What is the shape of the trajectory that a charged particle follows in a uniform magnetic field?

Which one the answer to this question

tangare [24]

The second bubble is the answer:)

4 0

3 years ago

The process called _____ destroys old oceanic crust at subduction zones.

ZanzabumX [31]

The process is called subduction

5 0

3 years ago

A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or sh

zalisa [80]

Complete question:

Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. In contrast, a passenger not wearing a seat belt or using an air bag decelerates over a distance of 5mm.

(a) A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.

Answer:

The net force on the person as the air bad deploys is -6750 N backwards

Explanation:

Given;

mass of the passenger, m = 60 kg

velocity of the car at impact, u = 15 m/s

final velocity of the car after impact, v = 0

distance moved as the front of the car crumples, s = 1 m

First, calculate the acceleration of the car at impact;

v² = u² + 2as

0² = 15² + (2 x 1)a

0 = 225 + 2a

2a = -225

a = -225 / 2

a = -112.5 m/s²

The net force on the person;

F = ma

F = 60 (-112.5)

F = -6750 N backwards

Therefore, the net force on the person as the air bad deploys is -6750 N backwards

4 0

3 years ago

A baseball is projected horizontally with an initial speed of 18.1 m/s from a height of 1.85 m. at what horizontal distance will

gtnhenbr [62]

1) The motion of the baseball consists of two separate motions along the horizontal (x) and vertical (y) axis. The position on the two directions at time t is given by:
$x(t)=v_0t$
$y(t)=h- \frac{1}{2}gt^2$
where the motion on the x-axis is a uniform motion with constant speed $v_0=18.1 m/s$ , while the motion on the y-axis is a uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$ , and initial height $h=1.85 m$ .

First of all, we can find the time t at which the ball reaches the ground by requiring $y(t)=0$ :
$0=h- \frac{1}{2}gt^2$
$t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(1.85 m)}{9.81 m/s^2} }= 0.61 s$

and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
$x(t)=v_0 t=(18.1 m/s)(0.61 s)=11.04 m$

2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one:
$v_x=18.1 m/s$
On the y-axis, the velocity is given by
$v_y(t)=gt$
If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
$v_y =gt=(9.81 m/s^2)(0.61 s)=6.0 m/s$
And the speed of the ball is the magnitude of the resultant of the two components:
$v= \sqrt{v_x^2+v_y^2}= \sqrt{(18.1 m/s)^2+(6.0 m/s)^2}=19.1 m/s$

8 0

3 years ago

Read 2 more answers

The force of gravity is twice as great on a 2-kg rock as on a 1-kg rock. Why then, dose the 2-kg rock not fall with twice the ac

Phoenix [80]

Answer:

because all objects fall at a rate of 9.8m/s²

8 0

3 years ago