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vitfil [10]
2 years ago
5

A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc

h mechanical energy was dissipated as this happened? The coefficient of kinetic friction between the tires and the road is 400.40un.
Physics
1 answer:
Harman [31]2 years ago
3 0

Answer:

E=88200\ J

Explanation:

Given:

  • mass of car, m=1500\ kg
  • distance of skidding after the application of brakes, d=15\ m
  • coefficient of kinetic friction, \mu_k=0.4

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

f=\mu_k.N

where N = normal reaction by ground on the car

f=0.4\ties 1500\times 9.8

f=5880\ N

<em>Now from the work-energy equivalence:</em>

E=f.d

E=5880\times 15

E=88200\ J is the dissipated energy.

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Answer:

Explanation:

If I assume that the wind did not cause the plane to chage its velocity.

The plane will have a velocity of vp =  (0*i + 100*j) km/h relative to ground

The cart has a velocity of vc = (0*i - 20*j) km/h relative to the plane

vc' = vc + vp

vc' = (0*i + 100*j) + (0*i - 20*j) = (0*i + 80*j) km/h relative to the ground.

If I assume that the wind move the plane:

The plane will have a velocity of vp =  (-40*i + 100*j) km/h relative to ground

The cart has a velocity of vc = (0*i - 20*j) km/h relative to the plane

vc' = vc + vp

vc' = (-40*i + 100*j) + (0*i - 20*j) = (-40*i + 80*j) km/h relative to the ground.

In reality the wind would move the plane a little, not to the full speed of the wind, somewhere between these two values, but without more data it cannot be calculated.

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2 years ago
Rod A and rod B are cylindrical rods made of the same metal. amd they differ only in size. Rod B has double the length and doubl
Sever21 [200]

Answer:

I believe the answer for this question is D

Explanation:

I hope this helps and is correct

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2 years ago
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Sara and Saba are identical twins who are the same in every way, including their weights. One day, Sara and Saba decided to go f
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As Saba was wearing high heels they are long from the bottom so they sank however Sana was wearing snow boots which means they were flat and so she didn’t sink.

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2 years ago
Th e heat capacity of air is much smaller than that of water, and relatively modest amounts of heat are needed to change its tem
Darina [25.2K]

Answer:

Q=1005 J

t= 0.67 sec

Explanation:

Lets take condition of room is 1 atm and 25°C.

Heat capacity ,c = 21 J /K.mol

If we assume that air is ideal gas that

P V = n R T

V= 5.5\times 6.5\times 3\ m^3

V=107.25\ m^3

V=107.25\times 1000 L

V= 107250 L

At STP number of moles given as

n=\dfrac{V}{V_{at\ S.T.P.}}

V=22.4 L at S.T.P.

n=\dfrac{107250}{22.4}\ moles

n=4787.94 moles

n= 4.784 Kmoles

So heat required to raise 10°C temperature

Q = n x c x ΔT

Q = 4.78794 x 21 x 10

Q=1004.64 J

Time t

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t = 1.004.64 /1.5

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2 years ago
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The air surrounding an airplane in flight exerts a drag force that acts opposite to the airplane's motion. When an Airbus A380 i
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Answer:

4.32\cdot 10^5 hp, 3.22\cdot 10^8 W

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The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.

Therefore, we can write:

4F_T - F_d = 0

where

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F_d is the drag force

Solving for Fd,

F_d = 4 F_T = 4(322,000)=1.288\cdot 10^6 N

The velocity of the jet is

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So, the rate at which the drag force does work (which is the power) is

P=F_d v

and substituting

F_d = 1.288\cdot 10^6 N\\v = 250 m/s

we find

P=(1.288\cdot 10^6)(250)=3.22\cdot 10^8 W

Converting into horsepower,

P=\frac{3.22\cdot 10^8}{746}=4.32\cdot 10^5 hp

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