Answer:
Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous planet increases linearly with the height of the atmosphere as measured from the top of a visible boundary layer, defined as 0 kilometers in altitude. The instruments on board can withstand a temperature of 601 K. At what altitude will the probe's instruments fail? A. 50 kilometers B. 80 kilometers C. 83 kilometers D. 100 kilometers E. 111 kilometers
Explanation:
A. 50 kilometers
Answer: 0.86 × 10^14
Explanation:
Given the following :
Radius of proton = 1.2 × 10-15 m
Radius of hydrogen atom = 5.3 × 10-11 m
Density of proton could be calculated thus:
Mass of proton = 1.67 × 10^-27 kg
Using the formula :
(4/3) × pi × r^3
(4/3) × 3.142 × (1.2 × 10^-15)^3 = 7.24 × 10^-45
Density = mass / volume
Density = (1.67 × 10^-27) / ( 7.24 × 10^-45)
= 0.2306 × 10^18
Density of hydrogen atom:
Mass of hydrogen atom= 1.67 × 10^-27 kg
Using the formula :
(4/3) × pi × r^3
(4/3) × 3.142 × (5.3 × 10^-11)^3 = 6.24 × 10^-31
Density = mass / volume
Density = (1.67 × 10^-27) / ( 6.24 × 10^-31)
= 0.2676 × 10^4
Ratio is thus:
Density of proton / density of hydrogen atom
0.2306 × 10^18 / 0.2676 × 10^4 = 0.8617 × 10^14
Answer:
The acceleration of the earth is 7.05 * 10^-25 m/s²
Explanation:
<u>Step 1:</u> Data given
mass of the apple = 0.43 kg
acceleration = 9.8 m/s²
mass of earth = 5.98 * 10 ^24 kg
<u>Step 2:</u> Calculate the acceleration of the earth
Following the third law of Newton F = m*a
F(apple) = F(earth) = m(apple)*a(apple)
F(apple) = 0.43 kg * 9.8 m/s² = 4.214 N
a(earth) = F(apple/earth)/m(earth)
a(earth) = 4.214N /5.98 * 10 ^24 kg
a(earth) = 7.05 * 10^-25 m/s²
The acceleration of the earth is 7.05 * 10^-25 m/s²
P=M(mass)G(Gravity)H(Height)
Gravity=9.8
M=1.5 G=9.8 H=35
so multiply all
=514.5 potential energy
Answer:
The change in velocity is 15.83 [m/s]
Explanation:
Using the Newton's second law we have:
ΣF = m*a
The force in the graph is 185 N, therefore:
![185=0.369*a\\Where\\a=acceleration made it by the force [m/s^2]](https://tex.z-dn.net/?f=185%3D0.369%2Aa%5C%5CWhere%5C%5Ca%3Dacceleration%20made%20it%20by%20the%20force%20%5Bm%2Fs%5E2%5D)
![a=501.35[m/s^2]](https://tex.z-dn.net/?f=a%3D501.35%5Bm%2Fs%5E2%5D)
Now using the following kinematic equation:
![V^{2}=Vi^{2} + 2*a*(x-xi) \\where\\V=final velocity [m/s]\\Vi= initial velocity [m/s] = 0 the hockey disk is in rest when receives the hit.\\ x = Final position [m] = 0.4 m\\xi = initial position [m] = 0.15m\\](https://tex.z-dn.net/?f=V%5E%7B2%7D%3DVi%5E%7B2%7D%20%2B%202%2Aa%2A%28x-xi%29%20%5C%5Cwhere%5C%5CV%3Dfinal%20velocity%20%5Bm%2Fs%5D%5C%5CVi%3D%20initial%20velocity%20%5Bm%2Fs%5D%20%3D%200%20the%20hockey%20disk%20is%20in%20rest%20when%20receives%20the%20hit.%5C%5C%20x%20%3D%20Final%20position%20%5Bm%5D%20%3D%200.4%20m%5C%5Cxi%20%3D%20initial%20position%20%5Bm%5D%20%3D%200.15m%5C%5C)
Now replacing the values:
![V^{2}=0 + 2*501.35*(0.4-0.15)\\ \\V= 15.83[m/s]](https://tex.z-dn.net/?f=V%5E%7B2%7D%3D0%20%2B%202%2A501.35%2A%280.4-0.15%29%5C%5C%20%5C%5CV%3D%2015.83%5Bm%2Fs%5D)
