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Arada [10]
3 years ago
13

Since monsoons are storms that usually occur during a specific time of year in certain regions, you could not compare them to th

understorms. Please select the best answer from the choices provided
a. True
b. False
Physics
2 answers:
Anarel [89]3 years ago
8 0

Answer:

FALSE

Explanation:

sweet [91]3 years ago
4 0
In the given statement: "<span>Since monsoons are storms that usually occur during a specific time of year in certain regions, you could not compare them to thunderstorms. </span>" is false. Therefore, among the given choices, the correct answer is B. False.
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As a moon follows its orbit around a planet, the maximum gravitational force exerted on the moon by the planet exceeds the minim
professor190 [17]

Answer:

rmax/rmin = √1.127

Explanation:

F = GmM / r²

As the masses can be assumed to be constant, the force between the two is proportional to the inverse of the square of the distance between them

(Fmax - Fmin) / Fmin = 0.127

           (Fmax - Fmin) = 0.127Fmin

       1/rmin² - 1/rmax² = 0.127(1/rmax²)

                      1/rmin² = 0.127(1/rmax²) + 1/rmax²

                      1/rmin² = 1.127(1/rmax²)

              rmax²/rmin² = 1.127

                 rmax/rmin = √1.127 ≈ 1.06160256...

8 0
3 years ago
Is gravity a force ?
alexandr402 [8]

Answer:

Gravity ( gravity , weight, traction force) is the force acting on any body on or near the surface of the celestial body directed at its center. Gravity, that is, the force with which the earth , moon , planets, and other celestial bodies or their systems act on other bodies, is theoretically exercised at any distance, but is practically examined on the surface of these bodies as well as at short distances.

6 0
2 years ago
Graph are blank a relationship
Maru [420]

I have no clue sorry maybe try applying the answers and questions then I'll answer for u

7 0
3 years ago
How is the voltage V across the resistor related to the current I and the resistance R of the resistor? (Use I for current and R
VladimirAG [237]

Answer:

This relationship is explained by Ohm's law

Explanation:

Ohm's law states that the current flowing through a circuit or a resistor is directly proportional to the voltage across the resistor and inversely proportional to the resistance. Where current is i, voltage is v and resistance is r, Ohm's law can be represented mathematically as

V= IR

8 0
3 years ago
A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

Magnitude of acceleration is -43200 ft/s²

5 0
3 years ago
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