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mihalych1998 [28]
3 years ago
14

An airplane pilot wishes to fly due west. a wind of 80.0 km> h (about 50 mi > h ) is blowing toward the south. (a) if the

airspeed of the plane (its speed in still air) is 320.0 km> h (about 200 mi> h ), in which direction should the pilot head?
Physics
1 answer:
diamong [38]3 years ago
6 0

let the plane is flying at some angle theta with west towards north

Now we can use the components

v_y = 320 sin\theta

v_x = 320 cos\theta

now since plane has to fly towards west so net speed in north must be equal to speed in south for air

So we can say

320 sin\theta = 80

sin\theta = \frac{1}{4}

\theta = 14.5 degree

so plane has to fly in direction 14.5 degree North of west

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A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h
Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

            \alpha_{1} = \frac{1}{4} \alpha_{2}

Thus, we can conclude that the new rotation is \frac{1}{4} times that of the first rotation rate.

8 0
3 years ago
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon’
Yuki888 [10]

Answer:

0.2 m

Explanation:

PHASE 1

First, we calculate the distance the tongue moved in the first 20 ms (0.02 secs). We use one of Newton's equations of linear motion:

s = ut + \frac{1}{2}at^2

where u = initial velocity = 0 m/s

a = acceleration = 250 m/s^2

t = time = 0.02 s

Therefore:

s = 0 + \frac{1}{2} * 250 * (0.02)^2\\\\\\s = 0.05 m

PHASE 2

Then, for the next 30 ms (0.03 secs), we use the formula:

distance = speed * time

This speed is the same as the final velocity of the tongue after the first 20 ms.

This can be obtained by using the formula:

v = u + at\\\\=> v = 0 + (250 * 0.02)\\\\\\v = 5 m/s

Therefore:

distance = 5 * 0.03 = 0.15 m

Therefore, the total distance moved by the tongue in the 50 ms interval is:

0.05 + 0.15 = 0.2 m

8 0
3 years ago
Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other.
Zielflug [23.3K]

Incomplete question.The complete question is attached below as screenshot along with figure

Answer:

F=6.00*10^{-6}N

Force is repulsive

Explanation:

Given data

Current I₁=5.00A

Current I₂=2.00A

Length L=1.20 m

Radius r=0.400m

To find

Force F

Solution

As the force is repulsive because currents are in opposite direction

From repulsive force we know that:

F=\frac{u_{o}I_{1}I_{2}L}{2\pi r}

Substitute the given values

F=\frac{u_{o}(5.00A)(2.00A)(1.20m)}{2\pi (0.400m)}\\ F=6.00*10^{-6}N

4 0
3 years ago
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kodGreya [7K]

Answer:

10 your absolutely beautiful, don't let anyone else tell you otherwise.

Explanation:

7 0
3 years ago
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a 905 - g meteor impacts the earth at a speed of 1623 m/s. if all of its energy is entirely converted to heat in the meteor, wha
nlexa [21]

Answer: 2859.78 k

Explanation: By using the law of conservation of energy, the kinetic energy of the meteor equals the heat energy.

Kinetic energy = 1/2mv^2

Heat energy = mcΔθ

Where m = mass of meteor , v = velocity of meteor = 1623 m/s

c = specific heat capacity of meteor (iron) = 460.548 j/kg/k

Δθ = change in temperature of meteor = ?

From law ofconservation of energy, we have that

1/2mv^2 = mcΔθ

By cancelling "m" on both sides, we have that

v^2/2 = cΔθ

v^2 = 2cΔθ

(1623)^2 = 2× 460.548 × Δθ

2634129 = 921.096 × Δθ

Δθ = 2634129 / 921.096

Δθ = 2859.78 k

6 0
3 years ago
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