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puteri [66]
3 years ago
7

Which temperature is the hottest? 98 F or 39 C or 303K? F= 1.8C + 32 C= (F-32)/1.8

Physics
1 answer:
sergejj [24]3 years ago
3 0

Answer:

The hottest temperature is  T_2 = 39^o C

Explanation:

From the question we are given

    T_1 =  98 F

  T_2 =  39^oC

  T_3 =  303 \  K

Generally converting T_3 to  Fahrenheit

    T_3' =  (T_3 -273 ) * \frac{9}{5}  + 32

=> T_3' =  (303 -273 ) * \frac{9}{5}  + 32

=> T_3' = 86 F

Converting  T_2 to  Fahrenheit

      T_2' =  T_2 * \frac{9}{5}  + 32

=> T_2' =  39 * \frac{9}{5}  + 32

=> T_2' =102.2 F  

Now comparing  the temperature  in Fahrenheit we see that T_2  is the hottest

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Explanation:

F=ma

7900=950a

a=7900/950=8.58m/s^2

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A river is flowing south at a rate of 3 m/s. Steven can roe directly across the river if he aims the raft 30 degrees. What rate
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Answer:

Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s

Explanation:

Given data,

The river flowing south at the rate, v = 3 m/s

To reach the other side directly across the river, he aims the raft, Ф = 30°

The speed of his raft across the river is given by the formula,

                                          V = v / Sin Ф

                                             = 3 / Sin 30°

                                              = 6 m/s

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Why the change of the pressure and temperature affect the velocity of the sound ​
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A permanent magnet has a magnetic dipole moment of 0.160 A · m^2. The magnet is in the presence of an external uniform magnetic
Elena L [17]

Answer:

the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

Explanation:

The torque is given by :

\bar {N} = \bar {m} * \bar {B}

where ;

m = 0.160 A.m²

B = 0.0800 T

θ = 35°

So the magnitude of the torque N = mBsinθ

N = (0.160)(0.0800)(sin 35°)

N = 0.007341

N = 7.34×10⁻³ Nm

Hence, the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

b) The potential energy \bar{U} = \bar{-m} * \bar{B}

U = -mBcosθ

U = (- 0.160)(0.0800)(cos 45)

U = -0.010485

U = -1.0485 ×10⁻² J

Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

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