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:
fact one
Explanation:
fact two is going slower and a longer distance than fact one so fact one will get there first.
hope this helps
The main danger is vehicles making u-turns or pulling out without signalling.
One of the main uses of this device is to test a beam of charged particles
What is cathode ray tube?
The cathode ray tube determines the charge flowing in a gas. When an electric field is set up using metal electrodes, the cathode ray tends to bend towards the positive electrode.
This concludes that a charge and the electrodes present helps determine the charge of the beam of charged particle.
Thus, one of the main uses of this device is to test a beam of charged particles.
Learn more about cathode ray tube.
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