The formula we can use here is:
g = G m / r^2
where g is gravity, G is gravitational constant, m is
mass, and r is radius
Since G is constant, therefore we can equate two
situations:
g1 m1 / r1^2 = g2 m2 / r2^2
(10 m/s^2) r1^2 / m1 = g2 * (1/2 r1)^2 / (1/8 m1)
<span>g2 = 5 m/s^2</span>
Answer:
The heat flux between the surface of the pond and the surrounding air is<em> 60 W/</em>
<em> </em>
Explanation:
Heat flux is the rate at which heat energy moves across a surface, it is the heat transferred per unit area of the surface. This can be calculated using the expression in equation 1;
q = Q/A ...............................1
since we are working with the convectional heat transfer coefficient equation 1 become;
q = h (
) ........................2
where q is the heat flux;
Q is the heat energy that will be transferred;
h is the convectional heat coefficient = 20 W/
.K;
is the surface temperature =
C 23°C + 273.15 = 296.15 K;
is the surrounding temperature =
C = 20°C + 273.15 = 293.15 K;
The values are substituted into equation 2;
q = 20 W/
.K ( 296.15 K - 293.15 K)
q = 20 W/
.K ( 3 K)
q = 60 W/
Therefore the heat flux between the surface of the pond and the surrounding air is 60 W/
Answer :
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Answer:
V(3) =14i^ -34j^ +8.57 k^
S(3) =(25,-45,3.97)
Explanation:
We know that
V =a dt
from t=0 to 3s
V = 4i - 6tJ + sin(.2t)k m/s² dt
V =4t i^ - 3t^2j^ - cos(2t)/2 k^ +C
So we have
V(0) =-1/2 k^ +C =2i -7j +8.4 k
C=2i -7j +8.9k^
V =4t i^ - 3t^2j^ - cos(2t)/2 k^ + 2i -7j +8.9k^
Then putting 3s
V(3) =14i^ -34j^ +8.57 k^
Also
S(t) =V(t) dt
S(t) = 4t i^ - 3t^2j^ - cos(2t)/2 k^ + 2i -7j +8.9k^] dt
S(t)= (2t^2 +2t)i^ - (t^3 +7t) j^ -[sin(2t)/4 - 8.9t] k^ +C =i+3j-5k
So at t= 0 we have
S(0) = C= i+3j-5k
So S(t) =(2t^2 +2t)i^ - (t^3 +7t) j^ -[sin(2t)/4 - 8.9t] k^ +i+3j-5k
When t= 3
S(3) =25i -45j + 3.97k
S(3) =(25,-45,3.97)