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3 years ago

A mosquito flying at 3 m/s that encounters a breeze blowing at 3 m/s in the same direction has a speed of

Physics

1 answer:

Vika [28.1K]3 years ago

4 0

Answer:

6m/s

Explanation:

The mosquito will have a speed of 6 m/s.

if the speed of the mosquito and breeze was in opposite direction, we will subtracted them from each other and the result will be the speed of mosquito moving in the direction of the higher speed (between the breeze and mosquito). while if they are both moving in the same direction the speed of the mosquito will be the sum of both the speed of the mosquito and breeze thus the mosquito and breeze are moving in the same direction so we just sum it up (3m/s + 3m/s = 6m/s).

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Answer:

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Explanation:

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2 years ago

A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreci

Simora [160]

Answer:

Explanation:

We shall first calculate the velocity at height h = 575 m .

acceleration a = 2.2 m /s²

v² = u² + 2 a s

u is initial velocity , v is final velocity , s is height achieved

v² = 0 + 2 x 2.2 x 575

v = 50.3 m /s

After 575 m , rocket moves under free fall so g will act on it downwards

If it travels further by height H

from the relation

v² = u² - 2 g H

v = 0 , u = 50.3 m /s

H = ?

0 = 50.3² - 2 x 9.8 H

H = 129.08 m

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4 0

2 years ago

Pls how to solve this problem. help would be appreciated

Bingel [31]

Answer:

80 m/s

Explanation:

Given:

a = -5 m/s²

v = 0 m/s

Δx = 640 m

Find: v₀

v² = v₀² + 2a(x − x₀)

(0 m/s)² = v₀² + 2(-5 m/s²) (640 m)

v₀ = 80 m/s

7 0

3 years ago

A 2.50 kg ball moving at 7.50 m/s is caught by a 70.0 kg man while the man is standing on ice. What is the common velocity of th

yarga [219]

Answer:

V = 2.5*7.0 / ( 2.5 + 70 )

8 0

2 years ago

What is the lift (in newtons) due to Bernoulli's principle on a wing of area 76 m2 if the air passes over the top and bottom sur

AveGali [126]

Answer:

So lift will be 30.19632 N

Explanation:

We have given area of the wing $a=76m^2$

We know that density of air $d=1.29kg/m^3$

Speed at top surface $v_2=290m/sec$ and speed at bottom surface $v_1=150m/sec$

According to Bernoulli's principle force is given by

$F=A\times d\times \frac{v_2^2-v_1^2}{2}=76\times 1.29\times \frac{290^2-150^2}{2}=3019632N$

4 0

3 years ago