Answer:
Force from the support closest = 79.8 N
Force from the support furthest = 61.9 N
Explanation:
Let's say the length of the beam is L. Let's say A is the near end of the beam and B is the far end of the beam.
Draw a free body diagram. There are four forces on the beam:
Reaction force Ra at the near end (0),
Reaction force Rb at the far end (L),
Weight force of the beam Mg at the center (L/2),
Weight force of the book mg at L/4 from A.
Sum of torques at A:
∑τ = Iα
Rb (L) − Mg (L/2) − mg (L/4) = 0
Rb (L) = Mg (L/2) + mg (L/4)
Rb = ½ Mg + ¼ mg
Rb = (½ M + ¼ m) g
Rb = (½ (10.8 kg) + ¼ (3.66 kg)) (9.8 m/s²)
Rb = 61.9 N
Sum of forces in the y direction:
∑F = ma
Ra + Rb − Mg − mg = 0
Ra = Mg + mg − Rb
Ra = (M + m) g − Rb
Ra = (10.8 kg + 3.66 kg) (9.8 m/s²) − 61.9 N
Ra = 79.8 N
<h2>
Answer: D. 70.0 J</h2>
Explanation:
3.00 m + 4.00 m = 7.00 m x 10.0 N = 70.0 J
Answer:
2 atoms of oxygen in carbon dioxide
Explanation:
Answer:
T1 = 417.48N
T2 = 361.54N
T3 = 208.74N
Explanation:
Using the sin rule to fine the tension in the strings;
Given
amass = 42.6kg
Weight = 42.6 * 9.8 = 417.48N
The third angle will be 180-(60+30)= 90 degrees
Using the sine rule
W/Sin 90 = T3/sin 30 = T2/sin 60
Get T3;
W/Sin 90 = T3/sin 30
417.48/1 = T3/sin30
T3 = 417.48sin30
T3 = 417.48(0.5)
T3 = 208.74N
Also;
W/sin90 = T2/sin 60
417.48/1 = T2/sin60
T2 = 417.48sin60
T2 = 417.48(0.8660)
T2 = 361.54N
The Tension T1 = Weight of the object = 417.48N
Answer:
False
Explanation:
Water contains ions that can conduct electricity and if touched it can cause harm (aka electrocution).
