<span>As we Know
ω=v/r
I2 = m*R² = 1.5*0.35² = 0.18375 kgâ™m²
I = Is + I2 = 4.28375 kgâ™m²
w = L/I = (v*R*m)/I = (2.8*0.35*1.5)/4.28375 = 0.343 rad/sec
So the answer is 0.343 rd/sec</span>
Whether a line is "steep" or "shallow" depends on the slope of the line. In the equation y=mx+b, m is the slope. In your first equation, 1/2x+4, the slope would be 1/2 which means there is 1 increase in the vertical, or y, value for each 2 increases in the horizontal, or x value. This line would be shallow.
Take a lamina with three holes near the periphery of the lamina, now suspend the lamina through them, one by one. Draw a line of equilibrium for each suspension point. The point of intersection of these three lines would be the centre of gravity.
Answer:
F = -49.1 10³ N
Explanation:
Let's use the kinematics to find the acceleration the acceleration of the bullet that they tell us is constant
² = v₀² + 2 a x
Since the bullet is at rest, the final speed is zero
x = 11.00 cm (1 m / 100 cm) = 0.110 m
0 = v₀² + 2 a x
a = -v₀² / 2 x
a = -1320²/(2 0.110)
a = -7.92 10⁶ m / s²
With Newton's second law we find the force
F = m a
F = 6.20 10⁻³ (-7.92 10⁶)
F = -49.1 10³ N
The sign means that it is the force that the tree exerts to stop the bullet
