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coldgirl [10]
2 years ago
8

Three-fourths of the elements on the

Physics
1 answer:
77julia77 [94]2 years ago
8 0

Answer:

b

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}

The man can jump a height of 18 m on the other planet.

5 0
2 years ago
Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

  $= 822486 \times 10^{-8}$

  $=0.822 \times 10^{-2} \ km/s$

 = 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

  $= 673552 \times 10^{-8}$

  $=0.673 \times 10^{-2} \ km/s$

 = 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

  $= 3371 \times 153.86 \times 10^{-8}$

  = 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

 = 3.64 m/s

       

3 0
3 years ago
Your mass is 65 kg. you stand on a bathroom scale in an elevator on earth. (a) what force would the scale exert when the elevato
erik [133]
I think you would use F = ma
F = 65*10
F = 650N
(The 10m/s is from acceleration due to gravity)
4 0
3 years ago
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