Answer:
diameter of largest orbit is 0.60 m
Explanation:
given data
isotopes accelerates KE = 6.5 MeV
magnetic field B = 1.2 T
to find out
diameter
solution
first we find velocity from kinetic energy equation
KE = 1/2 × m×v² ........1
6.5 × 1.6 × 
= 1/2 × 1.672 × 
×v²
v = 3.5 × 
m/s
so
radius will be
radius = 
........2
radius = 
radius = 0.30
so diameter = 2 × 0.30
so diameter of largest orbit is 0.60 m
Answer:
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Explanation:
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Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.
Answer:
umm section 2 and 4
Explanation:
because at section 2 it starts and at section 4 it moves again and stops at 3.
<span>118 C
The Clausius-Clapeyron equation is useful in calculating the boiling point of a liquid at various pressures. It is:
Tb = 1/(1/T0 - R ln(P/P0)/Hvap)
where
Tb = Temperature boiling
R = Ideal Gas Constant (8.3144598 J/(K*mol) )
P = Pressure of interest
Hvap = Heat of vaporization of the liquid
T0, P0 = Temperature and pressure at a known point.
The temperatures are absolute temperatures.
We know that water boils at 100C at 14.7 psi. Yes, it's ugly to be mixing metric and imperial units like that. But since we're only interested in relative pressure differences, it's safe enough. So
P0 = 14.7
P = 14.7 + 12.3 = 27
T0 = 100 + 273.15 = 373.15
And for water, the heat of vaporization per mole is 40660 J/mol
Let's substitute the known values and calculate.
Tb = 1/(1/T0 - R ln(P/P0)/Hvap)
Tb = 1/(1/373.15 K - 8.3144598 J/(K*mol) ln(27/14.7)/40660 J/mol)
Tb = 1/(0.002679887 1/K - 8.3144598 1/K ln(1.836734694)/40660)
Tb = 1/(0.002679887 1/K - 8.3144598 1/K 0.607989372/40660)
Tb = 1/(0.002679887 1/K - 5.055103194 1/K /40660)
Tb = 1/(0.002679887 1/K - 0.000124326 1/K)
Tb = 1/(0.002555561 1/K)
Tb = 391.3034763 K
Tb = 391.3034763 K - 273.15
Tb = 118.1534763 C
Rounding to 3 significant figures gives 118 C</span>
