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2 years ago

This problem, a squid at rest suddenly sees a predator coming toward it and needs to escape. Assume the following:______.

Physics

1 answer:

makkiz [27]2 years ago

8 0

Answer:

6.79 m/s

Explanation:

By applying the principle of conservation of momentum.

The total momentum = MV - mv = 0 (since the squid is beginning at rest)

the mass of the squid (M) in absence of water in its cavity = (6.5 - 1.75) kg

= 4.75 kg

speed of the squid (V) = 2.5 m/s

mass of the water expelled (m) = 1.75 kg

speed of the water (v) = ???

∴

4.75 × 2.5 = 1.75 × v

$v = \dfrac{4.75 \times 2.5}{1.75 }$

v = 6.79 m/s

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True or False?

olga nikolaevna [1]

Answer:

False

Explanation:

It is better to be logical when proving a point

6 0

3 years ago

The speed of sound in room temperature (20°C) air is 343 m/s; in room temperature helium, it is 1010 m/s. The fundamental freque

Lera25 [3.4K]

Answer: f = 927.55Hz

Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below

L = λ/4 where λ = wavelength.

speed of sound in air = v = 343m/s.

fundamental frequency of open closed tube = 315Hz

λ = 4L.

v = fλ

343 = 315 * 4L

343 = 1260 * L

L = 343/ 1260

L = 0.27m

In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.

The length of tube and wavelength are related by the formulae below

L = λ/4, λ=4L

λ = 4 * 0.27

λ = 1.087m.

v = fλ

1010 = f * 1.087

f = 1010/1.807

f = 927.55Hz

4 0

3 years ago

A physics student uses a spring loaded launcher to fire a 58 gram projectile horizontally. The projectile leaves the launcher at

timofeeve [1]

Answer:

no answer

Explanation:

the question does not sk what to find

3 0

3 years ago

The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol

Lesechka [4]

Answer:

$\Delta V = 0.053 A$

Explanation:

Electric field in a given region is given by equation

$E = \frac{A}{r^4}$

as we know the relation between electric field and potential difference is given as

$\Delta V = -\int E. dr$

so here we have

$\Delta V = - \int (\frac{A}{r^4}) .dr$

$\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}$

here we know that

$r_1 = 1.71 m$ and $r_2 = 2.89 m$

so we will have

$\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})$

so we will have

$\Delta V = 0.053 A$

8 0

3 years ago

The city is thinking about starting a program to add a new chemical to the public pool. Before they do, they want to find out if

Kitty [74]

The best and most correct answer among the choices provided by the question is the fourth choice.

The best people for advising is <span>the government agency that regulates these types of chemicals.</span>
I hope my answer has come to your help. God bless and have a nice day ahead!

4 0

3 years ago