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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.)

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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2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo

seropon [69]

Answer:

The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times60$

$\omega=376.9\ rad/s$

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

$B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}$

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

$B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}$

$B_{max}=2.901\times10^{-13}\ T$

Hence, The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

7 0

3 years ago

Which of the following statements explains one way that oceans influence the temperature of the tropics and poles?

pogonyaev

Answer:

c

Explanation:

5 0

2 years ago

The graph shows the progress of a car which traveled 300 km in 5 hrs. Assuming that its speed remained constant, how far had the

lisov135 [29]

If the car travels
300 km.........................5 hrs
?km ..............................2h
to find out we do (300*2)/5=600/5=120km
also
300km/5h=60km/h
in 2 hours
120km

6 0

3 years ago

1. Bone has a Young’s modulus of about

Blababa [14]

#1

As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

$strain = 9.33 \times 10^{-3}$

now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

$\Delta L = 5.04 \times 10^{-3} m$

$\Delta L = 5.04 mm$

#2

As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

#3

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

$B = 2.3 \times 10^9 N/m^2$

now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

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#4

as we know that pressure changes with depth as per following equation

$P = P_o + \rho g h$

here we know that

$P = 3 P_0$

now we will have

$3P_0 = P_0 + \rho g h$

$2P_0 = \rho g h$

$2(1.01 \times 10^5) = 1025 (9.81)(h)$

here we will have

$h = 20.1 m$

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

$mg = \rho_1V_1g + \rho_2V_2g$

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$46.6 = 43.89 - 922x + 1000x$

$x = 3.48 cm$

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