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svetlana [45]
2 years ago
8

Which of these is an appropriate treatment for a deep, bleeding wound?

Physics
2 answers:
kap26 [50]2 years ago
7 0

Answer:

Elevate the wound.

Hope this helped!

bija089 [108]2 years ago
4 0

Explanation:

A. Walk the person around

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An airplane flew 1000 m in 400 seconds. What is the airplane's speed?
weqwewe [10]

Answer:

speed =  \frac{distance}{time}  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   =  \frac{1000}{400}  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: = 2.5m {s}^{ - 1}

6 0
3 years ago
Read 2 more answers
A man pushes on piano with mass 170 kg; it slides at constant velocity down a ramp that is inclined at 20.0 ∘ above the horizont
nikdorinn [45]

Answer

given,                            

mass of the piano = 170 kg              

angle of the inclination = 20°                

moves with constant velocity hence acceleration = 0 m/s²    

neglecting friction                                  

so, force required to pull the piano                    

F = m g sin θ                                                      

F = 170 × 9.81 × sin 20°                                        

F = 570.39 N                                                    

so, force required by the man to push the piano is F = 570.39 N

4 0
3 years ago
Helium gas is compressed by an adiabatic compressor from an initial state of 14 psia and 50°F to a final temperature of 320°F in
iVinArrow [24]

Answer:

The value is  P_2 = 40.54 \ psla

Explanation:

From the question we are told that

 The initial pressure is  P_1 = 14\  psla

  The initial temperature is  T_1 =  50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283  \  K

   The final temperature is  T_2 =  320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433  \  K

Generally the equation for adiabatic process is mathematically represented as

         PT^{\frac{\gamma}{1- \gamma} } =  Constant

=>      P_1T_1^{\frac{\gamma}{1- \gamma} } =  P_2T_2^{\frac{\gamma}{1- \gamma} }

Generally for a monoatomic gas  \gamma =  \frac{5}{3}

So

           14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }

=>       14 * 283^{-2.5} =P_2 * 433^{-2.5}

=>       P_2 = 40.54 \ psla

8 0
2 years ago
The greater the mass is in an object, the higher resistance to a change in movement the object will have. Please select the best
Fofino [41]
This statement is true. The greater the mass is in an object, it is indeed the higher resistance to a change in movement the object will have. That only mean that the mass of an object and its resistance to change of movement is directly proportional.
3 0
3 years ago
The distance recorded for riding a motorcycle on its rear wheel without stopping is more than 320 km! Suppose the rider in this
antiseptic1488 [7]

Answer:

<h3>14.97m/s</h3>

Explanation:

Given

Initial velocity of the car u = 8m/s

Distance travelled by the rider S = 40m

Acceleration a = 2m/s²

Required

rider's velocity after the acceleration v

Using the equation of motion

v² = u²+2as

v² = 8²+2(2)(40)

v² = 64+160

v² = 224

v = √224

v = 14.97m/s

Hence the rider's velocity after the acceleration is 14.97m/s

5 0
3 years ago
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