The voltage drop across any component is

To determine an athlete's body fat, she is weighed first in air and then again while she's completely underwater. it is found th

il63 [147K]

The difference in weight is due to the displacement of water (the buoyancy of water is acting on the athlete thus giving her smaller weight).

The amount of weight displaced or the amount of buoyant force is:

Fb = 690 N - 48 N

Fb = 642 N

From newtons law, F = m*g. Using this formula, we can calculate for the mass of water displaced:

m of water displaced = 642N / 9.8m/s^2

m of water displaced = 65.5 kg

Assuming a water density of 1 kg/L, and using the formula volume = mass/density:

V of water displaced = 65.5kg / 1kg/L = 65.5 L

The volume of water displaced is equal to the volume of athlete. Therefore:

V of athlete = 65.5 L

The mass of athlete can also be calculated using, F = m*g

m of athlete = 690 N/ 9.8m/s^2

m of athlete = 70.41 kg

Knowing the volume and mass of athlete, her average density is therefore:

average density = 70.41 kg / 65.5 L

average density = 1.07 kg/L = 1.07 g/mL

5 0

3 years ago

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What is: the fastest car the fastest animal the fastest human the fastest aeroplane

seropon [69]

The Hennessey Venom GT is the fastest road car in the world.
The fastest land animal is the Cheetah
Usain Bolt, the World's fastest man.
The Lockheed SR-71 "Blackbird" the fastest airplane.

4 0

3 years ago

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A mass attached to a 50.0 cm long string starts from rest and is rotated 40 times in one minute before reaching a final angular

ch4aika [34]

To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

$\theta = \frac{1}{2}\alpha t^2 +\omega_0 t + \theta_0$

Where,

$\theta =$ Angular Displacement

$\alpha =$ Angular Acceleration

$\omega_0 =$ Angular velocity

$\theta_0 =$ Initial angular displacement

For this case we have neither angular velocity nor initial angular displacement, then

$\theta = \frac{1}{2}\alpha t^2$

Re-arrange for $\alpha,$

$\alpha = \frac{2\theta}{t^2}$

Replacing our values,

$\alpha = \frac{2(40rev*\frac{2\pi rad}{1rev})}{60^2}$

$\alpha = 0.139rad/s$

Therefore the ANgular acceleration of the mass is $0.139rad/s^2$

4 0

3 years ago

In a shipping yard, a crane operator attaches a cable to a 1,400 kg shipping container and then uses the crane to lift the conta

Molodets [167]

Answer:

Explanation:

Given

mass of crane $m=1400\ kg$

distance moved $d=40\ m$

Since it is moving with a constant velocity therefore net force on it is zero

Tension force=weight

T=mg

Work done by Tension T is

$W_T=T\cdot d$

$W_T=1400\times 9.8\cdot 40$

$W_T=548.8\ KJ$

Work done by Gravity will be equal in magnitude but opposite in sign and can be obtained by work energy theorem which states that change in kinetic energy of object is equal to work done by all the forces

$W_T+W_g=0$

$W_g=-548.8\ KJ$

4 0

3 years ago

2 Points

mezya [45]

The advantage is that we do not run out of resources and a disadvantage is that is dangerous when a “human” gets too close and gets sick by the radiation.

5 0

3 years ago