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2 years ago

Hurricane Katrina and hurricane Rita are similar in what way? (I will give brainliest to the correct answer) :)

Physics

2 answers:

scoundrel [369]2 years ago

5 0

Answer:

both same way

Explanation:

RoseWind [281]2 years ago

4 0

Answer:

Rita and Katrina both followed similar paths into the Gulf.

Explanation:

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Which organisms break down dead matter and waste into nonliving elements?

Alekssandra [29.7K]

Decomposers is the correct answer. ( I got your back bro)

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3 years ago

Read 2 more answers

A satellite has a mass of 5832 kg and is in a circular orbit 4.13 × 105 m above the surface of a planet. The period of the orbit

elena55 [62]

Answer:

W = 28226.88 N

Explanation:

Given,

Mass of the satellite, m = 5832 Kg

Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m

The time period of the orbit, T = 1.9 h

= 6840 s

The radius of the planet, R = 4.38 x 10⁶ m

The time period of the satellite is given by the formula

$T = 2\pi \sqrt{\frac{(R+h)^{3} }{R^{2} g} }$ second

Squaring the terms and solving it for 'g'

g = 4 π² $\frac{(R+h)^{3} }{R^{2}T^{2} }$ m/s²

Substituting the values in the above equation

g = 4 π² $\frac{(4.38X10^6+4.13X10^5)^{3} }{(4.38X10^6)^{2}X6840^{2}}$

g = 4.84 m/s²

Therefore, the weight

w = m x g newton

= 5832 Kg x 4.84 m/s²

= 28226.88 N

Hence, the weight of the satellite at the surface, W = 28226.88 N

8 0

3 years ago

At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

lbvjy [14]

Answer:

$F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

$|F_T|=2\sqrt{34}k\frac{Q^2}{L}$

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

$F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]$

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

$F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

and the magnitude is:

$|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}$

the direction is:

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

4 0

2 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

7 0

3 years ago

What is the name of the force that acts between any two objects because of<br>their masses?

svetoff [14.1K]

Answer:

Explanation:

The force of attraction between 2 masses.

4 0

3 years ago