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Solnce55 [7]
3 years ago
9

Key evidence that mars may once have had much more water than it does today-perhaps enough to form an ocean-comes from ________.

Physics
1 answer:
Mashutka [201]3 years ago
7 0
The answer to this question is precisely the "measurement of the atmospheric ratio of the given deuterium to the given ordinary hydrogen". Hence, if we complete the statement in the problem, we have it, the key evidence that Mars may once have had much more water than it does today- perhaps enough to form or accommodate an ocean comes from the measurement of the atmospheric ratio of the deuterium to the given ordinary hydrogen. 
You might be interested in
Classify the properties as extensive or intensive: mass density; color volume; total energy; temperature; melting point
m_a_m_a [10]

Answer:

Intensive properties

Density

Color

temperature

Melting point

Extensive properties

Mass

Volume

Total Energy

Explanation:

Intensive properties:  In Physics, Intensive properties which are not depend of the amount of matter in a sample, It only depends of the type of matter, some examples of intensive properties are:

1. Density: It is a intensive property. It can explain better with a example:  the water density is 1000 kg/m3, So if we have 1 liter or 1000 liters of water  the density will be the same for the two samples.

2. Color: Solid sodium chloride is white. If you have 2 samples the first recipient  with 2 kilograms of NaCl and the second with 10 kilograms of NaCl. The color of the substance does not depend on the amount of the substance.

As was mentioned before the same theory is applied to temperature and melting point concepts.

On the other hand,

Extensive properties are properties of the matter which depend on the amount of matter that is present in the system or sample. some examples are:

1. Mass: It is a property that measures the amount of matter that an object contains. For example, 10 kilograms of solid Copper contains a higher mass than 2 kilograms of the same metal.  

2. Volume: It is a property which measures the space occupied by an object or a substance. For example, the space occupied by a glass of milk is lower than the space occupied by a bottle of milk, Then the volume of the glass of milk is lower than the volume of the bottle of milk.

3. Finally the total energy is contained in molecules and atoms that constituted systems  so, if the amount of matter increases the number of molecules too, then the total energy will increase.

I hope it helps you.

6 0
3 years ago
The net torce on an object moving with constant speed in circular motion is in which direction?
aleksley [76]

The correct answer is C) towards the center of the circle.

Although the object is moving at a constant speed it is constantly accelerating due to the constant change in direction as it describes the circular path. This causes a constant change in velocity as velocity is a vector quantity.

For the object to maintain the circular path there has to be centripetal force acting on the object and this centripetal force is directed towards the center of the circle.

6 0
3 years ago
The tow spring on a car has a spring constant of 3,086 N / m and is initially stretched 18.00 cm by a 100.0 kg college student o
sdas [7]

Answer:

The velocity of the skateboard is 0.774 m/s.

Explanation:

Given that,

The spring constant of the spring, k = 3086 N/m

The spring is stretched 18 cm or 0.18 m

Mass of the student, m = 100 kg

Potential energy of the spring, P_f=20\ J

To find,

The velocity of the car.

Solution,

It is a case of conservation of energy. The total energy of the system remains conserved. So,

P_i=K_f+P_f

\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2+20

\dfrac{1}{2}\times 3086\times (0.18)^2=\dfrac{1}{2}mv^2+20

50-20=\dfrac{1}{2}mv^2

30=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{60}{100}}

v = 0.774 m/s

So, the velocity of the skateboard is 0.774 m/s.

7 0
3 years ago
San Jose to San Francisco is approximately 7200 meters away. If it takes you 3600 seconds to drive by car what is the velocity?
svetlana [45]
V=s/t=>7200/3600=2m/s
5 0
3 years ago
A common flashlight bulb is rated at 0.32 A and 4.3 V (the values of the current and voltage under operating conditions). If the
sleet_krkn [62]

Answer:

1176.01 °C

Explanation:

Using Ohm's law,

V = IR................. Equation 1

Where V = Voltage, I = current, R = Resistance when the bulb is on

make R the subject of the equation

R = V/I.................. Equation 2

R = 4.3/0.32

R = 13.4375 Ω

Using

R = R'(1+αΔθ)............................. Equation 3

Where R' = Resistance of the bulb at 20°, α = Temperature coefficient of resistivity, Δθ = change in temperature

make Δθ the subject of the equation

Δθ = (R-R')/αR'.................. Equation 4

Given: R = 13.4375 Ω, R' = 1.6 Ω, α = 6.4×10⁻³ K⁻¹

Substitute into equation 4

Δθ = (13.4375-1.6)/(1.6×0.0064)

Δθ = 11.8375/0.01024

Δθ = 1156.01 °C

But,

Δθ = T₂-T₁

T₂ = T₁+Δθ

Where T₂ and T₁ = Final and initial temperature respectively.

T₂ = 20+1156.01

T₂ = 1176.01 °C

5 0
3 years ago
Read 2 more answers
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