Both climate and weather measure conditions in the atmosphere.
Answer:
a) E = 2.00 10³ J
, b) I = 6.66 10⁻⁶ N s
, c) F = 1.66 10⁻⁶ N
Explanation:
a) The intensity is defined as the power per unit area
I = P / A
P = I A
Power is energy for time
P = E / t
We replace
E / t = I A
E = I A t
E = 1.0 10³ 2.0 1.00
E = 2.00 10³ J
b) The moment is
p = U / c
In the case of a reflection the speed is reversed, so the moment
Δp = 2 U / c
I = Δp
I = 2 U / c
I = 2.00 10³/3 10⁸
I = 6.66 10⁻⁶ N s
c) The defined impulse is
I = F t
F = I / t
For a time of 1 s
F = 6.66 10⁻⁶ / 1
F = 1.66 10⁻⁶ N
d) Suppose n small mass mirror m = 10 10⁻³ kg, we write Newton's second law
F = ma
a = F / m
a = 1.66 10⁻⁶ / 10 10⁻³
a = 1.66 10⁻⁴ m / s
We see that the acceleration is very small and attended to increase the mass of the mirror will be less and less, so the assumption of no twisting of the mirror is very reasonable
Answer: A difference in properties between two interacting systems as explained below.
Explanation: Planetary differentiation is the process of separation of different parts of a planetary body as a result of their physical or chemical behavior.
Causes for Physical Differentiation:
1. Melting and crystallization of source rock.
2. Differences in densities
3. Thermal diffusion
4. Collision of large bodies
Chemical Differentiation accounts for differences in chemical compositions of different materials.
The differentiation, or organization, of the Earth into layers led to the formation of a core, a crust, and eventually continents. The light elements were driven from the interior to form an ocean and atmosphere.
The difference between the frequencies of the piano key and the tuning fork gives the frequency of the beats.
When the tuning fork is 405 Hz, and no beats are heard, then the piano key is also 405 Hz.
When the piano key is 405 Hz and the tuning fork is 402 Hz, then 405 - 402 = 3 beats are heard.
The piano key is 405 Hz.
Answer:=14,160 kJ
Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below
Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase
C MPa m^3/kg kJ/kg kJ/kg
1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture
2 260 4.689 0.0422 2599 2797 1 Saturated Vapor
The mass initially contained in the tank is m1 = V/v1
m1 =0.85 m^3 /0.02993 m^3 /kg
= 28.4 kg
The mass finally contained in the tank is
m2 =V2/v
= 0.85 m^3 /0.0422 m^3 /kg
= 20.14 kg
The heat transfer is then
Qcv = m2u2 − m1u1 − he(m2 − m1)
Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ
