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3 years ago

HELP PLEASE!!!! What is the net force on this object?

Physics

2 answers:

Ede4ka [16]3 years ago

4 0

The 20N straight up exactly cancels the 20N straight down. So we can see that there's no net vertical force on the object, and if there's ANY net force on it at all, it's going to be horizontal.

The 14N force toward the left can't cancel the whole 22N toward the right. It can only cancel 14N of it, and then there's still another 8N more toward the right.

So the net force on the object is 8N toward the right.

rusak2 [61]3 years ago

3 0

the answer to this question is 8 newtons

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2 years ago

Read 2 more answers

Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen

kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

$\lambda = 436.1 \ nm$

or,

$=436.1\times 10^{-9} \ m$

Distance,

$d = 0.31 \ mm$

or,

$=0.31\times 10^{-3} \ m$

Distance between the 1st and 2nd dark fringes,

$(y_2-y_1) = 6\times 10^{-3} \ m$

As we know,

⇒ $\frac{d}{L} (y_2-y_1) = \lambda$

or,

⇒ $L=\frac{d(y_2-y_1)}{\lambda}$

By substituting the values, we get

$=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}$

$=\frac{0.31\times 6\times 10^3}{436.1}$

$=\frac{1860}{436.1}$

$=4.26 \ m$

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2 years ago

A 40W lamp wastes 34 J of energy every second by heating its surroundings.

Artemon [7]

Answer:

$15\%$ .

Explanation:

The efficiency of a machine is the percentage of energy input that was turned into useful energy.

The power rating of this lamp is $40\; \rm W$ (same as $40\; \rm J \cdot s^{-1}$ ,) meaning that $40\; \rm J$ of energy is supplied to this lamp every second.

The question states that $34\; \rm J$ out of that $40\; \rm J$ of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the $(40\; \rm J - 34\; \rm J) = 6\; \rm J$ of energy supplied to this lamp would be turned into useful energy output.

Thus, every second, this lamp would receive $40\; \rm J$ of energy input and would outputs $6\; \rm J$ of useful work. The efficiency of this lamp would be:

$\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}$ .

4 0

2 years ago