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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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STatiana [176]

The electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.

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Electric potential energy can be defined as the energy needed to move a charge against an electric field.

It is calculated using the formula;

U = Kq1 q2 ÷ r

Where Q = electric potential energy

k = Coulombs constant

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Electric potential energy is inversely proportional to the distance of separation of the charges.

If the distance of the charges changes from 3mm to 6mm, then the electric potential energy of the charges is reduced because it decreases with increase in the distance of the charges.

Therefore, the electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.

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1 year ago

a ultrasonic wave at 8x10^4 Hz is emitted into a vien where the speed of sound in blood is 1570 m/s. the wave reflects off the r

Aneli [31]

Answer: 0.392 m/s

Explanation:

The Doppler shift equation is:

$f'=\frac{V+V_{o}}{V-V_{s}} f$

Where:

$f=8(10)^{4} Hz$ is the actual frequency of the sound wave

$f'=8.002(10)^{4} Hz$ is the "observed" frequency

$V=1570 m/s$ is the speed of sound

$V_{o}=0 m/s$ is the velocity of the observer, which is stationary

$V_{s}$ is the velocity of the source, which are the red blood cells

Isolating $V_{s}$ :

$V_{s}=\frac{V(f'-f)}{f'}$

$V_{s}=\frac{1570 m/s(8.002(10)^{4} Hz-8(10)^{4} Hz)}{8.002(10)^{4} Hz}$

Finally:

$V_{s}=0.392 m/s$

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2 years ago

A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod.

mestny [16]

Answer:

a = 17.68 m/s²

Explanation:

given,

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angle made with vertical, θ = 61°

time to complete 1 rev, t = 1.25 s

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first we have to calculate the radius of the circle

R = L sin θ

R = 0.8 x sin 61°

R = 0.7 m

now, calculating at the angular velocity

$\omega =\dfrac{2\pi}{T}$

$\omega =\dfrac{2\pi}{1.25}$

ω = 5.026 rad/s

now, radial acceleration

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a = 0.7 x 5.026²

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hence, the radial acceleration of the ball is equal to 17.68 rad/s²

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3 years ago

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