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likoan [24]
3 years ago
8

Pls help me this is being timed.

Physics
1 answer:
GrogVix [38]3 years ago
5 0

Answer:

Increase air resistance

Explanation:

Gravity forces the parachute down but air resistance pushing up on the flat surface of the parachute causes it to fall slower to the ground.

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What two units of measure are necessary for describing speed
Shtirlitz [24]
Distance and time, distance because that's how far and time because that's how long
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Question 14 of 25
natali 33 [55]

the answer is sueist

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Refrigerant-134a enters a compressor at 180 kPa as a saturated vapor with a flow rate of 0.35 m3/min and leaves at 900 kPa. The
Alexus [3.1K]

Answer:

52.5°C

Explanation:

The final enthalpy is determined from energy balance where initial enthalpy and specific volume are obtained from A-12 for the given pressure and state

mh1 + W = mh2

h2 = h1 + W/m

h1 + Wα1/V1

242.9 kJ/kg + 2.35.0.11049kJ/ 0.35/60kg

=287.4 kJ/kg

From the final enthalpy and pressure the final temperature is obtained A-13 using interpolation

i.e T2 = T1 + T2 -T1/h2 -h1(h2 - h1)

= 50°C + 60 - 50/295.15 - 284.79

(287.4 - 284.79)°C

= 52.5°C

7 0
3 years ago
Determine the weight in newtons of a woman whose weight in pounds is 157. Also, find her mass in slugs and in kilograms. Determi
DerKrebs [107]

Answer:

Weight of the woman in Newton

   x =  698.6 \  N

 Mass of the woman in slug

  Mass =  4.86 \  slug

 Mass of the woman in kg

   Mass =  16 \  kg

  My weight in Newton

  W =  784 \  N

Explanation:

From the question we are told that

   The weight of the woman in pounds is  W = 157 \ lb

Converting to Newton

    1 N  =  0.22472 lb

    x N  =  157

=>   x =  \frac{157 *  1}{0.22472}

=>   x =  698.6 \  N

Obtaining the mass in slug

   Mass =  \frac{W}{g}

Here  g =  32.2 ft/s^2

So

     Mass =  \frac{157 }{32.2}

       Mass =  4.86 \  slug

Obtaining the mass in kilogram

     Mass =  \frac{W}{g}

Here  g =  9.8 \  m/s

So

   Mass =  \frac{157 }{9.8}

   Mass =  16 \  kg

Generally weight is mathematically represented as

     W =  m *  g

Given that my mass is   80 kg   then my weight is  

     W =   80 *9.8

      W =  784 \  N

6 0
3 years ago
A 4.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.650 kg ,
Maru [420]

Answer:

a) Coefficient of kinetic friction between block and surface = 0.12

b) Decrease in kinetic energy of the bullet = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = 0.541 J

Explanation:

Given,

Mass of bullet = 4.00 g = 0.004 kg

Initial velocity of the bullet = 400 m/s

Mass of wooden block = 0.65 kg

Initial velocity of the wooden block = 0 m/s (since it was initially at rest)

Final velocity of the bullet = 190 m/s

Distance slid through by the block after the collision = d = 72.0 cm = 0.72 m

Let the velocity of the wooden block after collision be v

According to the law of conservation of momentum,

Momentum before collision = Momentum after collision

Momentum before collision = (Momentum of bullet before collision) + (Momentum of wooden block before collision)

Momentum of bullet before collision = (0.004×400) = 1.6 kgm/s

Momentum of wooden block before collision = (0.65)(0) = 0 kgm/s

Momentum after collision = (Momentum of bullet after collision) + (Momentum of wooden block after collision)

Momentum of bullet after collision = (0.004×190) = 0.76 kgm/s

Momentum of wooden block after collision = (0.65)(v) = (0.65v) kgm/s

Momentum balance gives

1.6 + 0 = 0.76 + 0.65v

0.65v = 1.6 - 0.76 = 0.84

v = (0.84/0.65)

v = 1.29 m/s

The velocity of the wooden block after collision = 1.29 m/s

To obtain the coefficient of kinetic friction between block and surface, we will apply the work-energy theorem.

The work-energy theorem states that the work done in moving the block from one point to another is equal to the change in kinetic energy of the block between these two points.

The points to consider are the point when the block starts moving (immediately after collision) and when it stops as a result of frictional force.

Mathematically,

W = ΔK.E

W = workdone by the frictional force in stopping the wooden block (since there is no other horizontal force acting on the block)

W = -F.d (minus sign because the frictional force opposes motion)

d = Distance slid through by the block after the collision = 0.72 m

F = Frictional force = μN

where N = normal reaction of the surface on the wooden block and it is equal to the weight of the block.

N = W = mg

F = μmg

W = - μmg × d = (-μ)(0.65)(9.8) × 0.72 = (-4.59μ) J

ΔK.E = (final kinetic energy of the block) - (initial kinetic energy of the block)

Final kinetic energy of the block = 0 J (since the block comes to a rest)

(Initial kinetic energy of the block) = (1/2)(0.65)(1.29²) = 0.541 J

ΔK.E = 0 - 0.541 = - 0.541 J

W = ΔK.E

-4.59μ = -0.541

μ = (0.541/4.59)

μ = 0.12

b) The decrease in kinetic energy of the bullet

(Decrease in kinetic energy of the bullet) = (Kinetic energy of the bullet before collision) - (Kinetic energy of the bullet after collision)

Kinetic energy of the bullet before collision = (1/2)(0.004)(400²) = 320 J

Kinetic energy of the bullet after collision = (1/2)(0.004)(190²) = 72.2 J

Decrease in kinetic energy of the bullet = 320 - 72.2 = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = (1/2)(0.65)(1.29²) = 0.541 J

Hope this Helps!!!

4 0
3 years ago
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