Tornadoes are formed when wind travels into a low pressure area from opposite directions and creates a spinning storm. The tornado comes in contact to the earth’s ground and to the clouds (usually cumulus cloud) having a smaller cross sectional area at the ground than on the sky.
The rotation period that will provide normal gravity will be 57.12 seconds.
From the information given, the diameter of the cylinder is 1600m. Therefore, the radius will be:
= Diameter / 2.
= 1600 / 2
= 800m
Acceleration due to gravity = 9.8m/s²
Since a = rw², we'll use the subject of the formula to find w and this will be:
w = √g/✓r
w = ✓9.8 / √800
w = 0.11 rad/s
Therefore, the rotation period will be:
T = 2π/w
T = 2π/0.11
T = (2 × 3.142) / 0.11
T = 57.12 seconds
In conclusion, the correct option is 57.12 seconds.
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Answer
given,
diameter of the cylinder = 1600 m
radius = 800 m
acceleration due to gravity = 9.8 m/s²
a = r ω²
we know time period
T = 57.12 s
She can monitor her practice run times to see if they are decreasing.
Answer: a
Explanation:
The energy equation, E=12mvx2+12kx2=12kA2, is a useful alternative relationship between velocity and position, especially when energy quantities are also required. If the problem involves a relationship among position, velocity, and acceleration without reference to time, it is usually easier to use the equation for simple harmonic motion, ax=d2xdt2=−kmx (from Newton’s second law) or the energy equation above (from energy conservation) than to use the general expressions for x, vx, and ax as functions of time. Because the energy equation involves x2 and vx2, it cannot tell you the sign of x or of vx; you have to infer the sign from the situation. For instance, if the body is moving from the equilibrium position toward the point of greatest positive displacement, then x is positive and vx is positive.
IDENTIFY the relevant concepts
Energy quantities are required in this problem, therefore it is appropriate to use the energy equation for simple harmonic motion.
SET UP the problem using the following steps
Part A
The following is a list of quantities that describe specific properties of the toy. Identify which of these quantities are known in this problem.
Select all that apply.
Select all that apply.
maximum velocity vmax
amplitude A
force constant k
mass m
total energy E
potential energy U at x
kinetic energy K at x
position x from equilibrium
Part B
What is the kinetic energy of the object on the spring when the spring is compressed 5.1 cm from its equilibrium position?
Part C
What is the potential energy U of the toy when the spring is compressed 5.1 cm from its equilibrium position?
Answer:
Part A
Mass = 50g
Vmax = 3.2m/s
Amplitude= 6cm
Position x from the equilibrium= 5.1cm
Part B
Kinetic energy = 0.185J
Part C
Potential energy = 0.185J
Explanation:
Kinetic energy = 1/2mv×2
Vmax = wa
w = angular velocity= 53.33rad/s
Kinetic energy = 1/2mv^2×r^2 = 0.185J
Part c
Total energy = 1/2m×Vmax^2= 0.256J
1/2KA^2= 0.256J
K= 142.22N/m (force constant)
Potential energy = 1/2kx^2
=1/2×142.22×0.051^2
= 0.185J
To find the kinetic energy of the toy, we need to use the energy equation for simple harmonic motion and the relationship between velocity and position. We can then substitute the known values to calculate the kinetic energy.
In this problem, we are given the amplitude (A) of the oscillation and the maximum velocity (vmax) achieved by the toy. We need to find the kinetic energy (K) of the toy when the spring is compressed 5.1 cm from its equilibrium position.
To solve for the kinetic energy, we can use the energy equation for simple harmonic motion: K = 1/2mvx2, where m is the mass of the object and vx is the velocity of the object at position x. The mass of the object is given as 50 g, which is equal to 0.05 kg.
Since we know the maximum velocity (vmax = 3.2 m/s), we can use the relationship between velocity and position in simple harmonic motion to find the velocity (vx) at a displacement of 5.1 cm from the equilibrium position. The velocity and position in simple harmonic motion are related by vx = ±ω√(A2 - x2), where ω is the angular frequency of the motion.
Substituting the known values into the equations, we can calculate the kinetic energy of the toy.
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Knowing about distance between a gun fired and it's target is very useful for accurate shot.
It might be useful to know the distance between a gun fired and it's target because it gives us information about the angle at which the target has been hitting. If we do not know about distance between a gun fired and it's target, we can't shot the target accurately.
So we can conclude that knowing about distance between a gun fired and it's target is very useful for accurate shot.
Learn more about distance here: brainly.com/question/4931057
Answer: h =1.22 m
Explanation:
from the question we were given the following
mass of performer ( M1 ) = 77 kg
length of cable ( R ) = 3.65 m
mass of costar ( M2 ) = 55 kg
maximum height (h) = ?
acceleration due to gravity (g) = 9.8 m/s^2 (constant value)
We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy
work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )
initial velocity is zero in this case because the performer was at rest before swinging, therefore
work done = 1/2 x 77 x ( v^2 - 0)
work done = 38.5 x ( v^2 ) ......equation 1
work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have
m x g x R = 38.5 x ( v^2 )
77 x 9.8 x 3.65 = 38.5 x ( v^2 )
2754.29 = 38.5 x ( v^2 )
( v^2 ) = 71.54
v = 8.4 m/s ( velocity of the performer)
After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is
initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)
momentum = mass x velocity thereforethe equation above now becomes
(77 x 8.4) + (55 x 0) = (77 +55) x Vf
take note the the initial velocity of the costar is 0 before pick up because he is at rest
651.3 = 132 x Vf
Vf = 4.9 m/s
the performer and his costar is 4.9 m/s after pickup
to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0
initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy
where
kinetic energy = 1/2 x m x v^2
potential energy = m x g x h
after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes
initial kinetic energy = final potential energy
(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0
1584.7 = 1293 x h
h =1.22 m