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never [62]
3 years ago
8

A woman floats in a region of the Great Salt Lake where the water is about four times saltier than the ocean and has a density o

f about 1130 kg/m3. The woman has a mass of 52 kg, and her density is 943 kg/m3 after exhaling as much air as possible from her lungs.
1) Determine the percentage of her volume that will be above the waterline of the Great Salt Lake.
Physics
1 answer:
julsineya [31]3 years ago
3 0

To develop the problem it is necessary to use the concepts related to density and its definition with respect to mass and volume, as well

\rho= \frac{m}{V}

Therefore to calculate the volume of woman:

V = \frac{m}{\rho}\\V = \frac{52}{943}\\V = 0.0551 m^3

Now, volume of water of Great Salt Lake with the mass of 52 kg,

V_i = \frac{52}{1130} \\V_i = 0.0460 m^3

So, percentage of her volume changed will be,

\%=\frac{(V - Vi)}{V}*100\\\% = \frac{(0.0551 - 0.0460)}{0.0551}*100 \\\% == 16.51 \%\\

Therefore the percentage of her volume that will be above the waterline of the Great Salt Lake is 16.51%

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A 4.80 Kg watermelon is dropped from rest from the roof of an 18.0 m building. Calculate the work done by gravity on the waterme
Nonamiya [84]

Answer:

Work, W = 846.72 Joules

Explanation:

Given that,

Mass of the watermelon, m = 4.8 kg

It is dropped from rest from the roof of 18 m building. We need to find the work done by the gravity on the watermelon from the roof to the ground. It is same as gravitational potential energy i.e.

W = mgh

W=4.8\ kg\times 9.8\ m/s^2\times 18\ m

W = 846.72 Joules

So, the work done by the gravity on the watermelon is 846.72 Joules. Hence, this is the required solution.

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3 years ago
Hey! Can someone help with this question? Thx :)
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3 years ago
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
An object is placed on a surface. A student tries to apply various combinations of forces on the object. Which pair of forces wi
AVprozaik [17]

Answer:

See Explanation

Explanation:

The question is incomplete, as there are no diagrams or options to provide more information to the question.

The general explanation is as follows:

For the object not to move

(1): The forces acting on the object must opposite each other. i.e. if force A acts at the right (or positive direction), force B will act at the left (or negative direction).

(2) The two forces must be equal.

So, for instance:

If the pair of forces are 5N and 5N in opposite directions, the object wil not move.

However, if one of the forces is greater, the object will move towards the direction of the greater force.

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Visible electrical discharge from a storm cloud is called?
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id have to say its thunder

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