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pshichka [43]
4 years ago
14

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. By contrast, an unrestrained occupant keeps moving forward with no loss of speed (Newton's first law!) until hitting the dashboard or windshield. These are unyielding surfaces, and the unfortunate occupant then decelerates over a distance of only about 5 mm . Part APart complete 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.
Physics
1 answer:
gizmo_the_mogwai [7]4 years ago
7 0

Answer:

The net force is 1350 kN

Solution:

As per the question:

Mass of man, m = 60 kg

Initial speed of the car, v = 15 m/s

Final speed of the car, v' = 0 m/s

Distance covered by the person before coming to rest, d = 5 mm = 5\times 10^{- 3}\ m

By using the third eqn of motion:

v'^{2} = v^{2} + 2ad

0 = 15^{2} + 2\times a\times 5\times 10^{- 3}

a = - 22500\ m/s^{2}

Thus the force on the person can be given by:

F = ma = 60\times 22500 = 1350\ kN

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<span>A 67.0 kg crate is being raised by means of a rope. Its upward acceleration is 3.50 m/s2. What is the force exerted by the rope on the crate? 

</span>Newton's Second Law<span> of Motion states, “The force acting on an object is equal to the mass of that object times its acceleration.” We calculate as follows:
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F = ma = 67.0 kg (3.50 m/s^2) = 234.5 J
7 0
3 years ago
When light travels through a small hole, it appears to an observer that the light spreads out, blurring the outline of the hole.
Scilla [17]

wave theory

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4 0
3 years ago
A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder
QveST [7]

Answer:

\Delta \theta = 47.57^{\circ} C

Explanation:

given,

moles of air compressed, n = 1.70 mol

initial temperature, T₁ = 390 K

Power supply by the compressor, P = 7.5 kW

Heat removed = 1.3 kW

Angular frequency of the compressor, f = 110 rpm = 110/60 = 1.833 rps.

Time of compression = time of the hay revolution

             =\dfrac{1}{2}\ T

             =\dfrac{1}{2}\times \dfrac{1}{f}

             =\dfrac{1}{2}\times \dfrac{1}{1.833}

             =0.273 s

Using first law of thermodynamics

U = Q - W

now,

\dfrac{\Delta U}{\Delta t} = \dfrac{\Delta Q}{\Delta t}- \dfrac{\Delta W}{\Delta t}

Power supplied \dfrac{\Delta W}{\Delta t} = 7.5 kW

heat removed \dfrac{\Delta Q}{\Delta t} = 1.3 kW

now,

\dfrac{\Delta U}{\Delta t} = 7.5 -1.3

\dfrac{\Delta U}{\Delta t} = 6.2 kW

we know,

\dfrac{\Delta U}{\Delta t}=\dfrac{nC_v\Delta \theta}{\Delta t}

 C_v for air = 5 cal/° mol

                   = 5 x 4.186 J/mol°C  = 20.93 J/mol°C

now,

\Delta \theta = \dfrac{\Delta U}{\Deta t}\times \dfrac{\Delta t}{n C_v}

\Delta \theta = 6.2\times 10^3 \times \dfrac{0.273}{1.7\times 20.93}

\Delta \theta = 47.57^{\circ} C

the temperature change per compression stroke is equal to 47.57°C.

4 0
3 years ago
1 ) when a ball is projected upwords its time of rising is ...............the time of falling .
balu736 [363]

Answer:

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Explanation:

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1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down

correct answer C

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v₀ = 0

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5 0
3 years ago
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