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2 years ago

Salmon often jump waterfalls to reach their

Physics

1 answer:

PilotLPTM [1.2K]2 years ago

8 0

The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

height of the waterfall, h = 0.432 m
distance of the Salmon from the waterfall, s = 3.17 m
angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

$h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s$

The minimum velocity of the Salmon jumping at the given angle is calculated as;

$X = v_0_x t\\\\3.17 = (v_0\times cos(30.8)) \times 0.3\\\\10.567 = v_0\times cos(30.8)\\\\v_0 = \frac{10.567}{cos(30.8)} \\\\v_0 = 12.3 \ m/s$

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

Learn more here: brainly.com/question/20064545

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According to Oxford Dictionaries, a spit take is an act of suddenly spitting out liquid one is drinking in response to something

Tasya [4]

Answer:

The pressure is $p_1 = 4051.4 \ Pa$

Explanation:

From the question we are told that

The gauge pressure at the mouth is $p_1$

The radius of the column is $r_2 = 4 \ mm = 0.004 \ m$

The speed of the liquid outside the body is $v_2 = 3.1 \ m/s$

The area of the column is $A_2$

The area inside the mouth $A_1 = 10 A_2$

Generally according to continuity equation

$v_1 A_1 = v_2 A_2$

=> $v_ 1 = v_2 * \frac{A_2}{A_1}$

=> $v_ 1 = 3.1 * \frac{1}{10}$

=> $v_ 1 = 0.31 \ m/s$

$A_1 = 10A_2$

=> $\pi * r_1^2 = 10(\pi * r_2^2)$

=> $r_1 = 10 * r_2$

substituting values

$r_1 = 10 * 0.004$

$r_1 =0.04 \ m$

Now the height of inside the mouth is $h_1 = d = 2r_1 = 2* 0.04 = 0.08\ m$

Now the height of the column is $h_2 = d = 2r_2 = 2* 0.004 = 0.008\ m$

Generally according to Bernoulli's equation

$p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ]$

Now $\rho = 1000 \ kg m^{-3}$ which is the density of water

$p_2$ is the gauge pressure of the atmosphere which is zero

$p_1 = [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-$

$[(0.5 * 1000 * 0.31^2) + (0.08*1000 * 9.8)]$

$p_1 = 4051.4 \ Pa$

8 0

3 years ago

In which of these states is the age of consent 18 years old?

Andreas93 [3]

Answer:

all the ones that say

Explanation:

18.....

4 0

3 years ago

A goose with a mass of 2.0 kg strikes a commercial airliner with a mass of 160,000 kg head-on. Before the collision, the goose w

Goryan [66]

Answer:

The change in momentum of the goose during this interaction is 33.334 m/s

Explanation:

Given;

mass of goose, m₁ = 2.0 kg

mass of commercial airliner, m₂ = 160,000 kg

initial velocity of the bird, u₁ = 60 km/hr = 16.667 m/s

initial velocity of the airliner, u₂ = 870 km/hr = 241.667 m/s

Change in momentum is given as;

ΔP = mv - mu

where;

u is the initial velocity of the bird

v is the final velocity of the bird

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the final velocity of bird and airliner after collision;

(2 x 16.667) + (160,000 x 241.667) = v (2 + 160,000)

38,666,753.334 = 160,002v

v = 38,666,753.334 / 160,002

v = 241.664 m/s

Thus, the final velocity of the bird is negligible compared to final velocity of the airliner.

ΔP = mv - mu

ΔP = m(v - u)

ΔP = 2(0 - 16.667)

ΔP = -33.334 m/s

The negative sign implies a deceleration of the bird after the impact.

Therefore, the change in momentum of the goose during this interaction is 33.334 m/s

4 0

3 years ago

What is work please give ans

Fed [463]

Work= force x displacement :)

8 0

2 years ago

A phonograph record 0.15 m in its radius rotates 18 times per 90 seconds what is the frequency?

ioda

Answer:

The frequency of the phonograph record is 0.2 Hz

Explanation:

The frequency of an object moving in uniform circular motion is the number of completed cycles the object makes in a specified time period

The given parameters of the phonograph record are;

The radius of the record = 0.15 m

The number of times the phonograph record rotates, n = 18 times

The time it takes the phonograph record to rotate the 18 times, t = 90 seconds

The frequency of the phonograph record, f = (The number of times the phonograph record rotates) ÷ (The time it takes the phonograph record to rotate the 18 times)

∴ The frequency of the phonograph record, f = n/t = 18/(90 s) = 0.2 Hz

The frequency of the phonograph record = 0.2 Hz.

6 0

2 years ago