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s2008m [1.1K]
2 years ago
6

State Newton's first law of motion​

Physics
2 answers:
Ksivusya [100]2 years ago
6 0

Answer:

His first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia.

Explanation:

Leokris [45]2 years ago
5 0

Answer:

It states that: If a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it acted upon by a force

Explanation:

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5) A person holds a 1.7 kg bucket and lets it move down at
Elena L [17]

Answer:

12.5J

Explanation:

Given parameters:

Mass of bucket  = 1.7kg

Height  = 75cm  = 0.75m

Unknown;

Work done on the bucket by the person  = ?

Solution:

To solve this problem, we use the work done equation;

  Work done  = force x distance  = mgh

 m is the mass  

  g is the acceleration due to gravity

  h is the height

Now, insert parameters and solve ;

     Work done  = 1.7 x 9.8 x 0.75  = 12.5J

7 0
2 years ago
MATHPHYS CAN U HELP ME PLEASE
ludmilkaskok [199]

Explanation:

(1) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.041 kg) (2090 J/kg/°C) (0°C − (-11°C)) = 942.59 J

The heat added to melt the ice is:

q = mL = (0.041 kg) (3.33×10⁵ J/kg) = 13,653 J

The heat added to warm the water to 100°C is:

q = mCΔT = (0.041 kg) (4186 J/kg/°C) (100°C − 0°C) = 17,162.6 J

The heat added to evaporate the water is:

q = mL = (0.041 kg) (2.26×10⁶ J/kg) = 92,660 J

The heat added to warm the steam to 115°C is:

q = mCΔT = (0.041 kg) (2010 J/kg/°C) (115°C − 100°C) = 1236.15 J

The total heat needed is:

q = 942.59 J + 13,653 J + 17,162.6 J + 92,660 J + 1236.15 J

q = 125,654.34 J

(2) When the first two are mixed:

m C₁ (T₁ − T) + m C₂ (T₂ − T) = 0

C₁ (T₁ − T) + C₂ (T₂ − T) = 0

C₁ (6 − 11) + C₂ (25 − 11) = 0

-5 C₁ + 14 C₂ = 0

C₁ = 2.8 C₂

When the second and third are mixed:

m C₂ (T₂ − T) + m C₃ (T₃ − T) = 0

C₂ (T₂ − T) + C₃ (T₃ − T) = 0

C₂ (25 − 33) + C₃ (37 − 33) = 0

-8 C₂ + 4 C₃ = 0

C₂ = 0.5 C₃

Substituting:

C₁ = 2.8 (0.5 C₃)

C₁ = 1.4 C₃

When the first and third are mixed:

m C₁ (T₁ − T) + m C₃ (T₃ − T) = 0

C₁ (T₁ − T) + C₃ (T₃ − T) = 0

(1.4 C₃) (6 − T) + C₃ (37 − T) = 0

(1.4) (6 − T) + 37 − T = 0

8.4 − 1.4T + 37 − T = 0

2.4T = 45.4

T = 18.9°C

(3) Heat gained by the ice = heat lost by the tea

mL + mCΔT = -mCΔT

m (3.33×10⁵ J/kg) + m (2090 J/kg/°C) (30.8°C − 0°C) = -(0.176 kg) (4186 J/kg/°C) (30.8°C − 32.8°C)

m (397372 J/kg) = 1473.472 J

m = 0.004 kg

m = 4 g

4 grams of ice is melted and warmed to the final temperature, which leaves 128 grams unmelted.

(4) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.028 kg) (2090 J/kg/°C) (0°C − (-67°C)) = 3920.84 J

The heat added to melt the ice is:

q = mL = (0.028 kg) (3.33×10⁵ J/kg) = 9324 J

The heat added to warm the melted ice to T is:

q = mCΔT = (0.028 kg) (4186 J/kg/°C) (T − 0°C) = (117.208 J/°C) T

The heat removed to cool the water to T is:

q = -mCΔT = -(0.505 kg) (4186 J/kg/°C) (T − 27°C)

q = (2113.93 J/°C) (27°C − T) = 57076.11 J − (2113.93 J/°C) T

The heat removed to cool the copper to T is:

q = -mCΔT = -(0.092 kg) (387 J/kg/°C) (T − 27°C)

q = (35.604 J/°C) (27°C − T) = 961.308 J − (35.604 J/°C) T

Therefore:

3920.84 J + 9324 J + (117.208 J/°C) T = 57076.11 J − (2113.93 J/°C) T + 961.308 J − (35.604 J/°C) T

13244.84 J + (117.208 J/°C) T = 58037.418 J − (2149.534 J/°C) T

(2266.742 J/°C) T = 44792.58 J

T = 19.8°C

(5) Kinetic energy of the hammer = heat absorbed by ice

KE = q

½ mv² = mL

½ (0.8 kg) (0.9 m/s)² = m (80 cal/g × 4.186 J/cal × 1000 g/kg)

m = 9.68×10⁻⁷ kg

m = 9.68×10⁻⁴ g

(6) Heat rate = thermal conductivity × area × temperature difference / thickness

q' = kAΔT / t

q' = (1.09 W/m/°C) (4.5 m × 9 m) (10°C − 4°C) / (0.09 m)

q' = 2943 W

After 10.7 hours, the amount of heat transferred is:

q = (2943 J/s) (10.7 h × 3600 s/h)

q = 1.13×10⁸ J

q = 113 MJ

6 0
3 years ago
This is the si unit of current
skelet666 [1.2K]

Answer:

ampere

and it is denoted by A

3 0
2 years ago
Read 2 more answers
Two red and two overlapping balls in the center are surrounded by a green, fuzzy, circular cloud with a white line running throu
PolarNik [594]

Answer:

  • A. The line leading a bracket around the balls, labeled A, represents the nucleus of the atom.

  • B. The white line, labeled B, is signaling the first main energy level, as per the quantum model of the atom.

Explanation:

You are describing the model of an atom with two protons, two neutrons, the electron density (orbital), and two electrons.

Then, you want to know what the <em>labels A</em> and <em>B</em> represent.

The nucleus of the atom is at the center, where protons and neutrons are placed. This is supported by three models: Rutherford's model, Bohr's model, and the (currently accepted) quantum model.

Hence, the red balls and the two overlapping balls in the center are representing the protons and neutrons.

The <em>line leading a bracket around the balls</em>, labeled A, represents the nucleus of the atom, since it is signaling both protons and neutrons in the nucleus.

The green fuzzy circular cloud depicts the region where the electrons are: it is fuzzy in correspondence to the quantum model, which states the electrons are not in a fixed position but in a region around the atom. That region is named orbitals. As per the model you cannot tell the exact position of the electron, and that is what the fuzzy cloud means.

In the the Bohr's model that white line,  which should be circular, where the electrons are, depicts the orbit of the electrons, and the shell, which were idenfified with letters L, M, N, K. The first shell was the L shell.

In the quantum  model, that line corresponds to the principal quantum number n, which is the main energy level. Hence, the line leading to a bracket overlapping the white line depicts the main energy level.

The atom is neutral because it contains the same number of electrons aroud the nucleus as protons inside the nucleus.

3 0
2 years ago
Read 2 more answers
In comparison with other ocean basins, major sedimentary features such as continental rises and abyssal plains are relatively ra
Maksim231197 [3]

Answer:

Why are continental rises and abyssal plains relatively rare in the Pacific? This is because the extensive system of trenches along the active margins of the Pacific, trap much of the sediments flowing off the continents, preventing them from building the broad, flat abyssal plains typical of the Atlantic ocean basins.

3 0
2 years ago
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