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3 years ago

A car traveling with constant speed travels 150 km in 7200 s. What is the speed of the car?

Physics

2 answers:

Wewaii [24]3 years ago

6 0

75 km/h is the speed of the car

Akimi4 [234]3 years ago

5 0

.02 km/s
Hope this helped
(distance divided by time= speed)

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Climates on Earth get _____ as you move from the equator to the poles.

andreev551 [17]

The correct answer that would best complete the given statement above would be the last option: COLDER. Climates on Earth get colder <span>as you move from the equator to the poles. The places that are located near or on the equator experience the warmest or the hottest climates such as Africa. Hope this answer helps. </span>

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3 years ago

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When an object's final velocity is less than its initial velocity, however, it has ________________ acceleration.

algol [13]

Answer:

The body has negative acceleration PR a deceleration.

Explanation:

HOPE THAT THIS IS HELPFUL.

HAVE A GREAT DAY.

4 0

3 years ago

The components of vector A are:

Korvikt [17]

here as it is given that x component of the vector is positive while y component of the vector is negative so we can say the vector must inclined in Fourth quadrant.

So angle must be more than 270 degree and less than 360 degree

Now in order to find the value we can say that

$tan\theta = \frac{opposite\: side}{adjacent\: side}$

$tan\theta = \frac{8.6}{6.1}$

$\theta = tan^{-1}1.41$

$\theta = 54.65^0$

so it is inclined at above angle with X axis in fourth quadrant

Now if angle is to be measured counterclockwise then its magnitude will be

$\theta = 360 - 54.65 = 305.3^0$

so the correct answer will be 305 degree

3 0

3 years ago

A cooking pot is sitting in a kitchen for several hours unused. Why does it feel cold when you touch it?

masha68 [24]

Answer:

I would go with 2

Explanation:

But i would also not go with my answer. Lol

8 0

3 years ago

An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th

kvv77 [185]

Answer:

The plane would need to travel at least $8,\!580\; {\rm ft}$ ( $8.58 \times 10^{3}\; {\rm ft}$ .)

The $10,\!000\; {\rm ft}$ runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to $\rm ft\cdot s^{-1}$ :

$\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}$ .

Initial velocity of the plane: $u = 0\; {\rm ft \cdot s^{-1}}$ .

Take-off velocity of the plane $v =264\; {\rm ft\cdot s^{-1}}$ .

Let $x$ denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation $x = ((u + v) / 2) \, t$ .

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration $t$ times average velocity $(u + v) / 2$ .

The distance that the plane need to cover would be:

$\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}$ .

4 0

3 years ago