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ivolga24 [154]
2 years ago
11

How can hand tools help every technician for their safety in doing hard servicing activities​

Physics
1 answer:
Anni [7]2 years ago
5 0

Answer: Hand tools are necessary for daily activities.

Explanation:

The hand tools are useful for day to day activities they aid in pouring substances, holding things, mixing substances, cleaning and for other purposes. In a chemical laboratory the hand tools like test tube holders can help in holding the chemicals. In a medical laboratory the forceps can be useful in holding and cutting the sections without causing any harm to the hands. Hand tools avoid the direct contact with the microorganisms in medical laboratory thus protect against infections. Hand tools aid in surgical operations and they help in avoiding direct contact with the blood vessels and organs preventing contamination and infections.

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Hes ur clothes ur andwer is 27374747488

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Two trucks have the same velocity but different masses. Truck 1 has a mass of 200kg and truck 2 has a mass of 500kg . Which truc
Readme [11.4K]
Your answer would be Truck 2.
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3 years ago
What is carried by a wave?
Nata [24]
Waves carry energy from one place to another. Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals. ... Some types of waves need to be transmitted through matter, either a solid, liquid or a gas. For example, water waves have to travel in water.
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Read 2 more answers
You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they be
Alex

Answer:

The amplitude  is  A =  90.2 \ m

Explanation:

From the question we are told that

    The frequency of when sound is approaching observer is   f = 392 Hz

     The frequency as the move away from observer  is  f_ a =  330 \ Hz

    The time between the pitch are t =  10 \ s

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

       T =  2 t

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

          T = \frac{2 \pi}{w}

Where  w is the angular velocity

=>   \frac{2 \pi}{w}  =  2 * 10

=>   w =  0.314 \ rad/sec

Now using Doppler Effect,

   The source of the sound is approaching the observer

The

          f = f_o (\frac{v}{v- wA} )

         392  = f_o (\frac{v}{v- wA} )

Where A is the amplitude

    So when the source is moving away from the observer

         f_a =  f_o (\frac{v}{v+ wA} )  

        330  =  f_o (\frac{v}{v+ wA} )  

Here  f_o is the fundamental frequency

Dividing the both equation  we have

           \frac{392}{330}  =  \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}

           1.1878  = \frac{v+wA}{v-wA}

         1.1878 v -  1.1878 wA = v+wA

        1.1878 v = 2.1878 wA

=>     A =  \frac{(0.1878 * (330))}{(2.1878)* (0.314)}

         A =  90.2 \ m

7 0
3 years ago
A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to
Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa

On the lower part of the hatch, there is a pressure equal to

p_{bot}=p_{atm}=1.013\cdot 10^5 Pa

So, the net pressure acting on the hatch is

p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa

which acts from above.

The area of the hatch is given by:

A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N

8 0
3 years ago
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