3 years ago

How fast must a 350 gram softball be traveling in order to have a momentum of .600 kg

1 answer:

4 0

Momentum is mass times velocity. So here we can just substitute in our givens and solve for velocity.

.600kg*m/s/.350kg=1.71m/s

Hope this helps! Thank you!

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(d) What type of transformation is done by the following things?

mafiozo [28]

Answer:

human body is answer according to our studies

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2 years ago

Yub87654d HELP there is this person trying to kid,.,nap,./, kids so if you see an account called Hernyana report there questions

Morgarella [4.7K]

Answer:

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Explanation:

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2 years ago

elastic wire extend by 1.ocm when a load on 20g range from It, what additional load will it be required Cause the futher extensi

Citrus2011 [14]

Answer:

40g

Explanation:

20g range > 1.0cm

Therefore,

40g range > 2.0cm

6 0

1 year ago

g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses

statuscvo [17]

Answer:

v₁f = 0.5714 m/s (→)

v₂f = 2.5714 m/s (→)

e = 1

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)

v₁f = 0.5714 m/s (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s (→)

e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1

It was a perfectly elastic collision.

8 0

2 years ago

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a

Burka [1]

Solution :

a). B at the center :

$=\frac{u\times I}{2R}$

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

b). Also, the sum of the fields must be zero.

Therefore,

$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$