PLEASE HELP

A night-light uses 3 W.If the voltage is 120 V, what is the current? A.0.025A B.40 A C.360A D.0.25A

Tomtit [17]

The answer to this question is A.

6 0

3 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

3 years ago

Three-fourths of the area of a rectangular lawn 30 feet wide by 40 feet long is to be enclosed by a rectangular fence. If the en

satela [25.4K]

Answer:

The fence is 5feet less.

Explanation:

We need to determine

The less amount of fence required, if the enclosure has full width and reduced length, compared to full length and reduced width.

Approach & WorkingArea of lawn = 30 × 403/4th of the area of lawn = ¾(30 × 40) = 30 * 30

When full width will be fenced, and reduced length will be fenced.

Width = 30 feet30 * L = 30 * 30Hence, length = 30 feetLength of fence needed = 2(30 + 30) = 120 feet

When full length will be fenced, and reduced width will be fenced

Length = 40 feet40 * W = 30 * 30W = 22.5 feetLength of fence needed = 2(40 + 22.5) = 125 feet

Difference in length of fence needed = 125 – 120 = 5 feet.

4 0

4 years ago

Which of the materials would be able to scratch Quartz

Lelechka [254]

<span>anything harder than mohs scale 7 so eg Topaz, Corundum and diamond representing mohs scale 8 9 and 10 respectively.</span>

7 0

3 years ago

What is the net force acting on a 52 kg object that has a velocity of 8.0 m/s and is moving in a circle of radius 1.6 m? a. 4000

jenyasd209 [6]

Σf = m a
Σf = m v^2 / r
Σf = 52 8^2 / 1.6
Σf = 2080 N

7 0

3 years ago