M = 10.0 g, the mass of the iron sample
ΔT = 75 - 25.2 = 49.5°C, the decrease in temperature
c = 0.449 J/(g-°C), the specific heat of iron
The heat released is
Q = m*c*ΔT
= (10.0 g)*(0.449 J/(g-°C))*(49.5 C)
= 222.255 J
Answer: 222.3 J (nearest tenth)
Answer:
(a) 4.21 m/s
(b) 24.9 N
Explanation:
(a) Draw a free body diagram of the object when it is at the bottom of the circle. There are two forces on the object: tension force T pulling up and weight force mg pulling down.
Sum the forces in the radial (+y) direction:
∑F = ma
T − mg = m v² / r
v = √(r (T − mg) / m)
v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)
v = 4.21 m/s
(b) Draw a free body diagram of the object when it is at the top of the circle. There are two forces on the object: tension force T pulling down and weight force mg pulling down.
Sum the forces in the radial (-y) direction:
∑F = ma
T + mg = m v² / r
T = m v² / r − mg
T = (1.52 kg) (4.21 m/s)² / (0.676 m) − (1.52 kg) (9.8 m/s²)
T = 24.9 N
Answer:
Explanation:
So, we are looking for an expression of the amount of water that has been drained from the tub. The expression is in terms of v that represent the number of gallons of water drained since the plug was pulled. Since we are interested in the pounds of water that has been drained from the tub we need to take into account that for every gallon of water drained, 8.345 pounds have left the tub. Therefore, the expression for the weight of water Q that has been drained from the tub in terms of v is simply :
Where v is the amount of gallons that has been drained from the tub.
Have a nice day. let me know if I can help with anything else
Bending occurs when one side of the wave enters the new medium before the other side of the wave. ... The bending occurs because the two sides of the wave are traveling at different speeds.
