The area of the Earth (Ae) that is irradiated by is given by:
Ae = 4πRe^2, where Re = Distance from Sun to Earth
Substituting;
Ae = 4π*(1.5*10^8*1000)^2 = 2.827*10^23 m^2
On the Earth, insolation (We) = Psun/Ae
Therefore,
Psun (Rate at which sun emits energy) = We*Ae = 1.4*2.827*10^23 = 3.958*10^23 kW = 3.958*10^26 W
Answer:
497.00977 N
3742514.97005
Explanation:
= Density of water = 1000 kg/m³
C = Drag coefficient = 0.09
v = Velocity of dolphin = 7.5 m/s
r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m
A = Area
Drag force
![F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N](https://tex.z-dn.net/?f=F_d%3D%5Cfrac%7B1%7D%7B2%7D%5Crho%20CAv%5E2%5C%5C%5CRightarrow%20F_d%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%201000%20%5Ctimes%200.09%28%5Cpi%200.25%5E2%297.5%5E2%5C%5C%5CRightarrow%20F_d%3D497.00977%5C%20N)
The drag force on the dolphin's nose is 497.00977 N
at 20°C
= Dynamic viscosity = ![1.002\times 10^{-3}\ Pas](https://tex.z-dn.net/?f=1.002%5Ctimes%2010%5E%7B-3%7D%5C%20Pas)
Reynold's Number
![Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005](https://tex.z-dn.net/?f=Re%3D%5Cfrac%7B%5Crho%20vd%7D%7B%5Cmu%7D%5C%5C%5CRightarrow%20Re%3D%5Cfrac%7B1000%5Ctimes%207.5%5Ctimes%200.5%7D%7B1.002%5Ctimes%2010%5E%7B-3%7D%7D%5C%5C%5CRightarrow%20Re%3D3742514.97005)
The Reynolds number is 3742514.97005
Launch-capable countries
Order Country Satellite(s)
1 Soviet Union Sputnik 1
2 United States Explorer 1
3 France Astérix
4 Japan Ohsumi
10 more rows
Answer:
67
Explanation:
- The atomic number (Z) of an atom is equal to the number of protons in the nucleus
- The mass number (A) of an atom is equal to the sum of protons and neutrons in the nucleus
Therefore, calling p the number of protons and n the number of neutrons, for element X we have:
Z = p = 23
A = p + n = 90
Substituting p=23 into the second equation, we find the number of neutrons:
n = 90 - p = 90 - 23 = 67
Answer:
Explanation:
Given that,
Mass of the heavier car m_1 = 1750 kg
Mass of the lighter car m_2 = 1350 kg
The speed of the lighter car just after collision can be represented as follows
![m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}](https://tex.z-dn.net/?f=m_1u_1%2Bm_2u_2%3Dm_1v_1%2Bm_2v_2%5C%5C%5C%5Cv_2%3D%5Cfrac%7Bm_1u_1%2Bm_2u_2-m_1v_1%7D%7Bm_2%7D)
![v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s](https://tex.z-dn.net/?f=v_2%3D%5Cfrac%7B%281850%29%281.4%29%2B%281450%29%28-1.10%29-%281850%29%280.250%29%7D%7B1450%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B2590%2B%28-1595%29-%28462.5%29%7D%7B1450%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B2590-1595-462.5%7D%7B1450%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B532.5%7D%7B1450%7D%5C%5C%5C%5C%3D0.367m%2Fs)
b) the change in the combined kinetic energy of the two-car system during this collision
![\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))](https://tex.z-dn.net/?f=%5CDelta%20K.E%3D%28%5Cfrac%7B1%7D%7B2%7D%20m_1v_1%5E2%2B%5Cfrac%7B1%7D%7B2%7D%20m_2v_2%5E2%29-%28%5Cfrac%7B1%7D%7B2%7D%20m_1u_1%5E2%2B%5Cfrac%7B1%7D%7B2%7D%20m_2u_2%5E2%29%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%20%28m_1%28v_1%5E2-u_1%5E2%29%2Bm_2%28v_2%5E2-u_2%5E2%29%29)
substitute the value in the equation above
![=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%20%281850%28%280.250%29%5E2-%281.4%29%5E2%29%2B%281450%28%280.3670%29%5E2-%28-1.10%29%5E2%29%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%2811850%280.0625-1.96%29%2B%281450%280.1347%29-%281.21%29%29%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B2%7D%2811850%28-1.8975%29%29%2B%281450%28-1.0753%29%29%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%20%28-3510.375%2B%28-1559.185%29%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%20%28-5069.56%29%5C%5C%5C%5C%3D-2534.78J)
Hence, the change in combine kinetic energy is -2534.78J