All categories

9 months ago

Jan Baptista van Helmont's famous experiment incorrectly showed that the plants produce most of their mass from:

1 answer:

3 0

Jan Baptista van Helmont's famous experiment incorrectly showed that the plants produce most of their mass from water.

In physics, mass is used to express inertia, a property common to all matter. In essence, it is the resistance of a mass of matter to changing its course or speed in response to the application of a force. The more mass a body has, the less of a change an applied force makes. Using Planck's constant, the kilogram, the ISU's unit of mass, is equivalent to 6.62607015 1034 joule seconds (SI). One kilogram is multiplied by one square meter per second to produce one joule. Since the second and the meter have already been defined in terms of other physical constants, the kilogram is determined by precise measurements of Planck's constant.

To know more about mass visit : brainly.com/question/6782334

#SPJ4

You might be interested in

AYUDA!!!!!

ZanzabumX [31]

Answer:

a) El valor de la densidad es 0.79 $\frac{g}{cm^{3} }$ o 790 $\frac{g}{cm^{3} }$

b) El peso especifico es 7749.9 $\frac{N}{m^{3} }$

Explanation:

a) La densidad se define como la propiedad que tiene la materia, ya sean sólidos, líquidos o gases, para comprimirse en un espacio determinado. En otras palabras, la densidad es una magnitud que permite medir la cantidad de masa que hay en determinado volumen de una sustancia. Entonces, la expresión para el cálculo de la densidad es el cociente entre la masa de un cuerpo y el volumen que ocupa:

$d=\frac{m}{V}$

En este caso:

masa= 237 g= 0,237 kg (siendo 1000 g= 1 kg)
volumen= 300 cm³= 0,0003 m³ (siendo 1 cm³= 0,000001 m³)

Reemplazando:

$d=\frac{237 g}{300cm^{3} }$ → d=0.79 $\frac{g}{cm^{3} }$

$d=\frac{0,237 kg}{0,0003m^{3} }$ → d=790 $\frac{g}{cm^{3} }$

El valor de la densidad es 0.79 $\frac{g}{cm^{3} }$ o 790 $\frac{g}{cm^{3} }$

b) El peso específico es la relación existente entre el peso y el volumen que ocupa una sustancia en el espacio.

Entonces, en este caso, siendo el peso:

P= m*g= 0,237 kg* 9,81 $\frac{m}{s^{2} }$ = 2,32497 N

el peso especifico es calculado como:

$Pe=\frac{Peso}{Volumen}= \frac{2,32497N}{0,0003 m^{3} }$

Pe= 7749.9 $\frac{N}{m^{3} }$

El peso especifico es 7749.9 $\frac{N}{m^{3} }$

8 0

3 years ago

How to solve these two questions?

hammer [34]

1) See attached figure

The relationship between charge and current is:

$i = \frac{Q}{t}$

where

i is the current

Q is the charge

t is the time

Therefore, the current is the rate of change of the charge passing through a given point over time.

This means that for a graph of charge over time, the current is just equal to the slope of the graph.

For the graph in this problem:

- Between t = 0 and t = 2 s, the slope is

$\frac{50-0}{2-0}=25 C/s$

therefore the current is

i = 25 A

- Between t = 2 s and t = 6 s, the slope is

$\frac{-50-(50)}{6-2}=-25 C/s$

therefore the current is

i = -25 A

- Between t = 6 s and t = 8 s, the slope is

$\frac{0-(-50)}{8-6}=25 C/s$

therefore the current is

i = 25 A

The figure attached show these values plotted on a graph.

2) $15 \mu C$

The previous equation can be rewritten as

$Q = i t$

This equation is valid if the current is constant: if the current is not constant, then the total charge is simply equal to the area under a current vs time graph.

Here we have the current vs time graph, so we gave to find the area under it.

The area of the first triangle is:

$A_1 = \frac{1}{2}(0.001 s)(0.010 A)=5\cdot 10^{-6} C$