Answer: [tex]1.000\times 10^2m^2[/tex]
Explanation:
Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
The rule apply for the multiplication and division is :
The least precise number present determines the number of significant figures in the answer.
The rule apply for the addition and subtraction is :
The least precise number present after the decimal point determines the number of significant figures in the answer.
Area of square collector 1 =
Area of square collector 1 =[tex]6.00m\times 6.00m=36.0m^2[/tex]
Area of square collector 2 =[tex]length\times breadth[/tex]
Area of square collector 2 =[tex]8.000m\times 8.000m=64.00m^2[/tex]
Combined area of both collectors = Area of square collector 1 + Area of square collector 2 = [tex]36.0m^2+64.00m^2=100.0m^2=1.0\times 10^2m^2[/tex]
The combined area of both collectors is [tex]1.000\times 10^2m^2[/tex]
The combined area of both solar collectors is 1.0 × 10² m².
Explanation:To find the combined area of both solar collectors, simply add the areas of the two squares. The formula for the area of a square is length multiplied by width. For the first solar collector measuring 6.00 m by 6.00 m, the area is 6.00 m x 6.00 m = 36.00 m². To find the combined area of both solar collectors, simply add the areas of the two squares. The formula for the area of a square is length multiplied by width. For the first solar collector measuring 6.00 m by 6.00 m, the area is 6.00 m x 6.00 m = 36.00 m². For the second solar collector measuring 8.000 m by 8.000 m, the area is 8.000 m x 8.000 m = 64.00 m². Adding the areas, we get 36.00 m² + 64.00 m² = 100.00 m².The correct answer, using the correct number of significant figures, is therefore 1.0 × 10² m². This answer is represented by option b.For the second solar collector measuring 8.000 m by 8.000 m, the area is 8.000 m x 8.000 m = 64.00 m². Adding the areas, we get 36.00 m² + 64.00 m² = 100.00 m².
The correct answer, using the correct number of significant figures, is therefore 1.0 × 10² m². This answer is represented by option b.
At time t = 0 s, an object is observed at x = 0 m; and its position along the x-axis follows this expression: x = –4t + t2, where the units for distance and time are meters and seconds, respectively. What is the object's displacement between t = 1.0 s and t = 3.0 s?
The displacement of the object between t = 1.0 s and t = 3.0 s, using the equation x = –4t + t², is found to be 0 m.
Explanation:The displacement of the object between t = 1.0 s and t = 3.0 s can be found by substituting these time values into the given equation x = –4t + t2 and then subtracting the earlier position from the later one.
For t = 1.0 s; x1 = -4(1) + (1)² = -3 m
For t = 3.0 s; x2 = -4(3) + (3)² = -3 m
The displacement can be calculated as Δx = x2 - x1 = -3 - (-3) = 0 m. So, the displacement of the object between t = 1.0 s and t = 3.0 s is 0 m.
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If the total mass hanging on one end of the string that creates the tension in the string is 0.300 kg , what is the mass density (mass per unit length) μ of the string?
Answer:
The mass density of the string is (0.3/L)kg/m
Explanation:
Mass density of the string = Mass/Length
Mass = 0.3kg
The length of the string is unknown so it is assumed to be L meter(s)
Therefore, mass deny of the string = 0.3kg/Lm = (0.3/L)kg/m
Two vectors A and B are added together to form a vector C. The relationship between the magnitudes of the vectors is given by A + B = C. Which one of the following statements concerning these vectors is true?
Answer:
The question is incomplete, below is the complete question
"Two vectors A and B are added together to form a vector C. The relationship between the magnitudes of the vectors is given by A + B = C. Which one of the following statements concerning these vectors is true?
a. A and B must be displacements.
b. A and B must have equal lengths.
c.A and B must point in opposite direction.
d. A and B must point in the same direction.
e.A and B must be at right angles to each other."
answer:
d. A and B must point in the same direction.
Explanation:
a.false:From vector analysis, all forms of vector with the same unit can be added, and we add vector component by component. hence this defile option (a).
b. false: in addition of vectors, length is not a criteria to consider before carrying out the operation, hence vectors of different lengths can be added
c. false: this is against the claim as vectors in opposite direction can give rise to a new vector of negative value.
d. true: this is true as vector in the same direction add up to give rise to another vector.
e.false: this is not a true assumption
The relationship A + B = C only holds for vectors A and B that are collinear. Generally, the magnitude of the resultant vector C is derived using the triangle rule of vector addition, and is not a simple sum of the magnitudes of vectors A and B.
Explanation:The relationship between the magnitudes of vectors A, B, and C is not always expressed as A + B = C. This can only be true in the special case where the vectors A and B are collinear (aligned along the same line and direction).
In general, according to the triangle rule of vectors addition, if we add two vectors A and B (not necessary to be in the same direction), the resultant vector C can be represented by the diagonal of a parallelogram which is constructed using A and B as two adjacent sides. Meaning, |C|= √ (A² + B² + 2ABcosθ), where θ is the angle between A and B.
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Suppose you display a 1 kHz, 2 V sine wave in channel A and a DC, 1V signal in channel B with sensitivities set on 1V/div. You select 'Add' so that the two signals are combined and you readjust the position so the trace is in the middle of the screen. If you switch the signal in channel B from DC to AC, what will happen? (Draw a picture if it helps.)
Answer:
You are going to see the AC functions shifted down
Explanation:
Because you had a positive DC signal first, you had to readjust the position to the middle, otherwise, the sum of AC and DC would have been shifted up; because of this, your cursor has been moved down and ones the DC disappears from channel B the signal will be shifted down.
Note: Changing from DC to AC in channel B will kill all the DC signal
You are holding one end of an elastic cord that is fastened to a wall 3.0 m away. You begin shaking the end of the cord at 3.5 Hz, creating a continuous sinusoidal wave of wavelength 1.0 m. How much time will pass until a standing wave fills the entire length of string?
Answer:
1.714 s
Explanation:
The equation of a moving wave,
v = λf.................................. Equation 1
Where v = velocity of the wave, λ = wave length of the wave, f = frequancy of the wave,
Given: λ = 1.0 m, f = 3.5 Hz.
Substitute into equation 1
v = 1.0(3.5)
v = 3.5 m/s.
From motion,
velocity = distance/time.
v = d/t
t = d/v................................. Equation 2
Where distance of the elastic chord from the wall, t = time taken to travel to the other end.
Given: d = 3.0 m, v = 3.5 m/s
Substitute into equation 2
t = 3.0/3.5
t = 0.857 s.
Note: for the wave to form a standing wave, it has to travel twice.
Thus,
Time taken for standing wave to fill the entire length of the string = 2t
= 2×0.857
1.714 s
A Ping-Pong ball is given an upward initial velocity. The force of air resistance causes the times of ascent and descent to be unequal. Which time is greater?
Answer:
Time of ascent is greater than time of descent.
Explanation:
The gravitational force always acts in the downward direction. The air drag always opposes the motion.
During ascent, the gravitational force and air drag act in opposite direction to the motion where as during descent, only air drag acts in opposite direction to the motion of the ball while gravitational force acts in the same direction. Thus, the time of ascent and descent become unequal with time of ascent being greater than time of descent.
The normal boiling point of ethanol is 78.4 oC and its heat of vaporization is 38.56 kJ/mol. What is the vapor pressure of ethanol at 15 oC?
Answer: 0.05470atm
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:
[tex]ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})[/tex]
where,
[tex]P_1[/tex] = initial pressure at = 1 atm (standard atmospheric pressure
[tex]P_2[/tex] = final pressure at [tex]15^oC[/tex] = ?
[tex]\Delta H_{vap}[/tex] = enthalpy of vaporisation = 38.56 kJ/mol = 38560 J/mol
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex]= initial temperature = [tex]78.4^oC=273+78.4=351.4K[/tex]
[tex]T_2[/tex] = final temperature =[tex]15^oC=273+15=288K[/tex]
Now put all the given values in this formula, we get
[tex]\log (\frac{P_2}{1atm})=\frac{38560}{2.303\times 8.314J/mole.K}[\frac{1}{351.4K}-\frac{1}{288K}][/tex]
[tex]\log (\frac{P_2}{1atm})=-1.262[/tex]
[tex]P_2=0.05470atm[/tex]
Thus the vapor pressure of ethanol at [tex]15^0C[/tex] is 0.05470atm
Sirius, in Canis Major (the large hunting dog of Orion), is the brightest star in the winter night sky. It has a parallax angle of 0.38 seconds of arc. Orion's other hunting dog, Canis Minor, has as its brightest star Procyon, with a parallax angle of 0.29 seconds of arc. Vega, in Lyra the Lyre (Harp), is the brightest star in the summer night sky. It has a parallax angle of 0.13 seconds of arc. Which of these stars is farthest away? A.Procyon b.Vega c.It cannot be determined from this information alone d.Sirius
Answer:
Option b
b. Vega
Explanation:
The angle due to the change in position of a nearby object against the background stars it is known as parallax (see the image below).
The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth (as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km), is defined as 1 astronomical unit (AU).
[tex]\tan{p} = \frac{1AU}{d}[/tex]
Where d is the distance to the star.
[tex]p('') = \frac{1}{d}[/tex] (1)
Equation (1) can be rewritten in terms of d:
[tex]d(pc) = \frac{1}{p('')}[/tex] (2)
Equation (2) represents the distance in a unit known as parsec (pc).
For the case of Sirius ([tex]p('') = 0.38[/tex]):
[tex]d(pc) = \frac{1}{0.38}[/tex]
[tex]d(pc) = 2.63[/tex]
Hence, the object is 2.63 parsecs away from Earth.
For the case of Procyon ([tex]p('') = 0.29[/tex]):
[tex]d(pc) = \frac{1}{0.29}[/tex]
[tex]d(pc) = 3.44[/tex]
Hence, the object is 3.44 parsecs away from Earth.
For the case of Vega ([tex]p('') = 0.13[/tex]):
[tex]d(pc) = \frac{1}{0.13}[/tex]
[tex]d(pc) = 7.69[/tex]
Hence, the object is 7.69 parsecs away from Earth.
Therefore, Vega is the star farther away.
Summary:
Notice how a small parallax angle means that the object is farther away.
Last year a baseball player made 63 errors. This year he made 42. What percent decrease was there in the number of errors committed by the baseball player?
The percent decrease in the number of errors committed by the baseball player is 33.33%, calculated by finding the difference in errors between the two years and dividing by the number of errors in the first year.
Explanation:To calculate the percent decrease in the number of errors committed by the baseball player, we need to first find the difference in the number of errors between the two years and then divide that by the number of errors made in the first year. Finally, we'll multiply the result by 100 to get the percentage.
Here's the step-by-step calculation:
Find the difference in errors: 63 - 42 = 21 errors decreased.Calculate the percent: (21 ÷ 63) × 100 = 33.33%.Therefore, the baseball player had a 33.33% decrease in the number of errors committed compared to the previous year.
If stellar parallax can be measured to a precision of about 0.01 arcsec using telescopes on the Earth to observe stars, to what distance does this correspond in space?
Answer:
It corresponds to a distance of 100 parsecs away from Earth.
Explanation:
The angle due to the change in position of a nearby object against the background stars it is known as parallax.
It is defined in a analytic way as it follows:
[tex]\tan{p} = \frac{1AU}{d}[/tex]
Where d is the distance to the star.
[tex]p('') = \frac{1}{d}[/tex] (1)
Equation (1) can be rewritten in terms of d:
[tex]d(pc) = \frac{1}{p('')}[/tex] (2)
Equation (2) represents the distance in a unit known as parsec (pc).
The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth (as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km), is defined as 1 astronomical unit (1AU).
For the case of ([tex]p('') = 0.01[/tex]):
[tex]d(pc) = \frac{1}{0.01}[/tex]
[tex]d(pc) = 100[/tex]
Hence, it corresponds to a distance of 100 parsecs away from Earth.
Summary:
Notice how a small parallax angle means that the object is farther away.
Key terms:
Parsec: Parallax of arc second
A stellar parallax of 0.01 arcsecond measured using telescopes on Earth corresponds to a distance of 100 parsecs, or about 326 light-years.
Explanation:When measuring stellar parallax from Earth with a precision of about 0.01 arcsecond, we can calculate the corresponding distance in space using the concept that a parallax of 1 arcsecond is equivalent to a distance of 1 parsec. To find the distance for a parallax of 0.01 arcsecond, we use the formula: distance (parsecs) = 1 / parallax angle (arcseconds).
Thus, a parallax of 0.01 arcsecond corresponds to:
distance = 1 / 0.01 arcseconds = 100 parsecs. This means the star is 100 parsecs away, which is approximately 326 light-years, since 1 parsec equals 3.26 light-years.
Additional Context
Parallax measurements are vital for determining distances to the nearest stars in astronomy, and with ground-based telescopes, we can measure out to about 100 parsecs. However, due to atmospheric turbulence, accuracy diminishes beyond 40 parsecs, and ground-based parallax measurement becomes unfeasible beyond 100 parsecs. Space missions like Hipparcos have improved our ability to measure stellar parallax, pushing accurate distance measurements out beyond 500 parsecs.
A record turntable rotates through 5.0 rad in 2.8 s as it is accelerated uniformly from rest. What is the angular velocity at the end of that time?
Answer:
[tex]\omega_f = 3.584\ rad/s[/tex]
Explanation:
given,
turntable rotate to, θ = 5 rad
time, t = 2.8 s
initial angular speed = 0 rad/s
final angular speed = ?
now, using equation of rotational motion
[tex]\theta = \omega_i t + \dfrac{1}{2}\alpha t^2[/tex]
[tex]5 = 0+ \dfrac{1}{2}\alpha\times 2.8^2[/tex]
[tex]\alpha= \dfrac{10}{2.8^2}[/tex]
α = 1.28 rad/s²
now, calculation of angular velocity
[tex]\omega_f = \omega_i + \alpha t[/tex]
[tex]\omega_f =0 +1.28\times 2.8[/tex]
[tex]\omega_f = 3.584\ rad/s[/tex]
hence, the angular velocity at the end is equal to 3.584 rad/s
The angular velocity at the end of that time 10rad/s
In order to get the angular velocity [tex]\omega[/tex], we will use the equation of motion expressed as [tex]\omega = \omega_0 + \alpha t[/tex]
[tex]\alpha[/tex] is the angular acceleration
t is the time taken
[tex]\omega_0[/tex] is the initial angular velocity
Get the angular acceleration [tex]\alpha[/tex]
[tex]\theta = \omega_0 t+ \frac{1}{2} gt^2[/tex]
[tex]5=0+\frac{1}{2} \alpha (2.8)\\2\times5 = 2.8 \alpha \\10 = 2.8 \alpha\\\alpha = \frac{10}{2.8}\\\alpha = 3.57rad/s^2[/tex]
Get the angular velocity [tex]\omega[/tex]
[tex]\omega = \omega_0 + 3.57(2.8)\\\omega = 0+10\\\omega =10rad/s\\[/tex]
Hence the angular velocity at the end of that time 10rad/s
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In preparing to shoot an arrow, an archer pulls a bow string back 0.420 m by exerting a force that increases uniformly from 0 to 230 N. What is the equivalent spring constant of the bow? Answer in units of N/m.
Answer:
spring constant will be 547.619 N /m
Explanation:
We have given that force exerted if uniformly from 0 to 230 N
So exerted force F = 230 N
String is stretched by 0.420 m due to applying force
So x = 0.420 m
We have to find the spring constant in N/m
We know that stretched force is given by
[tex]F=Kx[/tex] , here K is spring constant and x is stretched length
So [tex]230=K\times 0.420[/tex]
[tex]K=\frac{230}{0.420}=547.619N/m[/tex]
So spring constant will be 547.619 N /m
The equivalent spring constant of the bow is 547.62 N/m.
Explanation:To determine the equivalent spring constant of the bow, we can use Hooke's law which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Given that the bow string is pulled back by 0.420 m and the force increases uniformly from 0 to 230 N, we can use the formula for spring force: F = kx, where F is the force, k is the spring constant, and x is the displacement.
Plugging in the given values, we have:
230 N = k * 0.420 m
Solving for k, we divide both sides of the equation by 0.420 m:
k = 230 N / 0.420 m = 547.62 N/m
Therefore, the equivalent spring constant of the bow is 547.62 N/m.
A 0.145-kg baseball pitched horizontally at 27.0 m/s strikes a bat and pops straight up to a height of 31.5 m. If the contact time between bat and ball is 2.5 ms, calculate the average force between the ball and bat during contact.
Answer:
The average force between the ball and bat during contact is 3006.72 N.
Explanation:
Given that,
Mass of the baseball, m = 0.145 kg
Initial speed of the ball, u = 27 m/s
Time of contact between bat and ball is 2.5 ms, [tex]t=2.5\times 10^{-3}\ s[/tex]
After striking the bat, it pops straight up to a height of 31.5 m. The final velocity of the ball is given by using third equation of motion as :
[tex]v^2-u^2=2as[/tex]
a = -g
And initially, u = 0
[tex]v=\sqrt{2gs}[/tex]
[tex]v=\sqrt{2\times 9.8\times 31.5}[/tex]
v = -24.84 m/s (as it pops straight up)
Let F is the average force between the ball and bat during contact. It is given by :
[tex]F=\dfrac{m(v-u)}{t}[/tex]
[tex]F=\dfrac{0.145\times (24.84 -(-27))}{2.5\times 10^{-3}}[/tex]
F = 3006.72 N
So, the average force between the ball and bat during contact is 3006.72 N. Hence, this is the required solution.
the repulsive force between two protons has a magnitude of 2.00 N. What is the distance between them?
A. 1.07 x 10^-14 m
B. 1.28 x 10^-38 m
C. 7.19 x 10^-10 m
D. 2.68 x 10^-4 m
Answer:
The answer to your question is letter A. r = 1.07 x 10⁻¹⁴ m
Explanation:
Data
F = 2 N
d = ?
q = 1.6 x 10 ⁻¹⁹ C
k = 8.987 Nm²/C²
Formula
[tex]F = K\frac{q1q2}{r^{2}}[/tex]
Solve for r
[tex]r = \sqrt{\frac{kq1q2}{F}}[/tex]
Substitution
[tex]r = \sqrt{\frac{8.987 x 10^{9}x1.6 x 10^{-19} x 1.6 x 10x^{-19}}{2}}[/tex]
Simplification
r = [tex]\sqrt{\frac{2.3 x 10^{-28}}{2}}[/tex]
r = [tex]\sqrt{1.15 x 10^{-24}}[/tex]
Result
r = 1.07 x 10⁻¹⁴ m
Answer:
A
Explanation:
Using F = kq1q2/r^2
r^2 = kq1q2/F
r = square root of (kq1q2/F)
r = square root of(8.99 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19/2)
r = square root of(1.15072 x 10^-28)
r = 1. 07 x 10^-14m
Fill in the blanks.(1) ______________________ consists of electric and magnetic disturbances, or waves, that travel through space. Human eyes see one form of this energy, called (2) ______________________. All forms of electromagnetic radiation, including X-rays and radio waves, make up the (3) ______________________. Each type of radiation can be classified in two ways. (4) ______________________ measures the distance between the peaks on a wave and (5) ______________________ is the number of waves that occurs each second. Scientists study radiation with (6) ______________________ , which collect and focus light. The (7) ______________________ the opening that gathers light in a telescope, the more light that can be collected. A(n) (8) ______________________ uses lenses to bring light to a focus, and a(n) (9) ______________________ uses mirrors to do the same thing. The process of linking several telescopes together so that they can act as one is called (10) ______________________.
Answer:
1. Electromagnetic Radiation
2. Visible light
3. Electromagnetic Spectrum
4. Wavelength
5. Frequency
6. Telescope
7. Bigger
8. Refractor telescope
9. Reflector telescope
10. Astronomical Interferometer
Explanation:
The light that we see and can not see is actually Electromagnetic radiation. It consists of various rays like X-rays, gamma, radio, IR etc. Out of the complete EM spectrum we can only see 7% i.e. the visible part of the spectrum. These waves are categorized basis their wavelength and frequency. Telescopes are used to study the EM spectrum. Bigger the opening of the telescope more light can be gathered and the image will be brighter and better. In refractor telescopes, convex lenses are used to collect the light while Reflector telescopes use concave mirrors are used to gather the light. When multiple telescopes, mirror segments are used in sync to work as one, the arrangement is called as Astronomical Interferometer.
The question is about the properties and tools used to study electromagnetic radiation. It introduces terms like electromagnetic spectrum, wavelength, frequency, telescopes, refracting telescope, reflecting telescope, and interferometry.
Explanation:(1) Electromagnetic radiation consists of electric and magnetic disturbances, or waves, that travel through space. Human eyes see one form of this energy, called (2) visible light. All forms of electromagnetic radiation, including X-rays and radio waves, make up the (3) electromagnetic spectrum. Each type of radiation can be classified in two ways. (4) Wavelength measures the distance between the peaks on a wave and (5) frequency is the number of waves that occur each second. Scientists study radiation with (6) telescopes, which collect and focus light. The (7) larger the opening that gathers light in a telescope, the more light that can be collected. A(n) (8) refracting telescope uses lenses to bring light to a focus, and a(n) (9) reflecting telescope uses mirrors to do the same thing. The process of linking several telescopes together so that they can act as one is called (10) interferometry.
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The space shuttle edeavor is launched to altitude of 500,000 m above the surface of earth. The shuttle travels at an average rate of 700 m/s.How long will it take for endeavor to reach its orbit
Answer:
714s
Explanation:
t=H/v=500000m/700m/s=714s
The space shuttle edeavor is launched to altitude of 500,000 m above the surface of earth. The shuttle travels at an average rate of 700 m/s. Time taken by space shuttle is 714.28 m/s to reach its orbit.
Given the data to find the time taken by space shuttle,
Distance covered by the space shuttle = 500,000 meter
Speed of space shuttle = 700 m/s
What is the time taken through the distance and speed?The distance covered by a particular body is equal to the sum of total path covered. It is equal to the total path traveled by an object during its entire journey.This quantity is always positive. It can't be 0 or a negative number.Distance is a scalar quantity.Mathematically, it can be calculated as follows :
distance = speed × time
As we have to find the time taken, the formula will be altered.
Computation:
Time taken = Distance / Speed
Time taken by space shuttle = Distance cover by space shuttle / Speed of space shuttle
Time taken by space shuttle = 500,000 / 700
Time taken by space shuttle = 714.28 m/s
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A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 7.0 s . Consider the merry-go-round to be a uniform disk of radius 6.0 m and mass 25,000 kg. Suppose that it is supported by bearings that produce negligible friction torque.
Answer
given,
initial speed of merry-go-round = 0 rad/s
final speed of merry-go-round = 1.5 rad/s
time = 7 s
Radius of the disk = 6 m
Mass of the merry-go-round = 25000 Kg
Moment of inertia of the disk
[tex]I = \dfrac{1}{2}MR^2[/tex]
[tex]I = \dfrac{1}{2}\times 25000\times 6^2[/tex]
I = 450000 kg.m²
angular acceleration
[tex]\alpha = \dfrac{\omega_f-\omega_0}{t}[/tex]
[tex]\alpha = \dfrac{1.5-0}{7}[/tex]
[tex]\alpha =0.214\ rad/s^2[/tex]
we know,
[tex]\tau= I \alpha[/tex]
[tex]\tau= 450000\times 0.214[/tex]
[tex]\tau=96300\ N.m[/tex]
The required value of frictional torque produced by the bearings is 96300 N-m.
Given data:
The final angular speed of merry-go round is, [tex]\omega_{2}=1.5 \;\rm rad/s[/tex].
The time interval is, t = 7.0 s.
The radius of merry-go round is, r = 6.0 m.
The mass of merry-go round is, m = 25,000 kg.
The frictional torque of an object undergoing rotational motion is the product of moment of inertia and angular acceleration. So,
[tex]T_{f} = I \times \alpha[/tex] ....................................................(1)
Here, I is the moment of inertia of merry-go round and its value is,
[tex]I =\dfrac{mr^{2}}{2}\\\\I =\dfrac{25000 \times (6.0)^{2}}{2}\\\\I =450000 \;\rm kg.m^{2}[/tex]
And angular acceleration is,
[tex]\alpha = \dfrac{\omega_{2}-\omega_{1}}{t}\\\\\alpha = \dfrac{1.5-0}{7}\\\\\alpha = 0.214 \;\rm rad/s^{2}[/tex]
Then the frictional torque is calculated from equation (1) as,
[tex]T_{f} = 450000 \times 0.214\\\\T_{f}=96300 \;\rm N.m[/tex]
Thus, we can conclude that the required value of frictional torque produced by the bearings is 96300 N-m.
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How does a rubber rod become negatively charged through friction?
a. It touches a negatively charged object, and protons move off of the rod.
b. It touches a positively charged object, and electrons move onto the rod.
c. It is rubbed with another object, and electrons move onto the rod.
d. It is rubbed with another object, and protons move off of the rod.
Answer:
c. It is rubbed with another object, and electrons move onto the rod.
Explanation:
A rubber rod is negatively charged through friction when gains electrons, this occurs when electrons are transferred from another material by simple contact between them. When these two materials are rubbed, the difference in electronic affinity causes one of the materials to be positively charged and the other negatively charged.
Despite the territorial gains west of the Appalachian Mountains, what decree by King George III forbid colonists from moving into this new territory?
Answer:
Royal proclamation was issued in 1763 by the king George III which forbade the american colonist from settling west of Appalachia. This proclamation was issued to avoid the conflicts between the native Americans and the new settlers.
Explanation:
A 20-kg fish swimming at 1 m/s swallows a 5-kg fish swimming toward it at a velocity that bring both to a halt. How fast was the smaller fish moving prior to being eaten?
Answer:
Swallowed fish speed=4 m/s
Explanation:
Given
m₁(fish mass)=20 kg
v₁(fish speed)=1 m/s
m₂(Swallowed fish mass)=5 kg
To find
v₂(Swallowed fish speed)
Solution
From law of conservation of momentum
[tex]m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v\\ (20kg)(1 m/s)+(5 kg)v_{2}=(20kg+5kg)0m/s\\20kg.m/s+5kgv_{2}=0\\v_{2}=(20kg.m/s)/5kg\\v_{2}=4m/s[/tex]
As I neglected the negative sign because it only shows that swallowed fish is moving in opposite direction
The smaller fish was moving at a velocity of 4 m/s towards the larger fish before being eaten
The question involves the principle of conservation of momentum to solve how fast a 5-kg fish was moving prior to being eaten by a 20-kg fish to bring both to a halt. In the scenario, the 20-kg fish is moving at 1 m/s towards the smaller fish. To find the speed of the smaller fish before being swallowed, we assume no external forces are acting on the system.
Momentum is calculated using the formula p = mv, where p is momentum, m is mass, and v is velocity. Initially, the momentum of the 20-kg fish is (20 kg)(1 m/s) = 20 kg·m/s, and let's say the smaller fish has a velocity v. Since they come to a halt, the final momentum is 0. By setting the total initial momentum equal to the final momentum, we have:
(20 kg)(1 m/s) + (5 kg)(-v) = 0
⇒ 20 - 5v = 0
⇒ v = 4 m/s
The amplitude of oscillation is the maximum distance between the oscillating weight and the equilibrium position. Determine the frequency of oscillation for several different amplitudes by pulling the weight down different amounts while still keeping the simulation within the top and botom boundaries. How does the frequency depend on the amplitude of oscillation
Answer:
In no case does the amplitude appear, so we would not have to change the period of the system when we change to the amplitude
Explanation:
In the harmonic movement when the amplitude of oscillation increases, body speed also increases, so the frequency remains constant.
When we solve the different case of harmonic movement
Simple pendulum 2pi f = Ra l / g
Spring mass 2pi f = RA k / m
2pi torsion pendulum f = RA I / k
In no case does the amplitude appear, so we would not have to change the period of the system when we change to the amplitude
Answer:
The frequency does not depend on the amplitude for any (ideal) mechanical or electromagnetic waves.
In electromagnetism we have that the relation is:
Velocity = wavelenght*frequency.
So the amplitude of the wave does not have any effect here.
For a mechanical system like an harmonic oscillator (that can be used to describe almost any oscillating system), we have that the frequency is:
f = (1/2*pi)*√(k/m)
Where m is the mass and k is the constant of the spring, again, you can see that the frequency only depends on the physical properties of the system, and no in how much you displace it from the equilibrium position.
This happens because as more you displace the mass from the equilibrium position, more will be the force acting on the mass, so while the "path" that the mass has to travel is bigger, the mas moves faster, so the frequency remains unaffected.
A tank of oil has a mass of 50 slugs.
A) Determine its weight in pounds and in Newtons at the earth’s surface.
B) What would be its mass (in slugs) and its weight (in pounds) if located on the moon’s surface where gravitational attraction is 1/6th that at the earth’s surface?
The weight of an oil tank with the mass of 50 slugs on the Earth's surface would be 1,610 pounds or 7164.2732 Newtons. On the moon, the weight would be 268.5 pounds, but its mass would remain the same (50 slugs) because mass is independent of location.
Explanation:To begin, the weight can be determined using the formula Weight = mass x gravitational force. On the Earth's surface, the gravitational force is approximately 32.2 ft/s2.
A) The weight of the oil tank on Earth can be calculated by multiplying its mass (50 slugs) by the gravitational force. Therefore, it would be 50 slugs * 32.2 ft/s2 = 1,610 pounds. To convert to Newtons, since 1 pound = 4.44822 Newtons, the weight is 1,610 pounds * 4.44822 = 7164.2732 Newtons.
B) On the moon, the gravitational force is a sixth of that on Earth, i.e., 32.2 ft/s2 / 6 = 5.37 ft/s2. The mass of the oil tank would remain the same, 50 slugs, because mass doesn't change with location. However, its weight on the moon would be 50 slugs * 5.37 ft/s2 = 268.5 pounds.
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To find the weight of 50 slugs on Earth, multiply by 32.17 ft/s^2 to get 1608.5 pounds and convert to 7157.48 Newtons. On the Moon, the weight is 1/6th, resulting in 268.083 pounds, with the mass still being 50 slugs.
To determine the weight of a tank of oil with a mass of 50 slugs on Earth's surface, we use the U.S. Customary system. The acceleration due to gravity (g) in this system is 32.17 feet per second per second. The formula to convert mass in slugs to weight in pounds is: Weight (lb) = mass (slugs) \\u00d7 acceleration due to gravity (ft/s^2).
A) On Earth, the weight in pounds is:
Weight (lb) = 50 slugs \\u00d7 32.17 ft/s^2 = 1608.5 pounds.
To convert this to Newtons, we use the conversion factor 1 pound = 4.44822 Newtons, resulting in:
Weight (N) = 1608.5 lb \\u00d7 4.44822 N/lb = 7157.48 Newtons.
B) On the Moon, where gravity is 1/6th that of Earth, the mass remains the same. Therefore, the weight in pounds is:
Weight (lb) = 50 slugs \\u00d7 (32.17 ft/s^2 / 6) = 268.083 pounds.
The mass in slugs remains the same at 50 slugs as mass is independent of gravity.
Write down the most general forms for the Work-Energy Theorem, the Conservation of Energy, and the relation between Forces and Potential Energies.
Answer:
Work energy theorem: [tex]W=\Delta KE=\frac{1}{2} m.v_f^2-\frac{1}{2} m.v_i^2[/tex]
Conservation of energy:
This theorem states that energy can neither be created nor be destroyed it can only get converted from one form to another.
[tex]PE=m.g.h[/tex]
Explanation:
Work energy theorem:
It states that the total work done by the sum of all the forces acting on a particle is equal to the change in kinetic energy of the particle.
Mathematically:
[tex]W=\Delta KE=\frac{1}{2} m.v_f^2-\frac{1}{2} m.v_i^2[/tex]
Conservation of energy:
This theorem states that energy can neither be created nor be destroyed it can only get converted from one form to another.
Relation between the Forces and potential energy:
Force here implies the force of gravitation. Potential energy in general means the energy which is contained in a body due to virtue of its position in a field of influence.
Here the field of influence is gravity. So,
[tex]PE=m.g.h[/tex]
where:]
m = mass of the body
g = acceleration due to gravity
h = height of the object from the earth surface
Consider a uniformly charged thin-walled right circularcylindrical shell having total charge Q, radiusR, and height h. Determine the electric field ata point a distance d from the right side of the cylinderas shown in the figure below. (Use k_e for ke,Q, d, R, and h as necessary.) Suggestion: Use the followingexpression and treat the cylinder as a collection of ring charges.
Answer:
[tex]\frac{k_eQ}{2h}[/tex]
Explanation:
The question is missing an image. I have added this as an attachment to my answer.
Given;
Q = total charge
R = radius of cylindrical shell
h = height of cylindrical shell
d = distance of point from the right side of the cylinder
Let the thickness of the cylindrical shell be [tex]dx[/tex] , and the charge [tex]\frac{Qdx}{h}[/tex],
Now, using the formula for finding the electric field due to a ring at a chosen point:
[tex]dE = \frac{k_ex}{(x^2 + R^2)^{\frac{3}{2}}} {\frac{Qdx}{h}i}[/tex]
where [tex]x[/tex] = center of the ring to the point
[tex]k_e[/tex] = electrostatic constant
We integrate on both sides from the limits [tex]d[/tex] to [tex]d + h[/tex] in order to determine the electric field at the point [tex]E[/tex]
[tex]\int\limits dE[/tex] = [tex]\int\limits^{d + h}_d {\frac{k_eQxdx}{h(x^2 + R^2)^{\frac{3}{2}}}i}[/tex]
[tex]E = \int\limits^{d + h}_d {\frac{k_eQdx}{h(x^2 + R^2)^{\frac{3}{2}}}i} = \frac{k_eQ}{2h}[/tex]
The electric field at a point a distance d from the right side of the cylinder is : E = [tex]\frac{K_{e}Q }{2h}[/tex]
Determine the electric field constantLet's' assume thickness of cylindrical shell = dx and charge = [tex]\frac{Qdx}{h}[/tex]
Next step : calculate the electric field due to a ring at a point
dE = [tex]\frac{kex}{(x^2+ R^2)^{\frac{3}{2} } } \frac{Qdx}{h} i[/tex] --- ( 1 )
where : x = centre of the ring to the specific point
ke = electrostatic constant
To determine the electric field constant by integrating equation within limits d to d + h
E = [tex]\int\limits^d_d {\frac{keQdx}{h(x^{2}+R^2)^{\frac{3}{2} } } } \, i = \frac{KeQ}{2h}[/tex]
Hence we can conclude that The electric field at a point a distance d from the right side of the cylinder is : E = [tex]\frac{K_{e}Q }{2h}[/tex].
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Suppose you are standing on the edge of a dock and jump straight down. If you land on sand your stopping time is much shorter than if you land on water. Using the impulse-momentum theorem as a guide, determine which one of the following statements is correct.
(a) In bringing you to a halt, the sand exerts a greater impulse on you than does the water.
(b) In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force.
(c) In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a smaller average force.
Answer:
In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force
Explanation:
Suppose you are standing on the edge of a dock and jump straight down. If you land on sand your stopping time is much shorter than if you land on water. Using the impulse-momentum theorem as a guide, the statement in bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force is correct.
What is Impulse momentum theorem?The impulse-theorem defined as the change in momentum of an object will be equal to the impulse exerted on it.
Mathematically, the theorem can be described as:
The impulse is defined as the product between the force exerted (F) and the duration of the collision Δt,
I = FΔt
The change in momentum is equal to the product between the mass of the object (m) and the change in velocity (Δυ):
Δp = mΔv
So, the theorem can be written as,
FΔt = mΔv
This theorem can also be proved using Newton's second law, we know that,
F = ma
where, F - Force,
m - mass
a - acceleration
Here, we have to alter it with the acceleration change.
a = Δv /Δt,
when we substitute this with the force,
F = mΔv /Δt,
FΔt =mΔv.
The above formula is the exact formula of impulse- momentum theorem.
By using this concept, according to the force and velocity, If you land on sand your stopping time is much shorter than if you land on water, in bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force.
Hence, Option B is the correct answer.
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A student throws a 120 g snowball at 7.5 m/s at the side of the schoolhouse, where it hits and sticks. Part A What is the magnitude of the average force on the wall if the duration of the collision is 0.18 s? Express your answer to two significant figures and include the appropriate units.
Answer:
5.00 N
Explanation:
Force: This is defined as the product of mass and acceleration. The S.I unit of Force is Newton ( N).
Mathematically,
F = ΔM/t...................... Equation 1
Where F = Force needed to stop the snowball, ΔM = change in momentum, t = time.
But
ΔM = mΔv
ΔM = m(v-u)......................... Equation 2
Where m = mass of the snowball, Δv = change n velocity of the snowball, u = initial velocity, v = final velocity.
Substituting equation 2 into equation 1
F = m(v-u)/t...................... Equation 3.
Given: m = 120 g = 120/1000 = 0.12 kg, v = 0 m/s, u = 7.5 m/s, t = 0.18 s
Substituting into equation 3,
F = 0.12(0-7.5)/0.18
F = -0.9/0.18
F = - 5.00 N.
From newtons third law of motion,
The Force need to stop the ball is equal and opposite to the force on the wall( Average force by the ball)
Therefore,
The average force on the wall = 5.00 N in opposite direction to the motion of the snowball
I've just discovered a new radioactive element. At 1:00 pm I measure 10 grams of the element, but at 1:24 pm I measure only 1.25 grams of the element (with 8.75 grams of daughter product). How long is each half life, in minutes? (Just enter the number).
Answer: Each half life is 8 minutes long.
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample = 24 minutes
a = initial amount of the reactant = 10 grams
a - x = amount left after decay process = 1.25 grams
[tex]24min=\frac{2.303}{k}\log\frac{10}{1.25}[/tex]
[tex]k=\frac{2.303}{24}\log\frac{10}{1.25}[/tex]
[tex]k=0.0866min^{-1}[/tex]
for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]t_{\frac{1}{2}}=\frac{0.693}{0.0866min^{-1}}=8.00min[/tex]
Thus half life is 8 minutes long
A proton is launched from an infinite plane of charge with surface charge density -1.10×10-6 C/m2. If the proton has an initial speed of 2.40×107 m/s, how far does it travel before reaching its turning point? 48.4 m 96.7 m 2.02×10-6 m 8.74×10-7 m
The distance covered by the proton is 48.4 m
Explanation:
The electric field produced by an electrically charged infinite plane is given by
[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]
where in this case,
[tex]\sigma = -1.10\cdot 10^{-6} C/m^2[/tex] is the surface charge density
[tex]\epsilon_0 = 8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity
Substituting,
[tex]E=\frac{-1.10\cdot 10^{-6}}{2(8.85\cdot 10^{-12})}=-5.65\cdot 10^4N/C[/tex]
And the direction is towards the plane (because the charge is negative).
The electric force on the proton due to this field is
[tex]F=qE[/tex]
where
[tex]q=1.6\cdot 10^{-19}C[/tex] is the proton charge
Substituting,
[tex]F=(1.6\cdot 10^{-19})(-5.65\cdot 10^4)=-9.0\cdot 10^{-15} N[/tex]
where the direction is toward the plane.
Now we can calculate the proton's acceleration using Newton's second law:
[tex]a=\frac{F}{m}[/tex]
where
[tex]m=1.67\cdot 10^{-27}kg[/tex] is the proton mass
Substituting,
[tex]a=\frac{-9.0\cdot 10^{-15}}{1.67\cdot 10^{-27}}=-5.4\cdot 10^{-12} m/s^2[/tex]
Now we can finally apply the following suvat equation for accelerated motion to find the distance travelled by the proton:
[tex]v^2-u^2=2as[/tex]
where
v = 0 is the final velocity
[tex]u=2.40\cdot 10^7 m/s[/tex] is the initial velocity
a is the acceleration
s is the distance covered
And solving for s,
[tex]s=\frac{v^2-u^2}{2a}=\frac{0-(2.4\cdot 10^7)^2}{2(-5.4\cdot 10^{12})}=53.3 m[/tex]
Therefore, the closest answer is 48.4 m.
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If you take two steps of different sizes, can you end up at your starting point (in other words, can two vectors with different magnitudes ever add to zero)? More generally, can three or more vectors with different magnitudes ever add to zero?
Final answer:
Two vectors with different magnitudes can add up to zero only if they are antiparallel. For three or more vectors of different magnitudes, they can sum to zero if they form a closed shape. The maximum distance achieved by adding two vectors is the sum of their lengths if they are in the same direction.
Explanation:
When dealing with vectors, it is possible for two vectors with different magnitudes to add up to zero if they are equal in magnitude but opposite in direction, making them antiparallel. This is due to the definition of vector addition, which considers both magnitude and direction.
For example, if you take a step of 5 units to the north (vector A) and then a step of 5 units to the south (vector B), you will end up back at your starting point as these two vectors cancel each other out. This can only happen if the two vectors are directly opposite to each other (antiparallel).
However, in the more general case with three or more vectors, they can add to zero even if they have different magnitudes, as long as they form a closed shape like a triangle or a polygon when placed tip-to-tail. The result of adding multiple vectors together in such a manner would be zero, bringing you back to your starting point.
Moreover, the maximum distance you can end up from the starting point when adding two vectors A and B is indeed the sum of their lengths, but only when the vectors are aligned and pointing in the same direction. When they point in opposite directions, the maximum distance will be the difference in their lengths.
The synthesis of a large collection of information that contains well-tested and verified hypotheses about certain aspects of the world is known as a scientific?
Answer:
The synthesis of a large collection of information that contains well-tested and verified hypotheses about certain aspects of the world is known as a scientific THEORY
A scientific theory is the synthesis of well-tested and verified hypotheses about certain aspects of the world.
Explanation:Scientific theory represents the pinnacle of scientific knowledge, signifying the meticulous synthesis of an expansive body of information that encompasses thoroughly tested and validated hypotheses pertaining to specific facets of our world. This robust and comprehensive explanation is not conjecture; instead, it is firmly rooted in a wealth of empirical evidence that has undergone rigorous scrutiny through a multitude of scientific experiments and meticulous observations.
A scientific theory, therefore, stands as an enduring framework that not only elucidates natural phenomena but also endows us with a profound comprehension of the intricate workings of the natural world. It serves as the bedrock upon which scientific understanding and further inquiry are built, fostering a deeper appreciation of the complexities that underlie our universe.
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