3 years ago

8. What current flows through a 1.15 kW electric fire at a potential difference of

Physics

1 answer:

Vilka [71]3 years ago

4 0

Answer:

I = 5[amp]

Explanation:

Electrical power is defined as the product of voltage by current.

$P=V*I$

where:

P = power = 1150 [W]

V = voltage = 230 [V]

I = current [amp]

Now replacing:

$1150=230*I\\I=1150/230\\I=5[amp]$

A 15 [amp] fuse must be used. Always the fuse must be larger than the operating current, to protect the equipment from very high currents. above 15 [amp]

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A company designed and sells an ultrasonic receiver, which detects sounds unable to be heard by the human ear. The receiver can

Kruka [31]

Answer:

the diameter of the outside edge of the receiver is $8\sqrt{18} \ \ \ inches$

Explanation:

From the schematic free body diagram illustrating what the question is all about below;

Let represent A to be the vertex where the receiver is being placed

S to be the focus

BP to be equal to r (i.e radius of the outer edge)

BC to be 2 r (i.e the diameter)

Given that AS = 4 in and AP is 18 in

Let AP be x- axis and AY be y -axis

A=(0,0)

S=(4,0) = (0,0)

So that the equation of the parabolic path of the receiver will be:

$y^2 =4 ax \\ \\ y^2 = 4*4*x \\ \\ y^2 = 16x$

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B = (18, r)

B lies y² = 16 x

r² = 16 x

r² 16 × 18

$r = \sqrt{16*18 } \\ \\ r = 4\sqrt{18}$

Diameter BC = 2r

$2* 4\sqrt{18} \\ \\= 8\sqrt{18 } \ \ \ inches$

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I believe the answer is C

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C:carry
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zavuch27 [327]

Answer:

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Explanation:

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