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Cerrena [4.2K]
3 years ago
10

A train moving with a velocity of 42.9 km/hour North, increases its speed with a uniform acceleration of 0.250 m/s^2 North until

it reaches a velocity of 160.0 km/hour North. What distance did the train travel while it was increasing its velocity, in units of meters? Give the answer as a positive number
Physics
1 answer:
ankoles [38]3 years ago
5 0

Answer:

3658.24m

Explanation:

Hello!

the first thing that we must be clear about is that the train moves with constant acceleration

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

\frac {Vf^{2}-Vo^2}{2.a} =X

Vf = final speed =160km/h=44.4m/s

Vo = Initial speed =42.9km/h=11.92m/s

A = acceleration =0.25m/s^2

X = displacement

solving

\frac {44.4^{2}-11.92^2}{2.(0.25)} =X\\X=3658.24m

the distance traveled by the train is 3658.24m

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7 0
3 years ago
A piece of plastic has a net charge of +9.9 μC. How many more protons than electrons does this piece of plastic have? (e = 1.60
Leto [7]

Answer:

The number of protons 6.19 more than electron.

Explanation:

Given that,

Charge Q=+9.9\times10^{-6}\ C

We know that,

formula of charge

Q = nc

Where,

Q = total charge

n = number of protons

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Put the value into the formula

n = \dfrac{Q}{e}

n=\dfrac{9.9\times10^{-6}}{1.6\times10^{-19}}

n=6.1875\times10^{13}\ C

According to statement of question

Divide the answer by 10^{13}

n=\dfrac{6.1875\times10^{13}}{10^{13}}

n=6.19\ C

Hence, The number of protons 6.19 more than electron.

6 0
3 years ago
What is the current when a typical static charge of 0.234~\mu\text{c}0.234 μc moves from your finger to a metal doorknob in 0.59
Black_prince [1.1K]
The current is defined as the quantity of charge Q that passes through a certain location in a time \Delta t:
I= \frac{Q}{\Delta t}
Using the data of the problem, we find:
I= \frac{Q}{\Delta t} = \frac{0.234 \mu C}{0.595 \mu s}=\frac{0.234\cdot 10^{-6} C}{0.595 \cdot 10^{-6} s}=0.39 A
4 0
3 years ago
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Answer:

D. 90

Explanation:

because even if there is a shadow of doubt the person should not be persecuted as it could lead to false imprisonment

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3 years ago
Can anyone tell me the ans of this question also please
Dmitry_Shevchenko [17]

Answer:

Hey there!

Stopwatch X recorded 40 seconds, and stopwatch Y recorded 50 seconds.

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Hope this helps :)

6 0
3 years ago
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