An object with high momentum _____

If a 0.5 kg pair of running shoes would weigh 11.55 Newton’s on Jupiter, what is the strength of gravity there?

OverLord2011 [107]

The weight of an object is given by
$F=mg$
where m is the mass of the object, while g is the strength of the gravity (which corresponds to the gravitational acceleration of the planet).

In our problem, the shoes have a mass of m=0.5 kg, and their weight is F=11.55 N. So, we can re-arrange the previous formula to find the value of g:
$g= \frac{F}{m}= \frac{11.55 N}{0.5 kg}=23.1 N kg^{-1}$
and this is the strength of the gravity on Jupiter's surface.

4 0

3 years ago

. An expedition in the arctic sea to study the e ect of climate change on deep-ocean ecosystemrequires precise measurement of te

Sergio039 [100]

Answer:

The complete question contains 2 parts such that

part a: Where the probe is submerged in water. In this case the wattage of heater is 90.47 W

part b: Where the probe is on a vessel. In this case the wattage of heater is So 121.34 W

Explanation:

Part a

For the first part when it is assumed that the probe is submerged in the water. In this case the heat loss is only due to the convective heat transfer which is given as

$Q=h \times A \times (T_{body}-T_{surr})$

Here

Q is the heat loss which is to be calculated

A is the area of the body which is given as $4 \pi r^2 \\$ . It is calculated as

$A=4 \pi r^2 \\A=4 \times 3.14 \times (0.2)^2 \\A=0.5026 m^2$

Tbody is 10 C
Tsurr is 0C
h is the convection coefficient of water which is 18 W/m^2K

$Q=h \times A \times (T_{body}-T_{surr})\\Q=18 \times 0.5026 \times (10-0)\\Q=90.47 W\\$

So the wattage of heater is 90.47 W

Part b

Now when the probe is above sea water now, the losses are both because of convection and radiation now the loss is given as

$Q_{con}=h_{air} \times A \times (T_{body}-T_{air})$

Here

Q_con is the heat loss due to convection which is to be calculated

A is the area of the body which is given as 0.5026 m^2

Tbody is 10 C
Tair is -10C
h is the convection coefficient of water which is 10 W/m^2K

$Q_{con}=h_{air} \times A \times (T_{body}-T_{air})\\Q_{con}=10 \times 0.5026 \times (10+10)\\Q_{con}=100.52 Watts$

Radiation loss is given as

$Q_{rad}=E_{emission}-E_{radiation}\\Q_{rad}=\epsilon A \tau (T_{body}^4-T_{sky}^4)\\$

Here

Q_rad is the heat loss due to radiation which is to be calculated

A is the area of the body which is given as 0.5026 m^2

Tbody is 10+273=283K
Tsky is 0+273=273K
ε is the emissivity of the shell which is 0.85

τ is the coefficient which is $5.67 \times 10^{-8}$

$Q_{rad}=E_{emission}-E_{radiation}\\Q_{rad}=\epsilon A \tau (T_{body}^4-T_{sky}^4)\\Q_{rad}=0.85 \times 0.5026 \times 5.67 \times 10^{-8} ((283)^4-(273)^4)\\Q_{rad}=20.823 Watts$