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3 years ago

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

2 answers:

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 SECOND 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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Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if a. The mass is doubled? b. The mass is halved? c. The amplitude is doubled? d. The spring constant is doubled?

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

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