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koban [17]
3 years ago
15

Convection is the type of heat transfer that occurs in which of the following?

Physics
1 answer:
katrin [286]3 years ago
5 0
C

Tfsdghjjjjjhvv (Ignore)
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What is the maximum eccentricity an ellipse can have
VashaNatasha [74]

Answer: 1

Explanation: The highest eccentricity an ellipse can have is '1', a straight line.

7 0
3 years ago
3 simple machines you find in a pinball machine
cestrela7 [59]

lever (The things that flick the ball around)

Inclined plane (The hill the ball rolls down)

Wedge (The bumps that stop the ball from rolling certain places)

6 0
2 years ago
The density of water is 1000 kg/m the pressure pf water at 10 m depth is about
puteri [66]

Answer:

pressure in liquids is given as:

P= hpg

where h is the depth

where p is the density

where g is 10

Explanation:

From the formula above

p = 10 X 1000 X 10

p = 100000N/m

6 0
2 years ago
A jet aircraft is traveling at 262 m/s in horizontal flight. The engine takes in air at a rate of 85.9 kg/s and burns fuel at a
Fittoniya [83]

Answer:

F_T=60132.52N

P=15814852.76W

Explanation:

From the question we are told that

Velocity of aircraft  V=263m/s

Engine air intake rate \triangle M_a=85.9kg/s

Fuel burn rate  \triangle M_f =3.92kg/s

Velocity of exhaust gas V_e =921m/s

Generally the Mass change rate of Rocket is mathematically given by

 \triangle M = \triangle M_a+\triangle M_f

 \triangle M= 85.9+3.92

 \triangle M=89.82kg/s

Generally the Trust of the rocket is given mathematically by

 F_T=(\triangle M *V_e)-(dM_a/dt)*(V)

 F_T=(89.82 *921)-(85.9)*(263)

 F_T=60132.52N

Generally the Rocket's delivered power is mathematically given by

Delivered power P

 P=V*F_T

 P=263*60132.52N

 P=15814852.76W.

8 0
3 years ago
A horizontal force of magnitude 46.3 n pushes a block of mass 4.14 kg across a floor where the coefficient of kinetic friction i
IrinaVladis [17]
A) Calling F the intensity of the horizontal force and d the displacement of the block across the floor, the work done by the horizontal force is equal to
W=Fd = (46.3 N)(4.25 m)=196.8 J

b) The work done by the frictional force against the motion of the block is equal to:
W_f =  -F_f d =- (\mu mg) d =-(0.609)(4.14 kg)(9.81 m/s^2)(4.25 m)=
=-105.1 J
Part of these 105.1 Joules of work becomes increase of thermal energy of the block (\Delta E_B), and part of it becomes increase of thermal energy of the floor (\Delta E_F). We already know the increase in thermal energy of the block (38.2 J), so we can find the increase in thermal energy of the floor:
\Delta E_F = 105.1 J - \Delta E_B = 105.1 J-38.2 J=66.9 J

c) The net work done on the block is the work done by the horizontal force F minus the work done by the frictional force (the frictional force acts against the motion, so we must take it with a negative sign):
W_{net}=W-W_f=196.8 J-105.1 J=91.7 J
For the work-energy theorem, the work done on the block is equal to its increase of kinetic energy:
W_{net} = \Delta K
So, we have \Delta K=+91.7 J


5 0
3 years ago
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