The radius of the sphere in meters is ,r = 
Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,
r/10 = tan [arctan(1/2)] = tan (1/2)
Since, tan (θ/2) = 1-cos(θ) / sin(θ)
we find that,
r/10 = 
Hence, r =
So, the radius of the sphere in meters is ,r =
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Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore,
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore,
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
Answer:
149 m
Explanation:
The distances across the lake is forming a triangle.
let the distance between the point and the left side be 'x'
and the distance between the point and the right be 'y'
and the distance across the lake be 'z' and the angle opposite to 'z' be 'Z' given:
∠Z = 83°
x = 105 m
y = 119 m
Now, applying the Law of Cosines, we get
z² = x² + y² - 2xycos(Z)
Substituting the values in the above equation, we get
z² = 105² + 119² - 2×105×119×cos(83°)
or
z = √22140.48
or
z = 148.796 m ≈ 149 m
The point is 149 m across the lake
Answer:
false
Explanation:
every object will always have the force of gravity acting upon it.
