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2 years ago

An object accelerates from rest and travels 53 m west in 5.2 s. Determine the acceleration

Physics

1 answer:

zmey [24]2 years ago

6 0

Answer:

20.4m/s²

Explanation:

Given parameters:

Initial velocity = 0m/s

Distance = 53m

Time = 5.2s

Unknown:

Acceleration = ?

Solution:

This is a linear motion and we use the right motion equation;

S = ut + $\frac{1}{2}$ at²

S is the distance

u is the initial velocity

a is the acceleration

t is the time

Insert the parameters and solve;

53 = (0x 5.2) + $\frac{1}{2}$ x a x 5.2

53 = 2.6a

a = $\frac{53}{2.6}$ = 20.4m/s²

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2 years ago

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svlad2 [7]

Answer:

F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.

Explanation:

The student wants to prove hooke's law which has the form

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2 years ago

An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro

Mashutka [201]

1) The distance travelled by the electron is $1\cdot 10^{17} m$

2) The time taken is $4.0\cdot 10^{10}s$

Explanation:

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

$v^2-u^2=2as$

where

v is the final velocity

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s is the distance travelled

For the electron in this problem,

$u=5\cdot 10^6 m/s$ is the initial velocity

v = 0 is the final velocity (it comes to a stop)

$a=-1.25\cdot 10^{-4} m/s^2$ is the acceleration

Solving for s, we find the distance travelled:

$s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m$

The total time taken for the electron in its motion can also be found by using another suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

$u=5\cdot 10^6 m/s$

v = 0

$a=-1.25\cdot 10^{-4} m/s^2$

And solving for t, we find the time taken:

$t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s$

Learn more about accelerated motion:

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