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Nadusha1986 [10]
3 years ago
15

An eagle is flying horizontally at a speed of 2.60 m/s when the fish in her talons wiggles loose and falls into the lake 4.70 m

below. Calculate the velocity (in m/s) of the fish just before it hits the water. (Assume that the eagle is flying in the x direction and that the y direction is up.)
Physics
2 answers:
patriot [66]3 years ago
4 0

Answer:

The speed of fish at the time of hitting the surface is 9.95 m/s.

Explanation:

Horizontal speed, u = 2.6 m/s

height, h = 4.7 m

Let the vertical velocity at the time of hitting to water is v.

Use third equation of motion

v^2 = u^2 - 2 gh \\\\v^2 = 0 + 2 \times 9.8\times 4.7\\\\v = 9.6 m/s

The net velocity with which the fish strikes to the water is

v' = \sqrt{9.6^2 + 2.6^2 }\\\\v' = 9.95 m/s

dimulka [17.4K]3 years ago
3 0

Answer:

Explanation:

The fish will have horizontal velocity of 2.6 m/s which is also the velocity of eagle. Additionally , he will have vertical velocity due to fall under gravity .

v² = u² + 2 g H .

v² = 0  + 2 x 9.8 x 4.7 m

= 92.12

v = 9.6 m /s

The fish's final velocity will have two components

vertical component = 9.6 m/s downwards

Horizontal component = 2.6 m /s  .

Resultant velocity = √ ( 9.6² + 2.6² )

= √ ( 92.16 + 6.76 )

= 9.9 m /s

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Newton’s Thrid Law, which states that for every reaction there is an opposite reaction.
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A 1000.0 kg truck accelerates from 20.0 m/s to 25.0 m/s over a distance of 300.0 m. What is the average net force on the truck?
choli [55]

Answer:

The average net force on the truck is 375 Newtons.

Explanation:

Using Newton's 3rd equation of motion, we have :

v^{2} - u^{2} = 2×a×s

where, v = final velocity = 25 m/s

u = initial velocity = 20 m/s

a = acceleration

s = distance traveled = 300 m

Using these values in the above equation, we get acceleration = 0.375 m/s^{2}

Using Newton's second law, we have:

F=m×a

where m = mass = 1000 kg

a= acceleration = 0.375 m/s^{2}

Putting values we have F=375 N

3 0
3 years ago
Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 30
dybincka [34]

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

<em><u>Support Cy:</u></em>

∑ M at Ay = 0

      (( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

                                      ( L ab + L bc )

Cy = 1.3 x 10³ k-N

<em><u>Support Ay:</u></em>

Since ∑ F = 0,           A + C - F - D = 0

                                   A = F  + D - C

                                  Ay = 200 k-N

4 0
3 years ago
Which of the following statements would NOT result in an eclipse? a. The Earth is positioned directly between the Sun and the Mo
vovangra [49]

c. The Moon is positioned directly between the earth and the sun is the statement that does not result in an eclipse.

Explanation:

  • The Sun is completely blocked in a solar eclipse because the Moon passes between Earth and the Sun.
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3 0
3 years ago
(8c7p26) During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with lar
likoan [24]

Answer: 1175 J

Explanation:

Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."

Given

Spring constant, k = 102 N/m

Extension of the hose, x = 4.8 m

from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m

Work done =

W = 1/2 k [x(i)² - x(f)²]

Since x(f) = 0, then

W = 1/2 k x(i)²

W = 1/2 * 102 * 4.8²

W = 1/2 * 102 * 23.04

W = 1/2 * 2350.08

W = 1175.04

W = 1175 J

Therefore, the hose does a work of exactly 1175 J on the balloon

7 0
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