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3 years ago

5

A fish (8 kg) is being yanked vertically upward out of the water and the fishing line breaks. If the line is rated to a maximum

tension of 160 N (~ 35 lb test), then what was the minimum acceleration of the fish (in m/s2)

2 answers:

kolezko [41]3 years ago

7 0

Answer:

The minimum acceleration of the fish is 10.2 m/s²

Explanation:

The force on string is:

F = 8 * 9.8 = 78.4 N

According the Newton second law:

T - W = ma

Where T = tension of the string = 160 N

W = weight of the fish = 78.4 N

m = mass of the fish = 8 kg

a = acceleration of the fish = ?

Clearing a:

a = (T- W)/m = (160 - 78.4)/8 = 10.2 m/s²

ki77a [65]3 years ago

5 0

Answer:

Explanation:

Given:

Mass of the fish, 8 kg

Tension of the line, F = 160 N

Tension of the line, F + normal force = 0

Normal force, Fn = Mass × acceleration

Acceleration = force/mass

= 160/8

= 20 m/s^2

Minimum acceleration = 20 m/s^2

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3 years ago

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uppose that white light strikes a flat piece of flint glass in air, coming in at an angle of 60 degrees to the surface (30 degre

Elodia [21]

Answer:

Explanation:

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As μ for blue light is greatest , angle of refraction for blue light will be least . Similarly , angle of refraction will be maximum for red light .

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7 0

3 years ago

The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the

nataly862011 [7]

Answer with Explanation:

Let rest mass $m_0$ at point P at distance x from center of the planet, along a line connecting the centers of planet and the moon.

Mass of moon=m

Distance between the center of moon and center of planet=D

Mass of planet=M

We are given that net force on an object will be zero

a.We have to derive an expression for x in terms of m, M and D.

We know that gravitational force= $\frac{GmM}{r^2}$

Distance of P from moon=D-x

$F_m$ =Force applied on rest mass due to m

$F_m$ =Force on rest mass due to mas M

$F_M=F_m$ because net force is equal to 0.

$F_m=F_M$

$\frac{Gm_0m}{(D-x)^2}=\frac{Gm_0M}{x^2}$

$\frac{m}{(D-x)^2}=\frac{M}{x^2}$

$\frac{x^2}{(D-x)^2}=\frac{M}{m}$

$\frac{x}{D-x}=\sqrt{\frac{M}{m}}$

Let $R=\sqrt{\frac{M}{m}}$

Then, $\frac{x}{D-x}=R$

$x=DR-xR$

$x+xR=DR$

$x(1+R)=DR$

$x=\frac{DR}{1+R}$

b.We have to find the ratio R of the mass of the mass of the planet to the mass of the moon when x= $\frac{2}{3}D$

Net force is zero

$F_m=F_M$

$\frac{Gm_0m}{(D-\frac{2}{3}D)^2}=\frac{Gm_0M}{\frac{4}{9}D^2}$

$\frac{m}{\frac{D^2}{9}}=\frac{9M}{4D^2}$

$\frac{M}{m}=4$

Hence, the ratio R of the mass of the planet to the mass of the moon=4:1

8 0

4 years ago

A physicist comes across an open container which is filled with two liquids. Since the two liquids have different density, there

Dominik [7]

Answer:

496.57492 kg/m³

Explanation:

$P_a$ = Atmospheric pressure = 101300 Pa

$\rho_w$ = Density of water = 1000 kg/m^3

$h_w$ = Height of water = 21.8 cm

$h_f$ = Height of fluid = 30 cm

g = Acceleration due to gravity = 9.81 m/s²

$\rho_f$ = Density of the unknown fluid

Absolute pressure at the bottom

$P_{abs}=P_a+\rho_wgh_w+\rho_fgh_f\\\Rightarrow \rho_f=\frac{P_{abs}-P_a-\rho_wgh_w}{gh_f}\\\Rightarrow \rho_f=\frac{104900-101300-1000\times 9.81\times 0.218}{9.81\times 0.3}\\\Rightarrow \rho_f=496.57492\ kg/m^3$

The density of the unknown fluid is 496.57492 kg/m³

6 0

3 years ago

(a) An ideal gas initially at pressure p0 undergoes a free expansion until its volume is 3.80 times its initial volume. What the

trapecia [35]

Answer:

a)P/Po=0.263

b)γ=1.33 So gas is triatomic

c) $\dfrac{KE_f}{KE_i}=1.56$

Explanation:

a)

initial pressure = Po

Initial volume = Vo

Final volume = 3.8 Vo

Lets take final pressure is P

we know that for free expansion process

PV= Constant

Po x Vo = P x 3.8 Vo

P=0.263 Po

So

P/Po=0.263

b)

Now gas is compressed in adiabatic manner

Final pressure = 1.56 Po

=1.56 Po

We know that for adiabatic process

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

$\dfrac{V_2}{V_1}=\left(\dfrac{P_1}{P_2}\right)^{\dfrac{1}{\gamma}}$

$0.263P_o(3.8V_o)^{\gamma}=1.56P_o\times V_o^{\gamma}$

γ=1.33 So gas is triatomic

c)

We know that average kinetic energy given as

$KE=\dfrac{3}{2}KT$

$\dfrac{KE_f}{KE_i}=\dfrac{T_f}{T_i}$ \

$\dfrac{KE_f}{KE_i}=\dfrac{P_fV_f}{P_iV_i}$

$\dfrac{KE_f}{KE_i}=\dfrac{1.56P_oV_o}{P_oV_o}$

$\dfrac{KE_f}{KE_i}=1.56$

3 0

3 years ago