How does inertia affect a person who is not wearing a seatbelt during a collision?

Answers

Answer 1

Answer:

A person who is not wearing a seatbelt during a collision will be thrown forward because it maintains forward motion

Further explanation

In Newton's law, it is stated that if the resultant force acting on an object of magnitude is zero, it can be formulated :

$\large{\boxed{\bold{\Sigma F = 0}}}$

then the object tends to defend itself from its state. So for objects in a state of movement, objects tend to move forever. Likewise, for objects in a state of rest, they tend to remain forever. The tendency of objects like this is called inertia

The size of inertia is proportional to mass, the greater the mass of the object, the greater the inertia of the object.

In objects with mass m that move translatively, the object will maintain its linear velocity

When we are in a vehicle that moves forward, then we will still maintain a state of forwarding motion. If our vehicle stops suddenly, then we keep moving forward so we will be pushed forward. From this point, the use of a safety belt serves to hold back our movements so that there are no fatal accidents or collisions.

Learn more

Newton's law of inertia

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example of Newton's First Law of inertia

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law of motion

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Keywords: inertia, Newton's First Law

Answer 2

Answer:

The person not wearing the seat-belt during a collision may hit the dashboard of the car very hard and this injury might be fatal.

Explanation:

The inertia of a body is a property contained by the body by virtue of its mass. Higher the mass of a body, higher will be the inertia.

According to the Newton's first law of motion i.e. law of Inertia, a body continues to be in its state of rest or motion until and unless an external force acts on the body to change its state.

When a car is moving on a road, the car is in motion and the body of the person in the car is also in the motion along with the car. Now if the car faces any collision, the car will suddenly come to rest.

But the body of the person sitting in the car continues to be in motion as there is no external force acting on the body. If the person is not wearing the seat-belt, the body moves forward and the person may hit the dashboard very hard.

The seat-belt stops the person's body to move ahead and hit the dashboard. So, the seat-belt prevents the person from injury by stopping the body.

Thus, The person not wearing the seat-belt during a collision may hit the dashboard of the car very hard and this injury might be fatal.

Learn More:

1. The component of a lever brainly.com/question/1073452

2. The kinetic energy of a body depends on brainly.com/question/137098

3. How long to walk 2000 miles brainly.com/question/3785992

Answer Details:

Grade: High School

Subject: Physics

Chapter: Laws of Motion

Keywords:

inertia, law of motion, newton's law, seat-belt, collision, wearing, dashboard, injury, external force, rest, motion.

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However, if the ray is within some suitable range of
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Answers

The ratio of output power to input power for a machine is called the machine's efficiency. It doesn't have to be expressed in percent.

Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on the other to be LaTeX: 1.35\times10^{-4}N1.35 × 10 − 4 N when they are 20 cm apart. Accidentally, one the the experimenters causes the balls to collide and then repositions them 20 cm apart . Now the repulsive force is found to be LaTeX: 1.406\times10^{-4}N1.406 × 10 − 4 N. What are the initial charges on the two metal balls?

Answers

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=(1)/(4\pi\spsilon_0)(q_1q_2)/(d^2)-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$(1)/(4\pi\spsilon_0)=9*10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35*10^(-4)$ N

So, from equation (i)

$1.35*10^(-4)=9*10^9(q_1q_2)/((0.2)^2)$

$\Rightarrow q_1q_2=6*10^(-16)\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$(q_1+q_2)/(2)$

d=20cm=0.2m, and $F_c=1.406*10^(-4)$ N

So, from equation (i)

$1.406*10^(-4)=9*10^9(\left((q_1+q_2)/(2)\right)^2)/((0.2)^2)$

$\Rightarrow (q_1+q_2)^2=2.50*10^(-15)$

$\Rightarrow q_1+q_2=\pm5* 10^(-8)$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5* 10^(-8)\;\cdots(iii)$

$\Rightarrow q_1=5* 10^(-8)-q_2$

The equation (ii) become:

$(5* 10^(-8)-q_2)q_2=6*10^(-16)$

$\Rightarrow -(q_2)^2+5* 10^(-8)q_2-6*10^(-16)=0$

$\Rightarrow q_2=3*10^(-8), 2*10^(-8)$

From equation (iii)

$q_1=2*10^(-8), 3*10^(-8)$

So, the magnitude of initial charges on both the sphere are $3*10^(-8)$ Coulombs $=0.03 \mu C$ and $2*10^(-8)$ Colombs or $0.02 \mu C$ .

Considerion the nature of charges too,

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

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Answers

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The given statements are :

Q1) Distance formula is displacement/time T/F?

- False

Q2)velocity formula is = displacement/time T/F?

- True

A compact disc has a radius of 6 centimeters.a. What is its circumference in meters?
b. If the cd rotates 4 times per second, what is the linear speed of a point on the outer edge of the cd? Give
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c. What is the linear speed of a point 3 centimeters from the center of the cd? (Assume the angular speed
has not changed).

Answers

Circumference is $2\pi{r}=37.7$ (approx.)
Speed is 4 times that. That is, 150.8 cm/s or 1.508 m/s.
If we get a point at distance 3 cm, we take the radius half of that. The circumference also gets half, so does the linear speed. We get 0.754 m/s or 75.398 cm/s.

Answer:75.389

Explanation:

stone thrown vertically upward with speed of 15.5m/s from edge of cliff 75.0m high. how much later does it reach bottom of cliff?

Answers

Answer:

Approximately $5.66\; {\rm s}$ , assuming that $g = 9.81\; {\rm m\cdot s^(-2)}$ and that air resistance is negligible.

Explanation:

Let upward be the positive direction. Under the assumptions, acceleration of the stone would be $a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}$ (negative since the stone is accelerating downward.)

The duration of the flight can be found in the following steps:

Find velocity right before landing given displacement, initial velocity, and acceleration.
Find duration of the flight from acceleration and the change in velocity.

In SUVAT equation $v^(2) - u^(2) = 2\, a\, x$ :

$v$ is the final velocity right before landing (needs to be found,)
$u = 15.5\; {\rm m\cdot s^(-1)}$ is the initial velocity,
$a = (-9.81)\; {\rm m\cdot s^(-2)}$ is acceleration, and
$x = (-75.0)\; {\rm m}$ is displacement (downward because the stone landed below where it was launched.)

Rearrange this equation to find $v$ :

$\begin{aligned}v &= \sqrt{u^(2) + 2\, a\, x} \n &= \sqrt{(15.5)^(2) + 2\, (-9.81)\, (-75.0)} \; {\rm m\cdot s^(-1)} \n &\approx -40.05\; {\rm m\cdot s^(-1)}\end{aligned}$ .

In other words, the velocity of this stone has changed from the initial value of $u = (15.5)\; {\rm m\cdot s^(-1)}$ to $v \approx (-40.05)\; {\rm m\cdot s^(-1)}$ during the flight. Divide the change in velocity by acceleration $a = (-9.81)\; {\rm m\cdot s^(-2)}$ (the rate of change in velocity) to find the duration of the flight:

$\begin{aligned}t &= (v - u)/(a) \n &\approx ((-40.05) - (15.5))/((-9.81))\; {\rm s} \n &\approx 5.66\; {\rm s}\end{aligned}$ .

In other words, the stone would be in the air for approximately $5.66\; {\rm s}$ .

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