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3 years ago

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A rescue helicopter is lifting a 78.9 kg man from a capsized boat by means of a cable and harness. What is the tension in the ca

ble when the man is given an initial upward acceleration of 1.8 m/s/s.

1 answer:

allsm [11]3 years ago

3 0

Answer:

the tension T = 916[N]

Explanation:

We need to analyze the sketch that is in the attached image, here we can find the free body diagram with the forces acting over him.

The weight is equal to the product of the mass by the gravity

w = m*g = 78.9[kg] * 9.81[m/s^2]

The acceleration given by the helicopter is 1.8 [m/s^2]

The sum of forces on the body must be equal to mass by acceleration.

The only forces acting are the tension of the cable and the weight of man, these forces are equal to the product of mass by acceleration.

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A 3.4kg aluminium ball has an apparent mass of 2.10kg when submerged in a particular liquid. Calculate the density of the liquid

Shalnov [3]

Answer:

1083.3kg/m ³

Explanation:

Given parameters:

Mass of aluminium ball = 3.4kg

Apparent mass of ball = 2.1kg

Unknown:

Density of the liquid = ?

Solution:

Density is is the mass per unit volume of any give substance;

Density = $\frac{mass}{volume}$

Now, we must understand that the apparent weight of the aluminium is the part of its weight supported by the fluid;

Pl = $\frac{Ml}{Vl}$

Pl = density of liquid

Ml = mass of liquid

Vl = volume of liquid

Mass of liquid = Mal - Mapparent

Mal = mass of aluminium

The volume of liquid displaced is the same as the volume of the aluminium according to Archimedes's principle;

Ml = Pl x Vl

Val = Vl

Val volume of aluminium

Vl = volume of liquid

****

Pal = $\frac{Mal}{Val}$

Val = $\frac{Mal}{Pal}$

Val = volume of aluminium

Mal = mass of aluminium

Pal = density of aluminium

*****

Since the Val = Vl

Ml = Pl x $\frac{Mal}{Pal}$

Since

Ml = Mal - Mapparent

Mal - Mapparent = Pl x $\frac{Mal}{Pal}$

Pal = density of aluminium = 2712 kg/m ³

3.4 - 2.1 = Pl x $\frac{3.4}{2712}$

1.3 = Pl x 0.00123

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One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

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0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

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