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belka [17]
3 years ago
5

Mr Johnson launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, how much time does it take

before the arrow hits the ground below (step 1 of a quesiton will need this answer for a future question)
a 2 seconds
b. 1 second
c.4 seconds
d 19.6​
Physics
1 answer:
vazorg [7]3 years ago
3 0

Answer:c

Explanation:

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Explanation:

Step 1: Data given

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After the motor has run for several hours the resistance rises to 58Ω.

Step 2: Calculate the new temperature

Formula: R = Rref(1 + α(T-Tref))

⇒with α = temperature coëfficiënt of Cupper at 20 °C = 0.00394/°C

⇒with Tref = reference temperature = 20°C

⇒with T = end temperature = TO BE DETERMINED

⇒with R = resistance at end temperature = 58Ω

⇒with Rref = resistance at reference temperature =  50 Ω

==> T = (R/Rref - 1)/α + Tref  

T = (58/50) - 1 )/ 0.00394 + 20

   T = 60.61 °C

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