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3 years ago

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An airplane, diving at an angle of 50.0° with the vertical, releases a projectile at an altitude of 554. m. The projectile hits

the ground 8.00 s after being released. (Assume a coordinate system in which the airplane is moving in the positive horizontal direction, and in the negative vertical direction.) (a) What is the speed of the aircraft?

1 answer:

Valentin [98]3 years ago

3 0

Answer:

Speed of the aircraft = 36.64 m/s

Explanation:

Consider the vertical motion of the projectile,

We have equation of motion s = ut+0.5at²

Let the velocity of plane be v.

Vertical velocity = vsin55 = Initial velocity of projectile in vertical direction = u

acceleration, a = 9.81 m/s²

displacement , s = 554 m

time, t = 8 s

Substituting,

554 = vsin55 x 8 +0.5 x 9.81 x 8²

v = 36.64 m/s

Speed of the aircraft = 36.64 m/s

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3 years ago

A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia

Musya8 [376]

Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

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3 years ago

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9966 [12]

a) 0 kg m/s

b) 0 kg m/s

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a)

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The principle of conservation of momentum states that when there are no external forces acting on a system, the total momentum of the system is conserved, so we can write:

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where

$p_i$ is the total momentum of the system before

$p_f$ is the total momentum of the system after

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$p_i = 0$

As we calculated in part a: this is because the total momentum of the two people before they push off each other is zero.

Therefore, according to the law of conservation of momentum,

$p_f = p_i = 0$

So the total momentum is zero also after they push off each other.

c)

The total momentum of the girl and the boy after they push off each other can be written as:

$p_f = m_1 v_1 + m_2 v_2$ (1)

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$p_f = 0$ (2)

By combining eq(1) and eq(2) we get

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d)

We can solve this part by applying the impulse theorem, which states that the change in momentum of an object is equal to the product between the force applied on it and the duration of the collision:

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where

$\Delta p$ is the change in momentum

F is the force

$\Delta t$ is the time during which the force is applied

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$\Delta t = 2.5 s$

For the boy, the change in momentum is:

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And since

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We have

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So, the force exerted between the boy and the girl is:

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3 years ago