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3 years ago

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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Let's take two balls, each of radius 0.1 meters and charge them each to 3 microCoulombs. Then, let's put them on a flat surface

salantis [7]

Explanation:

First, we will calculate the electric potential energy of two charges at a distance R as follows.

R = 2r

= $2 \times 0.1 m$

= 0.2 m

where, R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.

U = $\frac{k \times Q \times Q}{R}$

= $\frac{8.98 \times 10^{9} \times (3 \times 10^{-6} C)^{2}}{0.1}$

= 0.081 J

As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.

Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, $\frac{U}{2}$ .

Hence, K.E = $\frac{U}{2}$

= $\frac{0.081}{2}$

= 0.0405 J

Now, we will calculate the speed of balls as follows.

V = $\sqrt{\sqrt{\frac{2 \times K.E}{m}}$

= $\sqrt{\sqrt{\frac{2 \times 0.0405}{4 kg}}$

= 0.142 m/s

Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.

6 0

3 years ago

To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you

Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

β = 10 log $\frac{I}{I_o}$

let's write this expression for our case

β₁ = 10 log \frac{I_1}{I_o}

β₂ = 10 log \frac{I_2}{I_o}

β₂ -β₁ = 10 ( $log \frac{I_2}{I_o} - log \frac{I_1}{I_o}$ )

β₂ - β₁ = 10 $log \frac{I_2}{I_1}$

log \frac{I_2}{I_1} = $\frac{60 - 20}{10}$ = 3

$\frac{I_2}{I_1}$ = 10³

I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

I = P / A

P = I A

the area is of a sphere

A = 4π r²

the power of the sound does not change, so we can write it for the two points

P = I₁ A₁ = I₂ A₂

I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

I₁ r₁² = (10³ I₁ ) r₂²

r₁² = 10³ r₂²

r₂ = r₁ / √10³

we calculate

r₂ = $\frac{10.0}{\sqrt{10^3} }$

r₂ = 0.316 m

8 0

3 years ago

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Alex

Answer: the waves travel in an horizontal direction while the strings vibrate in a vertical direction.

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Science is made up of two things, what is it

rodikova [14]

Answer:

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Explanation:

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3 years ago

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Snezhnost [94]

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