A tin can has a radius of 0.0410m and is 185 m high if the air is removed from inside the can how much forces will the outside a
1 answer:
The force exerted on the bottle is 532 N
Explanation:
When the air is removed from inside the bottle, vacuum is created, therefore the external pressure (the atmospheric pressure) is no longer balanced, and it creates a net force downward on the can.
The pressure is related to the force by the equation:
where
p is the pressure
F is the force
A is the area of the can
Here we have:
is the atmospheric pressure
r = 0.0410 m is the radius of the can, so the area is
And solving for F, we can find the force on the can:
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Answer:
a
The total charge is
b
The change in potential energy is
c
The height is
Explanation:
From the question we are told that the
The magnitude of electric field is
The number of proton is
The distance between the plates is
The total charge can be mathematically represented as
Where e is the charge on one proton which has a value of
Substituting values
The change in potential energy is mathematically represented ads
where the negative sign shows that the work done by the electric force is against the electric field
Substituting values
The mass of the bead is given as 0.05kg
The change in potential due to gravity is mathematically given as
the negative sign is due to the fact that the height is decreasing
And
Making h the subject
Substituting values
Answer:
the block that starts moving first is block A , fr = 1.625 N , fr = 1.5 N
Explanation:
For this exercise we use Newton's second law, for which we take a reference system with the x axis parallel to the plane and the y axis perpendicular to the plane
X axis
fr- Wₓ = 0
fr = Wₓ
Axis y
N- = 0
N = W_{y}
Let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
Cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
Wₓ = 11 sin θ
W_{y} = 11 cos θ
The equation for friction force is
fr = μ N
We substitute
μ (W cos θ) = W sin θ
μ = tan θ
We can see that the system began to move the angle.
θ = tan⁻¹ μ
So the angles are
Block A θ = tan⁻¹ 0.15
θ = 8.5º
Block B θ = tan⁻¹ 0.26
θ = 14.6º
So the block that starts moving first is block A
The friction force is
Block A
fr = Wx = W sin θ
fr = 11 sin 8.5
fr = 1.625 N
Block B
fr = 6 sin 14.6
fr = 1.5 N