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3 years ago

What is the accleration of a car that travels 10 meters in 2seconds

1 answer:

5 0

If it happens to be cruising along at 5 meters per second (about
11 miles per hour), then it covers 10 meters in 2 seconds without
any acceleration at all.

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What is the main source of energy

jolli1 [7]

Sun is the main source of energy

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3 years ago

Read 2 more answers

If a car goes at a speed of 80 kph in 10 seconds, what is the distance

Blababa [14]

Answer:222 meters

Explanation:

Speed=80kph=(80x5/18)=(80x5)/18=22.2m/s

Time=10 seconds

distance=speed x time

distance=22.2 x 10

distance=222

distance=222 meters

3 0

3 years ago

Which of the following is a form of pollution created when vehicle exhaust interacts with sunlight?

SVEN [57.7K]

Photochemical smog is formed when primary air pollutants interact with sunlight.

Photochemical smog is the result of the reaction between pollutants like nitrogen oxides (NO), sunlight and volatile organic compound (VOC) in the atmosphere. The sources of NO are car exhaust, coal power plants, factory emissions, etc. This type of smog is also known by the name Los Angeles smog.

Air pollutants are the particles present dissolved in the air, which when inhaled by the organisms can cause serious health issues. These pollutants are :ozone, particulate matter, gaseous oxides, etc. These pollutants majorly affect the respiratory system of the humans.

Therefore, photochemical smog is a form of pollution created when vehicle exhaust interacts with sunlight.

To know more about photochemical smog, here: brainly.com/question/15728274

#SPJ4

8 0

1 year ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago

Suppose you wish to whirl a pail full of water in a vertical circle at a constant speed without spilling any of its contents (ev

Yanka [14]

Answer:

V = 2.87 m/s

Explanation:

The minimum speed required would be that at which the acceleration due to gravity is negated by the centrifugal force on the water.

Thus, we simply need to set the centripetal acceleration equal to gravity and solve for the speed V using the following equation:

Centripetal acceleration = V^2 / r

where r is the distance of water from the pivot or shoulder.

For our case, r will be 0.65 + 0.19 = 0.84 m

and solving the above equation we get:

9.81 = V^2 / 0.84

V^2 = 8.2404

V = 2.87 m/s

6 0

3 years ago