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Andreas93 [3]
3 years ago
7

How would I solve this? It's Newton's 2nd law

Physics
1 answer:
hammer [34]3 years ago
8 0

Answer:

5.65 m/s²

Explanation:

We'll begin by calculating the mass of PJ when in San Diego (i.e Earth). This can be obtained as follow:

Weight of PJ on Earth (Wₑ) = 545 N

Acceleration due to gravity (g) on Earth (gₑ) = 10 m/s²

Mass of PJ on Earth (mₑ) =.?

Wₑ = mₑ × gₑ

545 = mₑ × 10

Divide both side by 10

mₑ = 545 / 10

mₑ = 54.5 Kg

Thus, the mass of PJ on San Diego (i.e Earth) is 54.5 Kg

Finally, we shall determine the acceleration due to gravity of planet Koja. This can be obtained as follow:

Weight of PJ on Koja (Wₖ) = 308 N

Mass of PJ on Koja (mₖ) = mass of PJ on Earth (mₑ) because mass is constant irrespective of location.

Mass of PJ on Earth (mₑ) = 54.5 Kg

Mass of PJ on Koja (mₖ) = 54.5 Kg

Acceleration due to gravity of on Koja (gₖ) =?

Wₖ = mₖ × gₖ

308 = 54.5 × gₖ

Divide both side by 54.5

gₖ = 308 / 54.5

gₖ = 5.65 m/s²

Thus, the acceleration due to gravity on planet Koja is 5.65 m/s²

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a)   S = 2.35 10³   J/m²2 ,  

b)and the tape recorder must be in the positive Z-axis direction.

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c) the direction of the positive x axis

Explanation:

a) The Poynting vector or intensity of an electromagnetic wave is

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if we use that the fields are in phase

          B = E / c

we substitute

         S = E² /μ₀ c

let's calculate

        s = 941 2 / (4π 10⁻⁷  3 10⁸)

        S = 2.35 10³   J/m²2

 

b) the two fields are perpendicular to each other and in the direction of propagation of the radiation

In this case, the electro field is in the y direction and the wave propagates in the ax direction, so the magnetic cap must be in the y-axis direction, and the tape recorder must be in the positive Z-axis direction.

the answer is 5

C) The poynting electrode has the direction of the electric field, by which or which should be in the direction of the positive x axis

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3 years ago
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One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi
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Answer:

a)S_1=-9.65}\ J/K

b)S_2=71.18\ J/K

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

a)

The entropy change of reservoir 745 K

S_1=-\dfrac{7190}{745}\ J/K

Negative sign because heat is leaving.

S_1=-9.65}\ J/K

b)

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S_2=\dfrac{7190}{101}\ J/K

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c)

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S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

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