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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:

The amount of mass that needs to be converted to release that amount of energy is 1.122 X 10^{-7}  kg

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

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Making m the subject of the formula, we can find the change in mass to be

\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

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3 years ago
A proton is observed to have an instantaneous acceleration 11*10^11. what is the magnitude of e of the electric field at the pro
jek_recluse [69]

The magnitude of the electric field at the proton's location is 10,437.5 N/C.

<h3>What the magnitude of the electric field?</h3>

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6 0
2 years ago
A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre
Alisiya [41]

Answer:

0.546 \hat k

Explanation:

From the given information:

The force on a given current-carrying conductor is:

F = I ( \L  \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})

where the length usually in negative (x) direction can be computed as

\L ^ {\to }  = -x\hat i \\dL\limits ^ {\to }- dx\hat i

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})

F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)

F = I \int^3_1  - 9.0x^2 \ dx \hat k

F = I  (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k

F = I  (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ]  \hat k

where;

current I = 7.0 A

F = (7.0 \ A)  (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ]  \hat k

F = (7.0 \ A)  (9.0) \bigg [\dfrac{26}{3} \bigg ]  \hat k

F = 546 × 10⁻³ T/mT \hat k

F = 0.546 \hat k

4 0
3 years ago
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Answer:

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Explanation:

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