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3 years ago

a car is traveling at 26 m/s starts to decelerate steadily. It comes to a complete stop in 6 seconds. What is the acceleration?

Physics

2 answers:

torisob [31]3 years ago

5 0

I can’t see if anyone have gave you the answer yet

ad-work [718]3 years ago

4 0

-0.5 m/s^2

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A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. T

Nina [5.8K]

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= $\frac{1}{2}$ kx² = $\frac{1}{2}$ (530)(0.149)² = 5.88 J

b) Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mg $h_o$ = $\frac{1}{2}$ kx² - mgx

And, $h_o$ = ( $\frac{1}{2}$ kx² - mgx )/(mg) = $[\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)$ ]/(0.645x9.8)

$h_o$ = 0.78m

d) Now, if the initial initial height of block is 3 $h_o$

$h_o$ = 3 x 0.78 = 2.34m

then, $\frac{1}{2}$ kx² - mgx - mg $h_o$ =0

$\frac{1}{2}$ (530)x² - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum compression of the spring)

4 0

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

u is the initial velocity

a is the acceleration

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:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0

2 years ago

2. Using Graph 2, calculate the net force experienced by the particle between 4 and 6 seconds. The

Trava [24]

Using Newtons Second Law:

F = m×a

F = (0.25 kg)(-2 m/s²)

F = -0.5 N

<h2>The correct option is C</h2>

3 0

2 years ago

Which of the following is true of greenhouse gases?

Sedaia [141]

Answer:

A. They trap energy in the atmosphere.

Explanation:

Greenhouse gases are defined as gases that absorb and emit radiation within the infrared range. These gases allow the sun’s rays to pass through the ozone layer and warm the earth, but prevent this warmth from escaping atmosphere.

Greenhouse gases cause the greenhouse effect on planets. Examples of greenhouse gases in Earth's atmosphere are water vapor, carbon dioxide, methane, ozone etc.

4 0

2 years ago

What is the main cause of the seasons

irinina [24]

The main cause of the seasons is earths orbit since each time were in a new season the planet is facing the earth a different way which causes winter etc..

3 0

3 years ago