The linear speed of the ladybug is 4.1 m/s
Explanation:
First of all, we need to find the angular speed of the lady bug. This is given by:

where
T is the period of revolution
The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:
T = 1 s
Therefore, the angular speed is

Now we can find the linear speed of the ladybug, which is given by

where:
is the angular speed
r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation
Substituting, we find

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The number of heat units needed to raise the temperature of a body by one degree.
Answer:
A. Na2S + 2KCI - 2NaCl + KZS
Explanation:
Answer: The gravitational
Explanation: The student is pushing the box so u have to have gravitational force so it could move
Answer:
4.96×10¯¹⁰ N
Explanation:
The following data were obtained from the question:
Mass 1 (M1) = 300 Kg
Mass 2 (M2) = 300 Kg
Separating distance (r) = 110 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Gravitational force (F) =?
The gravitational force between the two goal posts can be obtained as follow:
F = GM1M2 / r²
F = 6.67×10¯¹¹ × 300 × 300 / 110²
F = 6.003×10¯⁶ / 12100
F = 4.96×10¯¹⁰ N
Therefore the gravitational force between the two goal posts is 4.96×10¯¹⁰ N