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UNO [17]
3 years ago
8

A car traveling at 26.8m/s hits the brakes to slow down before turning into a neighborhood. Determine how fast it is going after

slowing down at a rate of -8.2m/s2 for 14.3seconds.
Physics
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

The speed of the car after the break was applied is  90.46 m/s

Explanation:

Given;

initial velocity of the car, u = 26.8 m/s

accelertation of the car, a = -8.2 m/s²

time of motion, t = 14.3 s

final velocity, v = ?

Apply the following kinematic equation;

v = u + at

v = 26.8 + (-8.2 x 14.3)

v = 26.8 -  117.26

v = -90.46 m/s

magnitude of the final velocity = 90.46 m/s

Therefore, the speed of the car after the break was applied is  90.46 m/s

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Which circle passes through more continents antarctic or arctic
Alex787 [66]

Answer:Arctic circle

Explanation:

Arctic circle is one of the circles of latitude. It is located at the northern region if the equator and passes through more than one continent. These continent includes: North Asia,

Northern America, and

Europe.

In these continent, it passes through eight known counties. These countries includes: Canada, Russia, Norway, Sweden, Finland, Denmark, Alaska and Iceland. It covers 4% of the Earth surface and it's climatic conditions are extreme.

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3 years ago
Ok, so this question is probably really easy but I can't really be bothered to answer it, terrible I know, but I thank all usefu
Dvinal [7]
Without a bulb energy cant go through and it would be an open circuit blocking the energy from coming out.
3 0
3 years ago
. A ball is thrown downward at a speed of 20 m/s. Choosing
a_sh-v [17]

The final velocity is v = 20 - gt

The distance traveled by the ball at time t is y = y_o + (-20)t - \frac{1}{2} gt^2

The maximum distance traveled by the object is 0 = (20)^2 - 2g(y-y_0)

The given parameters;

initial velocity of the ball, u = 20 m/s

acceleration due to gravity, g = 9.8 m/s²

The final velocity can be calculate as;

v = 20 - gt

The distance traveled by the ball at time t;

y = y_o + (-20)t - \frac{1}{2} gt^2

The maximum distance traveled by the object is calculated as;

v^2 = u^2 - 2g(y - y_0)\\\\0 = (20)^2 - 2g(y-y_0)

Learn more here: brainly.com/question/16878713

3 0
2 years ago
Research the potential for a large earthquake off the coast of Oregon. What are two of the biggest concerns with a large earthqu
Sphinxa [80]

There is a 33% chance of a large earthquake off the coast of Oregon.

The two main concerns of a large earthquake are Gas fires and Tsunami .

Explanation:

Everytime a large earthquake hits in any other part of the world, We find a small earthquake hitting off the coast of Oregon.

So scientists feel that as United States is prone to Earthquakes, the people have to be prepared for such eventualities.

The main concern for this large earthquake is, that it might lead to gas leakages at homes that can start a gas fire, which might spread. So people have to be careful and check  on any gas leakages.

Such large earthquakes can also lead to tsunami waves. So people have to take precautions and evacuate, if they are near the coastline.

6 0
3 years ago
An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e
VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

r = \frac{8.26}{2} = 4.13m

And the distance 'd' is

d = lsin\theta

d = 1.14 sin 16.2\°

d = 0.318m

Through the free-body diagram the tension components are given by

Tcos\theta = mg

Tsin\theta = \frac{mv^2}{R}

Here we can watch that,

R = r+d

Dividing both expression we have that,

tan\theta = \frac{v^2}{Rg}

Replacing the values,

tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}

v = 4.83371m/s

PART B) Using the vertical component we can find the tension,

Tcos\theta = mg

T = \frac{mg}{cos\theta}

T = \frac{(10+26.2)(9.8)}{cos(16.2)}

T = 369.42N

6 0
3 years ago
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