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blagie [28]
4 years ago
5

Squids propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity

to force out the water through an opening. A 9 kg squid (including the water in the cavity) at rest suddenly sees a dangerous predator. If the squid expels 2 kg of water out of its body with a speed of 8 m/s, what would be its own escape speed
Physics
1 answer:
Musya8 [376]4 years ago
8 0

Answer:

v_squid = - 2,286 m / s

Explanation:

This exercise can be solved using conservation of the moment, the system is made up of the squid plus the water inside, therefore the force to expel the water is an internal force and the moment is conserved.

Initial moment. Before expelling the water

          p₀ = 0

the squid is at rest

Final moment. After expelling the water

         p_{f} = M V_squid + m v_water

         p₀ = p_{f}

          0 = M V_squid + m v_water

           c_squid = -m v_water / M

The mass of the squid without water is

            M = 9 -2 = 7 kg

let's calculate

           v_squid = 2 8/7

           v_squid = - 2,286 m / s

The negative sign indicates that the squid is moving in the opposite direction of the water

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3 years ago
Jack travelled 360 km at an average speed of 80 km/h. Elaine
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Answer:

Average speed of Elain = 60 km/h

Explanation:

Total Distance covered by Jack = 360km

Average Speed of Jack = 80 km/h

Time taken by Jack to complete his journey = Distance / Average speed = 360 km / 80 km/h

Time taken by Jack to complete his journey = 4.5 hours

As it is given the both Jack and Elain travelled the same amount of distance:

Total distance travelled by Elain = 360 km

It is given that Elain took 1.5 hourse more than Jack to cover the distance, so Time taken by Elain to cover the distance is = 4.5 hours + 1.5 hours = 6 hours

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3 years ago
Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa
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Answer:

The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

Explanation:

Given that,

First charge q_{1}= 2.9\mu C

Second chargeq_{2}= 2.9\mu C

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

E_{1}=\dfrac{kq}{r'^2}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}

E_{1}=52200=52.2\times10^{3}\ N/C

For E₂,

Using formula of electric field

E_{1}=\dfrac{kq}{r^2}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}

E_{2}=104400=104.4\times10^{3}\ N/C

We need to calculate the horizontal electric field

E_{x}=E_{1}\cos\theta

E_{x}=52.2\times10^{3}\times\cos45

E_{x}=36910.97=36.9\times10^{3}\ N/C

We need to calculate the vertical electric field

E_{y}=E_{2}+E_{1}\sin\theta

E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45

E_{y}=141310.97=141.3\times10^{3}\ N/C

We need to calculate the net electric field

E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

E_{net}=146038.69\ N/C

E_{net}=146.03\times10^{3}\ N/C

We need to calculate the direction of electric field

Using formula of direction

\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}

\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

\theta=75.36^{\circ}

Hence, The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

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