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3 years ago

Ixchelt burns her tongue when she takes a sip of hot coffee from her mug. Which part of this example represents heat?

Physics

2 answers:

bekas [8.4K]3 years ago

6 0

Answer:

Explanation:

"The thermal energy moving from her coffee to the tongue" represent the heat.

Here coffee is at high temperature while tongue is at low temperature, when Ixchelt tongue make contact with coffee then thermal energy of coffee is absorbed by tongue and tongue gets burned.

As heat always from high Potential to low that is why heat is absorbed by tongue.

meriva3 years ago

6 0

Answer:

B is the answer, if you dont have time to read through an explanation.

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MGS and later orbiters found spectral evidence for what minerals on the Martian surface? Check all that apply.

slamgirl [31]

Mars Global Surveyors (MGS) and later orbiters found the following minerals on the Martian surface;

Phyllosilicates

Carbonate
Sulfates
Iron oxide

The Mars Global Surveyors (MGS) and later orbiters suggest that the Martian crust contains a higher percentage of volatile elements such as Sulphur and chlorine than the Earth's crust does.

These scientists also conclude that the most abundant chemical elements in the Martian crust are those found in Igneous rock.

These elements include the following;

Silicon,
Oxygen,
Iron,
Magnesium,
Aluminum,
Calcium, and
Potassium.

They also, suggest that hydrogen is found in ice (water) while carbon is found in carbon dioxide and carbonates.

From the given options the minerals found in Martian surface include;

Phyllosilicates ------ these are sheet of silicate minerals

Carbonate
Sulfates
iron oxide

Learn more here: brainly.com/question/20470323

6 0

2 years ago

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial

elena55 [62]

Answer : The partial pressure of $SO_3$ is, 67.009 atm

Solution : Given,

Partial pressure of $SO_2$ at equilibrium = 30.6 atm

Partial pressure of $O_2$ at equilibrium = 13.9 atm

Equilibrium constant = $K_p=0.345$

The given balanced equilibrium reaction is,

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}$

Now put all the values of partial pressure, we get

$0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}$

$p_{SO_3}=67.009atm$

Therefore, the partial pressure of $SO_3$ is, 67.009 atm

6 0

2 years ago

the refractive index of cooking oil is 1.47, and the refractive index of water is 1.33. A thick layer of cooking oil is floating

VARVARA [1.3K]

When light travels from a medium with higher refractive index to a medium with lower refractive index, there is a critical angle after which all the light is reflected (so, there is no refraction).

The value of this critical angle can be derived by Snell's law, and it is equal to
$\theta_C = \arcsin ( \frac{n_2}{n_1} )$
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.

In our problem, n1=1.47 and n2=1.33, so the critical angle is
$\theta_C = \arcsin( \frac{1.33}{1.47} )=\arcsin (0.91)=65^{\circ}$

4 0

2 years ago

What is the acceleration of an object that goes from 45m/s to 10 m/s in 5 seconds?

kipiarov [429]

Answer:

$\boxed {\boxed {\sf a= -7 \ m/s^2}}$

Explanation:

Acceleration is the change in velocity over time.

$a= \frac {v_f-v_i}{t}$

The object accelerates <em>from</em> 45 meters per second <em>to </em>10 meters per second in 5 seconds. Therefore,

$v_f=10 \ m/s \\v_i= 45 \ m/s \\t= 5 \ s$

Substitute the values into the formula.

$a= \frac{ 10 \ m/s - 45 \ m/s}{5 \ s}$

Solve the numerator.

$a= \frac { -35 \ m/s}{5 \ s}$

Divide.

$a= -7 \ m/s/s$

$a= -7 \ m/s^2$

The acceleration of the object is -7 meters per square second. The acceleration is negative because the object's velocity decreases and the object slows down.

5 0

2 years ago

Two blocks of masses m and M are connected by a string and pass over a frictionless pulley. Mass m hangs vertically, and mass M

horsena [70]

Answer:

$sin\theta - \mu_k cos\theta = \frac{m}{M}$