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3 years ago

a Ferrari with an initial velocity of 10 m/s, comes to a complete stop in 5 seconds, what will its acceleration be?

Physics

2 answers:

Aleks04 [339]3 years ago

8 0

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) minus (speed at the beginning)

Change in speed = (zero) minus (10 m/s) = negative 10 m/s

Acceleration = (-10 m/s) / (5 sec)

Acceleration of the Ferrari = -2 m/s²

Democratically, it doesn't have to be a Ferrari. If my 16-yr-old Dodge is going 10 m/s and stops in 5 sec, its acceleration would be the same number.

11111nata11111 [884]3 years ago

6 0

10 m/s divided by 5 s

-2m/s/s

minus - deceleration

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A hovering mosquito is hit by a raindrop that is 50 times as massive and falling at 8.4 m/s, a typical raindrop speed. How fast

kenny6666 [7]

The final velocity after the collision is 8.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, if we consider the system to be isolated (=no external unbalanced forces), the total momentum of the raindrop+mosquito must be conserved before and after the collision.

If the collision is perfectly inelastic, moreover, the raindrop and the mosquito stick together and travel at the same velocity v after the collision.

Mathematically:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1$ is the mass of the first mosquito

$u_1 = 0$ is the initial velocity of the mosquito

$m_2 = 50 m_1$ is the mass of the raindrop

$u_2 = 8.4 m/s$ is the initial velocity of the raindrop

$v$ is the final combined velocity of the raindrop+mosquito

Re-arranging the equation and substituting, we find:

$m_1 u_1 + 50 m_1 u_2 = (m_1 + 50 m_1) v\\50 m_1 u_2 = 51 m_1 v\\50 u_2 = 51 v\\v=\frac{50}{51}u_2 = \frac{50}{51}(8.4)=8.2 m/s$

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3 years ago

What is the difference between celestial and terrestrial in planetary terms? Someone please answer

andriy [413]

"Celestial" = anything to do with the sky

("Cielo" ..... Spanish for "sky"
"Ceiling" ... that thing up over your head
"Caelum" .. Latin for "heaven")

"Terrestrial" = anything to do with the Earth

("Terra" ... Latin for "Earth")

5 0

3 years ago

A 25 kg child is riding on a swing. If the child travels 8.9 m/s at the bottom of their swing, how high into the air is the chil

Setler [38]

Answer:

h = 4.04 m

Explanation:

Given that,

Mass of a child, m = 25 kg

The speed of the child at the bottom of the swing is 8.9 m/s

We need to find the height in the air is the child is able to swing. Let the height is h. Using the conservation of energy such that,

$mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}$

Put all the values,

$h=\dfrac{(8.9)^2}{2\times 9.8}\\\\h=4.04\ m$

So, the child is able to go at a height of 4.04 m.

7 0

3 years ago

Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of

kondaur [170]

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

$m=\dfrac{F}{g}$

$m=\dfrac{160\ lb}{32\ ft/s^2}$

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

$m=\dfrac{F}{g}$

$m=\dfrac{1.9\ lb}{32\ ft/s^2}$

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

$W=2300\ kg\times 32\ ft/s^2=73600\ lb$

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

$W=0.022\ kg\times 32\ ft/s^2=0.704\ lb$

Hence, this is the required solution.

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3 years ago

A baseball player throws a baseball straight up into the air, the ball leaving his hand at time t = 0.0 s. The ball reaches maxi

laiz [17]

The ball should take twice as long to return to its original position as it took to reach its maximum height, so it should return to its original position at $t=2.8\,\rm s$ .

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3 years ago