Answer:
0.15
Explanation:
Assuming the rope is horizontal, sum the forces in the y direction:
∑F = ma
N − mg = 0
N = mg
Sum the forces in the x direction:
∑F = ma
F − Nμ = ma
Substitute:
F − mgμ = ma
mgμ = F − ma
μ = (F − ma) / (mg)
Plug in values:
μ = (8.0 N − 2.0 kg × 2.5 m/s²) / (2.0 kg × 9.8 m/s²)
μ = 0.15
Answer:
You wouldnt fall you would be sucked and you would lose all air supply and you lungs would pop
Explanation:
Answer:
.
Explanation:
When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.
The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.
Let 
denote the mass of this ball. Let 
denote the mass of water that this ball has displaced.
Let 
denote the gravitational field strength. The weight of this ball would be 
. Likewise, the weight of water displaced would be 
.
For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:
.
At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:
.
Therefore:
.
.
In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let 
denote the density of water. The volume of water that this ball should displace would be:
.
Given that 
while 
:
.
In other words, for this ball to stay afloat, at least of the volume of this ball should be under water. Therefore, the volume of this ball should be at least 
.
Answer:toward the ceiling
Explanation:
if you use the right hand rule it will point to the ceiling
Answer:
The force is 86.5×10^9 N towards the negative charge (to the right)
Explanation:
The electrostatic force on the charges is given by Coulomb's law;
F= Kq1q2/r^2
This an inverse square law.
F= electrostatic force on the charges
K= constant of Coulomb's law
q1 and q2= magnitude of the charges
Since K= 9.0×10^9Nm^2C^2
F= 9.0×10^9 × 5 × 3/(1.25)^2 = 135×10^9/1.56
F= 86.5×10^9 N
The force is 86.5×10^9 N towards the negative charge.
