Answer:
Explanation:
mass of ice skater, m = 55 kg
velocity, v = 4.07 m/s
radius, r = 0.808 m
The force is centripetal force, the formula for this force is
[tex]F_{c}=\frac{mv^{2}}{r}[/tex]
[tex]F_{c}=\frac{55\times 4.07\times 4.07}{0.808}[/tex]
Fc = 1127.56 N
Weight of the person, W = mg
W = 55 x 9.8 = 539 N
The ratio of force to he weight
[tex]\frac{F_{c}}{W}=\frac{1127.56}{539}[/tex]= 2.09
Suppose you move along a wire at the same speed as the drift speed of the electrons in the wire. Do you now measure a magnetic field of zero?
Answer:
False. Field is non-zero
Explanation:
If you were moving along with the electrons, they would appear stationary to you. You would measure a current of zero. However, the fixed positive charges in the wire seem to move backwards relative to you, creating the equivalent current as if you weren't moving. You would measure the same field, but the field would be caused by the 'backward' motion of positive particles.
Moving at the same speed as the drift speed of electrons does not result in a zero magnetic field. The movement of electrons in a wire creates a magnetic field, and this field would still be present even if you were moving at the same speed as the electrons.
Explanation:When moving along a wire at the same speed as the drift speed of the electrons, you will not measure a magnetic field of zero. The drift speed of electrons refers to the average velocity at which the electrons move in a conductor when an electric field is applied. This speed is generally very slow, but it does not mean that there is no magnetic field.
The movement of electrons in a wire creates a magnetic field around the wire, even if the drift speed is small. This is because the electric current generated by the movement of electrons is what produces the magnetic field.
So, even if you were to move at the same speed as the drift speed of the electrons in the wire, you would still measure a non-zero magnetic field because the movement of electrons in the wire is what generates the magnetic field.
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what will be the speed of a solid sphere of mass 2.0 kilograms and radius 15.0 centimeters when it reaches the bottom of a incline of length 5.0 meters. Assume the sphere starts from rest and rolls without slipping.
Explanation:
Below is an attachment containing the solution.
If you weigh 685 NN on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 25.0 kmkm ? Take the mass of the sun to be msmsm_s = 1.99×1030 kgkg , the gravitational constant to be GGG = 6.67×10−11 N⋅m2/kg2N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be ggg = 9.8 m/s2m/s2 .
Answer:
[tex] W = 5.94 \cdot 10^{15} N [/tex]
Explanation:
To calculate the weight on the surface of a neutron star we can use the following equation:
[tex] W = m*g [/tex]
Where:
W: is the weight of the person
m: is the mass of the person
g: is the gravity of the neutron star
Hence, first we need to find m and g. The mass is equal to:
[tex]m = \frac{W}{g} = \frac{685 N}{9.81 m/s^{2}} = 69.83 kg[/tex]
Now, the gravity of the neutron star can be found using the followig equation:
[tex]F = \frac{G*m*M}{r^{2}} = m*g \rightarrow g = \frac{G*M}{r^{2}}[/tex]
Where:
G: is the gravitational constant = 6.67x10⁻¹¹ m³ kg⁻¹ s⁻²
M: is the mass of the neutron star = 1.99x10³⁰ kg
r : is the distance between the person and the surface of the neutron star = 25/2 = 12.5 km
[tex] g = \frac{6.67 \cdot 10^{-11} m^{3}kg^{-1}s^{-2}*1.99 \cdot 10^{30} kg}{(12.5 \cdot 10^{3} m)^{2}} = 8.50 \cdot 10^{13} m/s^{2} [/tex]
Now, we can find the weight on the surface of the neutron star:
[tex]W = m*g = 69.83 kg * 8.50 \cdot 10^{13} m/s^{2} = 5.94 \cdot 10^{15} N[/tex]
I hope it helps you!
Your weight on a neutron star with the same mass as the Sun and a diameter of 25.0 km would be approximately 5.95 × 10¹⁴ N. This is due to the extremely high gravitational acceleration on the neutron star's surface. Such high gravity results from the star's compactness and mass.
To determine your weight on the surface of a neutron star, we need to calculate the gravitational acceleration on its surface and then use this to find the weight force.
Step-by-Step Solution:
Calculate the gravitational acceleration, gns, at the surface of the neutron star using the formula for gravitational acceleration: g = G * M / R², where:G is the gravitational constant, 6.67 × 10⁻¹¹N⋅m²/kg²M is the mass of the neutron star, which is equal to the mass of the Sun, 1.99 × 10³⁰ kgR is the radius of the neutron star, which is half of its diameter, so R = 25.0 km / 2 = 12.5 km = 1.25 × 10⁴ mSubstitute these values into the formula:gns = (6.67 × 10⁻¹¹N⋅m²/kg²) * (1.99 × 10³⁰ kg) / (1.25 × 10⁴ m)² = 8.51 × 10¹² m/s²To find your weight, use the weight formula: Weight = mass × gravitational acceleration.Your mass (m) can be found from your weight on Earth: WeightEarth = m × g, so:m = WeightEarth / g = 685 N / 9.8 m/s² = 69.9 kgNow, calculate your weight on the neutron star:Weightns = m × gns = 69.9 kg × 8.51 × 1012 m/s² ≈ 5.95 × 10¹⁴NSummary: Your weight on a neutron star with the same mass as the Sun and a diameter of 25.0 km would be approximately 5.95 × 10¹⁴N, which is significantly more than your weight on Earth.
energy conservation-problems 1. A slingshot fires a pebble from the top of a building at a speed of 14.0m/is. The building is 31.0m tall. Ignoring all frictional effects, find the speed with which the pebble strikes the ground
Answer:
Horizontal velocity is 14 m/s
Vertical velocity is 28.3 m/s
Explanation:
Hello dear friend, you have not mentioned the type of speed that you require for this problem.
In this case, there are three possibilities with which the slingshot can be fired i.e. Horizontally, Vertically straight up and vertically straight down. Below is the explanation / answer to all three possibilities
Fired horizontally:
initial conditions:
Vertical Velocity = 0 ; Horizontal Velocity = 14m/s
final conditions:
Vertical Velocity (v² = u² + 2gs) but initial vertical velocity is zero
v² = 2gs so v² = 2(9.8)(31) = 607
v = 24.6m/s
but Horizontal Velocity is still = 14m/s
Resultant velocity from these two velocity components (Pythagoras theorem)
V² = (v horizontal)² + (v vertical)² = 14² + 24.6²
V = 28.3m/s
angle = tan ⁻¹(24.3/14) = 60.1⁰
V = 28.3m/s at angle of 60.1⁰ to the horizontal
Fired Vertically Straight Up
distance before the pebble reaches maximum height from top of building
v² = u² + 2gs
where, v is zero at maximum height
g is minus for upward motion.
v² = u² + 2gs
0 = 14² - 2(9.8)s
s = 196/19.6 = 10.0m
totals distance from maximum height to the ground = 10.0 m + 31.0 m = 41.0m
v² = u² + 2gs
now u from maximum height is 0 and g is positive for downward motion
v² = 2gs
v² = 2(9.8)(41.0)
v = 28.3m/s
v = 28.3m/s vertically straight up
Fired Vertically Straight Down
v² = u² + 2gs
u = 14m/s, g = 9.8m/s², s = 31.0m
v² = 14² + 2(9.8)(31.0)
v = 28.3m/s
v = 28.3m/s vertically straight down
The astronomer who discovered the dwarf planet Eris suggests there might be another object far beyond the Kupier belt. If this Planet X exists, it would be about 10 times the mass of Earth and 2-3 times the size of Earth, putting it in the ice giant category, and have an orbit with a semi-major axis of 700 AU. You can read more about this object on NASA's page. If this object exists, what would we classify it as?
These objects would be classified as extreme trans Neptunian object (ETNO).
Explanation:
ETNO’s are the objects lying beyond the planet Neptune and orbiting the Sun. They follow a highly eccentric path which is tilted. ETNO has been grouped into three major according to their respective perihelia.
Within this region (beyond Neptune’s orbit), a hypothetical planet has been discovered. It was discovered following its gravitational effect on the other objects of Kuiper Belt (region beyond the orbit of the Neptune- the last planet of our Solar system)
The Planet is assumed to be around 2 times the Earth’s size and around 10 times heavier than Earth.
Planet X, if it exists, would likely be classified as a planet rather than a dwarf planet because it is much larger and more massive, suggesting it could clear its orbital neighborhood, which is one of the criteria that differentiate planets from dwarf planets.
The hypothetical object proposed by the astronomer beyond the Kuiper Belt, nicknamed Planet X, would be classified differently from Eris and other dwarf planets like Pluto, Makemake, and Haumea. Since it is speculated to be about 10 times the mass of Earth and 2-3 times its size, placing it within the ice giant category, it would resemble Uranus or Neptune rather than the smaller dwarf planets in the solar system.
Dwarf planets are generally smaller bodies that, while orbiting the Sun and having sufficient mass for their self-gravity to overcome rigid body forces, have not cleared their neighboring region of other objects. In contrast, Planet X, being significantly larger and more massive, would likely be considered a full-fledged planet if its existence were confirmed, primarily because of its mass, size, and potential to clear its orbit, aligning closer with the current criteria for a planet.
A centrifuge at a museum is used to separate seeds of different sizes. The average rotational acceleration of the centrifuge according to a sign is 30 rad/s2rad/s2. Part A If starting at rest, what is the rotational velocity of the centrifuge after 10 ss?
Answer:
[tex]\omega = 300 rad/s[/tex]
Explanation:
Given,
rotational acceleration = 30 rad/s²
initial angular speed = 0 m/s
time, t = 10 s
Final angular speed = ?
Using equation of rotation motion
[tex]\omega = \omega_o + \alpha t[/tex]
[tex]\omega =0+30\times 10[/tex]
[tex]\omega = 300 rad/s[/tex]
Rotational velocity after 10 s = 300 rad/s.
The velocity of a sky diver t seconds after jumping is given by v(t) = 80(1 − e−0.2t). After how many seconds is the velocity 65 ft/s? (Round your answer to the nearest whole number.)
Answer:
8 seconds
Explanation:
Given:
The velocity of the sky diver 't' seconds after jumping is given as:
[tex]v(t)=80(1-e^{-0.2t})[/tex]
The velocity is given as, [tex]v=65\ ft/s[/tex]
So, in order to find the time required to reach the above given velocity, we plug in 65 for 'v' in the above equation and solve for time 't'. This gives,
[tex]65=80(1-e^{-0.2t})\\\\\frac{65}{80}=1-e^{-0.2t}\\\\0.8125=1-e^{-0.2t}\\\\e^{-0.2t}=1-0.8125\\\\\textrm{Taking natural log on both sides, we get:}\\\\-0.2t=\ln(0.1875)\\\\t=\frac{\ln(0.1875)}{-0.2}\\\\t=8.4\ s\approx 8\ s(Nearest\ whole\ number)[/tex]
Therefore, the time taken to reach a velocity of 65 ft/s is nearly 8 seconds.
The velocity of the skydiver is 65 ft/s after approximately 9 seconds. This is found by solving the provided velocity function for the given speed.
To determine after how many seconds the velocity of the skydiver is 65 ft/s, we need to solve the equation
[tex]v(t) = 80(1 - e^-^0^.^2^t).[/tex]
Given v(t) = 65, we set up the equation:
[tex]65 = 80(1 - e^-^0^.^2^t)[/tex]
First, isolate the exponential term:
[tex]65/80 = 1 - e^-^0^.^2^t\\0.8125 = 1 - e^-^0^.^2^t[/tex]
Subtract 1 from both sides:
[tex]-0.1875 = -e^-^0^.^2^t[/tex]
Divide by -1:
[tex]0.1875 = e^-^0^.^2^t[/tex]
Take the natural logarithm (ln) of both sides to solve for t:
[tex]ln(0.1875) = -0.2t[/tex]
Solve for t:
[tex]t = ln(0.1875) / -0.2[/tex]
Using a calculator, we get:
[tex]t = 8.6 seconds[/tex]
Rounding to the nearest whole number, the velocity is 65 ft/s after about 9 seconds.
What happens to the direction of the magnetic field about an electric current when the direction of the current is reversed?
Answer:
The direction of the magnetic field is also reversed.
Explanation:
The direction of the magnetic field is also reversed when viewed form the same side if the direction of current is reversed. The direction of the magnetic field with respect to the direction of electric current is determined by the Maxwell's right-hand thumb rule.According to this rule we place our palm with the thumb pointing the direction of current flow and curling our finger in an action of gripping the wire. This position the the direction of curled fingers represents the direction of magnetic field.The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point (1, 2, 2) is 160°. (a) Find the rate of change of T at (1, 2, 2) in the direction toward the point (4, 1, 3). Incorrect: Your answer is incorrect. (b) Show that at any point in the ball the direction of greatest increase in temperature is given by a vector that
Answer:
The answers to the questions are as follows;
(a) The rate of change of T at (1, 2, 2) in the direction toward the point (4, 1, 3) is [tex]\frac{160\sqrt{11} }{33}[/tex]
(b) The direction of the gradient is in the direction of greatest increase and it is towards the origin.
Explanation:
To solve the question, we note that the shape of the ball is that of a sphere.
Therefore the distance of a point from the center is given by
f(x, y, z) = [tex]\sqrt{x^2+y^2+z^2}[/tex]
The temperature T in a metal ball is inversely proportional to the distance from the center of the ball
Therefore T ∝ [tex]\frac{1}{\sqrt{x^2+y^2+z^2}}[/tex] or T = [tex]\frac{C}{\sqrt{x^2+y^2+z^2}}[/tex]
Where
C = Constant of proportionality
x, y, and z are the x, y and z coordinates values
To find C, we note that at point (1, 2, 2), T = 160 °C.
Therefore 160 °C = [tex]\frac{C}{\sqrt{1^2+2^2+2^2}}[/tex] = [tex]\frac{C}{\sqrt{9}}[/tex] = [tex]\frac{C}{3}}[/tex]
Therefore C = 160 × 3 = 480 °C·(Unit length)
We therefore have the general equation as
T = [tex]\frac{480}{\sqrt{x^2+y^2+z^2}}[/tex]
The vector from points (1, 2, 2) to point (4, 1, 3) is given by
1·i + 2·j +2·k - (4·i + 1·j +3·k) = -3·i + j -k
From which we find the unit vector given by
u = [tex]\frac{1}{\sqrt{(-3)^2+1^2+(-)^2} } (-3, 1, -1)= \frac{1}{\sqrt{11} } (-3, 1, -1)[/tex]
From which we have the gradient equal to
∇T(x, y, z) = -480×(x²+y²+z²)[tex]^-{\frac{3}{2}}[/tex] in (x, y, z)
This gives D[tex]_u[/tex] = ∇T·u
= -480×(x²+y²+z²)[tex]^-{\frac{3}{2}}[/tex] in (x, y, z)·[tex]\frac{1}{\sqrt{11} } (-3, 1, -1)[/tex]
That is
[tex]-\frac{480}{\sqrt{11} }[/tex](x²+y²+z²)[tex]^-{\frac{3}{2}}[/tex] (-3·x + y - z)
From where D[tex]_u[/tex]Tat point (1, 2, 2) is = [tex]\frac{160\sqrt{11} }{33}[/tex]
(b) The direction of greatest increase in temperature is in the direction of the gradient and the direction of the gradient is opposite to the direction of {x, y, z}, which is away from the origin.
Hence the direction of the greatest increase in temperature is towards the origin.
An object of mass 0.77 kg is initially at rest. When a force acts on it for 2.9 ms it acquires a speed of 16.2 m/s. Find the magnitude (in N) of the average force acting on the object during the 2.9 ms time interval. (Enter a number.)
Answer: 4.3KN
Explanation:f=m(v-u)/t
m=0.77kg
t=0.0029s
s=16.2m/s
F= 0.77*16.2/0.0029
F=12.474/0.0029
F= 4301.38N
F=4.3KN
To find the magnitude of the average force acting on an object, one can use the derived form of the second law of motion, F = mΔv/Δt. Placing the given values into the equation, we calculate the force to be approximately 4307 N.
Explanation:To solve this question, we can use the equation F = mΔv/Δt which is derived from the second law of motion (Force = mass × acceleration), where F is the average force, m is the mass of the object, Δv is the change in velocity, and Δt is the time interval.
Substituting the given values, m = 0.77 kg, Δv = 16.2 m/s (the final velocity) - 0 m/s (the initial velocity) = 16.2 m/s and Δt = 2.9 ms = 2.9 × 10⁻³ s (as time should be converted to seconds).
Therefore, F = (0.77 kg × 16.2 m/s) / 2.9 × 10⁻³ s = 4306.9 N. Therefore, the magnitude of the average force acting on the object during the 2.9 ms time interval is approximately 4307 N.
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A box with its contents has a total mass of 40 kg. It is dropped from a very high building. After reaching terminal speed, what is the magnitude of the air resistance force acting upward on the falling box
Answer:
The magnitude of air = 392N
Explanation:
We use Newton's 2nd law. The sum of the vertical forces must be equal to zero because at terminal speed , the acceleration is zero. Solving for the air resistance force,F(air ) gives:
EFvertical = mg - F(air)= ma
F(air) = mg = 40 × 9.8 = 392N
Answer: 392N
Explanation:
Newton's second law of motion states that "The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object."
the sum of vertical forces has to be equal to zero because by the time the terminal speed has been attained, the acceleration is zero. Now, we solve for air resistance force.
summation of F(vertical) = mg - F(air) = ma
a = 0 m/s²
thus, F(air) = mg
F(air) = 40kg*9.8m/s²
F(air) = 392N
during a crash, an airbag inflates to stop a dummys forward motion. the dummy mass is 75 kg. if the net force is on the dummy is 825 N toward the rear of the car, what is the dumys deceleration
Answer:
11 m/s²
Explanation:
Deceleration: This can be defined as the rate of decrease of velocity. The S.I unit of deceleration is m/s².
From the question,
F = md ..................... Equation 1
Where F = Force acting on the dummy towards the rear of the car, m = mass of the dummy, d = deceleration of the dummy.
make d the subject of the equation
d = F/m............... Equation 2
Given: F = 825 N, m = 75 kg.
substitute into equation 2
d = 825/75
d = 11 m/s²
Answer:
11 m/s^2.
Explanation:
Given:
Mass = 75 kg
Force = 825 N
F = m × a
a = 825 ÷ 75
= 11 m/s^2.
Explain the differences between the geocentric theory of the universe and the heliocentric theory
Geocentric Theory
In astronomy, the geocentric model is a superseded description of the Universe with Earth at the center. Under the geocentric model, the Sun, Moon, stars, and planets all orbited Earth
Heliocentric Theory
Heliocentrism is the astronomical model in which the Earth and planets revolve around the Sun at the center of the Solar System. Historically, heliocentrism was opposed to geocentrism, which placed the Earth at the center
G-Theory is the earth is the center of the universe.
H-Theory is the sun is the center of the universe.
A block sliding along a horizontal frictionless surface with speed v collides with a spring and compresses it by 2.0 cm. What will be the compression if the same block collides with the spring at a speed of 2v?
Answer:
4.0 cm
Explanation:
For the compression of the spring, the kinetic energy of the mass equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² ⇒ x = (√m/k)v
Since m and k are constant since its the same spring x ∝ v
If our speed is now v₁ = 2v, our compression is x₁
x₁ = (√m/k)v₁ = (√m/k)2v = 2(√m/k)v = 2x
x₁ = 2x
Since x = 2.0 cm, our compression for speed = 2v is
x₁ = 2(2.0) = 4.0 cm
If the same block collides with the spring at a speed of 2v, the compression will be 4.0cm.
Given the data in the question;
Compression; [tex]x_1 = 2.0cm[/tex]Velocity 1; [tex]v_1 = v[/tex]Velocity 2; [tex]v_2 = 2v[/tex]Using conservation of energy:
Kinetic energy of the mass = Elastic potential energy of the spring
We have:
[tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2\\\\kx^2 = mv^2[/tex]
"v" is directly proportional to "x"
Hence,
[tex]\frac{x_1}{x_2} = \frac{v_1}{v_2}[/tex]
We substitute in our given values
[tex]\frac{2.0cm}{x_2} = \frac{v}{2v}\\\\x_2 = \frac{v(2.0cm*2)}{v} \\\\x_2 = (2.0cm*2)\\\\x_2 = 4.0cm[/tex]
Therefore, if the same block collides with the spring at a speed of 2v, the compression will be 4.0cm.
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A 25-kg iron block initially at 350oC is quenched in an insulated tank that contains 100 kg of water at 18oC. Assuming the water that vaporizes during the process condenses back in the tank, determine the total entropy change during this process.
Answer:
The value of total entropy change during the process
[tex]dS = 0.608 \frac{KJ}{K}[/tex]
Explanation:
mass of iron [tex]m_{iron}[/tex] = 25 kg
Initial temperature of iron [tex]T_{1}[/tex] = 350°c = 623 K
Mass of water [tex]m_{w}[/tex] = 100 kg
Initial temperature of water [tex]T_{2}[/tex] = 180°c = 453 K
When iron block is quenched inside the water the final temperature of both iron & water becomes equal. this is = [tex]T_{f}[/tex]
Thus heat lost by the iron block = heat gain by the water
⇒ [tex]m_{iron}[/tex] [tex]C_{iron}[/tex] ( [tex]T_{1}[/tex] - [tex]T_{f}[/tex] ) = [tex]m_{w}[/tex] [tex]C_{w}[/tex] ( [tex]T_{f}[/tex] - [tex]T_{2}[/tex] )
⇒ 25 × 0.448 × ( [tex]T_{1}[/tex] - [tex]T_{f}[/tex] ) = 100 × 4.2 × ( [tex]T_{f}[/tex] - [tex]T_{2}[/tex] )
⇒ [tex]( T_{1} - T_{f} ) = 37.5 ( T_{f} - T_{2} )[/tex]
⇒ [tex]( 623 - T_{f} ) = 37.5 ( T_{f} - 453 )[/tex]
⇒ [tex]( 623 - T_{f} ) = 37.5 T_{f} - 16987.5[/tex]
⇒ [tex]38.5 T_{f} = 17610.5[/tex]
⇒ [tex]T_{f} = 457.41 K[/tex]
This is the final temperature after quenching.
The total entropy change is given by,
[tex]dS = m_{iron}\ C_{iron} \ ln \frac{T_{f} }{T_{1} } + m_{w}\ C_{w} \ ln \frac{T_{f} }{T_{2} }[/tex]
Put all the values in above formula,
[tex]dS =[/tex] 25 × 0.448 × [tex]ln \frac{457.41}{623}[/tex] + 100 × 4.2 × [tex]ln \frac {457.41}{453}[/tex]
[tex]dS =[/tex] - 3.46 + 4.06
[tex]dS = 0.608 \frac{KJ}{K}[/tex]
This is the value of total entropy change.
. A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the bottom moving along the ground when the bottom of the ladder is 5 ft from the wall
Explanation:
Below is an attachment containing the solution.
A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.
Answer:
(a) I_A=1/12ML²
(b) I_B=1/3ML²
Explanation:
We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².
(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².
(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:
[tex]d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L[/tex]
Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:
[tex]x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L[/tex]
Finally, using the Parallel Axis Theorem, we calculate I_B:
[tex]I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}[/tex]
A) Moment of inertia about an axis passing through the point where the two segments meet : [tex]I_{A} = \frac{1}{12} ML^{2}[/tex]
B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends : [tex]Ix_{} = \frac{1}{3} ML^{2}[/tex]
A) The moment of inertia about an axis passing through the point where the two segments meet is [tex]I_{A} = \frac{1}{12} ML^{2}[/tex] given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.
B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends
First step: determine the distance between the ends ( d )
After applying Pythagoras theorem
d = [tex]\frac{\sqrt{2} }{2} L[/tex]
Next step : determine distance between the two axis ( x )
After applying Pythagoras theorem
x = [tex]\frac{\sqrt{2} }{4} L[/tex]
Final step : Calculate the value of Iₓ
applying Parallel Axis Theorem
Iₓ = Iₐ + Mx²
= [tex]\frac{1}{12} ML^{2}[/tex] + [tex]\frac{1}{4} ML^{2}[/tex]
∴ [tex]Ix_{} = \frac{1}{3} ML^{2}[/tex]
Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet : [tex]I_{A} = \frac{1}{12} ML^{2}[/tex], Moment of inertia passing through the point where the midpoint of the line connects its two ends : [tex]Ix_{} = \frac{1}{3} ML^{2}[/tex]
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What is the minimum diameter mirror on a telescope that would allow you to see details as small as 5.20 km on the moon some 384000 km away? Assume an average wavelength of 550 nm for the light received.
cm
Answer:
= 4.96cm
Explanation:
distance between objects of moon = 5.20km = 5.2 × 10³ m
Wavelength of light, λ = 550nm = 5.50 × 10⁻⁷m
distance of moon, L = 384000 km = 3.84 × 10⁸m
formula for resolving power of two objects
d = (1.22 × λ ×L) / D
D = (1.22 × λ ×L) / d
D = (1.22 × 5.50 × 10⁻⁷ ×3.84 × 10⁸) / 5.2 × 10³
D = 4.96cm
When charged particles are separated by an infinite dis- tance, the electric potential energy of the pair is zero. When 19. 7. the particles are brought close, the electric potential energy of a pair with the same sign is positive, whereas the electric potential energy of a pair with opposite signs is negative. Give a physical explanation of this statement.
Answer:
Explanation:
Potential energy Is given as.
U=kq1q2/r
Where k is a constant
q1 is charge
And q2 is also a charge
r is the distant between the two charges
When distance between the two charges is infinity then, the potential energy will b zero
U=kq1q2/∞
Therefore
U=0
Thus, the potential energy is zero when charged particles are separated by an infinite distance.
Let considered that both charges are positive.
Then U becomes
U=k(+q1)(+q2)/r
This shows that the energy is positive when the charges are positively charge. Also if the charges are also both negative charged the total energy will also be positive
U=k(-q1)(-q2)/r.
Both when their are opposite charges, the energy will be negative
U=k(-q1)(+q2)/r
U will be negative
With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you release the ball 1.4 mm above the ground? Solve this problem using energy.
Answer:
The initial velocity is 0.5114 m/s or 511.4 mm/s
Explanation:
Let the initial velocity be 'v'.
Given:
Mass of the ball (m) = 130 g = 0.130 kg [ 1 g = 0.001 kg]
Initial height of the ball (h₁) = 1.4 mm = 0.0014 m [ 1 mm = 0.001 m]
Final height of the ball (h₂) = 15 mm = 0.015 m
Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.
Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.
Change in kinetic energy is given as:
[tex]\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity[/tex]
As it just touches the 15 mm high roof, the final velocity will be zero. So,
[tex]v_f=0\ m/s[/tex].
Now, the change in kinetic energy is equal to:
[tex]\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2[/tex]
Change in gravitational potential energy = Final PE - Initial PE
So,
[tex]\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J[/tex] [ g = 9.8 m/s²]
Now, Change in KE = Change in PE
[tex]0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s[/tex]
Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s
When its 75 kW (100 hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m/s (150 m/min, or 500 ft/min). What fraction of the engine power is being used to make the airplane climb
Answer:
343/1500
Explanation:
Power: This can be defined as the product force and velocity. The S.I unit of power is Watt (w).
From the question,
P' = mg×v................. Equation 1
Where P' = power used to gain an altitude, m = mass of the engine, g = acceleration due to gravity of the engine, v = velocity of the engine.
Given: m = 700 kg, v = 2.5 m/s, g = 9.8 m/s²
Substitute into equation 1
P' = 700(2.5)(9.8)
P' = 17150 W.
If the full power generated by the engine = 75000 W
The fraction of the engine power used to make the climb = 17150/75000
= 343/1500
Before using a string in a comparison, you can use either the To Upper method or the To Lower method to convert the string to upper case or lower case, respectively, and then use the converted string in the comparison.1. True2. False
Answer:
True, check attachment for code
Explanation:
To convert java strings of text to upper or lower case, we can use and inbuilt methods To Uppercase and To lower case.
The first two lines of code will set up a String variable to hold the text "text to change", and then we print it out.
The third line sets of a second String variable called result.
The fourth line is where the conversion is done.
We can compare the string
We can compare one string to another. (When comparing, Java will use the hexadecimal values rather than the letters themselves.) For example, if we wanted to compare the word "Fat" with the word "App" to see which should come first, you can use an inbuilt string method called compareTo.
Check attachment for the code
A student with a mass of 80.0 kg runs up three flights of stairs in 12.0 sec. The student has gone a vertical distance of 8.0 m. Determine the amount of work done by the student to elevate his body to this height. Determine the power consumed by the student. Assume that his speed is constant.
Explanation:
Work = Force x Displacement
Force = Weight of student
Weight = Mass x Acceleration due to gravity
Mass, m = 80 kg
Acceleration due to gravity, g = 9.81 m/s²
Weight = 80 x 9.81 = 784.8 N
Displacement = 8 m
Work = 784.8 x 8 = 6278.4 J
The amount of work done by the student to elevate his body to this height is 6278.4 J
Power is the ratio of work to time taken
[tex]P=\frac{W}{t}\\\\P=\frac{6278.4}{12}\\\\P=523.2W[/tex]
Power is 523.2 Watts
The power consumed by the student is 523.2 watts.
The calculation is as follows:[tex]Work = Force \times Displacement[/tex]
Force = Weight of student
[tex]Weight = Mass \times Acceleration\ due\ to\ gravity[/tex]
Mass, m = 80 kg
Acceleration due to gravity, g = 9.81 m/s²
[tex]Weight = 80 \times 9.81 = 784.8 N[/tex]
Displacement = 8 m
[tex]Work = 784.8 \times 8 = 6278.4 J[/tex]
So,
Power consumed should be
[tex]= 6278.4 \div 12[/tex]
= 523.2
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A football punter accelerates a football from rest to a speed of 15 m/s during the time in which his toe is in contact with the ball (about 0.15 s). If the football has a mass of 0.44 kg, what average force does the punter exert on the ball?
Answer:
Force, F = 44 N
Explanation:
Given that,
Initial speed of the football, u = 0
Final speed, v = 15 m/s
The time of contact of the ball, t = 0.15 s
The mass of football, m = 0.44 kg
We need to find the average force exerted on the ball. It is given by the formula as :
[tex]F=ma\\\\F=\dfrac{mv}{t}\\\\F=\dfrac{0.44\times 15}{0.15}\\\\F=44\ N[/tex]
So, the average force exerted on the ball is 44 N. Hence, this is the required solution.
An elevator cab that weighs 27.8 kN moves upward. What is the tension in the cable if the cab's speed is (a) increasing at a rate of 2 1.22 / msand (b) decreasing at a rate of 2 1.22 / ms?
Answer:
a)T = 8.63 × 10 ⁴ N, b)T = -3.239 × 10 ⁴ N
Explanation:
Given:
W = 27.8 KN = 27.8 × 10 ³ N,
For upward motion: Fnet is upward, Tension T is upward and weight W is downward so
a) a=21.22 m/s² ( not written clearly the unit. if it is acceleration?) then
Fnet = T - W
⇒ T = F + W = ma + W
T = (W/g)a + W (W=mg ⇒m=W/g)
T = (27.8 × 10 ³ N / 9.8 ) 21.22 m/s² + 27.8 × 10 ³ N
T = 86,263.265 N
T = 8.63 × 10 ⁴ N
b) For Declaration in upward direction a = -21.22 m/s²
Fnet = T - W
⇒ T = F + W = ma + W
T = (W/g)a + W
T = (27.8 × 10 ³ N / 9.8 ) (-21.22 m/s²) + 27.8 × 10 ³ N
T = -32395.5 N
T = -3.239 × 10 ⁴ N
as Tension can not be negative I hope the value of acceleration and deceleration is correct.
An airplane needs to reach a velocity of 199.0 km/h to take off. On a 2000-m runway, what is the minimum acceleration necessary for the plane to take flight? Assume the plane begins at rest at one end of the runway.
Answer:
[tex]0.76m/s^2[/tex]
Explanation:
We are given that
Final velocity, v=199 km/h=[tex]199\times \frac{5}{18}=55.3m/s[/tex]
1km/h=[tex]\frac{5}{18}m/s[/tex]
Initial velocity, u=0
S=2000 m
We know that
[tex]v^2-u^2=2as[/tex]
Using the formula
[tex](55.3)^2=2a(2000)[/tex]
[tex]a=\frac{(55.3)^2}{2\times 2000}[/tex]
[tex]a=0.76m/s^2[/tex]
Hence, the minimum acceleration necessary for the plane to take flight=[tex]0.76m/s^2[/tex]
When two tuning forks are stuck simultaneously, 6 beats per second are heard. The frequency of one fork is 560 Hz. A piece of wax is placed on the 560 Hz fork to lower its frequency slightly. If the beat frequency is increased, what is the correct frequency for the second fork
Answer:
The correct frequenvy of the second fork is; ν2 = 566Hz
Explanation:
First of all, when two wave sources with slightly differing frequencies ν1 and ν2 generate waves at the same time and are superposed, then an interference effect will occur in time.
The intensity will be found to oscillate with time with a frequency (ν) called the beat frequency. It is depicted by:
ν = ± (ν1 −ν2).
In this question, there are two tuning forks, one with a frequency of let's say ν1=560Hz
The beat frequency is ν=6Hz
Therefore,
ν2=560 - 6 or ν2=560 + 6
i.e ν2=554 or ν2=566
For us to know the correct frequency, if the 560 Hz fork is loaded with wax, the increased inertia will lower its frequency.
Then it is found that the beat frequency increases. This can only mean that the other fork has a higher frequency. Hence the unkown frequency of the second fork must be,
ν2 = 566Hz
When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is 4.00 atm, is opened, what will be the final pressure in the two bulbs
Answer:
[tex]P_{C} = 3.2\, atm[/tex]
Explanation:
Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:
Bulb A (2 L, 2 atm) - Before opening:
[tex]P_{A} \cdot V_{A} = n_{A} \cdot R_{u} \cdot T[/tex]
Bulb B (3 L, 4 atm) - Before opening:
[tex]P_{B} \cdot V_{B} = n_{B} \cdot R_{u} \cdot T[/tex]
Bulbs A & B (5 L) - After opening:
[tex]P_{C} \cdot (V_{A} + V_{B}) = (n_{A} + n_{B})\cdot R_{u} \cdot T[/tex]
After some algebraic manipulation, a formula for final pressure is derived:
[tex]P_{C} = \frac{P_{A}\cdot V_{A} + P_{B}\cdot V_{B}}{V_{A}+V_{B}}[/tex]
And final pressure is obtained:
[tex]P_{C} = \frac{(2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L)}{5\,L}[/tex]
[tex]P_{C} = 3.2\, atm[/tex]
A current is established in a gas discharge tube when a sufficiently high potential difference is applied across the two electrodes in the tube. The gas ionizes; electrons move toward the positive terminal and singly charged positive ions move toward the negative terminal. What is the current in a hydrogen discharge tube in which 3.4 ✕ 1018 electrons and 1.4 ✕ 1018 protons move past a cross-sectional area of the tube each second? (Enter the magnitude.)
Explanation:
It is given that the number of electrons passing through the cross-sectional area in 1 s is [tex]3.4 \times 10^{18}[/tex]. Also, we know that charge on an electron is [tex]-1.60 \times 10^{-19} C[/tex], then negative charge crossing to the left per second is as follows.
I- = [tex]3.4 \times 10^{18} electrons \times -1.6 x 10^{-19} C/electrons[/tex]
I- = 0.544 A
As it is given that the number of protons crossing per second is [tex]1.4 \times 10^{18}[/tex], as the charge on the proton is [tex]+1.60 \times 10^{-19} C[/tex], then positive charge crossing to the right per second is calculated as follows.
I+ = [tex]1.4 \times 10^{18} electrons \times 1.6 \times 10^{-19} electrons/C[/tex]
I+ = 0.224 A
I = l I+ l + l I- l
So, I = 0.544 + 0.224
= 0.768 A
Thus, we can conclude that the current in given hydrogen discharge tube is 0.768 A.
A 497−g piece of copper tubing is heated to 89.5°C and placed in an insulated vessel containing 159 g of water at 22.8°C. Assuming no loss of water and a heat capacity for the vessel of 10.0 J/°C, what is the final temperature of the system (c of copper = 0.387 J/g·°C)?
Final answer:
To calculate the initial temperature of the copper piece, use the principle of energy conservation and the equations for heat gained and heat lost. The final temperature of the system is approximately 24.8 °C.
Explanation:
To calculate the initial temperature of the copper piece, we can use the principle of energy conservation. The heat gained by the water is equal to the heat lost by the copper. The heat gained by the water can be calculated using the formula:
Q = m * c * ΔT
Where Q is the heat gained, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The heat lost by the copper can be calculated using the formula:
Q = m * c * ΔT
Where Q is the heat lost, m is the mass of the copper, c is the specific heat capacity of copper, and ΔT is the change in temperature. Since the final temperature is known, we can rearrange the formulas to solve for the initial temperature of the copper:
Initial temperature of copper = (heat gained by water / (m * c)) + final temperature
Substituting the given values into the formulas, we get:
Heat gained by water = (159 g) * (4.18 J/g·°C) * (39.9 °C - 22.8 °C)
Heat lost by copper = (497 g) * (0.387 J/g·°C) * (final temperature - 89.5 °C)
Setting the two equations equal, we can solve for the final temperature:
(159 g) * (4.18 J/g·°C) * (39.9 °C - 22.8 °C) = (497 g) * (0.387 J/g·°C) * (final temperature - 89.5 °C)
Solving the equation, the final temperature of the system is approximately 24.8 °C.