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Rainbow [258]
3 years ago
10

Infant car seats are made to face the rear of the car. This is safer in a front end collision because of Newton's First law. New

ton's first law suggests A) Since a baby has a smaller mass it will feel a smaller force. B) The baby will move in the direction they are facing and not get throw forward. C) The baby will push on the car seat with a force equal and opposite to the own exerted on it. D) The baby will continue to move forward as the car slows and be push into the padded car seat.
Physics
2 answers:
Brut [27]3 years ago
8 0

Answer: D) The baby will continue to move forward as the car slows and be push into the padded car seat.

Explanation:

Newton's first law states that an object tends to be in state of rest or motion unless and until an unbalanced external force acts on it. This means, A body would continue to be in state of motion unless external force stops it. A body in the state of rest will remain at rest unless an external force moves it.

Infant car seat is made to face the rear of the car. This is because in case of front end collision, the car would come to sudden stop but the bodies inside the car are in state of motion and sudden halt will cause the body to push in the opposite direction.

Thus, for an infant in the car it would be a safer measure to place the seat facing the rear o the car because in this case it would cause the baby to safely collide with the padded seat of the car.

kicyunya [14]3 years ago
5 0
The answer is D hope it helps:)
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Answer

given,

V = 2 L

the left is an ideal gas at  P = 100 k Pa and T = 500 K

mass is constant

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\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}

Pressure is same because it's not changing due to process

\dfrac{V}{500} = \dfrac{2 V}{T_2}

T_2 = 1000\ K

\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}

\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})

m = \dfrac{P_1V_1}{RT_1}

m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}

m = 1.39 x 10⁻³ Kg

\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)

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spin [16.1K]

Answer:

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Explanation:

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I_{average}=\frac{cB_{0}^{2}}{2\mu_{0}} (1)

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I_{average}=\frac{3\cdot 10^{8}(1.5\cdot 10^{-10})^{2}}{2\cdot 1.26\cdot 10^{-6}}

I_{average}=2.68\cdot 10^{-6} W/m^{2}

Now, let's define the relationship between power (P) and average intensity (I).

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So we can calculate the power.

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Finally, energy is the product of P times time, so:

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I hope it helps you!

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