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.
60 RPM equals one hertz (i.e., one revolution per second, or a period of one second). The SI unit for period is the second.
Answer:
garden
Explanation: All the other habitats are aquatic
Supposing there's no air
resistance, horizontal velocity is constant, which makes it very easy to solve
for the amount of time that the rock was in the air.
Initial horizontal
velocity is: <span>
cos(30 degrees) * 12m/s = 10.3923m/s
15.5m / 10.3923m/s = 1.49s
So the rock was in the air for 1.49 seconds. </span>
<span>
Now that we know that, we can use the following kinematics
equation:
d = v i * t + 1/2 * a * t^2
Where d is the difference in y position, t is the time that
the rock was in the air, and a is the vertical acceleration: -9.80m/s^2. </span>
<span>
Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s
So:
d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2
d = 8.94 + -10.89</span>
d = -1.95<span>
<span>This means that the initial y position is 1.95 m higher than
where the rock lands. </span></span>
