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kompoz [17]
3 years ago
7

A temperature of 200°F is equivalent to approximately

Physics
2 answers:
Blizzard [7]3 years ago
8 0

93.3* C is correct, I just took the test and it was correct. Good luck

aev [14]3 years ago
7 0
<span>93.3°C
A temperature in Fahrenheit (°F) can be converted to Celsius (°C), using the formula
[°C] = ([°F] − 32) ×  5⁄9. Here we have to convert a temperature of 200°F in to Celsius. Thus Subtract 32 from Fahrenheit and multiply by 5 then divide by 9 . That is (200°F - 32) × 5/9=168 × 5/9
                                          =840/9
                                          =93.333333333°C
                                          = 93.3°C</span>
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A circular loop with radius R= 20 cm carrying a current, I= 5 A is placed in a uniform magnetic field B = 0.2 T. (a) Find the ma
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If Vx= 9.5 units and Vy= -6.4 units, determine the magnitude of V
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3 years ago
The position of a particle moving in an xy plane is given by with in meters and t in seconds. In unit-vector notation, calculate
Liula [17]

Answer:

a) r(@2sec) = (6.00i - 106j) m ; b) v = ( 19i - 224t³j) m/sec ; c) a = (24i - 336j) m/sec² ; d) θ = -85.1516°

Explanation:

Correct question statement:

The position r of a particle moving in an xy plane is given by r = (2.00t³ - 5.00t)° + (6.00 -7.00t 4 ) ĵ , with r in meters and t in seconds. In unit-vector notation, calculate (a) r , (b) v , find (c) a for t = 2.00 s. (d) What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t= 2.00 s?

Given Data:

r = (2t³ - 5t)i + ( 6 - 7t⁴)j ----------------equation (1)

a)

r(@2sec) = (2(2)³ - 5(2))i + ( 6 - 7(2)⁴)j

               = (6.00i - 106j) m

b)

By taking derivative of equation 1 once, we get

              v(t) = ( 6.00t²-5.00 )i - 28t³j ------------- equation (2)

By putting t = 2sec, we get

                 v = ( 19i - 224t³j) m/sec

c)

By differentiating equation 1 twice of equation 2 once, we get

              a(t) = 12ti - 84t²j

At t = 2 sec, we get

                 a = (24i - 336j) m/sec²

d)

θ = tan ⁻¹ (-224/19) = -85.1516° or 94.8483°

Answer:    θ = -85.1516° it is also equivalent to 274.8484°(counterclockwise from +x-axis), and it is in 4th quadrant.

8 0
3 years ago
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