Answer:
v = 0.42m/s
Explanation:
In order o calculate the linear speed of the point at the border of the wheel, you first take into account that the total acceleration of such a point is given by:
(1)
atotal: total acceleration = 1.2m/s^2
ar: radial acceleration of the wheel
at: tangential acceleration
The tangential acceleration is also given by:
(2)
r: radius of the wheel = (40cm/2 )= 20cm = 0.2m
α: angular acceleration = 4.0rad/s^2
You replace the expression (2) into the expression (1) and solve for the radial acceleration:
Next, you use the following formula for the radial acceleration and solve for the linear speed:
The linear speed of the point at the border of the wheel is 0.42m/s
A release of a large amount of energy
The kilogram is the Standard International System of Units unit of mass. It is defined as the mass of a particular international prototype made of platinum-iridium and kept at the International Bureau of Weights and Measures.
Answer:
1. Electromagnetic spectrum
2. Electromagnetic radiation
3. Radiant energy
4. Infared waves
5. Electricmagnetic radiation or Infared waves
6. Radio waves
7. Gamma rays
8. Radio waves
9. Ultraviolet rays
10. This is because it doesn’t contain water or a substance.
11. You should use sunscree and a hat to protect your skin from UV light that might cause skin cancer.
Hope this helps
Answer:
a) 3.7 m/s^2
b) 231.8 N
Explanation:
Let m1 be mass of the first object (m1 = 38.0 kg) and let m2 be the mass of the second object (m2 = 17.0 kg ). Let a be the acceleration of the two objects. Let F1 be the force of gravity exerted on m1 and F2 be the force of gravity exerted on m2. Let M = m1 +m2
a)
F1 = m1g and F2 = m2g
So Fnet = F1 + F2
Since the pulleys will move in different directions when accelerating...
Fnet = F1 - F2
M×a = m1g - mg2
M×a = g×(m1 -m2)
a = g×(m1 - m2)/M
a = 9.8×(38 - 17)/(38 + 17)
a = 3.7 m/s^2
b)
Looking at the part for m2
Fnet = T - m2g
-m2×a = T - m2g
T = m2(g - a)
T = 231.8 N
