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4 years ago

How large is the area of water found on Mars?

Physics

2 answers:

ehidna [41]4 years ago

8 0

More than five million cubic kilometers of ice have been identified.

Jlenok [28]4 years ago

7 0

Hello there!

More than five million cubic kilometers of ice have been identified at or near the surface of modern Mars, enough to cover the whole planet to a depth of 35 meters (115 ft).

You might be interested in

If you walk at an avaerage speed of 5 km/h for 30 min how far will you walk

PtichkaEL [24]

Answer:

2.5 km/30min

Explanation:

Divide 5 by 2 because 30 min is half of 1 hour. when you do that you get 2.5

7 0

3 years ago

The strong nuclear force holds together which two particles in an atom?

Yanka [14]

Protons and neutrons are held together

8 0

3 years ago

After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. when the box reaches position

KiRa [710]

As we know by work energy theorem

total work done = change in kinetic energy

so here we can say that wok done on the box will be equal to the change in kinetic energy of the system

$W_p = KE_f - KE_i$

initial the box is at rest at position x = x1

so initial kinetic energy will be ZERO

at final position x = x2 final kinetic energy is given as

$KE_f = \frac{1}{2}mv_1^2$

now work done is given as

$W_p = \frac{1}{2}mv_1^2 - 0$

so we can say

$W_p = \frac{1}{2}mv_1^2$

so above is the work done on the box to slide it from x1 to x2

3 0

4 years ago

Before going in for an annual physical, a 70.0-{\rm kg} person whose body temperature is 37.0{\rm ^{\circ} C} consumes an entire

Grace [21]

Answer:

A) The person's body temperature T_final after equilibrium is attained = 36.85°C

B) The change in the person's temperature after equilibrium is attained = 0.15°C

A high-quality medical thermometer can measure temperature changes as small as 0.1°C, hence, YES, it would detect the minute drop by 0.15°C too.

Explanation:

If we assume that the soft drink has the same density as water (since it is stated in the question that it is mostly water).

Density of water = 1 g/mL = 1 kg/L

Ignoring any heating by the person's metabolism,

A) So, heat lost by the human body = heat gained by the soft drink as it attains thermal equilibrium with the human body

Let the final temperature of the human body + soft drink set up be T

Heat lost by the human body = mCΔT

m = mass of the human body = 70.0 kg

C = Specific heat capacity of the human body = 3480 J/kg.K

ΔT = Temperature change of the human body = 37 - (Final temperature) = 37 - T

Heat lost by the body = 70 × 3480 × (37 - T)

= (9,013,200 - 243,600T) J

Heat gained by soft drink = mCΔT

m = mass of the soft drink = density × volume = 1 × 0.355 = 0.355 kg

C = specific heat capacity of the soft drink = specific heat capacity of the soft drink = 4182 J/kg.K

ΔT = (final temperature) - 12 = (T - 12)

Heat gained by the soft drink = 0.355 × 4182 × (T - 12) = (1,484.61T - 17,815.32) J

heat lost by the human body = heat gained by the soft drink as it attains thermal equilibrium with the human body

(9,013,200 - 243,600T) = (1,484.61T - 17,815.32)

9,013,200 + 17,815.32 = 1,484.61T + 243,600T

9,031,015.32 = 245,084.61T

T = (9,031,015.32/245,084.61)

= 36.8485614825 = 36.85°C

B) The change in the person's temperature = 37 - 36.85 = 0.15°C

A high-quality medical thermometer can measure temperature changes as small as 0.1°C, hence it would detect the minute drop by 0.15°C too.

Hope this Helps!!!

5 0

4 years ago

An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 1760 1760 × 103 seconds (a

Katen [24]

Answer:

$a_c=6.471\frac{m}{s^{2}}$

Explanation:

We can calculate the magnitude of the tangential velocity of the moon, using the equation:

$v=\frac{2\pi R}{\tau}$

(This works, since velocity is defined as distance over time; in this case the distance is 2πR and the time is the period τ)

Next, from the equation of the centripetal acceleration we have:

$a_c=\frac{v^{2} }{R}\\\\a_c=\frac{4\pi^{2}R}{\tau^{2} }$

Be careful, the radius of the orbit R is equal to the distance from the center of the moon to the center of the planet. So we have to sum the distance from the center of the moon to the surface of the planet and the radius of the planet to obtain R:

$R=285.0*10^{6} m+3.50*10^{6} m=288.5*10^{6}m$

Finally, plugging the given values into the centripetal acceleration formula, we have:

$a_c=\frac{4(3.141)^{2}(288.5*10^{6}m)}{(1760*10^{3}s)^{2}} =6.471\frac{m}{s^{2}}$

In words, the moon's radial acceleration a_c is 6.471m/s².

3 0

3 years ago