a). The initial vertical velocity of the cannonball is approximately 35.74 m/s, and the initial horizontal velocity is approximately 200.54 m/s.
b). The height of the cannonball at the top of its arc is approximately 72.664 m, and it takes approximately 3.65 seconds to reach that height.
a. To determine the initial vertical velocity (v1,y) and the initial horizontal velocity (v1,x) of the cannonball, we can use the given launch angle (θ) and the initial total velocity (v1).
The initial vertical velocity can be found using the following equation:
v1,y = v1 * sin(θ)
Substituting the given values: v1,y = 205 m/s * sin(10.0°)
v1,y ≈ 35.74 m/s
The initial horizontal velocity remains constant throughout the trajectory and can be found using:
v1,x = v1 * cos(θ)
Substituting the given values:
v1,x = 205 m/s * cos(10.0°)
v1,x ≈ 200.54 m/s
Therefore, the initial vertical velocity of the cannonball is approximately 35.74 m/s, and the initial horizontal velocity is approximately 200.54 m/s.
b. To determine the height (y2) at the top of the cannonball's arc and the time (t2) it takes to get there, we need to consider the vertical motion of the cannonball.
The vertical motion of the cannonball can be described using the following kinematic equation:
y2 = y1 + v1,y * t2 - (1/2) * g * t2^2
where y1 is the initial height of the cannonball, v1,y is the initial vertical velocity, t2 is the time taken to reach the top of the arc, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the given values:
y2 = 8.00 m + 35.74 m/s * t2 - (1/2) * (9.8 m/s^2) * t2^2
At the top of the arc, the vertical velocity becomes zero. Therefore, we can use this information to find t2.
v1,y - g * t2 = 0
35.74 m/s - 9.8 m/s^2 * t2 = 0
Solving for t2:
t2 = 35.74 m/s / 9.8 m/s^2
t2 ≈ 3.65 s
Substituting the value of t2 back into the equation for y2:
y2 = 8.00 m + 35.74 m/s * (3.65 s) - (1/2) * (9.8 m/s^2) * (3.65 s)^2
y2 ≈ 8.00 m + 130.511 m - 65.847 m
y2 ≈ 72.664 m
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1. A 2.08 m tall man stands 1.78 m from a lens with focal length 28.3 cm.
How tall (in m) is his image formed by the lens? Be sure to include the sign to indicate orientation!
2. When laser light of some unknown wavelength hits a diffraction grating with 20 slits per millimeter, it produces bright fringes separated by 27.2 mm on a screen that is 2.41 m away.
Given the pattern formed, what must be the wavelength of the light (in nm)?
1. The height of the image is 0.287 m.
2. The wavelength of the light is 563 nm.
1. The image distance, denoted as `i`, is determined by the lens formula: `1/f = 1/o + 1/i`, where `f` represents the focal length, `o` is the object distance, and `i` represents the image distance. Given `f = 28.3 cm` and `o = 1.78 m`, we need to convert the object distance from meters to centimeters: `o = 1.78 m = 178 cm`. Therefore, the image distance is calculated as follows:
i = (1/f - 1/o)^-1 = (1/28.3 - 1/178)^-1 = 24.53 cm.
The image height, denoted as `h'`, can be determined using the object height `h` and the magnification `m` relationship: `h' = m * h`. The magnification `m` is given by `m = -i/o`, where the negative sign indicates an inverted image. Thus,
m = -i/o = -(24.53 cm)/(178 cm) = -0.138.
The image height `h'` is obtained by multiplying `h` by `m`: `h' = m * h`, where `h = 2.08 m`. Therefore,
h' = (-0.138) * 2.08 = -0.287 m.
The negative sign signifies an inverted image. Hence, the height of the image is determined as `0.287 m`, and it is inverted.
2. Bright fringes are observed at angles `theta` satisfying the condition `d sin theta = m lambda`, where `d` represents the spacing between two slits, `m` is an integer indicating the fringe order, and `lambda` denotes the wavelength of light. In this case, given `d = 1/20 mm` and `m = 1`, the angle `theta` corresponding to the first bright fringe is given by `tan theta = x/L`, where `x` represents the separation between two fringes, and `L` is the distance from the grating to the screen. With `x = 27.2 mm` and `L = 2.41 m`, we can calculate:
tan theta = (27.2 mm)/(2.41 m) = 0.01126.
Therefore, `sin theta = tan theta = 0.01126`.
Consequently, the wavelength `lambda` is determined using the formula `lambda = d sin theta / m`, where `d = 1/20 x 10^-3 m`, `sin theta = 0.01126`, and `m = 1`:
lambda = (1/20 x 10^-3 m) x 0.01126 / 1 = 5.63 x 10^-7 m = 563 nm.
In summary:
1. The height of the image is 0.287 m.
2. The wavelength of the light is 563 nm.
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The low A string of a guitar is tightened to a tension of 62 N. The speed of the wave in the string is 143 m/s. If the string is 65 cm long, what is it’s mass in grams?
The mass of the string is 0.18655 g.The low A string of a guitar is tightened to a tension of 62 N, The speed of the wave in the string is 143 m/s, the string is 65 cm long.
The frequency of a guitar string is given by the relation:v = fλ,Where, v is the velocity, λ is the wavelength, f is the frequency. From the above relation, the wavelength of the string is given asλ = v/fWe know that the wave speed in a string is given asv = √(T/μ),Where, T is the tension force and μ is the mass per unit length.
Substituting the values given in the above equation, we have:143 = √(62/μ)Squaring on both sides, we getμ = 62/143²μ = 0.000287 kg/mThe total mass of the string m is given by:m = μlwhere l is the length of the string.Substituting the values, we have:m = 0.000287 × 0.65m = 0.00018655 kg = 0.18655 gB
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is a direct calculation question. You are in a spacecraft orbiting Venus. The mass of Venus is 4.87E+24 kg, and it's radius is 6.05E+3 km. What is the escape velocity from the surface? m/s. At a height of 11 radii above the surface of the Venus, what is the circular orbital velocity (orbital velocity, not escape velocity)? m/s Note: If your answer requires scientific notation, remember that OWL uses "e" notation: 1.1 x 105 is 1.1e5 to OWL.
The escape velocity from the surface of Venus is approximately 10.36 km/s. At a height of 11 radii above the surface of Venus, the circular orbital velocity is approximately 7.11 km/s.
To calculate the escape velocity from the surface of Venus, we can use the formula:
Escape velocity = √(2 * gravitational constant * mass of Venus / radius of Venus)
Plugging in the values, we get:
Escape velocity = √(2 * 6.67430e-11 m^3/kg/s^2 * 4.87e24 kg / (6.05e3 km * 1e3 m/km))
Calculating this expression, we find that the escape velocity from the surface of Venus is approximately 10.36 km/s.
To calculate the circular orbital velocity at a height of 11 radii above the surface of Venus, we need to consider the gravitational force between the spacecraft and Venus. The gravitational force provides the centripetal force required for circular motion. The formula for circular orbital velocity is:
Circular orbital velocity = √(gravitational constant * mass of Venus / (radius of Venus + 11 * radius of Venus))
Plugging in the values, we get:
Circular orbital velocity = √(6.67430e-11 m^3/kg/s^2 * 4.87e24 kg / (6.05e3 km * 1e3 m/km * (1 + 11)))
Calculating this expression, we find that the circular orbital velocity at a height of 11 radii above the surface of Venus is approximately 7.11 km/s.
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While targeting a wild ox, a stone-age hunter spins a rock around in his sling with a constant angular velocity of 4.5 rev/sec. If the sling has a radius of
0.55 m, what are the tangential velocity, tangential acceleration, and radial acceleration of the rock?
The tangential velocity of the rock in the sling is 2.475 m/s. The tangential acceleration is 0 m/s², and the radial acceleration is 11.137 m/s².
To calculate the tangential velocity, we can use the formula:
Tangential velocity = Angular velocity × Radius
Given that the angular velocity is 4.5 rev/sec and the radius is 0.55 m, we can substitute these values into the formula to find the tangential velocity.
Tangential velocity = 4.5 rev/sec × 0.55 m = 2.475 m/s
The tangential acceleration can be calculated using the formula:
Tangential acceleration = Angular acceleration × Radius
Since the problem does not provide information about the angular acceleration, we assume that it is constant and equal to zero. Therefore, the tangential acceleration is also zero.
The radial acceleration can be calculated using the formula:
Radial acceleration = (Tangential velocity)^2 / Radius
Substituting the values we already have, we can find the radial acceleration:
Radial acceleration = (2.475 m/s)^2 / 0.55 m ≈ 11.137 m/s^2
Therefore, the tangential velocity is 2.475 m/s, the tangential acceleration is 0 m/s^2, and the radial acceleration is approximately 11.137 m/s^2.
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(a) What is the radius of the sphere? \( \mathrm{mm} \) (b) What is the thickness of the shell wall? \( \mathrm{mm} \)
The radius of the sphere can be determined by dividing the diameter by 2, while the thickness of the shell wall can be found by subtracting the inner radius from the outer radius.
(a) To determine the radius of the sphere, we need more information. The radius is the distance from the center of the sphere to any point on its surface. If we have the diameter, we can divide it by 2 to find the radius. For example, if the diameter is 10 mm, then the radius would be 10 mm ÷ 2 = 5 mm.
(b) Similarly, to find the thickness of the shell wall, we need more details. The shell wall refers to the thickness of the hollow part of the sphere. If we know the outer radius and the inner radius of the shell, we can subtract the inner radius from the outer radius to find the thickness.
For example, if the outer radius is 8 mm and the inner radius is 6 mm, then the thickness of the shell wall would be 8 mm - 6 mm = 2 mm.
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Write the azimuth for the displayed bearing. Remember to enter
only a number, no spaces, letters, or punctuation.
The azimuth is a numerical representation of a bearing's direction, determined by determining the angle between the reference direction and the bearing's direction. It should be entered without spaces, letters, or punctuation.
The azimuth for a displayed bearing is a numerical representation of the direction. To write the azimuth, you should enter only a number without any spaces, letters, or punctuation.
To find the azimuth, you need to determine the angle between the reference direction (usually north) and the direction of the bearing.
Here's a step-by-step example:
1. Identify the reference direction, which is typically north.
2. Determine the direction of the bearing from the reference direction.
3. Measure the angle between the reference direction and the bearing direction.
4. Express the angle in degrees, ranging from 0 to 360.
For instance, if the bearing is 30 degrees east of north, the azimuth would be 30. If the bearing is 45 degrees west of north, the azimuth would be 315 (since it is 360 - 45).
Remember, when writing the azimuth, only enter the numerical value without any spaces, letters, or punctuation.
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when the central ray is angled, the structure situated _____ is projected the farthest.
When the central ray is angled, the structure situated **closest to the central ray** is projected the farthest.
A jetliner has a cruising air speed of 620mi/h relative to the air. Part A For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Beiative How long does it take this plane to fly round-trip trom San Francisco to Cricago, an east-west fight of 2000mi each way. velocity on the highway. Express your answer in hours. Part B How long does it take this plane to fly round tra from San Franciseo is Chicago an east-west fight of 2000 mi each waf. if the wind is blowing at 160mi/h from the west to the east? Express your answer in houns.
The time taken by the plane to fly round-trip from San Francisco to Chicago is 6.45 hours.the total time taken by the plane to fly round-trip is 6.91 hours.
Part AThe given the cruising air speed of the jetliner is 620 mi/h. Therefore, the time taken to fly 2000 miles in one direction is given by:Time = Distance/Speed
= 2000/620
= 3.225 hours. Therefore, the round-trip time taken is twice the time taken in one direction. So, the time taken by the plane to fly round-trip from San Francisco to Chicago is:Round-trip time = 2 × 3.225 = 6.45 hours. Hence, the required answer is 6.45 hours.
Part BThe cruising air speed of the jetliner is still 620 mi/h. However, now there is a wind blowing from west to east at 160 mi/h. This means that the plane will face a headwind while flying from west to east, and a tailwind while flying from east to west. The actual speed of the plane relative to the ground while flying in the two directions is given by:East to West Speed = 620 - 160
= 460 mi/h West to East Speed
= 620 + 160
= 780 mi/h. The time taken by the plane to fly 2000 miles in one direction is given by:Time taken to fly eastward = 2000/780
= 2.56 hours. Time taken to fly westward
= 2000/460
= 4.35 hours. Therefore, the total time taken by the plane to fly round-trip is:Total time taken = 2.56 + 4.35 = 6.91 hours. Hence, the required answer is 6.91 hours.
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Measuring m Measure the ma 4. Measuring force: Measure the weight of the 5. Measuring temperature: Using a digital thermometer: To alternate the unit between "C and "F press the button for this purpose. 1. Measure the temperature of cold water in the graduated cylinder in
∘
C and F. 2. Discard the cold water and put hot water in the graduated cylinder. Measure the temperature of the hot water in ⊙C and F. 6. Calculating average and percent error: 1.Set up a simple pendulum with an approximate length of 90 cm. Record the exact length. 2.Separate the metal sphere from vertical alignment by about 10 cm and release it to produce small and slow oscillations. 3. Measure the time t
10
that it takes the pendulum to make 10 oscillations and record its value. 4. Repeat the procedure four more times. 5. Calculate the average time tram and the average period T
av
=t
10m
/10. 6. Calculate the gravitational acceleration g using the formula: g=
T
av
2
4π
2
L
with if π=3.14 7. Calculate: % Erтor of g=
g
acpt
∣g−g
acpt
∣
=100% with g
eqpt
=9.8 m/s
2
By following these steps, you can measure force, temperature, and calculate the average time and percent error in the experiment involving the simple pendulum.
The provided instructions outline the steps to measure various physical quantities and calculate the average time and percent error. Here is a summary of the procedure:
Measuring force: Measure the weight of an object using a weighing scale.
Measuring temperature: Use a digital thermometer to measure the temperature. Press the button to alternate between Celsius (°C) and Fahrenheit (°F) units. Measure the temperature of cold water and hot water separately.
Calculating average and percent error:
Set up a simple pendulum with an approximate length of 90 cm and record the exact length.
Displace the metal sphere from its vertical alignment by about 10 cm and release it to observe small and slow oscillations.
Measure the time it takes for the pendulum to make 10 oscillations (t10) and record the value.
Repeat the procedure four more times to obtain a total of five measurements.
Calculate the average time (t10m) by summing up the five measurements and dividing by 5.
Calculate the average period (Tav) by dividing the average time by 10 (Tav = t10m/10).
Calculate the gravitational acceleration (g) using the formula: g = (4π²L) / Tav², where L is the length of the pendulum.
Calculate the percent error of g using the formula: % Error of g = (|g - gacpt| / gacpt) × 100, where gacpt is the accepted value of the gravitational acceleration (9.8 m/s²).
By following these steps, you can measure force, temperature, and calculate the average time and percent error in the experiment involving the simple pendulum.
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Determine the acceleration given this system (in m/s2 to the right). ]m/s2 (to the right) (b) Determine the tension in the cord connecting the 3.5 kg and the 1.0 kg blocks (in N ). ] N (c) Determine the force exerted by the 1.0 kg block on the 2.0 kg block (in N ). ∫N
(a) The acceleration of the system is approximately 2.25 m/s² to the right.
(b) The tension in the cord connecting the 3.5 kg and 1.0 kg blocks is approximately 23.556 N.
(c) The force exerted by the 1.0 kg block on the 2.0 kg block is approximately 4.50 N.
The system of two blocks, we need to find the acceleration, tension in the cord, and force exerted by the 1.0 kg block on the 2.0 kg block. Let's solve each part of the question.
(a) Acceleration (in m/s² to the right)
We know that the acceleration of the system is given by the formula:
F = ma
Where,
F is the net force on the system
m is the mass of the system
a is the acceleration of the system
The net force acting on the system is given by:
F = T - f
Where,
T is the tension in the cord connecting the blocks
f is the frictional force acting on the 2.0 kg block
So, the acceleration of the system is given by:
ma = T - f
Here, m = 3.5 kg
The frictional force f = μmg = 0.35 × 2.0 × 9.81 = 6.867 N (using μ = 0.35 as given)
Substituting the values, we get:
3.5a = T - 6.867
Also, for the second block, we know that:
F = ma
Here, F is the force exerted by the 1.0 kg block on the 2.0 kg block. We can write:
F = m₂a = 2.0a
Equating the two values of F, we get:
2.0a = T - 6.8673.5a = T - 6.867
Multiplying the second equation by 2 and adding it to the first equation, we get:
2(3.5a = T - 6.867) + 2.0a = T - 6.8677.0a = 3T - 6.867a = (3T - 6.867) / 7.0
Substituting the value of acceleration from the above equation, we get:
3.5[(3T - 6.867) / 7.0] = T - 6.867
Solving for T, we get:
T = 23.556 N (approx)
Therefore, the acceleration of the system is given by:
a = (3T - 6.867) / 7.0= (3 × 23.556 - 6.867) / 7.0= 2.25 m/s² (approx)
(b) Tension in the cord (in N)
From the above calculation, we get:
T = 23.556 N (approx)
Therefore, the tension in the cord connecting the 3.5 kg and 1.0 kg blocks is 23.556 N (approx).
(c) Force exerted by the 1.0 kg block on the 2.0 kg block (in N)
We know that the force exerted by the 1.0 kg block on the 2.0 kg block is given by:
F = m₂a
Here, m₂ = 2.0 kg, and a = 2.25 m/s² (as found in part (a)).
Therefore, the force exerted by the 1.0 kg block on the 2.0 kg block is:
F = m₂a= 2.0 × 2.25= 4.50 N (approx)
Therefore, the force exerted by the 1.0 kg block on the 2.0 kg block is 4.50 N (approx).
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If wind blows at 46 m/s over the roof of your housc, what is the pressure difference at the roof between the inside and outside air? Use an air density of 1,3 kg/m^3, A. 1.5×10 ^5Pa B. 7.5×10^5Pa C. 1.0×10^5Pa D. 1.4×10^3 Pa E. Orher 7. If the roof in problem 3 is 20 m×15 m what is the total force on the roof?
,The total force on the roof is 4,211,820 N.
Given:
Wind speed over the roof of the house, v1 = 46 m/s
Air density, ρ = 1.3 kg/m³
To find:
Pressure difference at the roof between the inside and outside air
Formula used:
Bernoulli's principle, which states that P1 + 0.5ρv1² = P2 + 0.5ρv2²
Where:
P1 = pressure at point 1
v1 = velocity at point 1
P2 = pressure at point 2
v2 = velocity at point 2
ρ = density of the fluid
Substituting the given values in the expression:
P1 + 0.5ρv1² = P2 + 0.5ρv2² ...(1)
Here, point 1 is outside the house and point 2 is inside the house.
Since the pressure inside the house is greater than the pressure outside the house, P2 > P1.
Given:
v1 = 46 m/s
v2 = 0 m/s (since the wind stops inside the house)
ρ = 1.3 kg/m³
Substituting these values in equation (1):
1.5P1 = P2
Since P2 > P1:
1.5P1 - P1 = 0.5P1
0.5P1 = 0.5 × 1.3 × 46²
P1 = 1403.94 Pa
Thus, the pressure difference at the roof between the inside and outside air is 1403.94 Pa. Hence, the correct option is D. 1.4×10³ Pa.
Now, let's find the total force on the roof using the formula:
F = P × A
Where:
F is the force
P is the pressure
A is the area of the roof
Substituting the given values in the above formula:
F = 1403.94 × 20 × 15
F = 4,211,820 N
Thus, the total force on the roof is 4,211,820 N.
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A ball is dropped from the top of a building and is
observed to hit the ground after 5.5 seconds. How far has it
traveled after 2.0s?
The ball has traveled 19.6 meters after 2.0 seconds. After 2.0 s, the distance traveled by the ball is 19.6 meters. We can use kinematic equations to solve this problem. The following equations are relevant in this scenario: vf = vi + atx
= vi(t) + 1/2(a)(t)^2
where vf = final velocity
= 0 m/s (since the ball has hit the ground, so its velocity is zero)
vi = initial velocity (unknown)
a = acceleration due to gravity = 5.5 s (time taken to hit the ground) In the vertical direction, we can write the second equation as follows: x = 150.125 m
This is the total distance traveled by the ball. However, we need to find the distance traveled by the ball in the first 2.0 s. We can use the same equation again, but with t = 2.0 s instead of 5.5 s:
x = 19.6 m Therefore, the ball has traveled 19.6 meters after 2.0 seconds.
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hockey player uses her stick to exert a force of 9.00 N on a stationary puck, which is resting on a nearly frictionless ice surface. The hockey puck has a mass of 180 g. What is the acceleration a of the hockey puck?
The hockey puck will experience an acceleration of [tex]50 m/s^2[/tex] when the hockey player exerts a force of 9.00 N on it.
For finding the acceleration of the hockey puck, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. The formula for calculating acceleration is
a = F/m,
where a is the acceleration, F is the force applied, and m is the mass of the object.
In this case, the force exerted by the hockey player is 9.00 N, and the mass of the puck is 180 g, which is equivalent to 0.180 kg.
Plugging these values into the formula,
a = 9.00 N / 0.180 kg.
Calculating the division, we find that the acceleration of the hockey puck is approximately [tex]50 m/s^2[/tex].
Therefore, the hockey puck will experience an acceleration of [tex]50 m/s^2[/tex] when the hockey player exerts a force of 9.00 N on it.
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Thermal neutrons are diffracted of (b.c.c) solid sample at diffraction angle 50.56 ^∘. If the diffraction take place of the plane (211) and the lattice constant of the unit cell is 2.868 A, what is the momentum of the incident neutron?
Thermal neutrons are diffracted of (b.c.c) solid sample at diffraction angle 50.56 ^∘. If the diffraction takes place of the plane (211) and the lattice constant of the unit cell is 2.868 A, what is the momentum of the incident neutron?
Thermal neutrons are slowed-down neutrons with energy lower than the energy of fast neutrons. They have a kinetic energy of about 0.025 eV (electron volts) or lower, which is roughly 2.2 kBT, where T is the temperature in kelvins. Thermal neutrons have a wavelength of about 0.2 nanometers, which is the same order of magnitude as the spacing between the atoms in a solid. As a result, when they collide with a solid, they can cause diffraction. When a neutron of momentum p and mass m is diffracted by a crystal, the diffraction angle θ is given by the Bragg equation:
2dsinθ = nλ,
where d is the spacing between the planes of atoms, λ is the wavelength of the neutron, n is an integer, and θ is the angle between the direction of the incident neutron and the plane of atoms.
In this problem, the diffraction angle is given as θ = 50.56°, and the spacing between the (211) plane of atoms is d = 2.868 Å.
The wavelength of the neutron is λ = h/p, where h is Planck's constant and p is the momentum of the neutron. Therefore, p = h/λ. Combining these equations gives:
2dsinθ = nλ
=> 2(2.868 Å)sin(50.56°) = n(h/p)
=> 2(2.868 × 10^−10 m)sin(50.56°) = n(6.63 × 10^−34 J s/p)
p=1.2 × 10^-24 kg m/s.
Approximately.
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In an old house, the heating system uses radiators, which are hollow metal devices through which hot water or steam circulates. In one room the radiator has a dark color (emissitivity =0.800 ). It has a temperature of 70.0
∘
C. The new owner of the house paints the radiator a lighter color (emissitivity =0.468 ). Assuming that it emits the same radiant power as it did before being painted, what is the temperature (in degrees Celsius) of the newly painted radiator?
Emissivity is a measure of how much radiation is emitted by a surface compared to a perfect black body at the same temperature. A perfect black body is one that absorbs all radiation that falls on it and does not reflect or transmit any of it. When a radiator is painted a lighter color, its emissivity changes, and its ability to emit radiation is reduced.
The equation used for the first radiator is given by the Stefan-Boltzmann equation: P = εσAT4
where P is power radiated, ε is emissivity, σ is Stefan-Boltzmann constant, A is surface area, and T is temperature. Since we are assuming that the power radiated remains the same and the surface area of the radiator does not change, we can equate the power equation for both radiators. Therefore: ε1σ[tex]A_1[/tex][tex]T_1^4[/tex] = ε2σ[tex]A_2T_2^4[/tex]
Since the surface area of the radiator does not change and the power radiated remains the same, ε1[tex]T_1^4[/tex] = ε2[tex]T_2^4[/tex]
Therefore: T2 = (ε1/ε2[tex])^{0.25[/tex][tex]T_1[/tex]
For the darker radiator,
[tex]T_1[/tex] = 70.0 °C
[tex]T_2[/tex] = (0.800/0.468)^0.25(70.0 °C)
[tex]T_2[/tex] = 79.6 °C
The temperature of the newly painted radiator is 79.6°C.
Radiators are common heating devices that can be found in old houses. Radiators are used to circulate hot water or steam throughout the building, providing warmth to the occupants. Radiators are hollow metal devices that emit heat through radiation, which is a process in which energy is transferred from one body to another in the form of electromagnetic waves. Radiators come in different sizes and shapes, and their heating efficiency depends on their surface area, temperature, and emissivity.
Emissivity is a measure of how much radiation is emitted by a surface compared to a perfect black body at the same temperature. A perfect black body is one that absorbs all radiation that falls on it and does not reflect or transmit any of it. When a radiator is painted a lighter color, its emissivity changes, and its ability to emit radiation is reduced. In this scenario, we are assuming that the power radiated by the radiator remains the same, and the surface area of the radiator does not change. Therefore, we can use the Stefan-Boltzmann equation to equate the power equation for both radiators. From this equation, we can solve for the temperature of the newly painted radiator, which is found to be 79.6 °C.
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A point charge Q creates an electric potential of +165 V at a distance of 18 cm. What is Q?
The point charge Q is approximately 3.30 x 10^-8 C, determined by using the formula for electric potential with the given values. The calculation involves multiplying the electric potential by the distance and dividing by the electrostatic constant.
To determine the value of the point charge Q, we can use the formula for electric potential:
V = k * Q / r
Where:
V is the electric potential (165 V)
k is the electrostatic constant (approximately 8.99 x 10^9 N m^2/C^2)
Q is the point charge (unknown)
r is the distance from the point charge (18 cm or 0.18 m)
Rearranging the formula, we can solve for Q:
Q = V * r / k
Substituting the given values into the equation:
Q = (165 V) * (0.18 m) / (8.99 x 10^9 N m^2/C^2)
Calculating the result:
Q ≈ 3.30 x 10^-8 C
Therefore, the value of the point charge Q is approximately 3.30 x 10^-8 Coulombs.
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1) Solve the points A, B, C and D (20 points) step by step
A). Coulomb's law is used to find the force of attraction or repulsion between
two point charges, which are at a certain distance. could you describe
why two charges attract or repel each other, use examples for your argument.
B). If two point charges are at a distance of 2.5 m and between them is exerted
a force equal to the day of your birthday (Keep in mind that force is measured in
Newton). Determine the value of the other load if one of them has a value of
3.2x10-6C.
Note: For example, if your birthday is February 19, then the force you use in the
the exercise would be 19 Newton.
C). Find the distance between two equal charges, which are equal to the
last digit of your identity document multiplied by ten to the minus three,
knowing that there is a repulsive force of 5.8 N.
Note: For example, if the last digit of your identity document is 5, the charges
will have the following value Q= 5x10-3
D). If we have two charges q1 and q2 that are at a distance # meters (# equals
at your age). Find the force exerted between them knowing that:
q1 is equal to 5x10-4
q2 is equal to 3.3x10-3
Comment, in your own words, in no less than 5 lines, why the
Name Coulomb's law.
Note: Keep in mind to attach with your answers the evidence of the values that you
use in each of the problems that require it.
A) Two charges attract or repel each other based on their electrical properties, B) Coulomb's constant k = 9x10^9 Nm^2/C^2, we can calculate the value of q2, c) Using the given values of the repulsive force F = 5.8 N, and the charge value Q = 5x10^(-3) C,D) Substituting the given values of q1 = 5x10^(-4) C, q2 = 3.3x10^(-3) C, and r = # meters (# equals your age), and using the value of the Coulomb's constant k = 9x10^9 Nm^2/C^2, we can calculate the force between the charges.
A) Two charges attract or repel each other based on their electrical properties. Like charges, such as two positive charges or two negative charges, repel each other because they have the same charge sign. This repulsion occurs because there is a force pushing them apart. On the other hand, unlike charges, such as a positive and a negative charge, attract each other because they have opposite charge signs. This attraction occurs because there is a force pulling them together. An example is the attraction between the positive and negative terminals of a battery, which allows for the flow of electric current.
B) Given that the force exerted between the two point charges is equal to the birthday value (19 Newton in this example), and one charge has a value of 3.2x10^(-6) C, we can use Coulomb's law to find the value of the other charge. Coulomb's law states that the force between two charges is given by the equation:
F = (k * q1 * q2) / r^2
Rearranging the equation, we can solve for q2:
q2 = (F * r^2) / (k * q1)
Substituting the given values, with F = 19 N, r = 2.5 m, q1 = 3.2x10^(-6) C, and using the value of the Coulomb's constant k = 9x10^9 Nm^2/C^2, we can calculate the value of q2.
C) To find the distance between two equal charges, given a repulsive force of 5.8 N, we can rearrange Coulomb's law equation to solve for distance (r):
r = sqrt((k * Q^2) / F)
Using the given values of the repulsive force F = 5.8 N, and the charge value Q = 5x10^(-3) C (assuming the last digit of the identity document is 5), and the Coulomb's constant k = 9x10^9 Nm^2/C^2, we can calculate the distance between the charges.
D) To find the force exerted between two charges q1 and q2 at a distance equal to the age in meters, we can use Coulomb's law:
F = (k * q1 * q2) / r^2
Substituting the given values of q1 = 5x10^(-4) C, q2 = 3.3x10^(-3) C, and r = # meters (# equals your age), and using the value of the Coulomb's constant k = 9x10^9 Nm^2/C^2, we can calculate the force between the charges.
The name Coulomb's law is given in honor of Charles-Augustin de Coulomb, a French physicist who formulated this law in the 18th century. Coulomb's law describes the fundamental relationship between electric charges and the force they exert on each other. It provides a mathematical framework for understanding and quantifying the forces between charged objects, making it a significant contribution to the field of electromagnetism.
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You have an object with a charge of negative 3 micro-Coulombs. How many extra electrons are on this object? 1.875×10
13
electrons B 5.33×10
14
electrons (C) 1.6×10
19
electrons D 6.25×10
18
electrons
The no. of extra electrons are on this object 1.875 x 10^13 electrons. The correct answer is A.
To determine the number of extra electrons on an object with a charge of -3 micro-Coulombs, we need to use the elementary charge (e) as a conversion factor.
The elementary charge represents the charge of a single electron and is approximately equal to 1.6 x 10^-19 Coulombs.
Given:
Charge of the object = -3 micro-Coulombs = -3 x 10^-6 Coulombs
To calculate the number of extra electrons, we divide the total charge by the elementary charge:
Number of extra electrons = (Charge of the object) / (Elementary charge)
Number of extra electrons = (-3 x 10^-6 C) / (1.6 x 10^-19 C)
Number of extra electrons ≈ -1.875 x 10^13 electrons
Since the charge of the object is negative, it indicates an excess of electrons. However, the magnitude of the charge is what determines the number of extra electrons, so we take the absolute value.
Therefore, the correct answer is A) 1.875 x 10^13 electrons.
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Short wavelength light has: A. High frequency and low energy B. High frequency and high energy C. Low frequency and low energy D. Low frequency and high energy E. None of the above 7. Which of the following statements describes the relationship between optical density of a material, refractive index of a material and the speed of light in a material? A. Optical density and refractive index are inversely proportional to each other and directly proportional to the speed of light in a material. B. Optical density, refractive index and the speed of light in a material are directly proportional to each other. C. Optical density is inversely proportional to the refractive index of a material and directly the speed of light in a material. D. Optical density and refractive index are directly proportional to each other
Optical density and refractive index are inversely proportional to each other and directly proportional to the speed of light in a material is the relationship between optical density of a material, refractive index of a material, and the speed of light in a material.
Short wavelength light has a High frequency and high energy.
This statement is true.
Optical density of a material, refractive index of a material, and the speed of light in a material are related to each other in the following manner.
Optical density is a measure of a material's ability to refract or bend light as it travels through it.
When light enters a material with a high refractive index, it bends more than when it enters a material with a lower refractive index.
The speed of light in a material is inversely related to its refractive index. When light enters a denser material, it slows down and bends.
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Construct the circuit.
•Setup a circuit with one battery, two resistors, 3 ammeters (one after each resistor and one after the battery), and wires, all in series.
The circuit consists of a battery, two resistors, and three ammeters connected in series.
To set up the circuit, first, connect the positive terminal of the battery to one end of the first resistor using a wire. Then, connect the other end of the resistor to the second resistor, creating a series connection.
Next, connect the remaining end of the second resistor to the positive terminal of the first ammeter. From the first ammeter, connect the negative terminal to the positive terminal of the second ammeter, and similarly, connect the negative terminal of the second ammeter to the positive terminal of the third ammeter.
connect the negative terminal of the third ammeter to the negative terminal of the battery, completing the series circuit.
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Larger worlds are geologically active longer than smaller worlds because larger worlds: have more volatile elements and higher gravity. have more angular momentum and are more primitive. have more out
The larger planets are geologically active longer than smaller planets because they have more angular momentum and are more primitive.
Larger worlds are geologically active longer than smaller worlds because they have more angular momentum.
Therefore, the correct option is "have more angular momentum and are more primitive.
Angular momentum is defined as the quantity of motion that an object possesses by virtue of its rotation or revolution. '
A planet has angular momentum since it is spinning around an axis while orbiting the sun.
If we compare a small world with a large world, the larger world has more mass.
As a result, it has a larger gravitational attraction.
Because the larger planet has more mass, it can retain more heat for a more extended period.
The larger planet's molten core takes longer to cool, and this molten core is responsible for generating the magnetic field, which is necessary for shielding the planet from radiation from space and solar winds.
In this way, the larger planets are geologically active longer than smaller planets because they have more angular momentum and are more primitive.
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A charge q
1
=3.80×10
−5
C is placed in an electric field of magnitude 5600 N/C. What is the electrostatic force that is exerted on charge q
1
? Force = Charge q
1
is replaced with a different charge (q
2
) at the same location (with the same electric fielc present). Charge q
2
experiences an electrostatic force of 0.30 N. What is the magnitude of charge q
2
? q
2
=×10
−5
C (Note the 10
−5
factor!)
The electrostatic force exerted on charge q1 is 0.2128 N, and the magnitude of charge q2 is 5.36 × 10^-5 C. The electrostatic force exerted on charge q1 can be calculated using the equation:
Force = q1 * E
where q1 is the charge and E is the electric field.
q1 = 3.80 × 10^-5 C
E = 5600 N/C
Plugging in the values into the equation:
Force = (3.80 × 10^-5 C) * (5600 N/C)
Force = 0.2128 N
To find the magnitude of charge q2, we can rearrange the equation:
Force = q2 * E
Force = 0.30 N
E = 5600 N/C
Plugging in the values into the equation:
0.30 N = q2 * 5600 N/C
Solving for q2:
q2 = (0.30 N) / (5600 N/C)
q2 = 5.36 × 10^-5 C
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The plane was going at 150 mph towards 270 degrees. I get into a wind zone and its speed combines with the wind speed and now it's going 250 mph towards 323.13 degrees. Which flow slows down the speed of the wind?
Answer: ____mi/h to ____degrees
When the plane is in the wind zone, the speed of the wind slows down the flow of the plane. As the wind is directly opposite to the plane, the wind resistance increases, and the speed of the plane decreases.
The original speed of the plane is 150 mph towards 270 degrees. After entering the wind zone, its speed combined with the wind speed, and it moved towards 323.
13 degrees at 250 mph. It is known that the wind resistance slows down the speed of the plane. So, the unknown answer would be less than 250 mph towards some unknown degree.
It cannot be determined without additional information. The question can't be fully answered, but the flow that slows down the speed of the plane is the wind.
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What is the slrengit of the electrie feed 3.0 cm from a amali plastic bead that has been charged to 7.4nC ? Express your answer to two significant figures and include the appropriate units. Part B What it me direction of the electric fiold 3.0 cm from a small plastic besd that has been charged to −7.AnC ?
To find the strength of the electric field at a distance of 3.0 cm from the charged plastic bead, we can use the equation:
Electric Field (E) = Charge (Q) / (4πε₀r²)
where E is the electric field, Q is the charge, ε₀ is the permittivity of free space (ε₀ ≈ 8.85 × 10^(-12) C²/(N·m²)), and r is the distance from the charge.
Given:
Charge (Q) = 7.4 nC = 7.4 × 10^(-9) C
Distance (r) = 3.0 cm = 3.0 × 10^(-2) m
Substituting these values into the equation, we get:
E = (7.4 × 10^(-9) C) / (4π(8.85 × 10^(-12) C²/(N·m²))(3.0 × 10^(-2) m)²)
Calculating this expression gives us:
E ≈ 7.89 × 10^7 N/C
Therefore, the strength of the electric field at a distance of 3.0 cm from the charged plastic bead is approximately 7.89 × 10^7 N/C.
Now, let's consider the direction of the electric field at the same distance but with a charge of -7.0 nC.
Given:
Charge (Q) = -7.0 nC = -7.0 × 10^(-9) C
Since the sign of the charge is negative, the direction of the electric field will be opposite to what it was in the previous case. Therefore, the direction of the electric field at 3.0 cm from the small plastic bead charged to -7.0 nC is in the opposite direction of the previous case.
Hence, the direction of the electric field is toward the charged plastic bead.
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A student in an undergraduate physics lab is studying Archimede's principle of bouyancy. The student is given a brass cylinder and, using a triple beam balance, finds the mass to be 1.25 kg. The density of this particular alloy of brass is 8.56 g/cm^3 . The student ties a massless string to one end of the cylinder and submerges it into a tank of water where there is an apparent reduction in the weight of the cylinder. With this information, calculate the volume, V, of the cylinder and the tension, T, in the string when it is submerged in the tank of water. The density of water is 1.00 g/cm3 , and the acceleration due to gravity is g=9.81 m/s ^2 V= cm^3 T= _____
The volume of the cylinder is 146 cm³ and the tension in the string when it is submerged in the tank of water is 10.83 N.
Given data:
Mass of the cylinder = 1.25 kg
Density of the alloy of brass = 8.56 g/cm³
Density of water = 1.00 g/cm³
The volume of the cylinder can be calculated using the formula:
Density = Mass / Volume
So, the formula for the volume of the cylinder is:
Volume = Mass / Density
= 1.25 kg / 8.56 g/cm³
= 1.460 × 10^5 mm³
= 146 cm³
The apparent reduction in the weight of the cylinder is equal to the buoyant force on the cylinder. So, the weight of the cylinder is equal to its actual weight minus the buoyant force. The weight of the cylinder is given by:
Weight = Mass × Acceleration due to gravity
= 1.25 kg × 9.81 m/s²
= 12.2625 N
To find the buoyant force on the cylinder, we need to know the volume of the cylinder.
The buoyant force is equal to the weight of the water displaced by the cylinder.
Since the density of water is 1.00 g/cm³, the weight of the water displaced by the cylinder is equal to its volume times the density of water times the acceleration due to gravity.
Thus, the buoyant force is given by:
Buoyant force = Weight of displaced water
= Volume of cylinder × Density of water × Acceleration due to gravity
= 146 cm³ × 1.00 g/cm³ × 9.81 m/s²
= 1.43226 N
Thus, the tension in the string when it is submerged in the tank of water is equal to the weight of the cylinder minus the buoyant force:
Tension = Weight - Buoyant force
= 12.2625 N - 1.43226 N
= 10.83 N
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How high above the ground does the bal go? A tennis player hits a ball at ground level, giving it an initial velocity of 20.0 m/s at 58.0
∘
above the horizontal. h= X Incorrect; Try Again: 9 attempts remaining Part c How lond does if take the ball to ceach ts maximin he jog? What is the ball's velocity at its highest point? v: Part E What is the ball's acceleration at its highest point? For how long a time is the ball in the air? t= Part G When this ball lands on the court, how far is it from the place where it was
Therefore, the ball lands 35.4 m away from the place where it was hit by the tennis player.
The question requires us to calculate the maximum height that the tennis ball attains, the time taken to reach the maximum height, velocity at its highest point, acceleration at the highest point, and the distance covered by the ball upon landing back on the ground.
We know that the initial velocity of the ball (u) = 20 m/s and the angle of projection (θ) = 58°.
The horizontal velocity[tex](ux) = u * cosθ= 20 * cos58° ≈ 10 m/s[/tex]
The vertical velocity [tex](uy) = u * sinθ= 20 * sin58° ≈ 17.32 m/s[/tex]
(a) To find the time taken to reach the maximum height:
At the maximum height, the vertical component of the velocity (uy) = 0m/s.
As per the third equation of motion, [tex]v^2 = u^2 + 2as[/tex] where v = final velocity (0 m/s)
u = initial velocity (17.32 m/s) a = acceleration (g = 9.8 m/s^2) s = displacement
Let’s calculate s. At the maximum height, the ball covers s distance.
[tex]uy^2 = u^2 + 2as0 = 17.32^2 - 2 × 9.8 × sts = (17.32^2) / (2 × 9.8) ≈ 15.20 m[/tex]
Time taken to reach the maximum height = [tex]t = (uy - 0)/g= 17.32/9.8 ≈ 1.77[/tex]s
(b) To find the maximum height:
The maximum height (h) attained by the ball can be found using the first equation of motion:
[tex]s = ut + (1/2)at^2[/tex], where s = h and u = uy = 17.32 m/s.
[tex]h = ut + (1/2)at^2= 17.32 × 1.77 - 0.5 × 9.8 × (1.77)^2= 30.1 m[/tex]
The maximum height attained by the ball is 30.1 m.
(c) To find the velocity at the highest point:
As the ball reaches the highest point, the vertical component of the velocity becomes zero.
Velocity at the highest point [tex](v) = √(ux^2 + uy^2)= √(10^2 + 0)= 10 m/s[/tex]
The velocity of the ball at the highest point is 10 m/s.
(d) To find the acceleration at the highest point:
As the ball reaches the highest point, the vertical component of the acceleration becomes equal to acceleration due to gravity. Hence, the acceleration at the highest point is equal to g = 9.8 m/s^2.
(e) To find the time for which the ball is in the air:
The ball is in the air for the time taken to reach the maximum height + time taken to land back on the ground.
[tex]t = 2 × 1.77≈ 3.54 s(f)[/tex]
To find the distance covered by the ball when it lands on the court:
Let the distance covered by the ball when it lands on the court = R (horizontal distance)
At the maximum height, the horizontal distance covered by the ball
[tex](Rx) = uxt= 10 × 1.77 ≈ 17.7 m[/tex]
When the ball lands on the court, the vertical distance covered by the ball
[tex](Ry) = h = 30.1 m[/tex]
The total horizontal distance covered by the ball is equal to the horizontal distance covered by the ball at the maximum height plus the horizontal distance covered by the ball while landing back on the court.
[tex]R = Rx + Rx= 2 × 17.7 ≈ 35.4 m[/tex]
Therefore, the ball lands 35.4 m away from the place where it was hit by the tennis player.
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A ball is thrown horizontally from the top of a 61.1 m building and lands 118.9 m from the base of the building. Ignore air resistance. (Assume the ball is thrown in the +x direction. Choose up to be +y direction from the top of the building.) a) How long (in s) is the ball in the air? b) What must have been the initial horizontal component of the velocity (in m/s )? (Indicate the direction of with the sign of your answer.) m/s c) What must have been the vertical component of the velocity (in m/s )just before the ball hits the ground? (Indicate the direction of with the sign of your answer.) m/s d) What is the final velocity vector in (m/s) of the ball just before it hits the ground? magnitude m/s direction degrees from +x direction (include the sign of the angle)
a. the time taken by the ball to reach the ground is 3.95 seconds.
b. the initial horizontal component of the velocity (in m/s) is 30.06 m/s.
c. the vertical component of the velocity just before the ball hits the ground is 38.81 m/s.
d. The ball is thrown in the +x direction, the sign of the angle is positive which is 51.88°.
Given that a ball is thrown horizontally from the top of a 61.1 m building and lands 118.9 m from the base of the building, we are to determine:
(a) How long (in s) is the ball in the air?
We can use the vertical motion equation to determine the time taken by the ball to reach the ground. We know that: y = vit + 1/2gt²
Where y = 61.1 m (height of the building)
vi = 0
t = ?
g = 9.8 m/s²
We can calculate t as follows:
61.1 = 0t + 1/2(9.8)t²
⇒ t = √(2 × 61.1/9.8)
≈ 3.95 seconds
Therefore, the time taken by the ball to reach the ground is 3.95 seconds.
(b) What must have been the initial horizontal component of the velocity (in m/s)?
Since the ball is thrown horizontally, the initial horizontal component of the velocity, vi, is constant throughout its motion. Therefore, the initial horizontal component of the velocity (in m/s) is:
vix = d/t
where d = 118.9 m and t = 3.95 s
Substituting the values, we get:
vix = 118.9/3.95
≈ 30.06 m/s
Therefore, the initial horizontal component of the velocity (in m/s) is 30.06 m/s.
(c) What must have been the vertical component of the velocity (in m/s) just before the ball hits the ground?
We can use the vertical motion equation to determine the vertical component of the velocity just before the ball hits the ground. We know that: v = u + gt
Where u = 0 m/s
t = 3.95 s
g = 9.8 m/s²
v = ?
Therefore, the vertical component of the velocity just before the ball hits the ground is:
v = gt
⇒ v = 9.8 × 3.95
≈ 38.81 m/s
(d) What is the final velocity vector in (m/s) of the ball just before it hits the ground?
The final velocity vector can be determined by combining the horizontal and vertical components of the velocity. We can use Pythagoras's theorem to find the magnitude of the final velocity. We know that:
v = √(v²x + v²y)
Where vx = 30.06 m/s
vy = 38.81 m/s
Substituting the values, we get:
v = √(30.06² + 38.81²)
≈ 49.19 m/s
Therefore, the magnitude of the final velocity is 49.19 m/s.
To find the direction, we can use the formula:
tanθ = vy/vx
Substituting the values, we get:
tanθ = 38.81/30.06
⇒ θ ≈ 51.88°
Therefore, the angle that the final velocity vector makes with the +x direction is 51.88°. Since the ball is thrown in the +x direction, the sign of the angle is positive.
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write an equation for the relationship between time and distance
The relationship between distance, speed, and time is given by the formula is Distance = Speed x Time
We can obtain an equation for the relationship between time and distance by modifying the above formula.
We can rearrange the formula by dividing both sides by speed, which yields the following equation:
Time = Distance / Speed
This equation shows that time is directly proportional to distance traveled, and inversely proportional to the speed of travel. If the speed is kept constant, then the time taken to travel a certain distance is directly proportional to the distance traveled.
For example, if a car travels at a speed of 50 km/hour, then the time taken to travel a distance of 150 km is given by:
Time = 150 / 50 = 3 hours.
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A small water pump for an aquarium delivers 14.0Lpm (litre per minute) of water at a net head of 1.5 m at its best efficiency point (BEP) with a motor that spins at 1200rpm. You are required to modify the system by attaching a different motor, for which the rpm is 1800rpm. If the pumps operate at homologous points (namely, at the BEP) for both cases,
6.1. Predict the volume flow rate and net head of the modified pump (4)
6.2. Calculate and compare the pump specific speed of both pumps (5)
The specific speed of the original pump (Ns1) is approximately 246.48, and the specific speed of the modified pump (Ns2) is approximately 245.52.
To predict the volume flow rate and net head of the modified pump, we can use the concept of pump affinity laws. These laws state that the flow rate and head developed by a pump are proportional to the speed of the pump.
Let's denote the subscript "1" for the original pump and "2" for the modified pump.
According to the pump affinity laws:
Flow Rate (Q2) = (N2 / N1) * (Q1)
Head (H2) = (N2 / N1)^2 * (H1)
where:
Q1 is the original flow rate (14.0 Lpm)
H1 is the original net head (1.5 m)
N1 is the original motor speed (1200 rpm)
N2 is the new motor speed (1800 rpm)
Let's calculate the values:
6.1. Predict the volume flow rate and net head of the modified pump:
Using the above formulas, we can calculate:
Q2 = (1800 rpm / 1200 rpm) * (14.0 Lpm)
= 21.0 Lpm
H2 = (1800 rpm / 1200 rpm)^2 * (1.5 m)
= 3.375 m
Therefore, the predicted volume flow rate of the modified pump is 21.0 Lpm, and the predicted net head is 3.375 m.
6.2. Calculate and compare the pump specific speed of both pumps:
The pump specific speed (Ns) is a dimensionless parameter that characterizes the geometry and performance of a pump. It is calculated using the following formula:
Ns = (N * Q^0.5) / H^0.75
where:
N is the pump speed (rpm)
Q is the flow rate (Lpm)
H is the head (m)
For the original pump:
Ns1 = (1200 rpm * (14.0 Lpm)^0.5) / (1.5 m)^0.75
For the modified pump:
Ns2 = (1800 rpm * (21.0 Lpm)^0.5) / (3.375 m)^0.75
Calculating the specific speeds:
Ns1 ≈ 246.48
Ns2 ≈ 245.52
Therefore, the specific speed of the original pump (Ns1) is approximately 246.48, and the specific speed of the modified pump (Ns2) is approximately 245.52.
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submitted in WebAssign.) velocity at end of engine burn time v=m/s engine burn time (b) Determine the maximum altitude of the rocket (in m ) and the total time (in s ) for the rocket to reach this altitude from ground level.
maximum altitude
time to reach maximum altitude
y=
t=
m
;
velocity just before ground impact v= total time of flight
Therefore, the maximum altitude of the rocket is given by(h) = (v² + 2 × m²) / (2 × m)and the total time for the rocket to reach the maximum altitude from the ground level is given by(t) = s.
Given that rocket is launched from ground level, the velocity of the rocket just before ground impact, the velocity at the end of engine burn time, and the engine burn time are given as:
v = 0 (velocity just before ground impact)
v1 = m/s (velocity at end of engine burn time)
b = s (engine burn time)
The acceleration of the rocket is given by the following relation:
a = (v1 - v) / b
Substituting the given values of v1, v and b in the above equation,
a = (v1 - v) / b
= (m/s - 0) / s
= m/s²
The maximum altitude of the rocket can be determined using the following relation:
v² - u² = 2as
Here,u = v1 = m/s (initial velocity)and a = m/s² (acceleration)
Let the maximum altitude be h.
Then,s = h - u
= h - m/s
Therefore, v² - u²
= 2asv² - (m/s)²
= 2 × m/s² × (h - m/s)v² - m²/s²
= 2 × m/s² × h - 2 × m²/s²v² - m²/s² + 2 × m²/s²
= 2 × m/s² × h
Therefore, h = (v² - m²/s² + 2 × m²/s²) / (2 × m/s²)
h = (v² + 2 × m²/s²) / (2 × m/s²)
h = (v² + 2 × m²) / (2 × m)
The total time for the rocket to reach the maximum altitude can be determined as follows:
t = (v - u) / a
Substituting the given values of v, u, and a in the above equation,
t = (m/s - m/s) / (m/s²)
= s
Therefore, the maximum altitude of the rocket is given by(h) = (v² + 2 × m²) / (2 × m)and the total time for the rocket to reach the maximum altitude from the ground level is given by(t) = s.
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