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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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English or metric <br> What is 16in to cm
il63 [147K]
40.64 cm is the answer to your question.
6 0
3 years ago
I need help on this question!!!
Irina-Kira [14]
2500
--------=3125
0.8

This can be can be viewed as compression due to the direction of the arrows.
7 0
3 years ago
A 1022kg Caprice car stopped at an intersection is rear-ended by a 1620kg ranger truck moving with a speed of 14.5m/s. If the ca
Alika [10]

Answer:

Explanation:

mass of car, m = 1022 kg

mass of truck, M = 1620 kg

initial velocity of truck, U = 14.5 m/s

initial velocity of car, u = 0 m/s

Let the final velocity of car is v and the final velocity of truck is V.

Collision is elastic, so the coefficient of restitution, e = 1

Use conservation of momentum

initial momentum of car + initial momentum of truck = final momentum of car + final momentum of truck

m x u + M x U = m x v + M x V

0 + 1620 x 14.5 = 1022 v + 1620 V

23490 = 1022 v + 1620 V ..... (1)

Use the formula of coefficient of restitution

e = \frac{V_{1}-V_{2}}{u_{2}-u_{1}}

1 (14.5 - 0) = v - V

14.5 = v - V

V = v - 14.5 .... (2)

Put in equation (1)

23490 = 1022 v + 1620 (v - 14.5)

23490 = 1022 v + 1620 v - 23490

46980 = 2642 v

v = 17.8 m/s

Put in equation (2)

V = 17.8 - 14.5

V = 3.3 m/s

Thus, the speed of car is 17.8 m/s and the velocity of truck is 3.3 m/s after collision.

8 0
3 years ago
The tension of a guitar string is increased by 40%. By what factor odes the fundamental frequency of vibration change? a. 1.13 b
bogdanovich [222]

Answer:

<h3> b. 1.18</h3>

Explanation:

The fundamental frequency in string is expressed as;

F1 = 1/2L√T/m .... 1

L is the length of the string

T is the tension

m is the mass per unit length

If the tension is increased by 40%, the new tension will be;

T2 = T + 40%T

T2 = T + 0.4T

T2 = 1.4T

The new fundamental frequency will be;

F2 = 1/2L√1.4T/m ..... 2

Divide 1 by 2;

F2/F = (1/2L√1.4T/m)/1/2L√T/m)+

F2/F = √1.4T/m ÷ √T/m

F2/F = √1.4T/√m ×√m/√T

F2/F = √1.4T/√T

F2/F = 1.18√T/√T

F2/F = 1.18

F2 = 1.18F

Hence the fundamental frequency of vibration changes by a factor of 1.18

8 0
3 years ago
The following equation is an example of decay?<br><br> 232/90 TH---4/2 HE +228/88 RA?
DaniilM [7]

Answer:

Alpha decay

Explanation:

  • Alpha decay is one of the three major types of decays, others being, beta decay and gamma decay.
  • <em><u>When a radioactive isotope undergoes alpha decay it emits alpha particles. An alpha particle is equivalent to the nucleus of Helium atom.</u></em>
  • <em><u>Therefore, an atom undergoing decay, its atomic mass is decreased by 4 and its atomic number is decreased by 2. </u></em>
  • Thus, since 232/90 Th, has undergone alpha decay its mass number is reduced by 4 to 228 and its atomic number by 2 to 88, and becomes 228/88 Ra.
5 0
3 years ago
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