Explanation:
Q= mc∆T
∆T= 5-24=- 19
Q= 0.5*4186*-19
Q= -39767 J
negative sign show heat releases
Explanation:
London dispersion forces will form between non-polar molecules(polar ) that are symmetrical like O₂, H₂, Cl₂ and noble gases.
- The attraction here is because non-polar molecules becomes polar due to the constant motion of its electrons.
- This lead to an uneven charge distribution at an instant.
- A temporary dipole or instantaneous dipole forms.
- The temporary dipole can induce neighboring molecules to be distorted and forms dipoles as well.
- This forms london dispersion forces.
Learn more:
Intermolecular forces brainly.com/question/10602513
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Answer:
<em>faster and at a higher luminosity and temperature.</em>
Explanation:
A protostar looks like a star but its core is not yet hot enough for fusion to take place. The luminosity comes exclusively from the heating of the protostar as it contracts. Protostars are usually surrounded by dust, which blocks the light that they emit, so they are difficult to observe in the visible spectrum.
A protostar becomes a main sequence star when its core temperature exceeds 10 million K. This is the temperature needed for hydrogen fusion to operate efficiently.
Stars above about 200 solar masses (Higher mass) generate power so furiously that gravity cannot contain their internal pressure. These stars blow themselves apart and do not exist for long if at all. A protostar with less than 0.08 solar masses never reaches the 10 million K temperature needed for efficient hydrogen fusion. These result in “failed stars” called brown dwarfs which radiate mainly in the infrared and look deep red in color. They are very dim and difficult to detect, but there might be many of them, and in fact they might outnumber other stars in the universe.
That is why higher mass protostars enter the main sequence at a <em>faster and at a higher luminosity and temperature.</em>
I am so sure it's 600,000 x 500 so you get 300,000,000
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
