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antoniya [11.8K]
3 years ago
12

Fuel is combusted in the _____ stroke of a four-stroke engine.

Physics
1 answer:
Margarita [4]3 years ago
3 0

Fuel is combusted in the <span>power </span>stroke of a four-stroke engine since this is where i<span>gnition of the fuel occurs and this stroke makes the substance gases to thrust in contact with the pistons.

</span>To add, the power stroke is the stage of the cycle of an internal combustion engine in which the piston is driven outward by the expansion of gases.

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Three collinear forces,F1=45N west,F2=63N east and an unknown force F3 are applied to an object.The resultant force of the three
Greeley [361]

Take east to be the positive direction. Then the resultant force from adding <em>F</em>₁ and <em>F</em>₂ is

<em>F</em>₁ + <em>F</em>₂ = (-45 N) + 63 N = 18 N

which is positive, so it's directed east.

To this we add a third force <em>F</em>₃ such that the resultant is 12 N pointing west, making it negative, so that

18 N + <em>F</em>₃ = -12 N

<em>F</em>₃ = -30 N

So <em>F</em>₃ has a magnitude of 30 N and points west.

6 0
3 years ago
If a nucleus decays by gamma decay to a daughter nucleus, which of the following statements about this decay are correct? (There
GarryVolchara [31]

Answer: Option (b) is the correct answer.

Explanation:

A gamma particle is basically a photon of electromagnetic radiation with a short wavelength.

Symbol of a gamma particle is ^{0}_{0}\gamma. Hence, charge on a gamma particle is also 0.

For example, ^{234}_{91}Pa \rightarrow ^{234}_{91}Pa + ^{0}_{0}\gamma + Energy

So, when a nucleus decays by gamma decay to a daughter nucleus then there will occur no change in the number of protons and neutrons of the parent atom but there will be loss of energy as a nuclear reaction has occurred.

Thus, we can conclude that the statement daughter nucleus has the same number of nucleons as the original nucleus., is correct about if  a nucleus decays by gamma decay to a daughter nucleus.

5 0
3 years ago
What force is needed to accelerate a 0.5kg football at a rate of 40m/s
GREYUIT [131]
Force=mass x acceleration
f= 0.5 x40
f=20N
6 0
3 years ago
A transformer is intended to decrease the value of the alternating current from 500 amperes to 25 amperes. The primary coil cont
EastWind [94]

Answer:

The number of turns in secondary coil is 4000

Explanation:

Given:

Current in primary coil I_{P} = 500 A

Current in secondary coil I_{S} = 25 A

Number of turns in primary coil N_{P} = 200

In case of transformer the relation between current and number of turns is given by,

     \frac{N_{S} }{N_{P}  } = \frac{I_{P} }{I_{S} }

For finding number of turns in secondary coil,

     N_{S} = \frac{I_{P} }{I_{S} }  N_{P}

     N_{S} = \frac{500}{25} \times 200

     N_{S} = 4000

Therefore, the number of turns in secondary coil is 4000

5 0
3 years ago
A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across
NARA [144]

Answer:

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:  

A=\frac{\pi }{4}d^{2}  \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\

a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper  </em>

<em>    wire: </em>

  L= 2.00 m

From Table  Copper Resistivity p= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

R=\frac{pL}{A}

   =0.0165Ω

The Potential difference across the copper wire is:  

V=IR

 =2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:  

R=\frac{pL}{A}

   =0.014Ω

The Potential difference across the Silver wire is:  

V=IR

 =1.75*10^-4

4 0
3 years ago
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