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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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the electric current in a wire is 1.5 A. How many electrons flow past a given point in a time of 2 s?​
34kurt

Answer:

The quantity of electrons that flows past a given point is 3.0 C.

Explanation:

An electric current (I) is the ratio of the quantity of charges (Q) that flows through a point to the time taken (t).

i.e            I = \frac{Q}{t}

It is measured in Ampere's by the use of an ammeter in the laboratory. The quantity of charge that flow through a given point is measured in Coulombs, while time is measured in seconds.

Given that;     I = 1.5A and t = 2s, find Q.

                          Q = It

                             = 1.5 × 2

                             = 3.0 C

The quantity of electrons that flows past a given point is 3.0 C.

6 0
3 years ago
A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows: ρ(r)=ρ0(1−r/r) for r
Nadusha1986 [10]

A)<span>
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² - r³/R) dr 
which when integrated from 0 to r is 
total charge = 4π * ρ0 (r³/3 + r^4/(4R)) 
and when r = R our total charge is 
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3 
and after substituting ρ0 = 3Q / πR³ we have 
total charge = Q ◄ 

B) E = kQ/d² 
since the distribution is symmetric spherically 

C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 - r/R)dr 
so 
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is 
and after substituting for ρ0 is 
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4)) 
which could be expressed other ways. 

D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that 
r = R for a min/max (and we know it's a max since r = 0 is a min). 

<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>

4 0
3 years ago
Magnetism/ magnetic field ana magnetic forces
katovenus [111]

Answer:

Magnetism is a physical phenomenon that manifests itself in a force acting between magnets or other magnetized or magnetisable objects, and a force acting on moving electric charges, such as in current-carrying cables. The force action takes place by means of a magnetic field, which is generated by the objects themselves or otherwise. There are natural and artificial magnets. All magnets have two poles called the north pole and the south pole. The north pole of one magnet repels the north pole of another magnet and attracts the south pole of another magnet; the same with south poles.  

6 0
3 years ago
A thin wire of length (2 m) and (1 mm2 ) cross-section area is clamped horizontally between two walls, a weight of (10 kg) is hu
allsm [11]

Answer:

Young modulus  = 9.8 × 10⁹ N/m²

Explanation:

From the information given:

Stress = F/A

Stress = (10 × 9.8) / 0.001²

Stress = 9.8× 10⁷ N/m²

Strain = increase in length / initial length of wire

Strain = 0.02/ 2

Strain = 0.01

Now;

The Young modulus (Y)= stress/strain

Young modulus  = (9.8 × 10⁷ N/m²) /  0.01

Young modulus  = 9.8 × 10⁹ N/m²

4 0
3 years ago
Is this correct and plz j answer the right one I can’t have an explanation rn I’m in a rush thx so much
vova2212 [387]
I pretty sure it’s C
6 0
3 years ago
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