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2 years ago

At what angle do the cars move after the collision?

Physics

1 answer:

liubo4ka [24]2 years ago

6 0

Answer:

his results in the final angle after the collision of 37.2 degrees basically what we did there is turn the vector into a right triangle. We use sohcahtoa to solve for the angle. Being.

Explanation:

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If you want to go to a community college and transfer to a four-year university, what classes are required to transfer? How many

bagirrra123 [75]

Typically, most four-year institutions will require the following classes to transfer:

2 English composition courses.

1 Speech course.

1 Critical thinking course.

1 Physical science class.

1 Biological science class.

1 Health class.

2 Physical education courses.

Completion of college algebra

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3 0

3 years ago

The time between a lightning flash and the following thunderclap may be used to estimate, in kilometers, how far away a storm is

Kruka [31]

Given Information:

Elapsed time = t = 6 seconds

Required Information:

Distance = d = ?

Answer:

Distance = d = 2.058 km

Explanation:

We know that the speed of sound in the air is given by

v = 343 m/s

The relation between distance, speed and time is given by

distance = speed*time

substituting the given values yields,

distance = 343*6

distance = 2058 m

There are 1000 meters in 1 km so

d = 2058/1000

d = 2.058 km

Therefore, the storm is about 2.058 km away when elapse time between the lightning and the thunderclap is 6 seconds.

3 0

3 years ago

What is the average horizontal component of force exerted on his feet by the ground during acceleration?

Arte-miy333 [17]

Well you have to think of it like electricity go through your answer closes to that and figure it out

7 0

3 years ago

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

<u>Given:</u>

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

5 0

3 years ago

A 75-m-long train begins uniform acceleration from rest. the the train has a speed of 23 m/s when it passes a railway worker tan

Lapatulllka [165]

Answer:

acceleration a = 1.04 m/s2

Explanation:

Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:

s = 75 + 180 = 255 m

We can use the following equation of motion to find out the distance traveled by the car:

$v^2 - v_0^2 = 2as$ where v = 23 m/s is the velocity of the car when it passes the worker, $v_0$ = 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.

$23^2 - 0^2 = 2*a*255$

$510a = 529$

$a = 529 / 510 = 1.04 m/s^2$

8 0

3 years ago