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Alex73 [517]
3 years ago
6

During a compression test, a cylinder has 40 percent of the specified compression reading. When the technician performs a wet te

st, the compression reading on this cylinder has 75 percent of the specified reading. The cause of the low compression reading could be:
Physics
1 answer:
Harrizon [31]3 years ago
8 0

Answer:

Worn out piston rings

Explanation:

Here there is increase in the pressure during the wet compression test. Since, the pressure reading from 40 percent of the specified compression reading raised up to 75% of the specified reading.

The cause of the low compression reading could be Worn out piston rings that is allowing the pressured to be released and hence the compression reading is low.

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Gravity largely depends on the comparison of two objects; it's why you have the equation F= (GMm)/r^2. On Earth, you have different altitudes that, with the formula, will give different results for gravity because the radius is different everywhere. This difference on calculations, however, are seen to be miniscule. We know gravity as 9.81 m/s^2 but it might be different by thousandths or hundreds of thousandths of a decimal.
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Nancy is sailing her boat toward Sam's boat at 5
Bas_tet [7]

Answer:

Nora's boat is moving towards Sam at 5 km/hr

Explanation:

The question says that Nora is few meters behind Nancy and is still with respect to her that means the relative velocity between Nora and Nancy is zero

Vrel = 5 - Vnora= 0 ⇒ Vnora = 5km/hr

Pictorially we can represent the given condition as:

Nora-------few meters------Nancy -----------------    Sam

5km/hr                                         5km/hr →

Hence,  Nora's boat is moving towards Sam at 5km/hr.

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2 years ago
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The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict
nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

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Answer:

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Explanation:

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Answer:

down below

Explanation:

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s = 50/25

s = 2

Therefore, the object was falling at a rate of 2 meters per second.

Best of Luck!

6 0
2 years ago
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