Answer:
sorry for you
The ratio of the man's moment of inertia in the first case (arms extended) to that in the second case (arms drawn in) is 3.2.
The relationship between the rotational frequency [tex](\(\omega\))[/tex] and moment of inertia (I) is given by the equation:
[tex]\[I_1\omega_1 = I_2\omega_2\][/tex]
where [tex]\(I_1\)[/tex]and [tex]\(I_2\)[/tex] are the moments of inertia in the two cases, and [tex]\(\omega_1\) and \(\omega_2\)[/tex] are the corresponding rotational frequencies.
Let's denote the moment of inertia in the first case (arms extended) as [tex]\(I_1\)[/tex] and in the second case (arms drawn in) as [tex]\(I_2\)[/tex]. The given rotational frequencies are [tex]\(\omega_1 = 0.25 \, \text{rev/s}\) and \(\omega_2 = 0.80 \, \text{rev/s}\)[/tex].
Using the equation [tex]\(I_1\omega_1 = I_2\omega_2\)[/tex], we can rearrange it to solve for the ratio of moments of inertia:
[tex]\[\frac{I_1}{I_2} = \frac{\omega_2}{\omega_1}\][/tex]
Substituting the given values, we have:
[tex]\[\frac{I_1}{I_2} = \frac{0.80 \, \text{rev/s}}{0.25 \, \text{rev/s}}\][/tex]
Simplifying the expression, we get:
[tex]\[\frac{I_1}{I_2} = 3.2\][/tex]
Therefore, the ratio of the man's moment of inertia in the first case (arms extended) to that in the second case (arms drawn in) is 3.2.
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If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k
Recall the definition of the cross product with respect to the unit vectors:
i × i = j × j = k × k = 0
i × j = k
j × k = i
k × i = j
and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)
Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have
a × b = (8i + j - 2k) × (5i - 3j + k)
a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)
… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)
… … … … + 8 (i × k) + (j × k) - 2 (k × k)
a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)
a × b = - 5k - 10j - 24k - 6i - 8j + i
a × b = -5i - 18j - 29k
Answer:
Explanation:
If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k
according to Newton's __ law, an object with no net force acting on it remains at rest or in motion with a constant veloctiy
Answer:
i think its his law of inertia
Explanation:
this law is about motion
A plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again. The apparent weight of the gas is the difference between these two masses. The gas is squeezed out of the bag to determine its volume by the displacement of water. What is the actual weight of the gas
The actual weight of the gas = apparent weight + weight.
The actual weight = [tex]W_{A}[/tex] + W
Given that a plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again.
If the apparent weight of the gas is the difference between these two masses, then let the apparent weight = [tex]W_{A}[/tex]
The gas is squeezed out of the bag to determine its volume by the displacement of water. Since
density = mass / volume
The density of water is 1000 kg/[tex]m^{2}[/tex]
we can get the mass of the gas by making m the subject of the formula.
W = mg
The actual weight of the gas = apparent weight + weight
That is,
The actual weight = [tex]W_{A}[/tex] + W
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TRUE OR FALSE! PLZ HELP
Magnitude is a measure of how strong a force is.
A. True
B. False
Answer:
True
Explanation:
Magnitude is the "value" the greater the value the greater the force is and vice versa
Answer:
TRUE
Explanation: I did the test
Hope this helps
a. Block on a smooth incline.
A block of mass m= 3.8 kg on a smooth inclined plane of angle 36° is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 3 kg hanging vertically. Take the positive direction up the incline and use 9.81 m/s2
for g.
What is the acceleration of each block to 1 decimal place?
Answer:
Explanation:
F = ma
3(9.81) - 3.8(9.81sin36) = (3 + 3.8)a
a = 1.10566...
a = 1.1 m/s²
the 3.8 kg mass will move up slope and the 3 kg mass will fall at that acceleration.
Throwing all over the place will___________the environment.
Answer:
pollute the environment
2. If a vehicle with four wheels is traveling in a linear path through soft dirt, how many and which tires would you expect to leave tire track impressions
Given our understanding of the situation, for this vehicle with four wheels is traveling in a linear path through soft dirt, we would expect for only two wheels to leave tire track impressions.
The vehicle has four wheels, and one might think initially that it makes sense for all four wheels to create tire track impressions on the dirt. This assessment is correct, all four tires will create impressions, However, not all four tires will leave a said impression.
This is due to the fact that as the vehicle advances, the tires in the back will erase the impressions left by the front tires. Therefore, only two tire track impressions will be left.
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Using the expression for the total energy of this system, it is possible to show that after the switch is closed, d2qdt2=−kq, where k is a constant. Find the value of the constant k.
The value of the constant K is [tex]\mathbf{K = \dfrac{1}{LC}}[/tex]
According to Kirchhoff's loop rule, the total algebraic sum of potential differences in any loop, combining voltage provided by voltage sources as well as resistive components, must equal zero.
Thus, the relation for Kirchhoff's loop rule can be expressed as:
[tex]\mathbf{\dfrac{q}{c}- L\dfrac{dI}{dt} = 0}[/tex]
We all know that the current in the nonconstant charge flow can be written as:
[tex]\mathbf{I = \dfrac{dq}{dt}}[/tex]
∴
Replacing the current (I) into Kirchhoff's loop rule, we have:
[tex]\mathbf{ L\dfrac{d}{dt} ( \dfrac{dq}{dt})= -\dfrac{q}{c}}[/tex]
[tex]\mathbf{ \dfrac{d^2q}{dt^2}= -\dfrac{q}{Lc} \ \ ---(1)}[/tex]
From the given question, when the switch is closed
[tex]\mathbf{ \dfrac{d^2q}{dt^2}= -kq\ \ ---(2)}[/tex]
Then, the charges on the capacitor start to b, resulting in the rise of the current in the circuit.
∴
By equating both equations (1) and (2);
[tex]\mathbf{K = \dfrac{1}{LC}}[/tex]
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help meee please please please
If it takes 50.0 seconds to lift 10.0 Newtons of books to a height of 7.0
meters, calculate the power required to lift it?
builder places a 3kg hammer on the top of a ladder, which is 4m above the ground. Calculate the gravitational potential energy of the hammer while on the ladder.
Answer:
[tex]E=mgh[/tex]
[tex]m=3kg[/tex]
[tex]h=4m[/tex]
[tex]g=9.8m/s^{2}[/tex]
[tex]E= 3*4*9.8=117.6J[/tex]
Explanation:
Only substitute amounts to formula.
Hope this helps ;)
Cheers :D
How does climate change?
Answer:
Con el cielo del agua
Explanation:
espero que te ayude
please.
I mark u brainliest answer
Answer:
destroy
Explanation:
cuz if anyhow throw rubbish, it will affect/destroy the world environment
Answer:
bathing regularly 2pollute 3 throwing waste in land water water pollution
PLEASEEEEEEEE HELPPPPPPP
Answer: 62.25
Explanation: F = ma
F = 7.5 * 8.3
F = 62.25
Thus, the answer is 62.25 Newtons.
Would appreciate brainliest <3
For a certain transverse standing wave on a long string, an antinode is at x = 0 and an adjacent node is at x = 0.20 m. The displacement y(t) of the string particle at x = 0 is shown in the figure, where the scale of the y axis is set by ys = 4.3 cm. When t = 0.90 s, what is the displacement of the string particle at (a) x = 0.30 m and (b) x = 0.40 m ? What is the transverse velocity of the string particle at x = 0.30 m at (c) t = 0.90 s and (d) t = 1.3 s?
The expressions for the traveling and standing wave to find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
The traveling wave is a disturbance in the medium that moves at constant speed, in the case of a transverse wave the expression for the perpendicular oscillation is:
y = A sin (kx - wt)
Where y is the oscillation perpendicular to the direction of the displacement, A the amplitude, k in wave number and w the angular velocity.
Standing waves are formed when a traveling wave collides with an obstacle and is reflected, in this case the sum of the two waves gives a wave that does not shift in time and fulfills the relationship
[tex]\frac{\lambda}{2} = \frac{L}{n}[/tex]
Where λ is the wavelength, L the distance between the reflection points and n the number of nodes.
Indicates that for the standing wave the distance between an antinode and the node is x = 0.20 m, therefore
[tex]\frac{\lambda}{4} = \frac{L}{1}[/tex]
λ = 4L
λ = 4 0.20
λ = 0.80 m
The wave number.
k = [tex]\frac{2\pi }{\lambda }[/tex]
k = [tex]\frac{2 \pi }{0.80 }[/tex]
k = 2.5π i m⁻¹
In the associated traveling wave, from the graph we can see that the period of the wave is:
T = 2.8 s
the angular velocity is related to the period.
[tex]w=\frac{2\pi}{T} \\w = \frac{2\pi }{2.8}[/tex]
w = 0.714π rad/s
indicate the maximum displacement that is the amplitude of the wave.
A = [tex]y_s[/tex]
A = 4.3 cm
Let's write the equation of the traveling wave.
y = 4.3 sin [π (2.5 x - 0.714 t)]
with this expression we can answer the questions.
a) the displacement of the particle for x = 0.30 m
y = 4.3 sin (π (2.5 0.30 - 0.714 t))
y = 4.3 sin π( 0.75 - 0.714 t(
Remember that the angles must be in radians. For time t = 0 the displacement is
y = 4.3 0.707
y = 3.04 cm
b) The displacement for x = 0.4m
y = 4.3 sin (π 2.5 0.4)
y = 0 cm
c) the transverse velocity of the wave at x = 0.30 m for the time of t = 0.90s
the speed of the wave is
[tex]v= \frac{dy}{dt} \\v= A w cos ( kx - wt)[/tex]
v = 4.3 0.714π cos π(2.5 0.3 - 0.714 t)
v = 9.65 cos π(0.75 - 0.714 t)
For time t = 0.90 s the velocity is:
v = 9.65 cos π(0.75 - 0.714 0.9)
v = 9.65 0.9436
v = 9.11 cm / s
d) The velocity for time t = 1.3 s
v = 9.65 cos π(0.75 - 0.714 1.3)
v = 9.65 0.9999
v = 9.65 cm / s
In conclusion, using the expressions for the traveling and standing wave, we can find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
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Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a horizontal surface of negligible friction, releasing the block, and measuring the velocity v of the block as it leaves the spring, as shown in Figure 1. The experiments indicate that as x increases, so does v in a linear relationship. The surface is now lifted so that the surface is at an angle θ above the horizontal. Which of the following indicates how the relationship between v and x changes?
Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.
The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.Reasons:
The energy given to the block by the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]
According to the principle of conservation of energy, we have;
On a flat plane, energy given to the block = [tex]0.5 \cdot k \cdot x^2[/tex] = kinetic energy of
block = [tex]0.5 \cdot m \cdot v^2[/tex]
Therefore;
0.5·k·x² = 0.5·m·v²
Which gives;
x² ∝ v²
x ∝ v
On a plane inclined at an angle θ, we have;
The energy of the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]
The force of the weight of the block on the string, [tex]F = m \cdot g \cdot sin(\theta)[/tex]
The energy given to the block = [tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)[/tex] = The kinetic energy of block as it leaves the spring = [tex]\mathbf{0.5 \cdot m \cdot v^2}[/tex]
Which gives;
[tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2[/tex]
Which is of the form;
a·x² - b = c·v²
a·x² + c·v² = b
Where;
a, b, and c are constants
The graph of the equation a·x² + c·v² = b is an ellipse
Therefore;
As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.Please find attached a drawing related to the question obtained from a similar question online
The possible question options are;
As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of xThe relationship is no longer linear and v will be more for the same value of xThe relationship is still linear, with lesser value of vThe relationship is still linear, with higher value of vThe relationship is still linear, but vary inversely, such that as x increases, v decreasesLearn more here:
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We assume that , the acceleration due to gravity on the surface of the earth, is 32 feet per second per second. If the pendulum is that of a clock that keeps good time when feet, how much time will the clock gain in 24 hours if the length of the pendulum is decreased to 3.97 feet
The period of the pendulum allows to find the result for how much it advances when the length is reduced is:
Δt = 1550 s
Simple harmonic motion is an oscillatory motion where the restoring force is proportional to the displacement.
In the case of the simple pendulum, this is fulfilled for small angles minus 15º, the angular velocity of the pendulum is
w = [tex]\frac{g}{L}[/tex]
Angular velocity and period are related
w = [tex]\frac{2pi}{T}[/tex]
We substitute
T = [tex]2\pi \ \frac{L}{g}[/tex]
They indicate that for the initial length L₀ and the pendulum marks the exact time, how much time changes if the length is 3.97 feet, therefore the initial length is L₀ = 3.90 feet.
[tex]T_o^2 = 4 \pi ^2 \ \frac{L_o}{g}[/tex]
The period for the reduced length is:
[tex]T'^2 = 4\pi ^2 \ \frac{L}{g}[/tex]
The relationship between the periods is:
[tex]( \frac{T}{T'}^2 = \frac{L_o}{L} \\T' = \sqrt{\frac{L_o}{L} } \ T[/tex]
Let's calculate
T ’= [tex]T \ \sqrt{\frac{3.97}{3.9} }[/tex]
T ’= T 1.01795
In the total time of a day.
T = 24 hours (3600 s / 2h) = 86 400 s
We calculate
T ’= 86400 1.01795
T ’= 87,950 s
Therefore the pendulum moves forward in a time of:
ΔT = T'- T
ΔT = 87950 - 86400
ΔT = 1550 s
In conclusion, using the period of the pendulum we can find the result for how much it advances when the length is change is:
ΔT = 1550 s
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A racecar can be slowed with a constant acceleration of -14 m/s2. If the car is going 75 m/s, how many meters will take to stop?
Answer:
Explanation:
v² = u² + 2as
s = (v² - u²)/2a
s = (0² - 75²) / (2(-14))
s = 200.8928
s = 200 m
Answer:
Identification
Explanation:
:-;..........
To become a healthy individual, a person must have a properly executed routine to achieve his/her fitness goals. On the contrary, in what instance does a person become unhealthy
A 260-kg object and a 560-kg object are separated by 3.10 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 49.0-kg object placed midway between them.
C is the correct answer:))))))))
Two ends of a steel wire of length 8m and 2mm radius are fixed to two rigid supports. Calculate the increase in tension when the temperature falls by 10°C. Given linear expansivity of steels = 12x10^_6 per kelvin and Young's modules for steel =2x10^11 n/m^2
The increase in tension on the steel wire is 8,484.75 N.
The given parameters;
original length of the wire, l = 8 mradius of the wire, r = 2 mmThe area of the steel wire is calculated as follows;
[tex]A = \pi r^2\\\\A = \pi \times (2\times 10^{-3})^2\\\\A = 1.257 \times 10^{-5} \ m^2[/tex]
The extension of the steel wire is calculated as follows;
[tex]\Delta l = \alpha \times l\times \Delta T\\\\\Delta l = (12\times 10^{-6}) \times (8) \times (10 + 273)\\\\\Delta l = 0.027 \ m[/tex]
The increase in tension on the steel wire is calculated as follows;
[tex]E = \frac{stress}{strain } = \frac{\ F/A}{\Delta l/l} \\\\E = \frac{F\times l}{A \times \Delta l} \\\\F = \frac{E\times A \times \Delta l }{l} \\\\F = \frac{(2\times 10^{11}) \times (1.257\times 10^{-5})\times 0.027}{8} \\\\F = 8,484.75 \ N[/tex]
Thus, the increase in tension on the steel wire is 8,484.75 N.
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What’s 1 + 1 many half a window
Answer:
1
+
1
is
2
Explanation:
because if you add one to a one it would end up being 2 <3
Answer: 2
Explanation: 1+1=2
This was THE hardest question I've ever decided to answer
This drawing of the field lines between these two charged plates is correct.
True or false
Answer:
True
Explanation:
Since positive and negative forces are both opposites of each other they attract. But if the forces are the same where both are positive or negative thye repel do to being the same charges in that case the picture would show lines moving in opposite directions.
The same type of engine is placed in iwo different airplanes The first airplane is twice as heavy as the second suchane Wrich
statement is correct about the mass of the airplane and its acceleration, assuming a constant not forest (1 point)
Increasing the acceleration of the airplane causes the enging to provide more force
Decreasing the mass of the airplane will result in the airplane accelerating loss
O Decreasing the mass of the airplane will result in the airplane accelerating more
increasing the acceleration of the airplane causes the engine to provide les force
Answer:
Explanation: If the engines in both of the airplanes are at the same Weight then The engines in both of the airplanes would stil have the same mass because mass stays the same everywhere
Since mass is constant everywhere, if the engines in both airplanes are at the same weight, the engines in both airplanes would still have the same mass.
What is Weight?Weight is the force an object exerts, mass is the amount of matter that makes up an object. Newton is the SI unit of weight while kilogram is the SI unit of mass.
Given that in question the same type of engine is placed in two different air-planes The first air-plane is twice as heavy as the second Suchane Wrich statement is correct about the mass of the air-plane and its acceleration, assuming a constant net force.
Increasing the acceleration of the air-plane causes the enging to provide more force.
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2. Match the term to the definition
1. Reversibility
a. A break between activities during a workout.
2. Overload
3. Progression
4 Duration
5. Rest
b. Length of an individual workout.
c. The Improvement seen when you are training will be lost if you stop
training.
d. Moving forward through a sequence to gradually build up the body.
e Number of training sessions per week.
f. Level of difficulty of your workout.
9. Improvements will only be seen in the areas that you work on.
h The time between workouts that allows your body to heal and adapt to
the stress from your workout.
6. Recovery
7. Frequency
8. Specificity
1. Increase the amount of stress you put on your body.
9. Intensity
the wheel of bicycle has a radius of 25cm. what will be the magnitudes of the angular displacement in radian and revolution respectively, when the wheel has rolled a distance of 350cm on straight level road?
Answer:
Explanation:
350 cm / 25 cm = 14 radians
14 rad / 2π rad/rev = 2.23 revolutions
A single paragraph can be selected by triple in the paragraph. ture or false
Answer:
The answer for the question should be True
II) A 0.40-kg ball, attached to the end of a horizontal ord, is rotated in a circle of radius 1.3 m on a friction- less horizontal surface. If the cord will break when the tension in it exceeds 60 N, what i~ the maximum speed the ball can have
Hi there!
In this instance, the object spinning in a horizontal circle will experience a net force in the horizontal direction due to tension.
The net force is equivalent to the centripetal force, so:
∑F = T
mv²/r = T
Solve for v:
v = √rT/m
v = 13.96 m/s
A car driving on level ground at 20.0 m/s slams on its brakes and skids to a halt. If the coefficient of kinetic friction between the car’s tires and the road is 0.750, how far does the car skid before stopping? How far would the car have skidded if it had been moving at 40.0 m/s?
The distance the car has skidded if it had been moving at 40.0 m/s is 27.2m
The linear force acting on the car is opposed by the frictional force. Hence;
[tex]F=F_f[/tex]
[tex]ma = -\mu R\\ma =-\mu mg[/tex]
m is the mass of the car
a is the acceleration
[tex]\mu[/tex] is the coefficient of friction
R is the normal force
Given the following parameters
[tex]a=-\mu g[/tex]
[tex]a =-0.75(9.8)\\a=-7.35m/s^2[/tex]
Get the distance the car has skidded if it had been moving at 40.0 m/s
[tex]v^2=u^2+2as[/tex]
[tex]0^2=20^2+2(-7.35)s\\0=400-14.7s\\s=\frac{400}{14.7}\\s= 27.2m[/tex]
Hence the distance the car has skidded if it had been moving at 40.0 m/s is 27.2m
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Calculate the altitude above the surface of Earth, in meters, at which the acceleration due to gravity is g
Answer:
By definition the acceleration due to gravity at the surface is g:
The altitude above the surface is zero for an acceleration of g.