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mr_godi [17]
3 years ago
8

Which is true about inelastic collisions: a. An inelastic collision does not obey conservation of energy. b. An inelastic collis

ion conserves kinetic energy. c. Objects will stick together upon collision. d. Momentum is not conserved in inelastic collisions..
Physics
1 answer:
Phoenix [80]3 years ago
5 0

Answer:

Option c is correct

Explanation:

There are two types of collisions-elastic collision and inelastic collision.

In elastic collision, both kinetic energy and total momentum are conserved. On the other hand, in inelastic collision, total momentum is conserved but kinetic energy is not conserved. Thus, option b and d are incorrect.

Total energy is always conserved in both types. Thus, option a is incorrect.

In a perfectly inelastic collision, objects stick together. This happens because maximum kinetic energy is dissipated and used in bonding of the two objects. Thus, correct option is c.

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A top-fuel dragster starts from rest and has a constant acceleration of 42.0 m/s2. What are (a) the final velocity of the dragst
disa [49]

Answer:

a)  Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) The displacement of the dragster at the end of 1.8 s = 68.04 m

d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

  v = u + at

  v  = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

  v = u + at

  v  = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 1.8 + 0.5 x 42 x 1.8²

    s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 3.6 + 0.5 x 42 x 3.6²

    s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

3 0
2 years ago
Points A (-9,2), B (2,-9), and C (-9,-9) are placed in three different quadrants of a Cartesian coordinate system. Convert each
slavikrds [6]

Answer:

answer

Explanation:

5 0
3 years ago
Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 45.0 mph . Beth leaves Los A
WARRIOR [948]

Answer:

a.Beth

b.2232 s

Explanation:

We are given that

Distance,d=400 mi

Speed of Alan,v=45 mph

Speed of Beth,v'=55 mph

a.Time =\frac{distance}{speed}

Using the formula

Time taken by Alan=\frac{400}{45}=8.89 hr

Time taken by Beth=\frac{400}{55}=7.27hr

Alan will reach San Francisco at 4:53 PM

Beth will reach San Francisco at 4:16 PM

Beth will reach before Alan.

b.Difference between time=8.89-7.27=1.62 hr

t=1.62 hr

1.62-1=0.62 hr

0.62 hr=0.62\times 60\times 60=2232 s

Hence, Beth has to wait 2232 s for Alan to arrive .

6 0
3 years ago
A 20 kg truck drives in a circle of radius 4 m at 10m/s. What causes
jek_recluse [69]

Answer:

Normal force

Explanation:

  • The force is centipetral force

Calculating :-

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  • Radius=r=4m
  • Velocity=10m/s=v

We know

\boxed{\sf F_c=\dfrac{mv^2}{r}}

\\ \sf\longmapsto F_c=\dfrac{20(10)^2}{4}

\\ \sf\longmapsto F_c=\dfrac{20(100)}{4}

\\ \sf\longmapsto F_c=\dfrac{2000}{4}

\\ \sf\longmapsto F_c=500N

6 0
2 years ago
I know this ain't right but can someone please correct the mistakes I have it for a test​
nikklg [1K]

Answer:

NICE HANDWRITING. I've done this before and I'm sure everything is correct. Although I would re-check to see if you had any mistakes with your calculations. Good job!

Explanation:

8 0
1 year ago
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