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marysya [2.9K]
3 years ago
7

A new mechanic foolishly connects an ammeter with 0.1 Ω resistance directly across a 12-V car battery with internal resistance o

f 0.01 Ω. What’s the power dissipation in the meter? (No wonder it gets destroyed!)
Physics
1 answer:
NemiM [27]3 years ago
8 0

Answer:

The power dissipated by the meter is 1188W

Explanation:

Here we have a circuit constituted with a power source and two resistors in series, we can calculate the power dissipated by the meter using the following formula:

P=I^2R_m

We first need to fin the current going through the circuit:

I=\frac{V}{R}\\where:\\V=voltage\\R=resistance

R=R_s+R_m

because they are connected in series. So:

I=\frac{12V}{(0.01+0.1)}=109A

P=(109)^2*0.1=1188W

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Plsss help I don’t understand this
nekit [7.7K]

B is the correct option.

1. Given eqn;

S(t) = 1/2t² - 4t + 8

2.Differentiate the above eqn with respect to t;

<u>d(S(t))</u> = t - 4

dt

When distance, S, is differentiated it results to velocity.

V = t - 4

at t = 10

V = 10 - 4

V = 6 feet/s

6 0
2 years ago
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If an object is being pulled by two forces, one 4 N to the left and the other 2 N to the right, what is the net
nika2105 [10]

Answer:

2N

Explanation:

subtract  rthe two forces to see which is greater

4-2=2

6 0
3 years ago
Describe what happens to water vapor when thermal energy is removed from it?
Marysya12 [62]

Explanation:

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3 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
4 years ago
Compare and contrast the front limb bones of a human and a bat.
riadik2000 [5.3K]

Answer:

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~i hope this helps~

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3 years ago
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