Answer: The velocity of the kid is 5.6 m/s
Explanation: Ok, the velocity of the kid will be:
v = w*r
where r is the radius, and w is the frequency.
We know that the diameter is 18m, and the diameter is equal to two times the radius. So r = 18m/2 = 9m
Now, we know that the circumference of a circle is equal to c = 2pi*r, so each revolution has this length, if the kid does 5.9 revolutions in one minute, then the kid spins at v = 5.9*2pi*9m/min
But we want to write this in meters per second, this means that we need to divide it by 60.
v = (5.9*2pi*9/60)m/s = 5.56 m/s
Velocity of the child= 20008.1 m/s
Explanation;
diameter= 18 m
the linear velocity v is given by
v= r ω
r= radius=18/2= 9 m
ω= 5.9 rev/ min = 5.9 rev/min* [2π rad/ 1 rev] *[60 s/ 1 min]=2223.1 rad/s
so V= 9 (2223.1)
V= 20008.1 m/s
(A) m3
(B) 1.8 m3
(C) 3.6 m3
(D) 6 m3
(E) 9 m3
Answer:
(C)
Explanation:
=
Since the object is a solid sphere, the equation for rotational inertia is:
The provided question seems to have a discrepancy as the calculated value of rotational inertia for a spherical object with a given mass-radius relationship is 4.5M³, which does not match any of the supplied answer choices.
The question is asking for the correct expression for the rotational inertia of a spherically shaped object with mass distribution given by the radius as a function of mass (r = km² where k = 3). The rotational inertia, or moment of inertia, for a solid sphere is given by the formula ⅒MR², where M is the mass of the sphere, and R is its radius. Considering that R is defined by r = km², we substitute R with km² in the formula:
I = ⅒M(km²)² = ⅒Mk²m⁴ = ⅒Mk²M²
Since k = 3, we further simplify the expression:
I = ⅒M(3M)² = ⅒(3²)M³ = ⅒ × 9M³ = 4.5M³
However, none of the options (A) to (E) match the value 4.5M³, which indicates there may be an error in the supplied options or an error within the initial assumptions or question parameters. It's important to recheck the given data and the calculation steps to ensure accuracy. If the question and the parameters are indeed accurate as stated, additional information or clarification would be necessary.
The correct option is option (1)
The faster movement of air on the upper surface of the paper creates lower pressure above the paper.
The movement of air is always from a region of higher pressure to a region of lower pressure.
As we blow air above the paper strip a low pressure is created above the strip due to the fast movement or high speed of the air. And the pressure below the strip is higher in comparison to the pressure above since the air below is not moving.
So, due to the pressure difference, a force is generated on the paper strip by the air from the lower surface to the upper surface.
Learn more about pressure difference:
This is happened because "the air" above "moves faster" and "the pressure" is "lower".
Option: 1
Explanation:
Air movement take place from the region where air pressure is more than the region where the pressure is low. When we "blow" air above the "paper strip" paper rises because "low pressure" is created above the strip as high speed winds always travel with reduced air pressure. Hence due to higher air pressure below the strip, it is pushed upwards. This difference in pressure results into fast air moves. This happen because "speed" of the wind depends on "the difference between the pressures" of the air in the two regions.
Answer:
Explanation:
designer
illusionist
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inventor
If f1= 300, F2= 60, and f2=260, find f1 to the nearest unit.
Answer:
F1 = 80
Explanation:
f1= f2 √ (F1/F2)
Where f1 = 300, f2 = 260 and F2 = 60
Putting in the above formula
300 = 260√(F1/60)
Dividing both sides by 260
=> 1.15 = √(F1/60)
Squaring both sides
=> 1.33 = F1/60
Multiplying both sides by 60
=> F1 = 80
Answer:
z_c = ⅜R
Explanation:
If we assume that the hemisphere has uniform density, we can express the centre of mass as;
z_c = (ρ/M)∫∫∫ z•dV
We know that density(ρ) = mass(M)/volume(V)
Thus, Vρ = M
And volume of hemisphere = 2πr³/3
Thus;
2Vρπr³/3 = M
So;
z_c = (ρ/(2Vρπr³/3))∫∫∫ z•dV
Where r = R in this case.
ρ will cancel out to give;
z_c = (3/(2πr³))∫∫∫_V (z•dV)
In spherical coordinates,
r is radius
Φ = angle between the point and the z − axis
θ = azimuthal angle
Therefore, the integral becomes what it is in the attached image.
I've completed the explanation as well in the attachment.
Answer:
Speed of ball just before it hit the surface is 31.62 m/s .
Explanation:
Given :
Mass of ball , m = 50 g = 0.05 kg .
Height from which it falls , h = 80 m .
Thermal energy , E = 15 J .
Now , Initial energy of the system is :
Here , initial velocity is zero .
Therefore ,
Now , final energy of the system :
Since , no external force is applied .
Therefore , total energy of the system will be constant .
By conservation of energy :
Therefore , speed of ball just before it hit the surface is 31.62 m/s .
Using the principle of conservation of energy, the speed of the ball just before hitting the Earth's surface is found to be 79.2 m/s after accounting for the 15 J increase in thermal energy.
This question is concerned with the concept of conservation of energy, specifically the principles of potential and kinetic energy. When the ball is 80 meters above the Earth's surface, the total gravitational potential energy is m*g*h = 50g*9.8m/s²*80m = 39200 J (where m is mass, g is gravity, and h is height), and the kinetic energy is 0.
As the ball falls, its potential energy gets converted into kinetic energy, but we also know that the total thermal energy of the ball and the air in the system increases by 15 J. That means that not all the potential energy is converted into kinetic energy, 15 J is lost to thermal energy. So, the kinetic energy of the ball when it hits the Earth is 39200 J - 15 J = 39185 J.
Finally, we know that kinetic energy equals (1/2)*m*v², where v is the speed of the ball. Rearranging this formula to solve for v we get, v = sqrt((2*kinetic energy)/m) = sqrt((2*39185 J)/50g) = 79.2 m/s. So, just before the ball hits the surface, its speed is 79.2 m/s.
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