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3 years ago

A lightbulb has a resistance of 195 Ω and carries a current of 0.62 A.

Physics

1 answer:

ki77a [65]3 years ago

8 0

Answer:

Explanation:

Power is current times voltage:

P = IV

Voltage is current times resistance:

V = IR

Therefore:

P = I²R

Given I = 0.62 A and R = 195 Ω:

P = (0.62 A)² (195 Ω)

P ≈ 75 W

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A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.

fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force $N= mg cos 22$

$\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N$

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3 years ago

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omeli [17]

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Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

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$Q_1 = Q_2$

$V_1A_1=V_2A_2$

Our values are given as,

$A_1=\frac{1}{2}^2*\pi=0.785 in^2$

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$\frac{V_1}{V_2}=\frac{A_2}{A_1}$

$\frac{V_1}{V_2}=\frac{1.227}{0.785}$

$\frac{V_1}{V_2}=1.56$

The second ratio of rates is

$\frac{V2}{V1}=\frac{A_1}{A2}$

$\frac{V2}{V1}=\frac{0.785}{1.227}$

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Lerok [7]

Answer:

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Explanation:

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v = x/t

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Answer:

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