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vagabundo [1.1K]
3 years ago
10

if large amount of heat are transferred to a solid, what will probably happen to the state of the solid?

Physics
2 answers:
Anna007 [38]3 years ago
3 0
The solid can melt down to a liquid like state, and if you keep applying heat, it would eventually become a gas state.
Korvikt [17]3 years ago
3 0
The solid will convert its state to liquid. It then may also convert to gas if the heat make it reaches boiling point.
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In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward
Vladimir79 [104]

Answer:

The answer is "1.01 \times 10^{-13}"

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

KE=U=\frac{kqq'}{r}\\\\

Calculating the closest distance:

\to r=\frac{kqq'}{KE}\\\\

=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\

=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}

5 0
3 years ago
A hummingbird can a flutter its wings 4800 times per minute what is the frequency of wing flutters per second
Scilla [17]

the answer i got is 80 i hope this helpss!!!!

7 0
4 years ago
Read 2 more answers
Which applied force will allow a 7.65 kg block of ice to begin sliding on a sheet of ice? The block of ice has a kinetic coeffic
Anni [7]

Answer:

force for start moving is 7.49 N

force for moving constant velocity 2.25 N

Explanation:

given data

mass = 7.65 kg

kinetic coefficient of friction = 0.030

static coefficient of friction = 0.10

solution

we get here first weight of block of ice that is

weight of block of ice = mass  ×  g

weight of block of ice = 7.65 × 9.8 = 74.97 N

so here Ff = Fa

so for force for start moving is

Fa = weight × static coefficient of friction  

Fa = 74.97 × 0.10

Fa =  7.49 N

and

force for moving constant velocity is

Fa =  weight × kinetic coefficient of friction

Fa = 74.97 × 0.030

Fa = 2.25 N

7 0
3 years ago
Read 2 more answers
Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia
sergejj [24]

Answer:

0.176m from the flagpole, westward.

Explanation:

Let the Eastward be the positive direction. So initially runner A is at position -6km, running with velocity of 9km/h while runner B is at position 5km running at a velocity of -8km/h. We can conduct the following equation for their distances over the same time t

s_A = -6 + 9t

s_B = 5 - 8t

When A an B meets, they are at the same position and at the same time. So

s_A = s_B

-6 +9t = 5 - 8t

17t = 5 + 6 = 11

t = 11/17 = 0.647 s

s_A = -6 + 9*0.647 = -0.176 m

So where they meet is 0.176m from the flagpole, westward.

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How is a warm front differnt from a cold front
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Bruh... but a warm front is different because it brings warm ness and sun, a cold front brings clouds , cloudy weather, and cold weather
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