All categories

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

You might be interested in

What is the pressure at the bottom<br> of a 2.50 m deep pool of water?

Grace [21]

Answer:

Pressure = Density x Gravity acceleration x Height

Density of water = 1000 Kgm^-3

Gravity acceleration = 9.8 ms^-2

So it is 4.50.

Hope this helps and please mark me brainliest!

8 0

3 years ago

If all else stays the same, which would cause an increase in the gravitational force on a space shuttle?

Alenkasestr [34]

Answer: Decreasing the distance of the space shuttle from Earth .

Explanation:

According to expression of gravitational force:

$F=\frac{G\times m_1\times m_2}{r^2}$

G = gravitational constant

$m_1, m_2$ = masses of two objects

r = Distance between the two objects.

F = Gravitational force

From the above expression we can say that gravitational force is inversely proportional to squared of the distance between the two masses.

$F\propto \frac{1}{r^2}$

So, in order to increase the gravitational force on space shuttle distance between the space space shuttle must be decreased.

Hence, the correct answer 'decreasing the distance of the space shuttle from Earth '.

8 0

3 years ago

Read 2 more answers

A power lifter performs a dead lift, raising a barbell with a mass of 305 kg to a height of 0.42 m above the ground, giving the

Ann [662]

Answer:

Explanation:

Before it hits the ground:

The initial potential energy = the final potential energy + the kinetic energy

mgH = mgh + 1/2 mv²

gH = gh + 1/2 v²

v = √(2g (H - h))

v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))

v ≈ 2.0 m/s

When it hits the ground:

Initial potential energy = final kinetic energy

mgH = 1/2 mv²

v = √(2gH)

v = √(2 * 9.81 m/s² * 0.42 m)

v ≈ 2.9 m/s

Using a kinematic equation to check our answer:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)

v ≈ 2.9 m/s

3 0

3 years ago

A proton and an alpha particle (q = +2e, m = 4 u) are fired directly toward each other from far away, each with an initial speed

stich3 [128]

Answer:

Distance of closest approach, $r=1.91\times 10^{-14}\ m$

Explanation:

It is given that,

Charge on proton, $q_p=e$

Charge on alpha particle, $q_a=2e$

Mass of proton, $m_p=1.67\times 10^{-27}\ kg$

Mass of alpha particle, $m_a=4m_p=6.68\times 10^{-27}\ kg$

The distance of closest approach for two charged particle is given by :

$r=\dfrac{k2e^2(m_p+m_a)}{2m_am_pv_p^2}$

$r=\dfrac{9\times 10^9\times 2(1.6\times 10^{-19})^2(1.67\times 10^{-27}+6.68\times 10^{-27})}{2\times 6.68\times 10^{-27}\times 1.67\times 10^{-27}(0.01\times 3\times 10^8)^2}$

$r=1.91\times 10^{-14}\ m$

So, their distance of closest approach, as measured between their centers $1.91\times 10^{-14}\ m$ . Hence, this is the required solution.

4 0

3 years ago

Motivate me for my exam

Brut [27]

Answer:

if i can pass, you can too :)

Explanation:

its super inspiring cuz im a dum‎‎b‎‎a‎‎ss

8 0

4 years ago