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Anarel [89]
3 years ago
6

What drives the movement of water through earths systems.

Physics
1 answer:
MrMuchimi3 years ago
8 0

Answer:

A. The Moon

Explanation:

I believe its the gravitational pool that effects the way out ocean tides moves

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Which of these black holes exerts the weakest tidal force on an object near its event horizon? Which of these black holes exerts
ratelena [41]

Answer:

THA ANSWER IS B

Explanation:

YEH ITS B

6 0
3 years ago
Kids are pelting a window with snowballs. On average, two snowballs of roughly 300-g mass hit the window each second, moving hor
hram777 [196]

Answer:

The average force exerted on the window due to two snowballs is 6 N

Explanation:

Given:

Mass of snowballs m = 300 \times 10^{-3} Kg

Velocity of snowball v = 10 \frac{m}{s}

For finding the average force,

Force is equal to the change in momentum,

   F = \frac{dP}{dt}

Here, final velocity is zero so we write,

 F = \frac{mv}{1}

Where dt = 1 sec

 F = 300 \times 10^{-30} \times 10

F = 3 N

Above value of force is due to one ball, but here given in question there are two ball,

F = 3 \times 2

F = 6 N

Therefore, the average force exerted on the window due to two snowballs is 6 N

5 0
3 years ago
The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (400-700 nm). The Arecibo radio telescope in Puert
stealth61 [152]

Answer:

y_{hubble} = 77\ \ m

y_{aceribo} = 1.1*10^6 \ \ m

Explanation:

what is the smallest crater that each of these telescopes could resolve on our moon?

For moon ;

s = 3.8 × 10 ⁸ m

y = 1.22 λs/D

where;

λ = 400 nm = 400× 10 ⁻⁹

D = 2.4 m

The smallest crater for the hubble space is calculated as follows:

y_{hubble} = 1.22*400*10^{-9}*3.8*10^8/2.4

y_{hubble} = 77\ \ m

For Aceribo ;

y = 1.22 λs/D

where :

λ = 75 cm = 0.75 m

D = 305 m

y_{acerbo} = 1.22*0.75 *3.8*10^8/305

y_{aceribo} = 1.1*10^6 \ \ m

5 0
2 years ago
A certain parallel-plate capacitor is filled with a dielectric for which Κ = 5.5 .The area of each plate is 0.034 m2 , and the p
Nesterboy [21]

Answer:

The maximum energy that can be stored in the capacitor is  6.62 x 10⁻⁵ J

Explanation:

Given that,

dielectric constant k = 5.5

the area of each plate, A = 0.034 m²

separating distance, d =  2.0 mm = 2 x 10⁻³ m

magnitude of the electric field =  200 kN/C

Capacitance of the capacitor is calculated as follows;

C = \frac{k \epsilon A}{d} = \frac{5.5*8.85*10^{-12}*0.034}{2*10^{-3}} = 8.275 *10^{-10} \ F

Maximum potential difference:

V = E x d

V = 200000 x 2 x 10⁻³ = 400 V

Maximum energy that can be stored in the capacitor:

E = ¹/₂CV²

E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²

E = 6.62 x 10⁻⁵ J

Therefore, the maximum energy that can be stored in the capacitor is  6.62 x 10⁻⁵ J

4 0
3 years ago
16. Kinematic equations can only be used if...​
Effectus [21]

the acceleration or the next force acting on the body is constant

7 0
2 years ago
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