An electromagnetic wave of a frequency between about 104<span> and 10</span>11<span> or 10</span>12<span> Hz, as used for long-distance communication</span>
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The answer is Tropical Cyclone. <span>A </span>tropical cyclone<span> with maximum sustained winds of 74 mph (64 knots) or higher. In the western North Pacific, hurricanes are called typhoons; similar storms in the Indian Ocean and South Pacific Ocean are called </span>cyclones<span>.</span>
Answer:
-209.42J
Explanation:
Here is the complete question.
A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?
Solution
The work done by a force W = ∫Fdx since our force is variable.
Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.
So, the workdone by the force on the cow is
W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx
= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx
= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹
= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]
= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J
= -209.415 J ≅ -209.42J
I believe this is it
The centripetal force is given by
F = mv^2 / r
When v' = v/2,
F' = mv'^2/r = m(v/2)^2/r = mv^2/4r = F/4.
So the centripetal force is divided by 4.