Dolphins communicate underwater using sound waves. This is called ________.
2 answers:
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Answer:
a. Energy lost, Q = 16,800 Joules.
b. Power = 28 J/s
c. Time, t = 2357.14 seconds
d. I assumed that the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.
Explanation:
<u>Given the following data;</u>
- Mass = 0.20 kg
- Initial temperature, T1 = 20°C
- Final temperature = 0°C
- Time = 10 minutes
a. To find the energy lost by the water as it cools to 0 degree celsius;
Mathematically, heat capacity is given by the formula;
Where;
- Q represents the heat capacity or quantity of heat.
- M represents the mass of an object.
- C represents the specific heat capacity of water.
- dt represents the change in temperature.
dt = T2 - T1
dt = 20 - 0
dt = 20°C
We know that the specific heat capacity of water is equal to 4200 J/kg°C
Substituting the values into the formula, we have;
<em>Energy lost, Q = 16,800 Joules.</em>
b. To find the average rate at which the water is losing energy in J/s by using the following formula;
First of all, we would have to convert the value of time in minutes to seconds.
<u>Conversion:</u>
1 minute = 60 seconds
10 minutes = X seconds
Cross-multiplying, we have;
X = 60 * 10
X = 600 seconds
Substituting the values into the formula, we have;
<em>Power = 28 J/s</em>
c. To estimate the time taken for the water at 0 degree celsius to turn completely into ice;
We know that the latent heat of fusion of water is equal to 3.3 * 10⁵ J/kg.
Mathematically, the latent heat of fusion is calculated by using the formula;
Energy, Q = ml = pt
Substituting the values into the formula, we have;
0.20 * 3.3 * 10⁵ = 28 * t
0.20 * 330000 = 28t
66000 = 28t
<em>Time, t = 2357.14 seconds.</em>
d. The assumption made is that, the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.
Answer:
Explanation:
We apply Newton's second law at the crate :
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
m=90kg : crate mass
F= 282 N
μk =0.351 :coefficient of kinetic friction
g = 9.8 m/s² : acceleration due to gravity
Crate weight (W)
W= m*g
W= 90kg*9.8 m/s²
W= 882 N
Friction force : Ff
Ff= μk*N Formula (2)
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W = 0
N = W
N = 882 N
We replace the data in the formula (2)
Ff= μk*N = 0.351* 882 N
Ff= 309.58 N
We apply the formula (1) in x direction:
∑Fx = m*ax , ax=0
282 N - 309.58 N = 90*a
a= (282 N - 309.58 N ) / (90)
a= - 0.306 m/s²
Kinematics of the crate
Because the crate moves with uniformly accelerated movement we apply the following formula :
vf²=v₀²+2*a*d Formula (3)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
v₀ = 0.850 m/s
d = 0.75 m
a= - 0.306 m/s²
We replace the data in the formula (3)
vf²=(0.850)²+(2)( - 0.306 )(0.75 )