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

Answers

Answer 1

Answer: The result of inertia due to not wearing a seatbelt will result inthe passenger becoming some sort of a missile .
The weight of the passenger plusethe speed of travel prior to the collision will be multiplied togetherto equal the force their body will endure.
For ex. the force aperson will be subjected to for a passenger weighing 100 pounds, in avehicle traveling at 60 mph, will be the equivalent of 6000 pounds hitting a brick wall.
I hope this helps:)

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You need to raise a heavy block by pulling it with a massless rope. You can either (a) pull the block straight up height h, or (b) pull it up a long, frictionless plane inclined at a 15∘ angle until its height has increased by h. Assume you will move the block at constant speed either way.

Answers

Final answer:

In both scenarios, the work done on the heavy block is the same, as it is determined by the change in the vertical height. However, pulling the block up the inclined plane may require less force because the work is distributed over a larger distance.

Explanation:

The subject of this question is based on the concept of work and energy in physics. When you pull the heavy block straight upwards (scenario a), the work done is equal to the force times the distance, or Work = mg*h, where m is the mass of the block, g is the acceleration due to gravity, and h is the height it needs to rise. For pulling the block up the inclined plane (scenario b), the work done still equals mg*h as the vertical distance it rises is the same.

This is because, according to the principle of work and energy, the work done on an object is equal to the change in its kinetic energy. Since the speed of the block remains constant in both scenarios, the kinetic energy does not change, meaning the work done on the block is the same in both scenarios.

However, pulling the block up the inclined plane may require less force because of the larger distance over which the work is done. But the overall work is the same in both cases.

Learn more about Work and Energy here:

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A long solenoid has 1400 turns per meter of length, and it carries a current of 4.9 A. A small circular coil of wire is placed inside the solenoid with the normal to the coil oriented at an angle of 90.0˚ with respect to the axis of the solenoid. The coil consists of 42 turns, has an area of 1.2 × 10-3 m2, and carries a current of 0.45 A. Find the torque exerted on the coil.

Answers

Answer:

The torque on the coil is $1.955* 10^(- 4)\ N-m$

Solution:

No. of turns per meter length, n = 1400 turns\m

Current, I = 4.9 A

Angle, $\theta = 90.0^(\circ)$

No. of turns of coil, N = 42 turns

Area, A = $1.2* 10^(- 3)m^(2)$

Current in the coil, I' = 0.45 A

Now,

To calculate the exerted torque on the coil:

The magnetic field, B produced inside the coil is given by:

$B = n\mu_(o)I$

$B = 1400* 4\pi times 10^(- 7)* 4.9 = 8.62* 10^(- 3)\ T$

Now, the torque exerted is given by:

$\tau = I'NAB$

$\tau = 0.45* 42* 1.2* 10^(- 3)* 8.62* 10^(- 3) = 1.955* 10^(- 4)\ N-m$

Answer:

$T\approx 1.95* 10^(-4) N.m$

Explanation:

Given:

A long solenoid having

no. of turns per meter, n =1400

current, I = 4.9 A

A small coil of wire placed inside the solenoid

angle of orientation with respect to the axis of the solenoid, $\theta=90\degree$ °

no. of turns in the coil, N = 42

area of the coil, $a= 1.2* 10^(-3) m^2$

current in the coil, $i =0.45 A$

We have for torque:

$T=n.i.a.B. sin\theta$ .......................(1)

∵ $B=\mu_(0) .n.I$ ................................(2)

where:

B= magnetic field

$\mu_0=$ The permeability of free space = $4\pi*10^(-7) T.m.A^(-1)$

Substitute B from eq. (2) into eq. (1) we have:

$T=n.i.a.(\mu_0.N.I ).sin\theta$

putting the respective values in above eq.

$T=42* 0.45* 1.2* 10^(-3)* 4\pi*10^(-7) * 1400* 4.9* sin 90^(\circ)$

$T\approx 1.95* 10^(-4) N.m$

A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck travels for 20.0 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00 s. (a) How long is the truck in motion? (b) What is the average velocity of the truck for the motion described?

Answers

You can use the equation
V_xf = V_xi + a_x(t)

V_xf = 20.0m/s
V_xi = 0m/s
ax = 2.0t

Thus, solve for t and get 10secondsand then take 5 seconds to break after 20 seconds of drivingso for

a) 10 + 20 + 5 = 35 seconds

for part b)
You can use the formula
Delta x/Delta t = average velocity

Need to find xf, knowing xi = 0

Thus, use the formula
x_f = x_i + V_xi(t) + (1/2)a_x(t)^(2)
x_f = 0 + 0(10) + (1/2)(2.0)(10)^(2)
x_f = 100m

so for the first 10 seconds the truck traveled 100msAt a speed of 20m/s

20m/s = xm/20s20*20 = x
x = 400

thus we have 100+400 = 500mthen it slows down from 500m to x_f

thus I use the equation
x_f = x_i + (1/2)(V_xf + V_xi)t
x_f = 500 + (1/2)(0 + 20)(5)x_f = 500 + 50
x_f = 550

therefore the total distance traveled is 550m

to calculate average velocity
550/35 = 16m/s

thusV_xavg = 16m/s

Where is a bibliography or works-cited list placed within a paper or report?in the introductory material
at the bottom of each page
at the end
in the appendix

Answers

Answer:

at the end of the page

Explanation:

leave a rating for my dying soul please. :)

Answer:

at the end

Explanation:

A 68 kg runner exerts a force of 59N. What is the acceleration of the runner?

Answers

Explanation:

Newton's First Law ,

F = ma

a = m/F

a = 68 / 59

a = 1.15 m/s2

How does size of an object affect its gravity

Answers

Answer:

the more particles packed together the faster it falls

Explanation:

the mass + the 1 constant g-force = the speed without adding air resistance

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