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N76 [4]
3 years ago
11

An unladen swallow that weighs 0.03 kg flies straight northeast a distance of 125 km in 4.0 hours. With the x x direction due ea

st and the y y direction due north, what is the average momentum of the bird (in unit vector notation)?
Physics
1 answer:
balandron [24]3 years ago
8 0

Answer:

The average momentum of the bird is 0.26 kgm/s

Explanation:

The formula to be used here is that of momentum which is

momentum (in kgm/s) = mass (in kg) × velocity (in m/s)

The velocity of the bird is

velocity (in m/s) = distance (in meter) ÷ time (in seconds)

distance in meters = 125km × 1000 = 125,000 m

time in seconds = 4 hrs × 60 × 60 = 14,400 secs

velocity = 125000/14400

velocity = 8.68 m/s

momentum (p) = 0.03 × 8.68

p = 0.26 kgm/s

The average momentum of the bird is 0.26 kgm/s

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A body weighs 10newton in air and 9.5 in water. How will it weigh in alchohol if density 0.8gcm-³​
Katen [24]

Answer:

The weight of object in alcohol is, W₀ = 9.6 N

Explanation:

Given,

The weight of body in air, Wₐ = 10 N

The weight of the body in water, Wₓ = 9.5 N

The density of the alcohol, ρ = 0.8 g/cm³

                                                = 800 kg/m³

The weight of body in water = weight of object in air - weight of water displaced

                        9.5 N = 10 N - weight of water displaced

                       Weight of water displaced = 0.5 N

The mass of the displaced water, m = 0.5 / 9.8

                                                            = 0.051 kg

Therefore the volume of the water displaced,

                                              V = m / ρ

                                                  = 0.051 / 1000

                                                   = 5.1 x 10⁻⁵ m³

<em>The volume of water displaced is equal to the volume of alcohol displaced</em>

Therefore, the mass of the alcohol displaced, = V x ρ

                                                                             = 5.1 x 10⁻⁵ x 800

                                                                             = 0.0405 kg

The weight of the alcohol displaced, w = 0.0405 x 9.8

                                                                    = 0.4 N

Therefore,

The weight of body in alcohol = weight of object in air - weight of alcohol displaced

                             W₀ = W - w

                                    = 10 N - 0.4 N

                                     = 9.6 N

Hence, the weight of object in alcohol is, W₀ = 9.6 N

4 0
3 years ago
A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re
lbvjy [14]

Answer:

The beam of light is moving at the peed of:

\frac{dy}{dt} = \frac{80\pi}{3} km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, \omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi    (1)

Now, in the right angle in the given fig.:

tan\theta' = \frac{y}{3}

Now, differentiating both the sides w.r.t t:

\frac{dtan\theta'}{dt} = \frac{dy}{3dt}

Applying chain rule:

\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}

sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}

Now, using tan\theta = \frac{1}{m} and y = 1 in the above eqn, we get:

(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}

Also, using eqn (1),

8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}

\frac{dy}{dt} = \frac{80\pi}{3}

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That makes no sense to me somehow
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nordsb [41]

Answer:

b.it depends on the distance it falls

8 0
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irinina [24]
Charge= Protons- electrons
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This Ion will have a charge of -2
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