Answer:

Explanation:
As per thermal radiation we know that rate is heat radiation is given as

here we know that
T = 34 degree C = 307 K

e = 0.557


now we have




A star chart<span> or </span>star map<span> is a </span>map<span> of the night </span>sky <span>Astronomers divide these into grids to use them more easil They are used to identify and locate astronomical objects such as stars constellations and galaxie it is in my text book </span>
Answer:
Approximately
, assuming that the acceleration of this ball is constant during the descent.
Explanation:
Assume that the acceleration of this ball,
, is constant during the entire descent.
Let
denote the displacement of this ball and let
denote the duration of the descent. The SUVAT equation
would apply.
Rearrange this equation to find an expression for the acceleration,
, of this ball:
.
Note that
and
in this question. Thus:
.
Let
denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is
, the net external force on this ball would be
.
Since
and
, the net external force on this ball would be:
.
Answer:
4*10^-2
Explanation:
for the scientific notation the first number must be between 1 and 10, so in this case it is 4. 4/100 is also equal to 0.04, and if we could the number of places before 4, there are two, therefore 4 times 10 to the power of -2
Gravitational force = G · (mass₁) · (mass₂) / (distance)
(distance²) = G · (mass₁) · (mass₂) / (Gravitational force)
G = 6.67 x 10⁻¹¹ n-m² / kg² (the "gravitational constant")
Distance² = (6.67 x 10⁻¹¹ n-m² / kg²) (28,500 kg) (2.2 x 10⁸ kg) / (39 N)
Distance² = (6.67 · 28,500 · 2.2 x 10⁻³ N-m²) / (39N)
Distance² = (418.209 N-m²) / (39N)
Distance² = 10.72 m²
<em>Distance = 3.275 meters</em>
An absurd scenario, but that's by golly what the math says with the numbers provided. I guess it's a teeny tiny planet orbiting 3.275 meters outside a teeny tiny black hole.