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2 years ago

What is the main way that scientists determine which conditions can support life

1 answer:

3 0

Bacteria they use bacteria there’s a microscope thing in the area and I see a lot of things and if the bacteria is some type away then they would find a medicine to treat direct her than hope everybody else

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A cat, walking along the window ledge of a new york apartment, knocks off a flower pot, which falls to the street 280 feet below

Harlamova29_29 [7]

H = 280 ft, the height of the flower pot.
g = 32 ft/s²

Neglect air resistance.
Note that 1 ft/s = 15/22 mi/h

The initial vertical velocity is zero.
Let v = the velocity with which the flower pot hits the ground.
Then
v² = 2gh
= 2*(32 ft/s²)*(280 ft)
= 17920 (ft/s)²
v = 133.866 ft/s

Also,
v = (133.866 ft/s)*(15/22 (mi/h)/(ft/s)) = 91.272 mi/h

Answer: 133.9 ft/s or 91.3 mi/h

5 0

2 years ago

types of plate boundaries where two plates separate or move apart?? does anyone know please help ??????

Shtirlitz [24]

Erosion i belive it is called

7 0

3 years ago

Read 2 more answers

imagine that 501 people are present in a movie theater of volume 8.00 x10^3 that is sealed shut so no air can escape. Each perso

Semenov [28]

Answer:

The temperature of air will increase by $\Delta T=41044.967\ K$

Explanation:

Given:

no. of person in a theater, $n=501$

volume of air in the theater, $V=8\times 10^3\ m^3$

rate of heat given off by each person, $P=110\ J.s^{-1}$

duration of movie, $t=2\ hr=7200\ s$

initial pressure in the theater, $p_i=1.01\times 10^5\ Pa$

initial temperature in the theater, $T_i=20+273=293\ K$

specific heat capacity of air at the given conditions, $c=1.0061\ J.kg^{-1}.K^{-1}$

The total quantity of heat released by the total people in the theater during the movie:

$Q=n.P.t$

$Q=501\times 110\times 7200$

$Q=396792000\ J$

Form the relation of heat capacity:

$Q=m.c.\Delta T$

∵ $p_i.V=m.R.T$

$Q=(\frac{p_i.V}{R.T}) \times c\times (T_f-T_i)$

$396792000=(\frac{1.01\times 10^5\times 8\times 10^3}{287\times 293}) \times 1.0061\times (T_f-293)$

$T_f=41337.967\ K$

Change in temperature of air:

$\Delta T=41044.967\ K$

8 0

2 years ago

When Emeka was first training his dog, Smooches, to heel he would give Smooches a treat when she stayed close during walks. Now

shepuryov [24]

Answer a. is correct. In this case the dog treats were positive reinforcers for staying close (the desirable behaviour). Repeated rewards for a certain behaviour work as positive reinforcers.

8 0

2 years ago

A 1.1-kg uniform bar of metal is 0.40 m long and has a diameter of 2.0 cm. When someone bangs one end of this bar, a 1.5 MHz sho

lyudmila [28]

Answer:

$\lambda = 2.22\times 10^{-3}\ m$

Explanation:

Given,

mass of the bar = 1.1 Kg

length of rod, l = 0.40 m

diameter of the rod, d = 2 cm

frequency, f = 1.5 MHz

time, t = 0.12 ms

wavelength of the shock wave = ?

Speed of the wave = $\dfrac{L}{t}=\dfrac{0.40}{0.12\times 10^{-3}}$

v = 3333.33 m/s

wavelength of the wave

$\lambda = \dfrac{v}{f}= \dfrac{3333.33}{1.5\times 10^6}$

$\lambda = 2.22\times 10^{-3}\ m$

6 0

2 years ago