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3 years ago

What are the top two gasses in the earth’s atmosphere ?

Physics

1 answer:

ch4aika [34]3 years ago

7 0

Answer:

Nitrogen and oxygen are by far the most common; dry air is composed of about 78% nitrogen (N2) and about 21% oxygen (O2). Argon, carbon dioxide (CO2), and many other gases are also present in much lower amounts; each makes up less than 1% of the atmosphere's mixture of gases.

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Which of these ideas was part of the earliest model of the atom?

babymother [125]

I think the answer is D

4 0

3 years ago

Read 2 more answers

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of

Serggg [28]

Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

τ = 26.58 Nm

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

τ = 18.79 Nm

3 0

2 years ago

A 0.2 kg block sliding on a horizontal table slows down from 25 m/s to 20 m/s. How much energy does the block lose due to fricti

Papessa [141]

Answer:

the kinetic energy lost due to friction is 22.5 J

Explanation:

Given;

mass of the block, m = 0.2 kg

initial velocity of the block, u = 25 m/s

final velocity of the block, v = 20 m/s

The kinetic energy lost due to friction is calculated as;

$\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 - \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J$

Therefore, the kinetic energy lost due to friction is 22.5 J

7 0

3 years ago

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The barometer of a mountain hiker reads 930 mbars at the beginning of a hiking trip and 780 mbars at the end. Neglecting the eff

lara31 [8.8K]

We could use the change of pressure to calculate for the height climbed by the mountain hiker. The change of pressure is given by

p = rho * g * h, where p is the change of pressure, rho is the air density, g is the acceleration due to gravity, and h is the height.

Using the conversion 1 mbar = 100 Pa,

(930 - 780)(100) = (1.20)(9.80)h
15000 = 1.20*9.80*h

h = 1.28 km

6 0

3 years ago

PLEASE HELP ME I HAVE BEEN DOING THIS FOR HOURS!! How do some carnivorous plants avoid beginning the digestion process from an a

konstantin123 [22]

Answer:

Once a carnivorous plant has procured an item for dinner, it has to have some way to turn it into fertilizer. What carnivorous plants do is very similar to what humans do with their dinner after they have eaten it. Most carnivorous plants have glands that secrete acids and enzymes to dissolve proteins and other compounds. The plants may also enlist other organisms to help with digestion. The plants then absorb the nutrients made available from the prey.

Drosera releases digestive juices through the glands at the tip of its tentacles and absorbs the nutrients through the tentacles, leaf surface, and sessile glands. In order to do this it bends its tentacles and rolls or bends the leaf to get as many tentacles as possible into contact with the prey for digestion and to make as much leaf surface available for absorption. Its relative Drosophyllum has differently structured, non moving tentacles and doesn't use them directly for digestion. Instead it has specialized glands on the surface of the leaf that release the digestive enzymes (see Carniv. Pl. Newslett. 11(3):66-73 ( PDF ) for drawings and discussion).

The sealed trap of Dionaea does digestion in a way similar to the leaf surface digestion carnivores—upon capture of a prey, digestive enzymes in mucous are released. The advantage of the sealed trap of Dionaea is rain won't wash away the nutrients as digestion proceeds.

The sealed trap carnivores Aldrovanda and Utricularia already have water in their traps so they only need to release enzymes. Utricularia appears to release the enzymes continuously into its traps.

The other carnivorous plants use either a mixed mode of digestive enzymes and partner organisms (Genlisea, Sarracenia, most Nepenthes, Cephalotus, some Heliamphora, Roridula) or other organisms exclusively for digestion (most Heliamphora, some Nepenthes, Darlingtonia). Part of the reason for partnering with other organisms is that the plants actually have little choice in the matter. This could also be a factor for the leaf surface and sealed trap digesters as well. The prey will have gut flora that are quite capable of digesting their host when it dies. In addition, insect larvae, frog tadpoles, and predacious protozoans will or will attempt to take up residence in water-filled traps. The plant releasing digestive enzymes and acids into the traps will help tip the nutrition balance to themselves, but there are limits.

Explanation:

6 0

2 years ago