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asambeis [7]
3 years ago
9

A 12 volt battery in a motor vehicle is capable of supplying the starter motor with 150 A. It is noticed that the terminal volta

ge of the battery drops to 10 V when the engine is cranked over with the starter motor. Determine the internal resistance of the 12 volt battery.
Physics
1 answer:
qwelly [4]3 years ago
4 0

Answer:0.0133 \Omega

Explanation:

Given

Voltage=12 V

Current(I)=150 A

V_{terminal}=10 V

r_{internal}=\frac{\Delta V}{I}

r_{internal}=\frac{12-10}{150}

r_{internal}=\frac{2}{150}=0.0133 \Omega

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A power plant burns fuel to convert water to steam. During the process, 83% of the heat produced is transferred. Of the heat car
9966 [12]

Answer: 33.2\%

Explanation:

Given

83% heat is transferred and out of this, only 40% is converted to the mechanical energy of spinning turbine.

The overall efficiency is the product of all the efficiency i.e.

\eta_t=0.83\times 0.4\\\eta_t=0.332\ \text{or}\ 33.2\%

6 0
3 years ago
The ball is launched with total initial energy Einitial = mgho. When it reaches the bottom of the hoops, it now has a potential
NeX [460]

Answer:

  K = m g (h₀ - h_plat)

Explanation:

Let's use energy conservation to solve this problem, write the energy at two points of interest

Initial. Higher

          E_initial = m g h₀

Final. Lower

          E_end = K + Ep

          E_end = ½ m v² + m g h_plat

Energy is conserved

          E_initial = E_end

          m g h₀ = K + m g h_plat

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4 0
3 years ago
KE=0.5.m.v2 or PE=m.g.h
LUCKY_DIMON [66]

Answer:

1. 37.8J

2. 18 Billion Joules, 18 Gigajoules

3. 9.81 Billion Joules, 9.81 Gigajoules

Explanation:

Use the formulas provided,

KE=(1/2)mv^2 and PE=mgh, noting that g=9.81

7 0
3 years ago
Read 2 more answers
3. When a magnetic sector instrument was operated with an accelerating voltage of 4.50*103 V, a field of 0.251 T was required to
Damm [24]

Answer:

The answer is "4,500 - 225 \ V".

Explanation:

Using formula for calculating the Voltage:

M_1=12.5\\\\M_2=250\\\\V_1=4,500 \\\\\bold{\text{Formula: }}\\\\\to \bold{\frac{m_1}{m_2}=\frac{V_2}{V_1}}\\\\\to \frac{12.5}{250}=\frac{V_2}{4,500}\\\\\to 0.05=\frac{v_2}{4,500}\\\\\to 0.05\times 4,500= V_2\\\\\to V_2=225\\\\

Hence the range of accelerating in voltage is 4,500 - 225 \ V

8 0
3 years ago
Help me please i need to get a good grade!
timama [110]
True. His third law is that there is an equal and opposite reaction for every action.
5 0
4 years ago
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