You have two points in a soil. Point A is at 75 cm; point B is at 25 cm above the reference. The capillary potential energy at p
1 answer:
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Answer:
The normal force exerted on the ant is 0.75 N.
Explanation:
Given;
diameter of the ball, D = 40 cm = 0.4m
radius of the ball, r = 0.2m
mass of the beach ball, m₁ = 300 g = 0.3 kg
mass of the ant, m₂ = 4 x 10⁻⁶ kg
speed of the ball, v = 4 m/s
The area of the ball, assuming spherical ball is given by;
A = 4πr²
A = 4π(0.2)² = 0.5027 m²
The drag force (resistance) experienced by the spherical ball is given as;
where;
C is the drag coefficient of the spherical ball = 0.45
ρ is density of air = 1.21 kg/m³
The downward force of the ball due to its weight and that of the ant is given by;
The net downward force experienced by the ball is given by;
This downward force experienced by the ball is equal to the normal reaction it exerts on the ant.
Thus, the normal force exerted on the ant is 0.75 N.
Answer:
a) Q = 111.6 cal
b) T2 = 415.15K = 142°C
Explanation:
a) To calculate the amount oh heat given to the block of silver, you use the following formula:
(1)
m: mass of the block = 40g
T2: final temperature = 80°C = 353.15k
T1: initial temperature = 30°C = 303.25K
c: specific heat of silver = 0.0558cal/g.K
You replace the values of the parameters in the equation (1):
hence, the amount of heat is 111.6 cal
b) The temperature is calculated by solving the equation (1) for T2:
hence, the final temperature when 250 cal of heat is geiven to the block of silver, is 415.15K