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2 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]2 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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A ball is dropped from rest at point O . It passes a window with height 3.8 m in time interval tAB = 0.02 s.Identify the correct

Taya2010 [7]

Answer:

VB − VA = g tAB & (VA + VB)/2 = h / tAB

Explanation:

s = h = Displacement

tAB = t = Time taken

VA = u = Initial velocity

VB = v = Final velocity

a = g = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{3.8-\frac{1}{2}\times 9.8\times 0.02^2}{0.02}\\\Rightarrow u=189.902\ m/s$

$v=u+at\\\Rightarrow v=189.902+9.8\times 0.02\\\Rightarrow v=190.098\ m/s$

$\frac{v+u}{2}=\frac{190.098+189.902}{2}=190\ s$

$\frac{h}{t}=\frac{3.8}{0.02}=190\ s$

Hence, the equations VB − VA = g tAB & (VA + VB)/2 = h / tAB will be used

3 0

3 years ago

What acceleration does the force of earth's gravity peoduce

maksim [4K]

Near Earth's surface, gravitational acceleration is approximately 9.81 m/s2, which means that, ignoring the effects of air resistance, the speed of an object falling freely will increase by about 9.81 metres per second every second.

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3 years ago

A truck is traveling east at 80 km/h. At an intersection 32 km ahead, a car is traveling north at 50 km/h. How long after this m

nirvana33 [79]

The time elapsed when the vehicles are closest to each other is 20 min.

The given parameters:

Speed of the truck, u = 80 km/h
Distance, d = 32 km
Speed of the car, v = 50 km/h

<h3>Principles of relative speed</h3>

The time elapsed when the cars are close to each other is calculated by applying the principles of relative speed.

$(V_r) t = d\\\\V_r^2 = 50^2 + 80^2\\\\V_r =\sqrt{50^2 + 80^2} \\\\V_r = 94.34 \ km/h$

$94.34 t = 32\\\\t = \frac{32}{94.34} \\\\t = 0.34 \ hr\\\\t \approx 20 \min$

Thus, the time elapsed when the vehicles are closest to each other is 20 min.

Learn more about relative velocity here: brainly.com/question/24430414

3 0

2 years ago

7. A force of 100 N acting on a body gives it a speed of 200 m/s in 2

alekssr [168]

Answer:

Choice a. 1 kg, assuming that all other forces on the object (if any) are balanced.

Explanation:

By Newton's Second Law,

$\displaystyle a = \frac{\Sigma F}{m}$ ,

where

$a$ is the acceleration of the object in $\text{m}\cdot\text{s}^{-2}$ ,
$\Sigma F$ is the net force on the object in Newtons, and
$m$ is the mass of the object in kilograms.

As a result,

$\displaystyle m = \frac{\Sigma F}{a}$ .

Assume that all other forces on this object are balanced. The net force on the object will be $100\;\text{N}$ . The net force is constant. Acceleration should also be constant and the same as the average acceleration in the two seconds.

<h3>What is the average acceleration of this object?</h3>

$\displaystyle \begin{aligned}\text{Acceleration} &= \text{Average Acceleration}=\frac{\text{Change in Velocity}}{\text{Time Taken}}\end{aligned}$ .

$\displaystyle {a} = \frac{200\;\text{m}\cdot\text{s}^{-1}}{2\;\text{s}}=100\;\text{m}\cdot\text{s}^{-2}$ .

<h3>Apply Newton's Second Law to find the mass of the object.</h3>

$\displaystyle m = \frac{\Sigma F}{a} = \frac{100\;\text{N}}{100\;\text{m}\cdot\text{s}^{-2}} = 1\;\text{kg}$ .

6 0

3 years ago

Read 2 more answers

How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

$T_{f}$ = (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

$T_{f}$ = 466.8 K

so change in temperature ΔT

ΔT = 466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x $_{i\\}$ ² - x $_{f\\}$ ² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q = 16 J

Therefore, the required heat energy is 16 J

5 0

2 years ago