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3 years ago

2. Determine the current in a 120-watt bulb plugged into a 120-volt outlet.

Physics

1 answer:

Sergio039 [100]3 years ago

8 0

Answer:

1 ampere

Explanation:

Power = VI

V = voltage

I = current

Given

Power = 120 watts

V = 120 volts

Therefore

120 = 120 x I

Divide both sides by 120

120/120 = 120/120 x I

1 = I

I = 1 ampere

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Find the area of the part of the plane 5x 3y z = 15 that lies in the first octant.

Lunna [17]

This part of the plane lies above a triangle with boundaries $x=0$ and $y=0$ along the coordinate axes, as well as the line

$z=0 \implies 5x + 3y = 15 \implies y = \dfrac{15 - 5x}3$

When $y=0$ , we have $15-5x=0\implies x=3$ . So this triangle is the set

$T = \left\{(x,y)\in\Bbb R^2 ~:~ 0\le x\le3 \text{ and } 0\le y\le\dfrac{15-5x}3\right\}$

Also, when $x=0$ , we have $y=\frac{15}3=5$ . So the triangle has length 3 and width 5, hence area 1/2•3•5 = 15/2.

Let $z=f(x,y) = 15 - 5x - 3y$ . Then the area of the plane over $T$ is

$\displaystyle \iint_T dA = \iint_T \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dx \, dy$

We have

$\dfrac{\partial f}{\partial x} = -5 \implies \left(\dfrac{\partial f}{\partial x}\right)^2 = 25$

$\dfrac{\partial f}{\partial y} = -3 \implies \left(\dfrac{\partial f}{\partial y}\right)^2 = 9$

$\implies\displaystyle \iint_T dA = \sqrt{35} \iint_T dx\,dy = \boxed{\frac{15\sqrt{35}}2}$

since the integral

$\displaystyle \iint_TdA$

is exactly the area of $T$ , 15/2.

8 0

1 year ago

Mention three features of the Constitution and a note about them

Dvinal [7]

Answer:

Explanation:(1) a constitution is a supreme law of the land, (2) a constitution is a framework for government; (3) a constitution is a legitimate way to grant and limit powers of government officials. Constitutional law is distinguished from statutory law.

7 0

2 years ago

When a can is crushed did it undergo a physical change

brilliants [131]

Because even though the object got crush and misshape it still has the same identity. the identity never change

4 0

3 years ago

What is the gravitational potential energy of a 15.0kg object that is 5.00m above the ground relative to a point 8.00m above the

Licemer1 [7]

Explanation:

an object's gravitational potential energy Eg is m×g×h where:

m=mass

g=9.8m/s²

h=height relative to the closest object below it (because it cannot potentially fall through it

so Eg = 15×9.8×5=735J

4 0

2 years ago

A- 1000 m/s2<br> Xi-0m<br> Xf-0.75m<br> Vf-?

sleet_krkn [62]

Answer:

The final velocity of the object is, $v_{f}$ = 27 m/s

Explanation:

Given,

The acceleration of the object, a = 1000 m/s²

The initial displacement of the object, $x_{i}$ = 0 m

The final displacement of the object, $x_{f}$ = 0.75 m

The initial velocity of the object will be, $v_{i}$ = o m/s

The final velocity of the object, $v_{f}$ = ?

The average velocity of the object,

v = ( $x_{f}$ - $x_{i}$ )/ t

= 0.75 / t

The acceleration is given by the relation

a = v / t

1000 m/s² = 0.75 / t²

t² = 7.5 x 10⁻⁴

t = 0.027 s

Using the I equation of motion,

$v_{f}$ = u + at

Substituting the values

$v_{f}$ = 0 + 1000 x 0.027

= 27 m/s

Hence, the final velocity of the object is, $v_{f}$ = 27 m/s

8 0

3 years ago