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Svet_ta [14]
3 years ago
10

Analyze the collision of a baseball with a bat. During which portion of the collision does the baseball’s velocity reach zero?

Physics
1 answer:
ehidna [41]3 years ago
3 0

Answer:During the collision

Explanation:

When a baseball collides with bat its velocity changes from Positive To negative i.e. ball changes its direction to exactly opposite of initial. During collision at some point in time velocity of ball reaches zero and then finally changes its direction associated with an increase in velocity of the ball. velocity after depends upon the amount of force acting on the ball and the time of contact of force.

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You run away from a plane mirror at 2.30 m/s. At what speed does your image move away from you?
ivanzaharov [21]

Answer:

4.60m/s

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5 0
4 years ago
Later, while taking your groceries back to the car, you accidentally let go of your cart. It rolls straight down a grassy slope
VladimirAG [237]

Car is moving on the glassy slope with constant speed

Now we know that

a = \frac{dv}{dt}

so acceleration is rate of change in velocity

as we know that velocity is constant here so acceleration is zero

so here

a = 0

now as we know by Newton's II law

F = ma

since a = 0

F = 0

so net force will be ZERO on it during this motion

6 0
4 years ago
The first antiparticle, the positron or antielectron, was discovered in 1932. It had been predicted by Paul Dirac in 1928, thoug
Taya2010 [7]

Answer:

Energy of Photon = 4.091 MeV

Explanation:

From the conservation of energy principle, we know that total energy of the system must remain conserved. So, the energy or particles before collision must be equal to the energy of photons after collision.

K.E OF electron + Rest Energy of electron + K.E of positron + Rest Energy of positron = 2(Energy of Photon)

where,

K.E OF electron = 3.58 MeV

Rest Energy of electron = 0.511 MeV

Rest Energy of positron = 0.511 MeV

K.E OF positron = 3.58 MeV

Energy of Photon = ?

Therefore,

3.58 MeV + 0.511 MeV + 3.58 MeV + 0.511 MeV = 2(Energy of Photon)

Energy of Photon = 8.182 MeV/2

<u>Energy of Photon = 4.091 MeV</u>

8 0
3 years ago
The collision between a hammer and a nail can be considered to be approximately elastic. estimate the kinetic energy acquired by
Setler [38]

Here we can use momentum conservation as in this type of collision there is no external force on it

m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}

now here we can say

m_1 = 10 g

v_{1i} = 0

m_2 = 550 g

v_{2i} = 3.5 m/s

now here we can say

10*0 + 550 * 3.5 = 10 v_{1f} + 550 v_{2f}

192.5 = v_{1f} + 55 v_{2f}

now by coefficient of restitution

for elastic collision we know that e = 1

v_{2f} - v_{1f} = e(v_{1i} - v_{2i})

v_{2f} - v_{1f} = 0 - 3.5

now by solving the two equation

56v_{2f} = 189

v_{2f} = 3.375 m/s

also we know that

v_{1f} = v_{2f} + 3.5 = 3.375 + 3.5 = 6.875 m/s

so final speed of the nail is 6.875 m/s


6 0
3 years ago
Read 2 more answers
What do the three variables (f,m, a) in the equation mean?
Anna [14]

Answer:

f=force m=mass and a=acceleration

7 0
3 years ago
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