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2 years ago

7

You wear lead to protect you from ____.

2 answers:

nata0808 [166]2 years ago

8 0

Answer would be x rays!

meriva2 years ago

6 0

Answer:

It's Uwu sisidijsisiidisd sussy Bakak uwuu s.m.d suiskskaksks uwu Sussy baka ksksksidijxjdk

Explanation:

isotonic because it us

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Two forces that are not equal in size are

Bumek [7]

They are unbalanced forces ..... Hope this helps :3

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2 years ago

. A huge pile of leaves was wrapped in a tarp in the middle of a lawn. The wrapped leaves weigh 580 newtons. The coefficient of

Rina8888 [55]

The force required is 319 N

Explanation:

The force of static friction is a force that acts an object on a surface, when this object is pushed by another force to put it in motion. The direction of the force of friction is opposite to the direction of the force of push, and its value increases as the force of push increases, up to a maximum value given by:

$F_f = \mu W$

where

$\mu$ is the coefficient of friction

W is the weight of the object

Therefore, in order to put the object in motion, the force applied must be greater than this value.

For the pile of leaves in this problem, we have:

$\mu = 0.55$ (coefficient of friction)

$W=580 N$ (weight of the leaves)

Substituting, we find:

$F=(0.55)(580)=319 N$

Learn more about force of friction:

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3 years ago

A typical adult can deliver about 12.5 N·m of torque when attempting to open a twist-off cap on a bottle. Assume that bottle cap

Nikitich [7]

Uhhhhhhhhh just tryna get a point so I can ask a question so eh I’m using ur question heheheheheh

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3 years ago

True or False. Where an element is located on the periodic table can help predict some of its properties.

lubasha [3.4K]

It’s true there are sections in the periodic table that define the elements.

7 0

3 years ago

Read 2 more answers

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

$I_2 = I_1 (\frac{r_1^2}{r_2^2})$

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

8 0

3 years ago