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3 years ago

Salmon often jump waterfalls to reach their

Physics

1 answer:

____ [38]3 years ago

5 0

Answer:

Using the range formula R = v^2 sin 2 theta / g

or v^2 = R * g / sin 86.4

v^2 = 3.14 m * 9.81 m/s2 / .998

v^2 = 30.9 m^2 / s^2

v = 5.56 m/s

This hasn't really proved the question - this would give

vy = 5.56 * sin 43.2 = 3.81 m/s

vx = 5.56 * cos 43.2. = 4.05 m/s

t = 1.57 / 4.05 = .387 sec to reach the waterfall

h = 3.81 * .387 - 4.9 (.387)^2 = .74 m well above the height of the falls

There seems another way to do this

vy / vx = tan 43.2 vy = .939 vx

h = vy t - 1/2 g t^2 and t = 1.57 / vx

h = 1.57 tan 43.2 - 4.9 (1.57 / vx)^2

Solving for vx I get vx = 3.26 m/s vy = 3.06 m/s v = 4.47 m/s

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Explanation:

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3 years ago

Starting from the orgin, a person walks 4 km East during the first day and 7 km west the next day. What is the net displacement

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Answer:

3km due West

Explanation:

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So;

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4km

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3 years ago

In July 2005, NASA's "Deep Impact" mission crashed a 372-kg probe directly onto the surface of the comet Tempel 1, hitting the s

eimsori [14]

Question:

(a) What change in the comet’s velocity did this collision produce? Would this change be noticeable? (b) Suppose this comet were to hit the earth and fuse with it. By how much would it change our planet’s velocity? Would this change be noticeable? (The mass of the earth is 5.97×1024kg.)

Answer:

The answers to the question are;

(a) The change in the comet’s velocity produced by the collision is 2.86 × 10⁻⁶km/h or 7.944 × 10⁻⁷ m/s

(b) It would change our planet’s velocity by 6.70× 10⁻⁸ km/h or 1.86× 10⁻⁸ m/s

Change is too small to be noticeable

Explanation:

We not that the question is about conservation of liner momentum

Therefore we have, by listing out the known parameters

m₁ = Mass of "Deep Impact" = 372 kg

m₂ = Mass of Tempel 1 comet = (0.1 to 2.5) × 10¹⁴ kg,

v₁ = Vaelocity of "Deep Impact" = 37000 km/h

v₂ = Velocity of Tempel 1 comet = 40000 km/h

From the principle of linear momentum, we have, for both bodies moving in opposite direction;

m₁×v₁ + m₂×v₂ = m₁×v₃ + m₂×v₃ since it was a crash, it is assumed that they both have the same final velocity

This gives

372 kg ×37000 km/h - 0.1 × 10¹⁴ kg × 40000 km/h = (372 kg + 0.1 × 10¹⁴ kg )×v₃

13764000 kg·km/h - 4.0 × 10¹⁷ kg·km/h = 10000000000372×v₃

v₃ = (-399999999986236000 kg·km/h)/10000000000372 = -39999.999997 km/h ≈ - 40000 km/h in the direction of Deep Impact

Change in comet velocity

= 40000 km/h - 39999.999997 km/h

= 2.86 × 10⁻⁶km/h = 7.944 × 10⁻⁷ m/s

(b) If the colission is with earth, we have

m₃ = Mass of earth

From the principle of conservation of linear momentum, we have

m₂v₂+m₃ v₃ = (m₂ + m₃) v₄

v₃ = Initial velocity of Earth = 0 km/h

m₃ = Mass of Earth = 5.97 × 10²⁴ kg

Therefore, pluggin in the vaalues gives

0.1 × 10¹⁴ kg × 40000 km/h + 5.97 × 10²⁴ kg × 0 km/h = (0.1 × 10¹⁴ kg + 5.97 × 10²⁴ kg) × v₄

Therefore,

v₄ = (4.0 × 10¹⁷ kg·km/h + 0 kg·km/h)/ (5970000000010000000000000 kg)

= 6.70× 10⁻⁸ km/h = 1.86× 10⁻⁸ m/s

Change is too small

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4 years ago

Enterprising students set an enormous slip-n-slide (a plastic sheetcovered in water to reduce friction) on flat ground. If the s

Klio2033 [76]

Answer:

a = -0.05 m/s² (negative sign shows deceleration)

Explanation:

In order, to find out the minimum average acceleration for a student starting at 5 m/s to slide to the end, we can use 3rd equation of motion. 3rd equation of motion is given as follows:

2as = Vf² - Vi²

where,

a = minimum acceleration required = ?

s = minimum distance covered = 250 m

Vf = Final Speed = 0 m/s (for minimum acceleration the student will barely cover 250 m and then stop)

Vi = Initial Velocity = 5 m/s

Therefore,

2a(250 m) = (0 m/s)² - (5 m/s)²

a = - (25 m²/s²)/(500 m)

<u>a = -0.05 m/s²</u> (negative sign shows deceleration)

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3 years ago