<h2>
Answer: Toward the center of the circle.</h2>
This situation is characteristic of the uniform circular motion , in which the movement of a body describes a circumference of a given radius with constant speed.
However, in this movement the velocity has a constant magnitude, but its direction varies continuously.
Let's say is the velocity vector, whose direction is perpendicular to the radius of the trajectory, therefore
the acceleration is directed toward the center of the circumference.
Taking ratio of W & w. ≈ 6 . w = 1/6 W. Therefore , Weight of an object on the moon is 1/6 of its weight on the earth.
Answer:
Explanation:
a )
Depth of hole from surface of water d = .50 m - .03 m = .47 m
velocity of efflux v = √ 2gd
v = √ (2 x 9.8 x .47 )
v = 3.03 m /s
b )
Volume flow rate = π R² v where R is radius of hole at the bottom .
= 3.14 x ( .005 ) ² x 3.03 m/s
= 2.378 x 10⁻⁴ m³ /s
c )
Volume of water collected in 60 s
= 2.378 x 10⁻⁴ x 60
= 1.4268 x 10⁻² m³
If height attained in collecting container be h
π R² h = 1.4268 x 10⁻² m³ where R is radius of container
3.14 x ( .1 )² x h = 1.4268 x 10⁻²
h = .4544 m .
Pressure at the bottom of container = hρ g
where h is height of water , ρ is density of water
Pressure = .4544 x 1000 x 9.8 N /m²
= 4453.12 N /m²
Answer:
2.3687599 m/s
0.91106 m/s
0.617213012 J
Explanation:
f = Frequency =
A = Amplitude = 0.13 m
k = Spring constant
m = Mass of object = 0.22 kg
Angular velocity is given by
Velocity is given by
Speed when it passes the equilibrium point is 2.3687599 m/s
Frequency is given by
x = Displacement = 0.12 m
In this system the energies are conserved
The speed when it is 0.12 m from equilibrium is 0.91106 m/s
The energy in the system is given by
The total energy of the system is 0.617213012 J
Answer:
The ballon will brust at
<em>Pmax = 518 Torr ≈ 0.687 Atm </em>
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Explanation:
Hello!
To solve this problem we are going to use the ideal gass law
PV = nRT
Where n (number of moles) and R are constants (in the present case)
Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.
--- (*)
Our initial state is:
P1 = 754 torr
V1 = 3.1 L
T1 = 294 K
If we consider the final state at which the ballon will explode, then:
P2 = Pmax
V2 = Vmax
T2 = 273 K
We also know that the maximum surface area is: 1257 cm^2
If we consider a spherical ballon, we can obtain the maximum radius:
Rmax = 10.001 cm
Therefore, the max volume will be:
Vmax = 4 190.05 cm^3 = 4.19 L
Now, from (*)
Therefore:
Pmax= P1 * (0.687)
That is:
Pmax = 518 Torr