Answer:
C. Up, equal to the can's weight
Explanation:
You are camping in the breathtaking mountains if Colorado. You spy an unopened diet soda can floating motionless below the surface of a lake. What is the direction and amount of force the water exerts on it?
A. Zero
B. Down, equal to the can's weight
C. Up, equal to the can's weight
D. Not enough information is given
from the principle of flotation which states that a
When a body displaces a weight of water equal to its own weight, it floats. : A floating object displaces a weight of fluid equal to its own weight. ... Archimedes' principle equates the buoyant force to the weight of the fluid displaced.
the upthrust (this is the upward vertical force exerted on an object in fluid)in the water equals the weight of the body in water it floats.
Answer:
The only incorrect statement is from student B
Explanation:
The planet mercury has a period of revolution of 58.7 Earth days and a rotation period around the sun of 87 days 23 ha, approximately 88 Earth days.
Let's examine student claims using these rotation periods
Student A. The time for 4 turns around the sun is
t = 4 88
t = 352 / 58.7 Earth days
In this time I make as many rotations on itself each one with a time to = 58.7 Earth days
#_rotaciones = t / to
#_rotations = 352 / 58.7
#_rotations = 6
therefore this statement is TRUE
student B. the planet rotates 6 times around the Sun
t = 6 88
t = 528 s
The number of rotations on itself is
#_rotaciones = t / to
#_rotations = 528 / 58.7
#_rotations = 9
False, turn 9 times
Student C. 8 turns around the sun
t = 8 88
t = 704 days
the number of turns on itself is
#_rotaciones = t / to
#_rotations = 704 / 58.7
#_rotations = 12
True
The only incorrect statement is from student B
The moons volume is that of 2 percent of the earth.
Answer:
Explanation:
Given a particle of mass
M = 1.7 × 10^-3 kg
Given a potential as a function of x
U(x) = -17 J Cos[x/0.35 m]
U(x) = -17 Cos(x/0.35)
Angular frequency at x = 0
Let find the force at x = 0
F = dU/dx
F = -17 × -Sin(x/0.35) / 0.35
F = 48.57 Sin(x/0.35)
At x = 0
Sin(0) =0
Then,
F = 0 N
So, from hooke's law
F = -kx
Then,
0 = -kx
This shows that k = 0
Then, angular frequency can be calculated using
ω = √(k/m)
So, since k = 0 at x = 0
Then,
ω = √0/m
ω = √0
ω = 0 rad/s
So, the angular frequency is 0 rad/s
Working of a Half wave rectifier
The diode is connected in series with the secondary of the transformer and the load resistance RL. The primary of the transformer is being connected to the ac supply mains. The ac voltage across the secondary winding changes polarities after every half cycle of the input wave.
