Answer:
Explanation:
Applied force, F = 18 N
Coefficient of static friction, μs = 0.4
Coefficient of kinetic friction, μs = 0.3
θ = 27°
Let N be the normal reaction of the wall acting on the block and m be the mass of block.
Resolve the components of force F.
As the block is in the horizontal equilibrium, so
F Cos 27° = N
N = 18 Cos 27° = 16.04 N
As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .
The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)
The vertically downward force acting on the block is mg - F Sin 27°
= mg - 18 Sin 27° = mg - 8.172 ... (2)
Now by equating the forces from equation (1) and (2), we get
mg - 8.172 = 6.42
mg = 14.592
m x 9.8 = 14.592
m = 1.49 kg
Thus, the mass of block is 1.5 kg.
Answer:
what was the answer
Explanation:
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Answer options:
A) Some connections are lost because critical stages of development were missed.
B) Some connections are lost so others can be strengthened.
C) Some connections are lost because of experimentation with drugs or alcohol.
D) Some connections are lost so that children become less dependent and are less tied to their parents.
Answer:
B) Some connections are lost so others can be strengthened.
Explanation:
Synaptic pruning refers to a natural process that occurs in the brain between childhood and adulthood. Extra synaptic connections are eliminated in order to increase the efficiency of neuronal communications. It is thought to be the brain’s way of removing connections that are no longer needed.
Answer:
F = 789 Newton
Explanation:
Given that,
Speed of the car, v = 10 m/s
Radius of circular path, r = 30 m
Mass of the passenger, m = 60 kg
To find :
The normal force exerted by the seat of the car when the it is at the bottom of the depression.
Solution,
Normal force acting on the car at the bottom of the depression is the sum of centripetal force and its weight.



N = 788.6 Newton
N = 789 Newton
So, the normal force exerted by the seat of the car is 789 Newton.
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