1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w
Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, a=3\ m/s^2

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.  

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

m=\dfrac{W}{g}

m=\dfrac{150}{9.8}

m=15.3\ kg

Frictional force is given by :

F=ma

F=15.3\times 3

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0
3 years ago
Hi i have homework can you help me
soldier1979 [14.2K]

Explanation:

The particle will be at rest when its velocity v(t) is equal to zero. Recall that the velocity is simply the derivative of the position x(t) with respect to time:

v(t) = \dfrac{dx}{dt}

Since x(t) = t^2 - 2t + 2

then

v(t) = \dfrac{d}{dt}(t^2 - 2t + 2) = 2t - 2 = 0

Solving for t, we find that the particle will be at rest at

t = 1\:\text{s}

6 0
2 years ago
A body of mass 2.9 kg is suspended from a string of length 2.5 metres and is at rest. A bullet of mass 100 gram moving horizonta
givi [52]

Heres your answer to ur question

6 0
2 years ago
What evidence did Wegener make use of to develop the theory of continental drift? A The climate of Siberia has always been the s
Neko [114]
The answer is D. Mountains in North America appear to be the same as those in Northern Europe
3 0
3 years ago
Read 2 more answers
A ball of mass 2 kg moving at 6 m/s collides directly with another ball of mass 3 kg travelling in the same direction at 4 m/s.
nata0808 [166]

Answer:

v₁ = 3.9 m/s

v₂ = 5.4 m/s

The loss in the kinetic energy = 1.05 J

Explanation:

Given:

mass, m₁ = 2 kg

m₂ = 3 kg

initial speed of mass m₁, u₁ = 6 m/s

Initial speed of the mass m₂, u₂ = 4 m/s

coefficient of restitution, e = (3/4)

let, the final speed of mass m₁ and m₂ be v₁ and v₂ respectively

Now,

e=\frac{v_2-v_1}{u_1-u_2}\\

on substituting the values, we get

\frac{3}{4}=\frac{v_2-v_1}{6-4}\\

or

1.5 = v₂ - v₁  

or

v₂ = 1.5 + v₁

also,

from the conservation of momentum, we have

( 2 × 6 ) + ( 3 × 4 ) = ( 2 × v₁ ) + ( 3 × v₂ )

or

24 = 2 × v₁ + ( 3 × ( 1.5 + v₁ ) )

or

24 = 2 × v₁ + 4.5 + 3 × v₁

19.5 = 5 × v₁

or

v₁ = 3.9 m/s

and

v₂ = 1.5 + v₁

or

v₂ = 1.5 + 3.9

or

v₂ = 5.4 m/s

now,

the loss in the kinetic energy = initial kinetic energy - Final kinetic energy

or

= (\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2)-(\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2)

or

= (\frac{1}{2}\times2\times6^2+\frac{1}{2}\times3\times4^2)-(\frac{1}{2}\times2\times3.9^2+\frac{1}{2}\times3\times5.4^2)

or

the loss in the kinetic energy = 1.05 J

6 0
3 years ago
Other questions:
  • At 9:13AM a car is traveling at 35 miles per hour. Two minutes later, the car istraveling at 85 miles per hour. Prove that are s
    15·1 answer
  • 1)
    11·1 answer
  • Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement
    9·1 answer
  • A tradesman sharpens a knife by pushing it with a constant force against the rim of a grindstone. The 30-cm-diameter stone is sp
    13·1 answer
  • Which of the following is true statement about isotopes?
    9·2 answers
  • An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude
    14·1 answer
  • PLEEEEASE HELP ME PLEASE WILL GIVE BRAINLIEST
    8·1 answer
  • During sublimation the particles in a solid?
    7·1 answer
  • I don't understand help​
    11·1 answer
  • A 375-pound concrete cylinder has a base area of 144 square inches. with the cylinder resting on its base, the pressure exerted
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!