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ioda
3 years ago
13

Why might it be necessary to ignore some of the data points just before and just after the collision?

Physics
1 answer:
krek1111 [17]3 years ago
7 0
We have no idea. We need to examine the experimental set-up. You've given us no information, except that there may have been some sort of collision.
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I have a big test coming up and these two questions i don't understand: Choose the best answer: A device that transforms mechani
Jlenok [28]
1. B- electric generator

Although its really sometimes called an AC generator.

- i hoped i helped u :)





3 0
3 years ago
15) What is the frequency of a pendulum that is moving at 30 m/s with a wavelength of .35 m?
____ [38]

A pendulum is not a wave.

-- A pendulum doesn't have a 'wavelength'.

-- There's no way to define how many of its "waves" pass a point
every second.

--  Whatever you say is the speed of the pendulum, that speed
can only be true at one or two points in the pendulum's swing,
and it's different everywhere else in the swing.

-- The frequency of a pendulum depends only on the length
of the string from which it hangs.


If you take the given information and try to apply wave motion to it:

             Wave speed = (wavelength) x (frequency)

             Frequency  =  (speed) / (wavelength) ,

you would end up with

             Frequency = (30 meter/sec) / (0.35 meter) = 85.7 Hz

Have you ever seen anything that could be described as
a pendulum, swinging or even wiggling back and forth
85 times every second ? ! ?     That's pretty absurd. 

This math is not applicable to the pendulum.

6 0
2 years ago
An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much
cupoosta [38]

Answer:

time will elapse before it return to  its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × 10^{7} m/s

uniform electric field E = 1.18 × 10^{4} N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × 10^{-31} kg ×a = 1.18 × 10^{4} × 1.602 × 10^{-19}

a = 20.75 × 10^{14} m/s²

so acceleration is 20.75 × 10^{14} m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × 10^{7} = 0 + 20.75 × 10^{14} (t)

t = 11.80 × 10^{-9} s

so time will elapse before it return to  its staring point is

time = 2t

time = 2 ×11.80 × 10^{-9}

time is 23.6 × 10^{-9} s

time will elapse before it return to  its staring point is 23.6 ns

7 0
2 years ago
a hammer drops from a height of 8 meters. calculate the speed with which it hits the ground. show work
ioda

Answer:

12.5 m/s

Explanation:

The motion of the hammer is a free fall motion, so a uniformly accelerated motion, therefore we can use the following suvat equation:

v^2-u^2=2as

Where, taking downward as positive direction, we have:

s = 8 m is the displacement of the hammer

u = 0 is the initial velocity (it is dropped from rest)

v is the final velocity

a=g=9.8 m/s^2 is the acceleration of gravity

Solving the equation for v, we find the final velocity:

v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(8)}=12.5 m/s

So, the final speed is 12.5 m/s.

3 0
2 years ago
Help ASAP plsssss:///
kenny6666 [7]

Answer:

c

Explanation:

8 0
3 years ago
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