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3 years ago

15

The diagram below shows a circuit with a battery that provides an unknown potential difference. Work out the potential differenc

e provided by the battery.

1 answer:

Gnesinka [82]3 years ago

5 0

Answer:

no diagram attached :/

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A chair of mass 15.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force of

Neko [114]

Answer: 185.5672566

Explanation: The friction is not relevant

Normal reaction is the force perpendicular to the surface.

this force resists the downwards forces applied which are gravity and a component of the applied force.

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3 years ago

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3. A microwave oven draws 12 A of current on a 110 V household circuit. What is its power

steposvetlana [31]

Answer:

W = 1320Watts

Explanation:

W = I*V

W = 12A*110V

W = 1320Watts

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3 years ago

If the acceleration of a motorboat is 4.0 m/s2, and the motorboat starts from rest, what is its velocity after 6.0 s?

umka2103 [35]

Answer:

The velocity of the motorboat after 6s is 24 m/s.

Explanation:

Given;

acceleration of the motorboat, a = 4.0 m/s²

initial velocity of the motorboat, u = 0

time of motion of the motorboat = 6s

Apply the following kinematic equation to determine the velocity of the motorboat after 6 ;

v = u + at

v = 0 + (4 x 6)

v = 24 m/s

Therefore, the velocity of the motorboat after 6s is 24 m/s.

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3 years ago

Is it possible for a distance versus-time graph to be a vertical line?

Feliz [49]

If time is the x axis and distance is the y axis then yes, in the case that time is going by but distance remains the same.

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3 years ago

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When one person shouts at a football game, the sound intensity level at the center of the field is 58.4 dB. When all the people

Tanzania [10]

Answer:

The number of people at game are approximately 22909

Explanation:

Given data

When one person shout $\beta _{1}=58.4dB$

When n number of person shout together $\beta _{n}=102dB$

The sound intensity level during one person shout is given by:

$\beta _{1}=10log(\frac{I_{1}}{I_{o}} )\\58.4=10log(\frac{I_{1}}{I_{o}} )\\5.84=log(\frac{I_{1}}{I_{o}} )\\\frac{I_{1}}{I_{o}} =10^{5.84}\\I_{1}=10^{5.84}*I_{o}$

The sound intensity level during n number of person shout is given by:

$\beta _{n}=10log(\frac{I_{n}}{I_{o}} )\\102=10log(\frac{I_{n}}{I_{o}} )\\10.2=log(\frac{I_{n}}{I_{o}} )\\\frac{I_{n}}{I_{o}}=10^{10.2}\\I_{n}=10^{10.2}*I_{o}$

Since each person generates same sound intensity and hence total number of persons can be determined as

$=\frac{I_{n}}{I_{1}}\\ =\frac{10^{10.2}I_{o}}{10^{5.84}I_{o}} \\=22909$

Hence

The number of people at game are approximately 22909

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3 years ago

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