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frez [133]
3 years ago
8

Convert 125 mL to L

Physics
2 answers:
muminat3 years ago
6 0

Answer:

0.125

Explanation:

divide by 1 000 to convert mL to liters

Alexandra [31]3 years ago
3 0

Answer:

1 L = 1000 mL

125 mL = 125/1000 = 0.125 L

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Your car's 30.0 W headlight and 2.50 kW starter are ordinarily connected in parallel in a 12.0 V system. What power (in W) would
Tatiana [17]

Answer:

<h3>The power of headlight in series connection is 29.64 W</h3>

Explanation:

Given :

Power of headlight P_{1} = 30 W

Power of starter P_{2} = 2500 W

Voltage of headlight and starter V = 12 V

From equation of power,

 P = \frac{V^{2} }{R}

 R = \frac{V^{2} }{P}

For finding the resistance of headlight and starter,

⇒ For headlight,

 R_{1}  = \frac{144}{30} = 4.8 Ω

⇒ For starter,

R_{2} = \frac{144}{2500} = 0.057 Ω

Since equivalent resistance,

R_{eq} = R_{1} + R_{2} + ........

R_{eq} = 4.8 +0.057 = 4.857 Ω

So power in series is given by,

 P_{s } = \frac{V^{2} }{R_{eq} }  = \frac{144}{4.857}

 P_{s } = 29.64 W

8 0
3 years ago
For a reaction to occur what must happen to the energy in order to break the chemical bond
Andrej [43]
Energy needs to realease
6 0
3 years ago
Calculate the average velocity of a dancer who moves 5 m toward the left of the stage over the course of 15 s. ** Velocity = dis
Alex73 [517]

Answer:

B

Explanation:

Velocity=disp/time

V=5m/15s

V=1/3 m/s

3 0
3 years ago
A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
Mila [183]

Answer:

v=6.65m/sec

Explanation:

From the Question we are told that:

Mass m=97.6

Coefficient of kinetic friction  \mu k=0.555

Generally the equation for Frictional force is mathematically given by

 F=\mu mg

 F=0.555*97.6*9.8

 F=531.388N

Generally the  Newton's equation for Acceleration due to Friction force is mathematically given by

 a_f=-\mu g

 a_f=-0.555 *9.81

 a_f=-54455m/sec^2

Therefore

 v=u-at

 v=0+5.45*1.22

 v=6.65m/sec

4 0
3 years ago
Consider a Carnot heat-engine cycle executed in a closed system using 0.025 kg of steam as the working fluid. It is known that t
ArbitrLikvidat [17]

Answer:

The temperature of the steam during the heat rejection process is 42.5°C

Explanation:

Given the data in the question;

the maximum temperature T_H in the cycle is twice the minimum absolute temperature T_L in the cycle

T_H  = 0.5T_L

now, we find the efficiency of the Carnot cycle engine

η_{th = 1 - T_L/T_H

η_{th = 1 - T_L/0.5T_L

η_{th = 0.5

the efficiency of the Carnot heat engine can be expressed as;

η_{th = 1 - W_{net/Q_H

where W_{net is net work done, Q_H is is the heat supplied

we substitute

0.5 = 60 / Q_H

Q_H = 60 / 0.5

Q_H = 120 kJ

Now, we apply the first law of thermodynamics to the system

W_{net = Q_H - Q_L

60 = 120 -  Q_L

Q_L = 60 kJ

now, the amount of heat rejection per kg of steam is;

q_L = Q_L/m

we substitute

q_L = 60/0.025

q_L = 2400 kJ/kg

which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)

q_L = h_{fg = 2400 kJ/kg

now, at  h_{fg  = 2400 kJ/kg from saturated water tables;

T_L = 40 + ( 45 - 40 ) ( \frac{2400-2406.0}{2394.0-2406.0}\\} )

T_L = 40 + (5) × (0.5)

T_L = 40 + 2.5

T_L = 42.5°C  

Therefore, The temperature of the steam during the heat rejection process is 42.5°C  

4 0
3 years ago
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