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yanalaym [24]
3 years ago
12

if a ball is rolling at a velocity of 1.5 m/sec and has the momentum of 10.0 kg m/sec, what is the mass of the ball?

Physics
1 answer:
PolarNik [594]3 years ago
6 0
Okay. There is a simple formula to go with this where:

p = mv

P: Momentum.
M: Mass.
V: Velocity

Sub the numbers in and solve for M.

10.0 = m(1.5)
10.0/1.5 = m
6.67 kg = m

Therefore the mass of the ball is 6.67kg.
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A current of 1.8 A delivers 2.5 C of charge. How much time was required? 0.70 s 0.72 s 1.4 s 4.5 s
Natalka [10]

Answer:

1.4 s

Explanation:

Given the following data;

Quantity of charge, Q = 2.5 C

Current = 1.8 A

To find the time required;

Mathematically, the quantity of charge passing through a conductor is given by the formula;

Quantity of charge, Q = current * time

Substituting into the formula, we have;

2.5 = 1.8 * time

Time = 2.5/1.8

Time = 1.4 s

5 0
3 years ago
Read 2 more answers
Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep
SIZIF [17.4K]

Answer:

1) iii i= 1m, 2)  iii and iv, 3)  i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

              f = r / 2

              f = 2/2

             f = 1 m

The builder's equation

           1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

              1 / f = 0 + 1 / i

              i = f

              i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

              R₂ = infinity

              R₁ > 0

Cas2 Flat-concave (divergent) lens

             R₂ = infinity

              R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

          1 / f₂ = 1 / (L-f₁) + 1 / i

          1 / i = 1 / f₂ - 1 / (L-f₁)

           1 / i = (L-f₁-f₂) / f₂ (L-f₁)

           i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

6 0
4 years ago
You are classifying an organism that is unfamiliar to you.
MatroZZZ [7]

Answer:

I know that we can identify it

Explanation:

C. structure

4 0
3 years ago
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A 2.00 kg, frictionless block s attached to an ideal spring with force constant 550 N/m. At t = 0 the spring is neither stretche
gladu [14]

Answer:

a)    A = 0.603 m , b) a = 165.8 m / s² , c)  F = 331.7 N

Explanation:

For this exercise we use the law of conservation of energy

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    Em₀ = K = ½ m v²

End Point with fully compressed spring

    Em_{f} = K_{e} = ½ k x²

    Emo = Em_{f}

    ½ m v² = ½ k x²

    x = √(m / k)    v

    x = √ (2.00 / 550)   10.0

    x = 0.603 m

This is the maximum compression corresponding to the range of motion

     A = 0.603 m

b) Let's write Newton's second law at the point of maximum compression

    F = m a

    k x = ma

    a = k / m x

    a = 550 / 2.00 0.603

    a = 165.8 m / s²

With direction to the right (positive)

c) The value of the elastic force, let's calculate

    F = k x

    F = 550 0.603

   F = 331.65 N

5 0
3 years ago
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Answer:

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Explanation:

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4 0
3 years ago
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