Answer:
what I show me a image plz mark me as brainleast
Water evaporates at 100⁰C
So change in temperature = 100-20 = 80⁰C
Amount of water to be evaporated = 1 liter = 1L*1kg/liter = 1 kg
Specific heat of water is 1 calorie/gram ⁰C = 4.186 joule/gram =4186 J/kg
So heat required E = mcΔT = 1 * 4186 *80= 334880 J =334.88 kJ
So amount of heat require to evaporate water = 334.88 kJ
You can eliminate the answer A because the moon is super cold
For answer B, atmosphere is contained of gasses, not just oxygen alone
Definitely not C
The answer is D because the moon's gravity isn't strong enough to hold the gasses, as a result, only a small amount of gasses has an attraction to it ( the moon has a little atmosphere though) but not enough to be considered
Answer: 4 s
Explanation:
Given
The ball leaves the hand of student with a speed of 
When the hand is 
above the ground
Using the equation of motion we can write
Substitute the values
![\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]](https://tex.z-dn.net/?f=%5CRightarrow%202.5%3D-19t%2B0.5%5Ctimes%209.8t%5E2%5C%5C%5CRightarrow%204.9t%5E2-19t-2.5%3D0%5C%5C%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B19%5Cpm%20%5Csqrt%7B%28-19%29%5E2-4%5Ctimes%204.9%5Ctimes%20%28-2.5%29%7D%7D%7B2%5Ctimes%2019%7D%5C%5C%5CRightarrow%20t%3D4.0049%5Cquad%20%5B%5Ctext%7BNeglecting%20the%20negative%20value%20of%20%7Dt%5D)
Thus, the ball will take 4 s to hit the ground.
Answer:
B meet A 0.01 km east of flagpole
Explanation:
given data
distance A = 5.7 km west
velocity V1 = 8.9 km/h
distance B = 4.5 km east
velocity V2 = 7 km/h
to find out
How far runners from the flagpole, when paths cross
solution
we know A and B are 5.7 + 4.5 = 10.2 km apart
and we consider here B will run distance x km for meet
so time will be for B is
time B = distance / velocity
time B = x / 7 ...................1
and
for A distance for meet = ( 10.2 - x ) km
so time A = distance / velocity
time A = ( 10.2 - x ) / 8.9 .............2
now equating equation 1 and 2
time A = time B
x / 7 = ( 10.2 - x ) / 8.9
x = 4.490
so distance of B run for meet is 4.490 km
so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km
so B meet A 0.01 km east of flagpole
