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JulsSmile [24]
3 years ago
8

In a collision, the __________ collision is when an unsecured driver strikes the inside of the vehicle.

Physics
2 answers:
zaharov [31]3 years ago
6 0
In a collision, the second collision is when an unsecured driver strikes the inside of the vehicle. It is a collision that happens between an occupant of a vehicle and the vehicle he is riding during the impact. The first collision would be the collision of the vehicle and the other object.
Leokris [45]3 years ago
5 0

Answer:

In a collision, the <em><u>SECOND</u></em> collision is when an unsecured driver strikes the inside of the vehicle.

Explanation:

When driver in the car is unsecured then it means that the driver is not having his safety seat belt.

Now when collision occurs between the car and some other object then due to the property of inertia of driver it has tendency to move in the direction of his motion.

So the driver will continue his motion and collide with other parts of the car and due to this he may got injured.

So in any car collision there are two types of collision

first is the collision of car with other objects

second collision is between car and the driver inside the car

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Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper
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Explanation:

Expression to calculate thermal resistance for iron (R_{I}) is as follows.

             R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}  

where,   L_{I} = length of the iron bar

             k_{I} = thermal conductivity of iron

             A_{I} = Area of cross-section for the iron bar

Thermal resistance for copper (R_{c}) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where,  L_{c} = length of copper bar

             k_{c} = thermal conductivity of copper

            A_{c} = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

           Q = \frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

 Putting the given values into the above formula as follows.

       Q = \frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

  = \frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}

           = 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

                 P = \frac{Q}{T}

Here, T is 1 second so, power conducted is equal to heat transferred.

So,           P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

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3 years ago
When do you turn your wheels sharply left before backing slowly in a parallel parking maneuver?
Alex_Xolod [135]

Answer:

What is C - when your front bumper is even with the front vehicles back bumper.

Explanation:

Good Luck

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2 years ago
A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang
koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × 10^{-4}  Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux  is express as

flux Ф = \int \vec{B}.d\vec{A}   .................1

and Ф = BA cos∅    ............2

so B = \frac{\phi }{Acos\theta }    

and we know

A = ab

so

B = \frac{\phi }{abcos\theta }    ..............3

put here value

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solve we get

B = 0.692 T

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Answer:

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F=ke

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