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tia_tia [17]
3 years ago
7

Which of the following is not an example of centripetal acceleration?

Physics
1 answer:
Amanda [17]3 years ago
5 0
An apple falling to the ground is not an example of centripetal acceleration.
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The main purpose of an air bag is to stop a passenger during a car accident in a greater amount of time than if the air bag were
Simora [160]

Answer:

a) 45571 N  

b) 22786 N

c) 4557 N

Explanation:

  • Since the goal of the airbag is helping the person to stop after the collision in a greater time, this means that the change in momentum must finish when this is just zero.
  • In other words, the change in momentum, must be equal to the initial one, but with opposite sign.

       \Delta p = - p_{o} = -m*v = -55 kg*29m/s = -1595 kgm/s (1)

  • Now, just applying the original form of  Newton's 2nd Law, we know that this change in momentum must be equal to the impulse needed to stop the person:

       \Delta p = F* \Delta t  (2)

  • So, as we know the magnitude of Δp from (1) and we have different Δt as givens, we can get the different values of F (in magnitude) required to stop the person for each one of them, as follows:

       F_{1} = \frac{\Delta p}{\Delta t_{1}} = \frac{1595kgm/s}{0.035s} = 45571 N (3)

       F_{2} = \frac{\Delta p}{\Delta t_{2}} = \frac{1595kgm/s}{0.07s} = 22786 N (4)

       F_{3} = \frac{\Delta p}{\Delta t_{3}} = \frac{1595kgm/s}{0.35s} = 4557 N (5)

4 0
3 years ago
if a gas has a gage pressure of 156 kpa its absolute pressure is approximately .....A... 56 kPa ....B....100 kPa.......C......25
NeX [460]

256 kPa because p-guage + p-absolute + p-atmospheric = 256

7 0
3 years ago
Earth’s polar ice caps contain about 2.3 × 1019 kg of ice. This mass contributes essentially nothing to the moment of inertia of
sp2606 [1]

Answer:

Explanation:

Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass  2.3 x 10¹⁹ Kg

Final moment of inertia I₂ = 2/5 M R² + 2/3  x 2.3 x 10¹⁹ x R²

For change in period of rotation we shall apply conservation of angular momentum law

I₁ ω₁  = I₂ ω₂  ,  ω₁ and   ω₂ are angular velocities initially and finally .

I₁ / I₂     =  ω₂ / ω₁

I₁ / I₂     =  T₁ / T₂  , T₁ , T₂ are time period initially and finally .

T₂ / T₁ = I₂ / I₁

(2/5 M R² + 2/3  x 2.3 x 10¹⁹ x R²) / 2/5 MR²

1 + 5 / 3  x 2.3 x 10¹⁹ / M

= 1 + 5 / 3  x 2.3 x 10¹⁹ / 5.97 x 10²⁴

= 1 + .0000064

T₂ = 24 (1 + .0000064)

= 24 hours + .55 s

change in length of the day = .55 s .

3 0
2 years ago
Where are you likely to find a cooling plant located in the ductwork? A. A medium-sized office B. A large factory C. A single fa
Rufina [12.5K]

Answer:

i think it would be B, a large factory

Explanation:

8 0
2 years ago
A) Show that the surface temperature of a star can be inferred from measurements of blackbodyfluxes at two different frequencies
Roman55 [17]

Answer:

The answers to the questions have been solved in the attachment.

Explanation:

The answers to part a to e are all contained in the attachment. For answer part b, temperature and frequency were assumed to be fixed or constant. V² is directly proportional to T telling us that variation in T gives us a square in the frequency variation. This tells us why it is difficult when both frequencies are on this side of the black body.

3 0
3 years ago
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