Answer:
h = 4T/dgr
Explanation:
Using Laplace law for spherical bubble, the pressure difference P'- P = 4T/r where T = surface tension and r = radius of sphere.
Now, the pressure difference on the hollow sphere P' - P = dgh where d = density of liquid, g = acceleration due to gravity and h = maximum depth to which sphere must be immersed.
So dgh = 4T/r
h = 4T/dgr
Answer:
Upward: Negative
Downward: Positve
Explanation:
Let upward be positive
We want to find the work done on the object upwards.
The equation of work done due to gravity is
It is read as force of gravity times displacement times cos of the angle between the force and displacement.
Since gravity is happening down of the center of mass and the displacement is upwards, the angle is 180° so
Cosine of 180 is -1
so
If we go downwards, the sign will be positive because cos(0)=1
Answer:
1800W
Explanation:
Power=Current×Potential difference
=15A×120V
=1800W
Answer:
Explanation:
According to the described situation we have the following data:
Horizontal distance between lily pads:
Ferdinand's initial velocity:
Time it takes a jump:
We need to find the angle at which Ferdinand jumps.
In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:
(1)
(2)
(3)
On the other hand, the x-component of the velocity is expressed as:
(4)
Substituting (3) in (4):
(5)
Clearing :
This is the angle at which Ferdinand the frog jumps between lily pads
Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m