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Setler [38]
3 years ago
15

If two balls collide with each other, they will move apart at the same speed if

Physics
2 answers:
denis-greek [22]3 years ago
7 0
A.
if you have seen a newton's cradle this will make sense.

in order for both of them to travel at the same speed, the balls need to have the same mass and the speed to begin with tocontinue to travel at the same speed because mass can affect the impact of the force on the balls by each other, causing each ball to have different speeds.
Kryger [21]3 years ago
6 0
Answer a. would be correct. Since momentum is conserved, Δmv must be constant.
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Choose all that apply. Solids, liquids, and gases can be distinguished by their:
vodomira [7]
I believe its by there shape
8 0
3 years ago
Read 2 more answers
Initially sliding with a speed of 1.9 m/s, a 1.8 kg block collides with a spring and compresses it 0.35 m before coming to rest.
Alika [10]
Let k =  the force constant of the spring (N/m).

The strain energy (SE) stored in the spring when it is compressed by a distance x=0.35 m is
SE = (1/2)*k*x²
     = 0.5*(k N/m)*(0.35 m)²
     = 0.06125k J

The KE (kinetic energy) of the sliding block is
KE = (1/2)*mass*velocity²
     = 0.5*(1.8 kg)*(1.9 m/s)²
     = 3.249 J

Assume that negligible energy is lost when KE is converted into SE.
Therefore
0.06125k = 3.249
k = 53.04 N/m

Answer:  53 N/m  (nearest integer)

3 0
3 years ago
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lidiya [134]

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C-SUPPLY

D-DEMAND

E-EQUILIBRIUM POINT

Explanation:

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7 0
3 years ago
The speed of light in a solid is 1.24 x 108 m/s. <br> Calculate the index of refraction
Dahasolnce [82]

Answer:

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7 0
3 years ago
A 0.49-kg cord is stretched between two supports, 7.8m apart. When one support is struck by a hammer, a transverse wave travels
katovenus [111]

To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,

v = \sqrt{\frac{T}{\mu}}

Here,

v = Velocity

\mu= Linear density (Mass per  unit length)

T = Tension

Rearranging to find the Period we have that

T = v^2 \mu

T = v^2 (\frac{m}{L})

As we know that speed is equivalent to displacement in a unit of time, we will have to

T = (\frac{L}{t}) ^2(\frac{m}{L})

T = (\frac{7.8}{0.83})^2 (\frac{0.49}{7.8})

T = 5.54N

Therefore the tension is 5.54N

8 0
3 years ago
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