Answer:
A. We have that radius r = 4.00m intensity I = 8.00 W/m^
total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W
b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2
c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J
Answer:
Explanation:
The relation between time period of moon in the orbit around a planet can be given by the following relation .
T² = 4 π² R³ / GM
G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .
Substituting the values in the equation
(.3189 x 24 x 60 x 60 s)² = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)
759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )
M = .06424 x 10²⁵
= 6.4 x 10²³ kg .
Answer:
B is the best answer for the question
Answer:
a) 25.5°(south of east)
b) 119 s
c) 238 m
Explanation:
solution:
we have river speed 
=2 m/s
velocity of motorboat relative to water is 
=4.2 m/s
so speed will be:
a) 
=+
solving graphically
=4.7 m/s
Ф=
=25.5°(south of east)
b) time to cross the river: t=
=
=119 s
c) d=
=(2)(119)=238 m
note :
pic is attached
Answer:
a transverse (sort of a plot of a sine or cosine graph, basically)
b longitudinal
c Electromagnetic (an electric wave and a magnetic wave travelling together at right angles to each other)
Explanation:
