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2 years ago

Un'automobile di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Physics

1 answer:

sesenic [268]2 years ago

3 0

Answer:

e di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiute di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiut

Explanation:

e di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiute di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiut

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A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the

irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

$x=V_{o} cos \theta t$ (1)

$y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2}$ (2)

Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

$\theta=0\°$ since the cannonball was shoot horizontally

$t$ is the time

$y=0$ is the final height of the cannonball

$y_{o}=1.96 m$ is the initial height of the cannonball

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $t$ from (2):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (3)

$t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}}$ (4)

$t=0.63 s$ (5)

Substituting (5) in (1):

$x=(23 m/s) cos(0\°) 0.63 s$ (6)

Finally:

$x=14.49 m$

5 0

3 years ago

I need help!! i’ll give 20 points

larisa86 [58]

Answer:

780 m to travel north

Explanation:

6 m over = 750

53 degree so it will take about 2 min to reach the destination

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3 years ago

Blizzards and hurricanes are examples of contrasting observations evens. Truth or False

noname [10]

The correct answer should be True

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3 years ago

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The weight of a standard object defined as having a mass of exactly 2.9 kg is measured to be 28.449 n. in the same laboratory, a

PtichkaEL [24]

Mass of the object m = 2.9 kg
Force F1 = 28.449 N
F1 = m1 x a => a = F / m => 28.449 / 2.9 => a = 9.81, which is gravitational acceleration.
In the same lab, a = g = 9.81, second object F2 = 48.7N = m2 x a
m2 = F2 / a => 48.7 / 9.81 => m2 = 4.96 kg
Mass of the second object m2 = 4.96 kg

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3 years ago

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Mendeleev gave the name eka-aluminum to an A. compound containing aluminum. B. mixture of aluminum and an unknown element. C. un

PolarNik [594]

Answer: C. unknown element he predicted would have properties similar to those of aluminum.

Explanation: Mendeleev arranged the elements in order of increasing atomic masses. He left spaces for the elements that were predicted to exist, but were not discovered.

One space was left below Aluminum where an element of mass 70 was predicted to exist with properties similar to Aluminum. It was referred to as eka-Aluminum.

Later it was named as gallium.

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3 years ago

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