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3 years ago

PLEASE HELP ME PLEASE! BRAINLEST

Physics

1 answer:

Vadim26 [7]3 years ago

6 0

Answer:

10kg

Explanation:

Weight is "how much does gravity drag this down".

Mass is "how much matter is there here".

The relation is:

$F_g = mg$

where $F_g$ is the weight, $m$ is the mass and $g$ is the gravitational acceleration (roughly equal to 10N/kg on Earth).

From the task we know that:

$F_g = 100N\\g = 10\frac{N}{kg}$

So let's input it into the relation:

$100N = m\cdot 10\frac{N}{kg}\\10N = m \cdot 1\frac{N}{kg}\\10N \cdot \frac{kg}{N} = m\\~\\m = 10kg$

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A sound is recorded at 19 decibels. What is the intensity of the sound?

sp2606 [1]

$1 \times 10^{-10.1} \mathrm{Wm}^{-2}$ is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about $1 \times 10^{-12} \mathrm{Wm}^{-2}$ ). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB). Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

$\text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)$

Where,

I = Intensity of the sound produced

$I_{0}$ = Standard Intensity of sound of 60 decibels = $1 \times 10^{-12} \mathrm{Wm}^{-2}$

So for 19 decibels, determine I as follows,

$19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9$

When log goes to other side, express in 10 to the power of that side value,

$\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}$

$I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}$

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3 years ago

A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in

katen-ka-za [31]

Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

Displacement,

d = 300 m

Now,

The boat is travelling,

= $10+1.50$

= $11.5 \ m/s$

Travelling such distance for 300 m will be:

⇒ $v = \frac{d}{t} \ sot \ t$

$=\frac{d}{v}$

On putting the values, we get

$=\frac{300}{11.5}$

$=26.08 \ s$

Throughout the opposite direction, when the boat seems to be travelling then,

= $10-1.50$

= $8.5 \ m/s$

Travelling such distance for 300 m will be:

⇒ $v=\frac{v}{t} \ sot \ t$

$=\frac{d}{v}$

On putting the values, we get

$=\frac{300}{8.5}$

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hence,

The time taken by the boat will be:

= $26.08+35.29$

= $61.37 \ s$

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2 years ago

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We may be positive that an object is in mechanical equilibrium if it is not rotating and experiences no acceleration.

<h3>What is mechanical equilibrium?</h3>

There are numerous other definitions for mechanical equilibrium that are all mathematically comparable in addition to the definition in terms of force. A system is in equilibrium in terms of momentum if the component motions are all constant. If velocity is constant, the system is in equilibrium in terms of velocity. When an item is in a state of rotational mechanical equilibrium, its angular momentum is preserved and its net torque is zero. More generally, equilibrium is reached in conservative systems at a configuration space location where the gradient of the potential energy concerning the generalized coordinates is zero.

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6 0

1 year ago

PLEASE HELP ME PLEASE! BRAINLEST ​

PLEASE HELP ME PLEASE! BRAINLEST