<span>An observation of the red shift of galaxies suggests that the universe is expanding. The correct option among all the options that are given in the question is the first option or option "a". I hope that this is the answer that you were looking for and it has actually come to your help.</span>
For the first question it is the fourth option. Cryosphere is a term for the portions of earth that are covered in water when the water is in solid form. this includes both snow and ice.
For the second question the answer is a delta is formed at the mouth of the river a sediment is carried down stream. The hydrosphere refers to all water on earth.
Assume no air resistance, and g = 9.8 m/s².
Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity
The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s
With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0
Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°
The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s
If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)
Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s
The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m
Answer: h = 110.4 m
Answer:
The time taken by the satellite to orbit earth at its surface, t = 1.66 hr
Explanation:
Given data,
The velocity of the satellite, v = 15000 miles/hr
The distance of travel, d = 24901 miles,
It is equal to the circumference of earth,
So the time taken by the satellite to orbit earth at its surface,
v = d/t
t = d/v
Substituting the given values in the above equation,
t = 24901 / 15000
= 1.66 hr
Hence, the time taken by the satellite to orbit earth at its surface, t = 1.66 hr
