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Komok [63]
3 years ago
7

A police officer is called to the scene of a car accident in his accident he sketches the scene and describes it according to hi

s description the car went off of the road and hit a tree right after a bend the driver claimed that a second car ran them off a road by hitting them from behind using his observations and his knowledge of physics the police officer determined that the driver was not telling the truth
Physics
1 answer:
abruzzese [7]3 years ago
7 0

Answer:

The driver was not telling the truth because it is not possible for a car to hit another car from behind and generate a force to the sides that deflects it from its path.

Explanation:

First, we analyze the driver's statement.

The driver when arriving at the curve, is collided from behind by another car and deviates from his path and crashes into a tree. For the car to go to the tree there must be a force towards the tree.

The net force that causes the car to deviate must be formed by the sum of the motion vector of the first car plus the force that is directed towards the tree.

Here we verify that a car hitting from behind will not generate a force to the sides, but will generate a force in the same direction that the car moves, forward.

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Electrons and protons travel from the Sun to the Earth at a typical velocity of 3.83 ✕ 105 m/s in the positive x-direction. Thou
Leona [35]

Answer:

F=2.84*10^{-26}N  & -y direction

F=2.84*10^{-26}N & +y direction

Explanation:

From the question we are told that:

Speed of electron V_e=3.83 * 10^5 m/s +x direction

Earths magnetic field B_e=3.04 * 10^-^8 +z direction

a)

Generally the equation for magnetic force F_m is mathematically given by

F=q(V_e*B_e)

where

q=1.6*10^{-19}c\\\=i*\=z=-\=j

F=1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)

F=1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)

F=-2.84*10^{-26}N \=j

Magnitude & Direction

F=2.84*10^{-26}N  -y direction

b)

Generally the equation for magnitude and direction of the magnetic force on an electron. is mathematically given by

\=F'=-1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)

\=F'=-2.84*10^{-26}N \=j

Magnitude & Direction

F=2.84*10^{-26}N & +y direction

5 0
3 years ago
Of the following states of motion, the one which requires balanced forces is:
Tanzania [10]
I believe it would be D a change in direction of motion
3 0
3 years ago
Read 2 more answers
PLSS HELP ASAP
Feliz [49]
40 seconds I’m pretty sure sorry if I’m wrong
6 0
3 years ago
Read 2 more answers
A package is dropped from a helicopter moving upward at 15 m/s
daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

#SPJ4

7 0
2 years ago
Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air?
aniked [119]
An applied force<span> is a </span>force<span> that is </span>applied<span> to an object by a person or another object.
An attractive force is a force of an attraction (where object are attracted by each other). Gravitation is an example of attractive force.
</span>Normal force<span> is the component, perpendicular to the surface (surface being a plane) of contact.
</span><span>The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences attractive force from the air it passes through. It also experiences a downward pull because of the normal force.
Solution A.</span>
4 0
3 years ago
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