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kompoz [17]
3 years ago
14

Heat energy always moves from a _ _ _ place to a _ _ _ _ _ _ place. The three methods of heat transfer are?

Physics
1 answer:
Viefleur [7K]3 years ago
8 0

Answer:Heat can travel from one place to another in three ways: Conduction, Convection and Radiation.

Explanation:

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Two forces Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts and Upper FSubscript Upper B Baseline Overscript
Pavel [41]

Answer:

Part a)

F_A = 4.59 N

Part B)

F_B = 1.28 N

Explanation:

As we know that when both the forces are acting on the object in same direction then we will have

F_A + F_B = ma

as we know that

a = 0.554 m/s^2

m = 10.6 kg

now we will have

F_A + F_B = 10.6(0.554)

F_A + F_B = 5.87 N

Now two forces are in opposite direction then we have

F_A - F_B = 10.6(0.313)

F_A - F_B = 3.32 N

Part A)

Now we will have from above two equation

F_A = 4.59 N

Part B)

Similarly for other force we have

F_B = 1.28 N

5 0
2 years ago
What information and measurements would you need to calculate the rate of movement?
aliina [53]
When you talk about rate, you will expect that it will be in terms of a time unit. It measures how fast it is going. So, you would expect that the denominator is in time units. For the movement, you can measure this with either distance or velocity.

So, for the first variety, you would need distance and time to measure the rate of how far you go at a certain time. It is also called as velocity. For the second variety, you would need velocity and time to measure the rate of how fast you are going at a certain interval. It is also called as acceleration.
8 0
3 years ago
Could someone summarize these into a paragraph? i'll give brainliest<br> DUE TODAY!!
slamgirl [31]

Answer:

I can't see the picture

Explanation:

Im sorry can you right whats on the picture down in the comments plz.

6 0
2 years ago
Read 2 more answers
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
2 years ago
Emboldened by the success of their late night keg pull in Exercise 61 above, our intrepid young scholars have decided to pay hom
alexdok [17]

Answer is answer

XD                                ssssss

5 0
3 years ago
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