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Feliz [49]
4 years ago
12

I need help plz help me​

Physics
1 answer:
kirza4 [7]4 years ago
8 0

Answer:ghhg

Explanation:

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Over a span of 6.0 seconds, a car changes it's speed from 89 km/h to 37 km/h. What is its average acceleration in meters per sec
scoundrel [369]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) - (speed at the beginning)

change in speed = (37 km/hr) - (89 km/hr) = -52 km/hr

Acceleration = (-52 km/hr) / (6 sec)

Acceleration = (-26/3) km/(hr·sec)

Units: (1/hr·sec) · (hr/3600 sec) = 1 / 3600 sec²

(-26/3) km/(hr·sec) = (-26/3) km/(3600 sec²)

= -26,000/(3 · 3600) m/s²

<em>Acceleration = -2.41 m/s²</em>

3 0
4 years ago
A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or sh
zalisa [80]

Complete question:

Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. In contrast,  a passenger not wearing a seat belt or using an air bag decelerates over a distance of 5mm.

(a) A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.

Answer:

The net force on the person as the air bad deploys is -6750 N backwards

Explanation:

Given;

mass of the passenger, m = 60 kg

velocity of the car at impact, u = 15 m/s

final velocity of the car after impact, v = 0

distance moved as the front of the car crumples, s = 1 m

First, calculate the acceleration of the car at impact;

v² = u² + 2as

0² = 15² + (2 x 1)a

0 = 225 + 2a

2a = -225

a = -225 / 2

a = -112.5 m/s²

The net force on the person;

F = ma

F = 60 (-112.5)

F = -6750 N backwards

Therefore, the net force on the person as the air bad deploys is -6750 N backwards

4 0
3 years ago
If a sound is 30 dB and its absolute pressure was 66 x 10-9 Pa, what must have been the reference pressure?
Ksenya-84 [330]

Given:

I = 30dB

P = 66 × 10^{-9} Pa

Solution:

Formula used:

I = 20\log_{10}(\frac{P}{P_{o}})           (1)

where,

I = intensity of sound

P = absolute pressure

P_{o} = reference pressure

Using Eqn (1), we get:

30 = 20\log _{10}\frac{66\times 10^{-9}}{P^{o}}

P_{o} = \frac{66\times 10^{-9}}{10^{1.5}}

P_{o} = 2.08 × 10^{-9} Pa

4 0
4 years ago
Um carro viaja de uma cidade A a uma cidade B, distantes 200km.seu percurso demora 4 horas ,pois decorrida uma hora de viagem ,o
Aleksandr-060686 [28]

Answer: B

Explanation:

5 0
3 years ago
HEEEEELLLPPPPPP!!!!!!!!!!! PPPPPPPPPPPLLLEAASSEE!!!
o-na [289]

Answer:

The arrows always start at the magnet's north pole and point towards its south pole. When two like-poles point together, the arrows from the two magnets point in OPPOSITE directions and the field lines cannot join up. So the magnets will push apart (repel).

4 0
3 years ago
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