Ask question

Ask question

All categories

3 years ago

15

What is antoine lavoisier conservation of mass equation

1 answer:

ratelena [41]3 years ago

5 0

Answer:

the mass of any one element at the beginning of a reaction will equal the mass of that element at the end of the reaction.

Explanation:

The antoine lavoiser conservation of mass equation can be equal to the mass of the element

You might be interested in

A string is waved up and down to create a wave pattern with a wavelength of 0.5 m. If the waves are generated with a frequency o

Pachacha [2.7K]

Answer:

1m/s

Explanation:

Speed of a wave is expressed as;

Speed = frequency × wavelength

Given

Frequency = 2Hz

Wavelength = 0.5m

Required

Speed of the wave

Substitute

Speed = 2×0.5m

Speed = 1.0m/s

Hence the speed is 1.0m/s

6 0

2 years ago

Billiard ball A of mass mA = 0.125 kg moving with speed vA = 2.80 m/s strikes ball B, initially at rest, of mass mB = 0.140 kg .

lianna [129]

Answer:

V=1.309

β= -41.997

Explanation:

Law Newton's conservation motion

Axis x

$m_{1}*v_{x1}+m_{2}*v_{x2}=m_{1}*v_{fx1}+m_{2}*v_{fx2}\\v_{x1}=2.8\frac{m}{s}\\v_{x2}=0 \frac{m}{s}\\m_{1}=0.125kg\frac{m}{s}\\m_{2}=0.140kg\frac{m}{s}\\0.125kg*2.8\frac{m}{s}+0.14kg*0=0.125kg*2.10\frac{m}{s}*cos(30) +0.14kg*v_{fx2}\\0.35 \frac{kg*m}{s} =0.125kg*1.81\frac{m}{s}+0.14kg*v_{fx2}\\v_{fx2}=\frac{0.12\frac{kg*m}{s} }{0.14kg} \\v_{fx2}=0.876 \frac{m}{s}$

Axis y

$m_{1}*v_{y1}+m_{2}*v_{y2}=m_{1}*v_{fy1}+m_{2}*v_{fy2}$

$v_{y1} =0\\v_{y2} =0$

$0=m_{1}*v_{fy1} +m_{2}*v_{fy2} \\v_{fy2}=-\frac{m_{1}*v_{fy1} }{m_{2}}\\ v_{fy2}=-\frac{0.125kg*2.10*sen(30)\frac{m}{s}}{0.14kg}\\v_{fy2}= -0.937\frac{m}{s}$

So the velocity $v_{f2}$

$v_{f2}=\sqrt{v_{fx2}^{2} +v_{fy2}^{2} } \\v_{f2}=\sqrt{0.876^{2} +0.983^{2} } \\v_{f2}=1.309\frac{m}{s}$

The angle can be find using both velocity factors

$\alpha =tanx^{-1}*\frac{v_{fx2}}{v_{fy2}} \\\alpha =tanx^{-1}*\frac{0.876}{-0.973}\\ \alpha =tanx^{-1}*\\ \alpha =-41.997$

Check:

$m_{1}*v_{x1}+m_{2}*v_{x2}=m_{1}*v_{fx1}+m_{2}*v_{fx2}$

$0.125*2.80=0.125*2.1*cos(30)+0.14*1.03*cos(-41.997)\\0.35=0.227+0.107\\$

0.35≅0.3489

4 0

3 years ago

Im not going to lie, this is a physics Q right here<br> plz guys im being serious i need help

Anastaziya [24]

Answer:

to much to small

Explanation:

i cant zoom in and i cant see

5 0

3 years ago

A scientist is subjected to a dose of ionizing radiation in his laboratory. without the help of radiation detection instruments,

vesna_86 [32]

<span>Health problems that develop later</span>

5 0

3 years ago

A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai

shutvik [7]

Answer:

d) None of the above

Explanation:

$v_{o}$ = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m

t = time of travel

using the equation

$y=v_{oy} t+(0.5)a_{y} t^{2}$

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

$v_{ox}$ = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

$a_{x}$ = acceleration along the horizontal direction = 0 m/s²

t = time of travel = 2.1 s

Using the kinematics equation

$x =v_{ox} t+(0.5)a_{x} t^{2}$

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

$y_{o}$ =initial vertical position at the time of launch = 20 m

$y$ = vertical position at the maximum height = 20 m

$v_{fy}$ = final velocity along vertical direction at highest point = 0 m/s

using the equation

${v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})$

$0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)$

$y$ = 20.02 m

h = height above the starting height

h = $y$ - $y_{o}$

h = 20.02 - 20

h = 0.02 m

7 0

3 years ago