The three parts of the Earth are Atmosphere, Hydrosphere and Lithosphere.
Atmosphere is the blanket of air that surrounds the earth. It is densest close to the surface and thins out as one moves higher. Atmosphere of Earth contains mainly Nitrogen, followed by Oxygen and small amounts of water vapor, Carbondioxide and other gases.
Lithosphere is the outer most part of the earth's surface. The Earth's crust and the mantle form Lithosphere.
Hydrosphere is the part of the Earth that has water. The Oceans, seas, rivers, lakes and other water bodies constitute the Hydrosphere.
Stratosphere, Mesosphere and Ionosphere are different layers of atmosphere.
Hence, for the study of the Earth, one needs to consider earth to be made of three parts- atmosphere, Lithosphere and Hydrosphere.
The safety items such as eye washing station, fire blanket, and the class shower.
Answer:
a. 5 × 10¹⁹ protons b. 2.05 × 10⁷ °C
Explanation:
Here is the complete question
A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is 0.42 A. (a) How many protons strike the target in 19 seconds? (b) Each proton has a kinetic energy of 6.0 x 10-12 J. Suppose the target is a 17-gram block of metal whose specific heat capacity is 860 J/(kg Co), and all the kinetic energy of the protons goes into heating it up. What is the change in temperature of the block at the end of 19 s?
Solution
a.
i = Q/t = ne/t
n = it/e where i = current = 0.42 A, n = number of protons, e = proton charge = 1.602 × 10⁻¹⁹ C and t = time = 19 s
So n = 0.42 A × 19 s/1.602 × 10⁻¹⁹ C
= 4.98 × 10¹⁹ protons
≅ 5 × 10¹⁹ protons
b
The total kinetic energy of the protons = heat change of target
total kinetic energy of the protons = n × kinetic energy per proton
= 5 × 10¹⁹ protons × 6.0 × 10⁻¹² J per proton
= 30 × 10⁷ J
heat change of target = Q = mcΔT ⇒ ΔT = Q/mc where m = mass of block = 17 g = 0.017 kg and c = specific heat capacity = 860 J/(kg °C)
ΔT = Q/mc = 30 × 10⁷ J/0.017 kg × 860 J/(kg °C)
= 30 × 10⁷/14.62
= 2.05 × 10⁷ °C
Answer:
When the air pressure in the throat and outside the body is less than the air pressure in the middle ear, barotrauma occurs.
Explanation:
Ear barotrauma is a medical condition that describes discomfort in the ear which is caused by pressure differences in the inner and outer ear drum.
Usually, the air pressure in the middle ear is the same as the air pressure in the throat and outside the body.
When we swallow, the eustachian tube opens up and air flows out of and into the middle ear, this balances the pressure. But if the eustachian tube is blocked, the air pressure in the throat and outer body become different from the air pressure in the middle ear.
