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2 years ago

Three-fourths of the elements on the

Physics

1 answer:

77julia77 [94]2 years ago

8 0

Answer:

Explanation:

because the metalloids are the thing in the middle

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What are the independent and dependant variables in Plate Tectonics Lab Report?

tresset_1 [31]

Answer:

I don't know the exact answer but I think this is it.

The independent variable is the variable the experimenter changes or controls and is assumed to have a direct effect on the dependent variable. The dependent variable is the variable being tested and measured in an experiment, and is 'dependent' on the independent variable.

5 0

3 years ago

Read 2 more answers

How long does it take electrons to get from the car battery to the starting motor? Assume the current is 143 A and the electrons

Ostrovityanka [42]

Answer

the concentration of electron/ volume (n) = 8.5 × 10²⁸ /m³

the drift velocity of electron

$v = \dfrac{I}{neA}$

$v = \dfrac{143}{8.5\times 10^{28}\times 1.6\times 10^{-19}\times 40.2 \times 10^{-6}}$

v = 2.615 × 10⁻⁴ m/s

time require is

$t=\dfrac{L}{v}$

$t=\dfrac{0.845}{2.615\times 10^{-4}}$

t = 3231.35 s

5 0

3 years ago

A man can jump 1.5 m on earth. calculate the approximate

Olegator [25]

Answer:

18 m

Explanation:

G = Gravitational constant

m = Mass of planet = $\rho V$

$\rho$ = Density of planet

V = Volume of planet assuming it is a sphere = $\dfrac{4}{3}\pi r^3$

r = Radius of planet

Acceleration due to gravity on a planet is given by

$g=\dfrac{Gm}{r^2}\\\Rightarrow g=\dfrac{G\rho V}{r^2}\\\Rightarrow g=\dfrac{G\rho \dfrac{4}{3}\pi r^3}{r^2}\\\Rightarrow g=\dfrac{4G\rho\pi r}{3}$

So,

$g\propto \rho r$

Density of other planet = $\rho_p=\dfrac{1}{4}\rho_e$

Radius of other planet = $r_p=\dfrac{1}{3}r_e$

$\dfrac{g_e}{g_p}=\dfrac{\rho_e r_e}{\rho_p r_p}\\\Rightarrow \dfrac{g_e}{g_p}=\dfrac{\rho_e r_e}{\dfrac{1}{4}\rho_e\times \dfrac{1}{3}r_e}\\\Rightarrow \dfrac{g_e}{g_p}=12\\\Rightarrow g_p=\dfrac{g_e}{12}\\\Rightarrow g_p=\dfrac{9.8}{12}$

Since the person is jumping up the acceleration due to gravity will be negative.

From kinematic equations we have

$v^2-u^2=2g_es\\\Rightarrow u^2=v^2-2g_es\\\Rightarrow u^2=0-2\times -9.8\times 1.5\\\Rightarrow u^2=2\times 9.8\times 1.5$

On the other planet

$v^2-u^2=2g_ps\\\Rightarrow s=\dfrac{v^2-u^2}{2g_p}\\\Rightarrow s=\dfrac{0-(2\times 9.8\times 1.5)}{2\times -\dfrac{9.8}{12}}\\\Rightarrow s=18\ \text{m}$