tires is 0.102.
The formula for minimum stopping distance (wet road) is given by: d = (v²/2gμ) + v²/2a
Where
v = initial velocity = 50 miles/hour = (22/15)*50 m/s = 73.33 m/s
μ = coefficient of static friction = 0.102
g = acceleration due to gravity = 9.81 m/s²
a = acceleration = gμ = (9.81)(0.102) = 1.00062 m/s²
Substituting the values in the formula,
d = (73.33²/2*9.81*0.102) + 73.33²/2*1.00062= 52.37 + 1983.19= 2035.56 m
(b) Dry RoadWhen the road is dry, the coefficient of static friction between the road and the tires is 0.595.
The formula for minimum stopping distance (dry road) is given by: d = (v²/2gμ) + v²/2a
Where
v = initial velocity = 50 miles/hour = (22/15)*50 m/s = 73.33 m/s
μ = coefficient of static friction = 0.595
g = acceleration due to gravity = 9.81 m/s²
a = acceleration = gμ = (9.81)(0.595) = 5.83995 m/s²
Substituting the values in the formula,
d = (73.33²/2*9.81*0.595) + 73.33²/2*5.83995= 15.28 + 1344.28= 1359.56 m
Thus, the minimum stopping distance (in m) when the road is wet is 2035.56m and when the road is dry is 1359.56m.
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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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If a charge, Q, is located at the center of a spherical volume of radius R
0
=3.65 cm and the total electric flux through the surface of the sphere is Φ
0
=−6.66
C
Nm
2
, what is the total flux through the surface if the radius of the sphere is changed to R=14.1 cm (in
C
Nm
2
).
Therefore, the total flux through the surface if the radius of the sphere is changed to R = 14.1 cm (in C Nm²) is: Phi = \frac{-6.66}{4.95 × 10^{-10}
Phi = -1.35 × 10^7 C Nm^{-2}
The total electric flux through a surface of a sphere when a charge, Q, is located at the center of a spherical volume of radius R₀ is given by the expression;Phi_0 = \frac{Q}{4πε_0R_0^2}
Where;Phi_0 = the total electric flux through the surface of the sphere Q =the charge located at the center of the spherical volume
ε_0 = the permittivity of free spaceR_0 =the initial radius of the spherical volume.
When the radius of the sphere is changed to R, the total electric flux through the surface of the sphere is given by the expression; Phi = \frac{Q}{4πε_0R^2}
Where; Phi = the total electric flux through the surface of the sphere (in C Nm²)Q = the charge located at the center of the spherical volumeε_0 = the permittivity of free space R =the new radius of the spherical volume.
Thus, the total flux through the surface if the radius of the sphere is changed to R = 14.1 cm is given by; Phi = \frac{Q}{4πε_0R^2} Phi = frac{Q}{4π(8.85 × 10^{−12} N^{-1}m^{-2}) (0.141 m)^2} Phi = \frac{Q}{4.95 × 10^{-10}}
Therefore, the total flux through the surface if the radius of the sphere is changed to R = 14.1 cm (in C Nm²) is; Phi = \frac{-6.66}{4.95 × 10^{-10}}
Phi = -1.35 × 10^7 C Nm^{-2}$$
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Air at 20°C and at a atmospheric pressure flows over a flat plate at a velocity of 3 m/s. If the plate is 30 cm long and at a temperature of 60°C, calculate: (a) the thickness of velocity and thermal boundary layers at 20 cm. (b) the average heat transfer coefficient. (c) total drag force on the plate, per unit width. Take the following properties of air: = P = 1.18 kg/m³, kinematic viscosity = 17 x 10-6 m²/s, k = 0.0272 W/m-K, Cp = 1.007 kJ/kg K
The thickness of velocity and thermal boundary layers is 0.0567 m and 0.0347 m respectively. The average heat transfer coefficient is 57.11 W/m² K. The total drag force on the plate is 0.05677 N/m.
According to the given problem, the properties of air are: ρ = 1.18 kg/m³, Kinematic viscosity (μ) = 17 × 10⁻⁶ m²/s, Thermal conductivity (k) = 0.0272 W/m-K, Specific heat (Cp) = 1.007 kJ/kg K, Reynolds number (Re)
Re = ρVxδvx / μ
= (1.18 × 3 × 0.2 × 0.017) / 0.000017 = 2222.4
Prandtl number (Pr)
Pr = Cp μ / k
= (1.007 × 0.000017) / 0.0272
= 0.00064
Nusselt number (Nu)
Nu = 0.332 × Re1/2 Pr1/3
= 0.332 × 2222.4 1/2 × 0.00064 1/3
= 73.324
Average heat transfer coefficient:
h = k Nu / δtx
= 0.0272 × 73.324 / 0.0347 = 57.11 W/m² K
The average skin friction coefficient is:
cf = 0.664 / Re1/2 = 0.664 / 2222.4 1/2 = 0.01575
Total drag force per unit width:
Fx = 0.5 ρ Vx³ Cf
L = 0.5 × 1.18 × 3³ × 0.01575 × 0.3
= 0.05677 N/m
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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 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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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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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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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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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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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?
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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A runner hopes to completo the 10,000−m fun in less than 30.0 min. After running at constant speed for exactly 25.0 min, there are still 1900 II to go. The runner must then accelerate at 0.19 m/s
2
for how many seconds in order to achieve the desired time Express your answer using two significant figures. * Incorrect; Try Again; 3 attempts remaining
The runner needs to accelerate for approximately 368 seconds in order to achieve the desired time of completing the 10,000m run in less than 30.0 minutes.
To find the time needed to accelerate, we can use the formula:
d = v_i * t + (1/2) * a * t^2
Where:
d = distance to go after running for 25 minutes (1900m)
v_i = initial velocity (unknown)
t = time to accelerate (unknown)
a = acceleration (0.19 m/s^2)
Since the runner is running at a constant speed for the first 25 minutes, the initial velocity is equal to the average velocity during this time. We can calculate it using the formula:
v_i = d / t
Substituting the given values, we have:
v_i = 1900m / 25min
Now, we can use the equation for distance with the known values to solve for t:
1900m = (v_i * t) + (1/2) * (0.19 m/s^2) * t^2
Simplifying the equation, we get:
1900m = (1900m/25min) * t + 0.095t^2
Rearranging the equation, we have:
0.095t^2 + (1900m/25min) * t - 1900m = 0
Solving this quadratic equation for t, we find:
t ≈ 368 seconds
Therefore, the runner needs to accelerate for approximately in order to achieve the desired time of completing the 10,000m run in less than 30.0 minutes.
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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 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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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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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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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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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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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 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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Suppose a 52-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.275 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V ) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T?
Faraday's law of electromagnetic induction is used to compute induced EMF, abbreviated as e. It expresses the relationship between the EMF generated and the magnetic flux's rate of change, abbreviated as φ.
The induced EMF in the coil with 52 turns that lies in the plane of the page in a uniform magnetic field of 1.60 T that is directed out of the page is calculated as follows:
Given values are as follows:52-turn coilInitial area, A1 = 0.275 m²Final area, A2 = 0m² (as it is stretched to have no area in 0.100s)Time, t = 0.100 s
Strength of the uniform magnetic field, B = 1.60 T
We need to calculate the magnitude (in V) and direction (as seen from above) of the average induced EMF, e.We know that the flux is defined as φ = B.A.
Therefore, we can write:[tex]B * A1 = B * A2 + [(ΔB / Δt) * A2][/tex]
By substituting the given values in the above formula,
we get: (1.60 T)(0.275 m²)
= [tex](1.60 T)(0 m²) + [(ΔB / Δt) * 0 m²]ΔB / Δt[/tex]
= [tex][(1.60 T)(0.275 m²)] / (0 m²)(0.100 s)ΔB / Δt[/tex]
= [tex]4.40 T/s[/tex]
Now, by using Faraday's law of electromagnetic induction, we can calculate the induced EMF.
[tex]e = -N (Δφ / Δt).[/tex]
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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
