Answer:
W = 28226.88 N
Explanation:
Given,
Mass of the satellite, m = 5832 Kg
Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m
The time period of the orbit, T = 1.9 h
= 6840 s
The radius of the planet, R = 4.38 x 10⁶ m
The time period of the satellite is given by the formula
second
Squaring the terms and solving it for 'g'
g = 4 π² 
m/s²
Substituting the values in the above equation
g = 4 π² 
g = 4.84 m/s²
Therefore, the weight
w = m x g newton
= 5832 Kg x 4.84 m/s²
= 28226.88 N
Hence, the weight of the satellite at the surface, W = 28226.88 N
Answer:
L = 41.09 Kg m2 / s The angular momentum does not depend on the time
Explanation:
The definition of angular momentum is
L = r x p
Where blacks indicate vectors
Let's apply this definition our case. Linear momentum
p = m v
Let's replace
L = m r x v
The given function is
x = 6.00 i ^ + 4.15 t j
^
We look for speed
v = dx / dt
v = 0 + 4.15 j ^
To evaluate the angular momentum one of the best ways is to use determinants
![L = m \left[\begin{array}{ccc}i&j&k\\6&4.15t&0\\0&4.15&0\end{array}\right]](https://tex.z-dn.net/?f=L%20%3D%20m%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C6%264.15t%260%5C%5C0%264.15%260%5Cend%7Barray%7D%5Cright%5D)
L = m 6 4.15 k ^
The other products give zero
Let's calculate
L = 1.65 6 4.15 k ^
L = 41.09 Kg m2 / s
The angular momentum does not depend on the time
Answer:
-30°C
Explanation:
F-32/180 =C-0/100
or, -22-32/180=C/100
or, -54/180*100=C
or, -0.3*100=C
therefore, C= -30
-22°F = -30°C
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Explanation :
It is given that,
Diameter of the coil, d = 20 cm = 0.2 m
Radius of the coil, r = 0.1 m
Number of turns, N = 3000
Induced EMF, 
Magnitude of Earth's field, 
We need to find the angular frequency with which it is rotated. The induced emf due to rotation is given by :
So, the angular frequency with which the loop is rotated is 159.15 rad/s. Hence, this is the required solution.
Answer:
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