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3241004551 [841]
3 years ago
5

How can liquid pressure phenomena be used in daily life. Give one example with explanation of working​?

Physics
1 answer:
N76 [4]3 years ago
8 0

Answer:

An example in which liquid pressure phenomena can be used in daily life is in Water blasting

Explanation:

Water blasting refers application of pressurized water to remove materials from the surface of objects.

There are different varieties of water blasting, including;

Hydrocleaning; Cleaning enabled by the use of high pressure water

Hydrodemolition; Demolition or removal of concrete using pressurized water

Hydrojetting; The spraying of water under pressure on surfaces in order to remove surface contaminants.

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Most new jobs in the United States will be in the _____.
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in the service producing sector

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you are doing an experiment outside on a sunny day you find the temperature of some sand is 28 degress Celsius you also find the
OLga [1]
E the temperature of a substance. Water has a very high specific heat. That means it needs to absorb a lot of energy before its temperature changes. Sand , on the other hand, have lower specific heats. This means that their temperatures change more quickly. When the summer sun shines down on them, they quickly become hot.
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3 years ago
How long will it take a person walking at 2.1 m/s to travel 13 m?
MrRissso [65]

Answer:

I gonna give you the number so but you need to round 6.19047619048

Explanation:

  • This is a speed formula so you would use the formula speed=distance/time
  • You need to rearrange it to time=distance/speed
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7 0
3 years ago
Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio
Alecsey [184]

Answer:

The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

Explanation:

Given that,

First charge q_{1}= 20\mu C

second charge q_{2}= 8\mu C

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

E_{1}= \dfrac{kq_{1}}{r^2}

Put the valueinto the formula

E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}

E_{1}=18\times10^{5}\ N/C

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

E_{2}= \dfrac{kq_{2}}{r^2}

Put the valueinto the formula

E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}

E_{2}=7.2\times10^{5}\ N/C

Direction of electric field along AO

We need to calculate the net electric field at midpoint

E_{net}=E_{1}-E_{2}

E_{net}=(18-7.2)\times10^{5}\ N/C

E_{net}=10.8\times10^{5}\ N/C

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

8 0
3 years ago
How many cm3 are there in 1 dm3?
Verdich [7]

The answer is in the picture.

8 0
2 years ago
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