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2 years ago

The real advantage to hydrostatic weighing is that it __________.

Physics

2 answers:

katovenus [111]2 years ago

7 0

The real advantage to hydrostatic weighing is that it gives one of the most accurate measurements of body fat.

KengaRu [80]2 years ago

7 0

Answer:

yup its B as in bagel

Explanation:

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If an oscillating mass has a frequency of 1.25 Hz, it makes 100 oscillations in

KatRina [158]

Answer:

Time, t = 80 seconds

Explanation:

Given that,

The frequency of the oscillating mass, f = 1.25 Hz

Number of oscillations, n = 100

We need to find the time in which it makes 100 oscillations. We know that the frequency of an object is number of oscillations per unit time. It is given by :

$f=\dfrac{n}{t}$

$t=\dfrac{n}{f}$

$t=\dfrac{100}{1.25\ Hz}$

t = 80 seconds

So, it will make 100 oscillations in 80 seconds. Hence, this is the required solution.

4 0

2 years ago

The primary source of most of the moisture for the earths atmosphere is

SpyIntel [72]

I think is ocean but I'm not sure

6 0

2 years ago

According to the Heisenberg uncertainty principle, if the uncertainty in the speed of an electron is 3.5 × 103 m/s, the uncertai

GREYUIT [131]

Explanation:

It is given that,

Uncertainty in the speed of an electron, $\Delta v=3.5\times 10^3\ m/s$

According to Heisenberg uncertainty principle,

$\Delta x.\Delta p=\dfrac{h}{4\pi}$

$\Delta x$ is the uncertainty in the position of an electron

Since, $\Delta p=m\Delta v$

$\Delta x=\dfrac{h}{4\pi.m \Delta v}$

$\Delta x=\dfrac{6.6\times 10^{-34}}{4\pi\times 9.1\times 10^{-31}\times 3.5\times 10^3}$

$\Delta x=1.64\times 10^{-8}\ m$

So, the uncertainty in its position is $1.64\times 10^{-8}\ m$ . Hence, this is the required solution.

6 0

2 years ago

An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole drilled normal to the surface of a large block of

slega [8]

Answer:

$T_{1}=94.9^{o}C$

Explanation:

Given data

length=100mm

Diameter=5mm

Thermal conductivity=5 W/m.K

Power=50 W

Temperature=25°C

The temperature of heater surface follows from the rate equation written as:

$T_{1}=T_{2}+\frac{q}{kS}$

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

$S=\frac{2\pi L}{ln(\frac{4L}{D} )} \\$

Substitute the given values

$S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m$

The temperature of heater is then:

$T_{1}=25^{o}C+\frac{50W}{5W/m.K*0.143m} \\T_{1}=94.9^{o}C$

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.

$T_{1}=94.9^{o}C$

5 0

2 years ago

a star has a distance of 30 parsecs and a apparent magnitude of 3 --what would its absolute magnitude be?

Alborosie

The absolute magnitude of the star would be +5.

8 0

1 year ago