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bonufazy [111]
3 years ago
12

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

When it reaches the highest point of its trajectory, its speed is150 m/s. In a second trial with the same projectile, the initialspeed is the same but the angle is now 37 degree with thehorizontal. At its highest point in this trajectory, the velocityof the projectile would be what?
Physics
1 answer:
ivann1987 [24]3 years ago
8 0

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

H_{max}= \dfrac{v^2 sin^2(\theta)}{g}

now,

H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}

H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}

now, equating both the equations

\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}

\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}

   v₂² = 31061.79

   v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

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Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.
ozzi

Answer:

The thrust is 6\times 10^6\ N

Explanation:

Given that,

Mass of gas, m=1.5\times 10^3\ kg

The rate at which the gas is expelling, \dfrac{dv}{dt}=4\times 10^{3}\ m/s

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

F=\dfrac{p}{t}

Also, p = mv

F=\dfrac{mv}{t}

So,

F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N

So, the thrust is 6\times 10^6\ N

3 0
3 years ago
Two pebbles, Pebble A and Pebble B, are thrown horizontally with the same force. Pebble A's mass is 3 times the
Hitman42 [59]

Answer:

Pebble A has 1/3 the acceleration as pebble B.

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F = m×a

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Same starting force, F

m_a = mass of a

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a_a = acceleration of a

a_b = acceleration of b

F = m_a × a_a = m_b × a_b

3 × m_b × a_a = m_b × a_b

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zhannawk [14.2K]
Average speed= total distance/total time =12km/h
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3 years ago
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