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o-na [289]
2 years ago
6

The temperature of an air parcel and the kinetic energy of an air parcel are ___________ related. this means that as the tempera

ture of the air parcel increases, the kinetic energy ________________.
Physics
1 answer:
torisob [31]2 years ago
8 0

The temperature of an air parcel and the kinetic energy of an air parcel are directly related. this means that as the temperature of the air parcel increases, the kinetic energy increases.

<h3>What is temperature?</h3>

Temperature is the measure of degree of hotness or coldness of a body.

Temperature is also the measure of the average kinetic energy of a system.

When the heat is applied to body, its temperature increases as the body gains heat.

Thus, the temperature of an air parcel and the kinetic energy of an air parcel are directly related. this means that as the temperature of the air parcel increases, the kinetic energy increases.

Learn more about temperature here: brainly.com/question/25677592

#SPJ1

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a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
3 years ago
Which type of radiation travels at the speed of light and penetrates matter easily?
Alina [70]
I think the answer is Gamma Rays
5 0
3 years ago
If you travel 1.7 km north from your house at noon, and at 6:00 PM you travel 5.4 km south, what is your displacement? 3.7 km no
Yakvenalex [24]

<u>Answer</u>

3.7 Km south


<u>Explanation</u>

The definition of displacement is the distance traveled in a specific direction. It is the vector quantity. We add displacements like the way we add vectors.

Taking the direction towards North to be positive (+1.7 Km), the distance towards south would be negative (-5.4 Km).

Now lets add the two values.

(+1.7) + (-5.4) = 1.7 - 5.4

                    = - 3.7 Km      But negative was towards south.

∴ Answer = 3.7 Km south.


6 0
3 years ago
How much work must be done to stop a 1100-kg car traveling at 112 km/h?(Hint: You will need to convert the speed first.)Answer:
zimovet [89]

According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:

W=\Delta K

Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:

K=\frac{1}{2}mv^2

Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:

\begin{gathered} K=\frac{1}{2}(1100kg)\left(112\frac{km}{h}\times\frac{1\frac{m}{s}}{3.6\frac{km}{h}}\right)^2=532,345.679...J \\  \\ \therefore K\approx532,346J \end{gathered}

Therefore, the answer is: 532,346 J.

5 0
1 year ago
Introduction to Forces<br> Warm-Up Active<br> How do forces affect the motion of an object?
In-s [12.5K]

Answer:

Forces can affect an object.

Balanced forces allow an object to continue moving at a constant motion (law of inertia).

Unbalanced forces cause a change in motion.

4 0
3 years ago
Read 2 more answers
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