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ELEN [110]
3 years ago
9

A projectile is fired in such a way that its horizontal range is equal to 13.5 times its maximum height. What is the angle of pr

ojection?
Physics
2 answers:
hjlf3 years ago
8 0

Answer:

Maximum height: h  = v0² sin² α / 2 g

Horizontal range: x = v0² sin 2 α / g

13.5 v0² sin² α /2 g  = v0² sin 2 α /g          / :  v0²/g

6.75 sin² α = 2 sin α cos α    / : sin α

6.75 sin α = 2 cos α    / : cos α

6.75 tan α = 2

tan α = 0.2963

α = tan^(-1) 0.2963

α = 16.504° ≈ 16.5°

The angle of projection is 16° 30`.

Read more on Brainly.com - brainly.com/question/1596989#readmore

Explanation:

True [87]3 years ago
6 0
Maximum height: h  = v0² sin² α / 2 g
Horizontal range: x = v0² sin 2 α / g
13.5 v0² sin² α /2 g  = v0² sin 2 α /g          / :  v0²/g
6.75 sin² α = 2 sin α cos α    / : sin α
6.75 sin α = 2 cos α    / : cos α
6.75 tan α = 2
tan α = 0.2963
α = tan^(-1) 0.2963
α = 16.504° ≈ 16.5°
The angle of projection is 16° 30`.
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Answer:

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Explanation:

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Q(A) = caΔT

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ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

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where;

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The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

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T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

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Answer:

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