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2 years ago

What is the resistance of a 15 ampere current with 8 volts of potential difference?

Physics

1 answer:

Alex Ar [27]2 years ago

8 0

Answer:

Resistance in circuit = 0.53 ohm (Approx.)

Explanation:

Given:

Flow of current in circuit = 15 amp

Potential difference = 8 Volts

Find:

Resistance in circuit

Computation:

In an electrical system, resistance is a stopper of a material to electric current.

Resistance in circuit = Potential difference / Flow of current in circuit

Resistance in circuit = 8 / 15

Resistance in circuit = 0.53 ohm (Approx.)

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What two energies does a bulldozer use

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2 years ago

An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17

Softa [21]

Answer:

distance cover is = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

$t_1 = 3.32 sec$

$t_2 = 5.08 sec$

from equation of motion we know that

$d_1 = vt_1 + \frac{1}{2} gt_1^2$

where d_1 is distance covered in time t1

so $d_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2$ =

$d_1 = 110.78 m$

$d_2 = vt_2 + \frac{1}{2} gt_2^2$

where d_2 is distance covered in time t2

$d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2$

$d_2 = 213.31 m$

distance cover is = 213.31 - 110.78 = 102.53 m

3 0

3 years ago

A 52 kg child on a swing is travelling at 6 m/s . What is his gravitational potential energy if he has 1000 J of the mechanical

DiKsa [7]

Answer:

The correct answer is "64 J".

Explanation:

The given values are:

Mass,

m = 52 kg

Velocity,

v = 6 m/s

Mechanical energy,

= 1000 J

Now,

The gravitational potential energy will be:

⇒ $P.E=1000-\frac{1}{2}mv^2$

$=1000-\frac{1}{2}\times 52\times (6)^2$

$=1000-26\times 36$

$=1000-936$

$=64 \ J$

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2 years ago

Plot StartRoot 1.5 EndRoot and StartRoot 1.9 EndRoot on the number line to find which inequalities are true. Check all that appl

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Answer:

A, B, C, E

Explanation:

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2 years ago

Read 2 more answers

A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.

densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (5*10^{-3})(300)^2$

$KE = 225J$

On the other hand the required Energy to heat up t melting point is

$Q_1 = mC_p \Delta T$

$Q_2 = L_f m$

Where,

m = Mass

$C_p =$ Specific Heat

$\Delta T =$ Change at temperature

$L_f =$ Latent heat of fussion

Heat required to heat up to melting point,

$Q = Q_1+Q_2$

$Q = mC_p \Delta T+L_f m$

$Q = 5*0.128*(327-20) + 5*24.7$

$Q = 310J$

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

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3 years ago