The initial temperature of the bar is 25. To get to the t temperature you need to add (t-25) degrees Celsius.
for 1 degree................... 7 Joules
y given degree........ p Joules
p=7y
In our case y=(t-25) .
h(t) = 7(t-25) which is the final answer.
Complete question:
A diver is 10 m below the surface of water. Calculate the pressure the fluid exerted on the diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is 1000 kg/m3. Answer in units of Pa. Show your work.
Answer:
Tthe pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa
Explanation:
Given;
density of water, ρ = 1000 kg/m³
diver's position below the surface of the water, h = 10 m
acceleration due to gravity, g = 9.8 m/s²
Let the atmospheric pressure, P₀ = 101325 Pa
The pressure 10 m below the surface of the water is calculated as;
P = P₀ + ρgh
P = 101325 Pa + (1000 x 9.8 x 10)Pa
P = 199325 Pa
P = 1.99 x 10⁵ Pa.
Therefore, the pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa
Answer:
The increase in thermal energy results in an increase in pressure.
Explanation:
- The increase in entropy is directly related to the increase in temperature. So the thermal energy in heat engine increases the temperature of that surrounding.
- Higher temperature means the kinetic energy of particles is also higher, their vibration is increasing. So it increases the pressure (ideal gas law).
- In this way the increment in the thermal energy in heat engine moves piston by increasing the pressure.
The problem seems to be incomplete because there is no question. However, from the problem description, the logical question is to find he acceleration needed by the jet to land on the airplane carrier. The working equation would be:
2ad = v₂² - v₁²
Since the jet stops, v₂ = 0. Substituting the values:
2(a)(95 m) = 0² - [(240 km/h)(1000 m/1 km)(1h/3600 s)]²
Solving for a,
<em>a = -23.39 m/s² (the negative sign indicates that the jet is decelerating)</em>
Answer:isotopes
Explanation:Isotopes form when the number of neutrons and atomic number change except for the protons don't change