Equilibrium force is the force that will keep the small
mass in place, hence no movement must be made. So we know that 32 N of force is
acted towards the positive direction so +32 N. Which is counteracted by 26 N
force so:
32 N – 26 N = 6 N (positive)
Since positive 6 is left, therefore this must be acted by
an equilibrant negative 6 N.
Answer:
<span>- 6 N </span>
Answer:
a)15 N
b)12.6 N
Explanation:
Given that
Weight of block (wt)= 21 N
μs = 0.80 and μk = 0.60
We know that
Maximum value of static friction given as
Frs = μs m g = μs .wt
by putting the values
Frs= 0.8 x 21 = 16.8 N
Value of kinetic friction
Frk= μk m g = μk .wt
By putting the values
Frk= 0.6 x 21 = 12.6 N
a)
When T = 15 N
Static friction Frs= 16.8 N
Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.
Friction force = 15 N
b)
When T= 35 N
Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N
Friction force = 12.6 N
Answer:
22.36 rad
Explanation:
Applying,
ω = θ/t.............. Equation 1
Where ω = angular velocity, θ = angular displacement of the baseball, t = time
make θ the subject of the equation
θ = ωt............... Equation 2
From the question,
Given: ω = 350 rev/min = 350(0.10472) = 36.652 rad/s, t = 0.61 s
Substitute these values into equation 1
θ = 0.61(36.652)
θ = 22.36 rad
Hence the angular displacement of the baseball is 22.36 rad
0.495 m/s
Explanation
the formula for the terminal velocity is given by:
![\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2mg%7D%7B%5Csigma%20AC%7D%7D%20%5C%5C%20%5Ctext%7Bwhere%7D%20%5C%5C%20%20%5Cend%7Bgathered%7D)
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient
then
Step 1
Let's find the mass
now, replace
![\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2mg%7D%7B%5Csigma%20AC%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2%280.002kg%29%289.81%5Ctext%7B%20%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%7D%7B%282%5Ccdot10%5E3%5Cfrac%7B%5Coperatorname%7Bkg%7D%7D%7Bm%5E3%7D%29%280.0001m%5E2%290.8%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.03924%5Cfrac%7B%5Coperatorname%7Bkg%7Dm%7D%7Bs%5E2%7D%7D%7B0.16%5Cfrac%7B%5Coperatorname%7Bkg%7D%7D%7Bm%5E%7B%7D%7D%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B0.2452%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%7D%20%5C%5C%20v%3D0.495%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
hence, the answer is 0.495 m/s
