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Tatiana [17]
3 years ago
5

Conduction, convection, and radiation are ways by which,

Physics
2 answers:
Amiraneli [1.4K]3 years ago
6 0

the answer is THERMAL

mrs_skeptik [129]3 years ago
5 0

Answer:

Conduction, convection, and radiation are ways by which thermal energy can be transferred.

Explanation:

just answered it

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How do the meetings of the words exercise and fitness differ?
ICE Princess25 [194]
Exercise is the activity and fitness is a lifestyle and done with time
7 0
2 years ago
(URGENT) A ball rolls off a ledge. Its velocity is 7.70m/s in a horizontal direction. It falls on the floor 1.60m below the ledg
elena-s [515]

Answer:

the horizontal distance is 4.355 meters

Explanation:

The computation of the horizontal distance while travelling in the air is shown below:

Data provided in the question is as follows

Velocity = u = 7.70 m/s

H = 1.60 m

R = horizontal direction

Based on the above information

As we know that

R = u × time

where,

Time = \sqrt{\frac{2H}{g} }

So,

= 7.70\times \sqrt{\frac{2\times 1.60}{10} }

= 4.355 meters

hence, the horizontal distance is 4.355 meters

6 0
3 years ago
Which biome has the most extreme conditions
Lelu [443]
Well this all depends on the region you would like to know about. One biome would be The Tundra. This biome is a very bitter cold. Some times the temperature can drop to -45f! So your answer more than likely would be Tundra.

Have a wonderful day user!
3 0
3 years ago
The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ
prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

          \nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})

                           = (30)^{2} + 2(-0.25(412 - 0))

                           = 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

          y = 200 e^{\frac{x}{1000}}

      \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}

Now, second derivative of the train is calculated as follows.

         \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}      

       \frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}    

Radius of curvature of the train is as follows.  

   \rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}

               = \frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}

              = 3808.96 m

Now, we will calculate the normal component of the train as follows.

            a_{n} = \frac{\nu^{2}_{B}}{\rho}

                        = \frac{(26.34)^{2}}{3808.96}

                        = 0.1822 m/s^{2}

The magnitude of acceleration of train is calculated as follows.

            a = \sqrt{(a_{t})^{2} + (a_{n})^{2}}

               = \sqrt{(-0.25)^{2} + (0.1822)^{2}}

              = 0.309 m/s^{2}

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is 0.309 m/s^{2}.

6 0
3 years ago
Atom in the ground state is said to be​
koban [17]

Answer:

Ground-state atom

Explanation:

When an atom is not excited, it is in its ground-state, which we refer as "standard" or "normal" state.

(Hopefully that helped you!)

GOOD LUCK

Astrophysicist Dr. D

3 0
3 years ago
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