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Crank
4 years ago
5

for a freely falling object dropped from rest, what is the acceleration at the end of the fifth second of fall?

Physics
1 answer:
lara31 [8.8K]4 years ago
3 0

acceleration due to gravity is always 9.8 m/s/s (on earth)
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John took 0.75 hours to bicycle to his grandmother's house, a distance of 4 km. What is his velocity?
ddd [48]

Answer:

5.3Km/hr

Explanation:

Velocity=Displacement/Time

D=4km;T=0.75hr

V=4/0.75=5.33..

6 0
3 years ago
Uma chapa metálica, com um furo central de diâmetro "d", é aquecida dentro de um forno. Com o aumento da temperatura, podemos af
Tatiana [17]

Answer:  O furo diminui enquanto a chapa aumenta a sua dimensão.

(The diameter decreases and the dimensions of the metal increases)

Explanation:

I will answer in English:

Here we have a piece of metal with a small hole on it, that is heated in a hoven.

We know that when a piece of metal is heated, it will expand (the density decreases, so the dimensions of the piece of metal increase).

Now, particularly, the hole will also expand but inwards, as each "extreme" of the piece of metal, will expand into where it has less resistance, so it will expand in the region where there is no material. This "inwards" expansion is why the diameter of the hole will decrease.

4 0
3 years ago
Calculate the energy needed to change 200.0 g of ice from -20.0°C to water at 35.0°C.
oksano4ka [1.4K]

it's C  24.94KCAL

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3 0
3 years ago
Which resource would be the best choice to learn more information about studying martial arts?
zhuklara [117]

Answer:

Dance studio

Explanation:

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3 0
3 years ago
From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up
NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

7 0
3 years ago
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