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sergij07 [2.7K]
3 years ago
9

When is the pressure of the man on the ground more – lying position or standing

Physics
1 answer:
Wittaler [7]3 years ago
7 0

Hi,

<u>The man on the ground in standing position has more pressure</u>. This is because when he stands, only his legs are in contact with the ground. While lying, his body is more in contact with the ground, therefore, he exerts less pressure.

To the point, a man standing position on the ground had more pressure.

More is the area of contact, less is the pressure efforted.

Thank you...

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For a particular pipe in a pipe-organ, it has been determined that the frequencies 296 Hz and 370 Hz are two adjacent natural fr
user100 [1]

Answer:

fundamental frequency of pipe will be equal to 74 Hz

Explanation:

We have given for a particular organ pipe two adjacent frequency are 296 Hz and 370 Hz

Speed of the sound in air is 343 m/sec

We have to find the fundamental frequency for the pipe

Fundamental frequency will be equal to difference of the two adjacent frequency

So fundamental frequency = 370 - 296 = 74 Hz

So fundamental frequency of pipe will be equal to 74 Hz

8 0
3 years ago
A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if
Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by \sqrt{2}. Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>

We have a block-spring system, whose angular frequency (\omega) is defined by the following formula:

\omega = \sqrt{\frac{k}{m} } (1)

Where:

k - Spring constant, measured in newtons per meter.

m - Mass, measured in kilograms.

And the period (T), measured in seconds, is determined by the following expression:

T = \frac{2\pi}{\omega} (2)

By applying (1) in (2), we get the following formula:

T = 2\pi\cdot \sqrt{\frac{m}{k} }

a) If the mass is doubled, then the period is increased by \sqrt{2}. Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

8 0
3 years ago
Which of the 3 states of matter is easier to increase the temperature by one degree; in a liquid, in a solid, or in a gas?
Mashcka [7]

Answer:

gas

Explanation:

this is bcz they have higher kinetic energy compared to liquids and gases

4 0
3 years ago
A scientist that applies the laws of science to the needs of communities is called _____.
s2008m [1.1K]

Answer:

The experimental scientist

5 0
2 years ago
12)A black body is heated from 27°C to 127° C. The ratio of their energies of radiations emitted will be
Nat2105 [25]

Answer:

81:256.

Explanation:

Let T denote the absolute temperature of this object.

Calculate the value of T before and after heating:

T(\text{before}) = 27 + 273 = 300\; \rm K.

T(\text{after}) = 127 + 273 = 400\; \rm K.

By the Stefan-Boltzmann Law, the energy that this object emits (over all frequencies) would be proportional to T^4.

Ratio between the absolute temperature of this object before and after heating:

\displaystyle \frac{T(\text{before})}{T(\text{after})} = \frac{3}{4}.

Therefore, by the Stefan-Boltzmann Law, the ratio between the energy that this object emits before and after heating would be:

\displaystyle \left(\frac{T(\text{before})}{T(\text{after})}\right)^{4} = \left(\frac{3}{4}\right)^{4} = \frac{81}{256}.

4 0
3 years ago
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