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sladkih [1.3K]
3 years ago
13

A bullet is fired vertically upward from a gun with an initial velocity of 225 m/s. How high does the bullet go in meters? [Igno

re air resistance]
Physics
1 answer:
Alchen [17]3 years ago
7 0

Answer:

h= 2582.9 m

Explanation:

bullet kinematics:  bullet moves with uniformly accelerated movement in freefall  

The negative sign in the equation indicates that the acceleration (g) is in the direction opposite to the movement of the  bullet.

v f₁²=v₀²-2g*h₁  Equation (1))

h: hight in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

Known data

v₀=  225 m/s

g= 9.8 m/s² : acceleration due to gravity

Problem development

When the  bulletl reaches the maximum height the final speed is equal to zero. vf=0

We replace data in the equation (1)

v f₁²=v₀²-2g*h₁

0 =  (225)²-2*9.8*h

2*9.8*h=(225)²

h= (225)²/ (2*9.8)

h= 2582.9 m

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The mechanical energy for the first and the second ball is

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The maximum height reached by the first ball is,

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The angle made by the cannonball while being fired from the cannon = 90 °

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Therefore, the total mechanical energy for the first and the

\:second \:  cannonball \:  is  \: 10 ^{7}  \:joules.

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In static friction, the frictional force resists force that is applied to an object, and the object remains at rest until the force of static friction is overcome. In kinetic friction, the frictional force resists the motion of an object. ... The frictional force itself is directed oppositely to the motion of the object.

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A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕
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Explanation:

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c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

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