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3 years ago

How many years would take to terraform for Mars

Physics

1 answer:

dybincka [34]3 years ago

8 0

Some say one year and a half

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Since the noble gases are stable (VERY unreactive), oxygen would like to look like one in terms of the number and arrangement of

iogann1982 [59]

Answer:

oxygen gains two electron to attain electron configuration like neon. Oe-²

Explanation:

3 0

3 years ago

What current flows through a 2.60-cm-diameter rod of pure silicon that is 20.5 cm long, when 1.00 ✕ 103 V is applied to it? (Suc

Virty [35]

Answer:

1.13 mA

Explanation:

Length of wire L = 20.5 cm = 0.205m

Radius of wire r = 2.60/2 = 1.3cm = 0.0130m

Voltage V = 1 × 10³ V

Resistivity of pure silicon p = 2300 Ohms • m

Cross sectional area of the wire

A = pi × r² = pi × (0.013)² = 5.307 × 10 ^-4 m²

Resistance of the material

R = p• L/A

= 2300 • 0.205/5.307 × 10^-4 = 0.888 × 10⁶ Ohms

Using Ohms Law

R = V/ I

I = V/R

I = 10³/0.888 × 10⁶

= 0.001126 A

= 1.13 mA

5 0

4 years ago

Kathy had a heavy metal ball and Rich had a light wooden ball they drop the balls from the same height. they were surprised to d

Colt1911 [192]

Answer: I scored a 9/10

Explanation:

I agree with Kathy gravity must be acting the same on both balls just because one ball is heavier doesn’t mean they will hit different times if the wooden ball was drop closer to the ground then the wooden ball would hit the ground first

3 0

3 years ago

Through what media does mechanical waves move

Aleks [24]

A mechanical wave moves through all matter

6 0

3 years ago

The average weight of a particular box of crackers is 26.0 ounces with a standard deviation of 0.5 ounce. The weights of the

Olenka [21]

Answer:

(a) 99.865%

(b) 0.135%

Explanation:

Given that the weight of the boxes are normally distributed.

The average weight of the particular box,

$\mu=26.0$ ounce

The standard deviation of weight,

$\sigma=0.5$ ounce.

Let $z$ be the standard normal variable,

$z=\frac{x-\mu}{\sigma}$

And, the probability of the boxes having weight x ounces is

$P(z)=\frac{1}{2\pi}e^{-\frac{z^2}{2}}$

For $x=24.5$ ,

$z=\frac{x-\mu}{\sigma}=\frac{24.5-26}{0.5}=-3$

(a) For the boxes having weight more than 24.5 ounces: $z>-3$

So, the probability of boxes for $z>-3$ is

$P(z>-3)=\int_{-3}^{\infty}\left(\frac{1}{2\pi}e^{-\frac{z^2}{2}}\right)dx$

=0.99865

So, the percent of the boxes weigh more than 24.5 ounces is 99.865%.

(b) For the boxes having weight less than 24.5 ounces: z<-3

So, the probability of boxes for z<-3 is

$P(z-3}=1-0.99865$

=0.00135

So, the percent of the boxes weigh more than 24.5 ounces is 0.135%.

8 0

4 years ago