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3 years ago

The state of matter that keeps its shape and volume when it is placed in a different container

Physics

1 answer:

RUDIKE [14]3 years ago

7 0

Is this what you need physical property because if you move it somewhere it is still in its physical property.

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what was the mass of a cannkn ball whose velocity is 200m/s if it were shot from 1000kg that recoils a 2m/s. Solve step by step

Whitepunk [10]

The famous Newton’s Third Law states that “For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.”

By using this,

10grams or 0.01kg of bullet with speed 400 m/sec and 5kg gun recoil with speed suppose ‘v’.

0.01×400=5×v

4/5=v

v=0.8m/sec ANSWER.

6 0

3 years ago

What description refers to fog?

wolverine [178]

I think the answer is D

4 0

2 years ago

bagirrra123 [75]

Explanation:

The reading on the scale is

W = m(g + a)

= (77 kg)(9.8 m/s^2 + 2 m/s^2)

= 908.6 N

7 0

2 years ago

Pls help pls pls pls pls

Elza [17]

1.cool down
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6 0

3 years ago

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,

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

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

5 0

3 years ago