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vampirchik [111]

3 years ago

12

Se lanza una pelota de béisbol desde la azotea de un edificio de 25 m de altura con velocidad inicial de magnitud 10 m/s y dirig

ida con un ángulo de 63.1° sobre la horizontal. A) ¿Qué rapidez tiene la pelota justo antes de tocar el suelo? Use métodos de energía y desprecie la resistencia del aire.

1 answer:

MissTica3 years ago

6 0

Answer:

v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

Starting point. On the roof of the building

Em₀ = K + U = ½ m v₀² + m g y₀

Final point. On the floor

Em_f = K = ½ m v_f²

Emo = Em_g

½ m v₀² + m g y₀ = ½ m v_f²

v_f² = v₀² + 2 g y₀

let's calculate

v_f = √(10² + 2 9.8 25)

v_f = 24.3 m / s

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This solid layer of the earth is made of mostly iron and nickel.

Studentka2010 [4]

Answer

Explanation:

Yes, it's true that the solid layer of the earth is known as the most dense part as it is made up of the heavy metals like iron and nickel. Inner part is the hotter part due to the high pressure and temperature. It has the temperature of about 5,200°C and the pressure of 3.6 million atm but still the iron and nickel are present there in the solid form as they withstand such high temperature and pressure values.

5 0

3 years ago

A rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s. (a) Find the recoil speed of the rifle. m/s (b) If a

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Answer:

The recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$ .

Explanation:

The expression for the force in terms of mg is as follows;

F=mg

Here, m is the mass and acceleration due to gravity.

Rearrange the expression for mass.

$m=\frac{F}{g}$

Calculate the combined mass of the man and rifle.

$m_{man,rifle}=\frac{650+25}{g}$

Put $g=9.8 ms^{-2}$ .

$m_{man,rifle}=\frac{650+25}{9.8}$

$m_{man,rifle}=68.88 kg$

The expression for the conservation of momentum is as follows as;

$m_{man}u_{man}+m_{bullet}u_{bullet}=m_{man}v_{man}+m_{rifle}v_{man,rifle}$

Here, $m_{man,rifle}$ is the mass of the man and rifle, $m_{rifle}$ is the mass of the rifle, $u_{man},u{bullet}$ are the initial velocities of the man and bullet and $v_{man},v{man,rifle}$ are the final velocities of the man and rifle and rifle.

It is given in the problem that a rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s.

Convert mass of rifle from gram to kilogram.

$m_{bullet}=4.5 g$

$m_{bullet}=.0045 kg$

Put $m_{bullet}=.0045 kg$ , $m_{man,rifle}=68.88 kg$ , $u_{man,rifle}=0$ , $v_{bullet}= 240 ms^{-1}$ and $u_{bullet}=0$ .

$m_{man}(0)+m_{bullet}(0)=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+1.08$

$(68.88)v_{man,rifle}=\frac{-1.08}{68.88}$

$v_{man,rifle}=-0.016 ms^{-1}$

Therefore, the recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$ .

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noname [10]

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Harrizon [31]

Answer:

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Answer:

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Explanation:

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