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Korvikt [17]
3 years ago
8

A 0.5 kg ball traveling at 1 m/s to the right hits a second ball of equal mass that is initially at rest. After the perfectly el

astic collision, the second ball travels at 0.543 m/s at a 57.2 degree angle upward from the point of contact.
Draw a before and after diagram of this collision

Physics
1 answer:
ruslelena [56]3 years ago
8 0

Comment if u dont understand

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Convert to the fractional equivalent and reduce 21.12
nignag [31]

The decimal point is placed after two digits starting from the end. For each decimal place, we can write the number divided by 100.

21.12 can be written as \frac{2112}{100}.

Divide the numerator and denominator by 2:

\frac{2112}{100}= \frac{156}{50}

The numerator and denominator can be divided by 2 again:

\frac{78}{25}

There is no other common factor between numerator and denominator other than 1. Hence, it is the reduced form.




3 0
3 years ago
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PLEASE TRY TO ANSWER AS MANY QUESTIONS AS YOU CAN !
suter [353]
Good luck with solving this
3 0
3 years ago
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A 70 ft rope hangs from a helicopter above this room. The rope has a mass per unit length of 2 lb/ft. In order to be rescued fro
Mrac [35]

Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope  = 2 lb/ft x 70 ft

                                         = 140 lbs.

Total mass to be pulled to the helicopter, M = 120 lbs  + 140 lbs  

                                                                       = 260 lbs

The work done is calculated from work-energy theorem as follows;

W = Mgh

where;

g is acceleration due gravity = 32.17 ft/s²

h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s²  x 70 ft

W = 585494 lb.ft²/s²

1 lb.ft²/s² = 0.0421 J

W = 585494 lb.ft²/s²  = 2.47 X 10⁴ J.

Therefore, the work done to get you safely away from the test is  2.47 X 10⁴ J.

4 0
2 years ago
Why do racing cars have wider wheels?​
tamaranim1 [39]

Answer:

So they can last longer and have more grip than normal on-road cars. They need that in order for them to run well

Explanation:

3 0
3 years ago
Read 2 more answers
By how much do the critical angles for red (660 nm) and violet (410 nm) light differ in a diamond surrounded by air?
Stella [2.4K]

the critical angle for red light in diamond is 24.51°

critical angle for violet light in diamond is 24.01°.

given:

wavelength of red light λr =660nm

wavelength of violet light λv =410nm

refractive index for air=1

as the refractive index of red and violet colour in diamond is

Refractive index of red color nr= 2.407

Refractive index of violet color nv= 2.451

as it is mentioned that angle is critical angle,thus θr =90°

so,from Snell's law

n1sinθc=n2 sin 90°

Here,

n1 is the refractive index for incident medium

n2 is the refractive index for refractive medium

θc is the critical angle after which light starts reflecting internally

The critical angle for red light is here,n1=2.407, n2=1

n1sinθc=n2 sin 90°

2.407×sin θc=1×1

sin θc =(1/2.407)

θc=sin-1 0.415

θc=24.51°

Thus,the critical angle for red light in diamond is 24.51°

The critical angle for violet light is,

here,n1=2.451,n2=1

n1sinθc=n2 sin 90°

2.407×sin θc=1×1

sin θc= (1/2.451)

θc=sin-1 0.407

=24.01°

Thus,the critical angle for violet light in diamond is 24.01°.

learn more about Snell's law from here: brainly.com/question/2273464

#SPJ4

3 0
2 years ago
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