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algol [13]
3 years ago
12

A cannonball is shot off a cliff with a speed of 77.02 m/s and an angle of 34 degrees above the horizontal. What is the horizont

al components of the initial velocity?
Physics
1 answer:
spin [16.1K]3 years ago
3 0

The horizontal component would be

(77.02 m/s) cos(34°) ≈ 63.85 m/s

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V=at and a=F/m

140/.070 = 2000m/s^2

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Lower in an arrangement

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Se dispone de dos vasos con agua a 20 grados y queremos calentarlo hasta alcanzar 50 grados. Si el primero contiene 0,5l y el se
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Answer:

The cup with 0.5L

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c_{water}=4186\frac{J}{kg\°C}

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4 0
3 years ago
A small wooden block with mass 0.775 kg is suspended from the lower end of a light cord that is 1.50 m long. The block is initia
jeka94

Answer:

34.83 m/s

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mv₀ = (m + M)V

V = mv₀/(m + M)

where m = mass of bullet = 0.0120 kg, v₀ = initial momentum of bullet, M = mass of block = 0.775 kg, V = final velocity of block + bullet.

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gh = -1/2(V₁² - V²)   (2)

Now, we obtain V₁ from

F = (m + M)V₁²/R since a centripetal force acts on the block + bullet at height 0.725 m. F = tension in chord = 4.88 N and R = length of cord = 1.50 m.

V₁ = √[FR/(m + M)]

Substituting V and V₁ into (2) above, we get

gh = -1/2(FR/(m + M) - [mv₀/(m + M)]²)

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substituting the values of the variables into v₀ we have

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= √(5.76 + 8.80)/0.012 kg

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So the initial speed v₀ = 34.83 m/s

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