Answer:
600 KPa.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V1) = 0.075 m³
Final volume (V2) = 0.45 m³
Final pressure (P2) = 100 KPa
Initial pressure (P1) =?
Temperature = constant
The initial pressure can be obtained by using the Boyle's law equation as shown below:
P1V1 = P2V2
P1 × 0.075 = 100 × 0.45
P1 × 0.075 = 45
Divide both side by 0.075
P1 = 45 / 0.075
P1 = 600 KPa.
Thus, the initial pressure in the balloon is 600 KPa.
Answer:
The electromagnetic wave that travels the fastest through space is gamma ray
Explanation:
Electromagnetic wave is a type of wave which does not require material medium for its propagation. Examples of electromagnetic waves according to increasing frequency of the waves are gamma ray, x-ray, ultra violet ray, infra red, visible light, micro wave and radio waves.
Frequency of a wave is inversely proportional to period of oscillation of the wave. The higher the frequency of a wave, the shorter the period of oscillation. Gamma ray has the highest wave frequency in electromagnetic spectrum and shorter period of oscillation, thereby causing it to have the highest penetration power.
Metal ores
Explanation:
in an area where subduction has occurred in times past, metal ores are likely to be found.
Metallic ores find subduction zone regions very favorable to crystallize out of a magma.
- Ores have different modes of formation.
- Typically, they are found in hydrothermal vents and black smokers of igneous intrusives.
- These are igneous terrains where metallic sulfides and other minerals crystallize out of magmatic body.
- Metals in magma usually have large sizes and do not partition easily in the melt.
At a subduction zone, partial melting of the subducting plate forces magma into nearby country rock as an intrusive and to the ocean floor where they form black smokers.
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Answer:
we can not use the suns energy too effectively in power cells and with human power we can generate more energy
Explanation:
According to the law of conservation of energy
,
Potential energy = kinetic energy
I = 
![mgh = [\frac{1}{2} \times mv^{2}] + [\frac{1}{2} \times (\frac{mr^{2}}{2}) \frac{v^{2}}{r^{2}}]](https://tex.z-dn.net/?f=mgh%20%3D%20%5B%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20mv%5E%7B2%7D%5D%20%2B%20%5B%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%28%5Cfrac%7Bmr%5E%7B2%7D%7D%7B2%7D%29%20%5Cfrac%7Bv%5E%7B2%7D%7D%7Br%5E%7B2%7D%7D%5D)
mgh = ![[\frac{1}{2} \times mv^{2}] + [\frac{1}{4} \times mv^{2}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20mv%5E%7B2%7D%5D%20%2B%20%5B%5Cfrac%7B1%7D%7B4%7D%20%5Ctimes%20mv%5E%7B2%7D%5D)
v = 7.4 m/s
thus, we can conclude that the translational speed of the cylinder when it leaves the incline is 7.4 m/s.
