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Mariulka [41]
3 years ago
15

What is the momentum of a baseball with a mass of 4 kg being thrown at a velocity of 84 m/s towards the hitter

Physics
2 answers:
Setler79 [48]3 years ago
8 0
<h2><u>Given</u> :</h2>
  • A mass of 4 kg being thrown at a velocity of 84 m/s towards the hitter
<h2><u>To Find</u> :</h2>
  • Momentum of a baseball
<h2><u>Solution</u> :</h2>

<u>We have</u>,

  • Mass = 4 kg
  • Velocity = 84 m/s

<u>We know that</u>,

<h3>→ Momentum = Mass × Velocity</h3>

<u>Substituting the values we get</u>,

\impliesMomentum = 4 × 84

\impliesMomentum = 336 kgm/s

∴ The momentum is 336 kgm/s.

Delvig [45]3 years ago
7 0

Answer:

<h3>The answer is 336 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 4 kg

velocity = 84 m/s

We have

momentum = 4 × 84

We have the final answer as

<h3>336 kgm/s</h3>

Hope this helps you

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Answer:

Real images are formed where rays of light actually converge, whereas virtual images occur with they are perceived to converge. Real images can be produced by passing light through converging lenses or with a concave mirror of some sort.

Explanation:

hope that helped a little :)

6 0
3 years ago
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A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x
DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is \lambda

Total charge, Q = 7 nC = 7\times 10^{- 9} C

At x = 2L,

Electric field, \vec{E_{2L}} = 500N/C

Coulomb constant, K = 8.99\times 10^{9} N.m^{2}/C^{2}

Now, we know that:

\vec{E} = K\frac{Q}{x^{2}}

Also the line charge density:

\lambda = \frac{Q}{L}

Thus

Q = \lambda L

Now, for small element:

d\vec{E} = K\frac{dq}{x^{2}}

d\vec{E} = K\frac{\lambda }{x^{2}}dx

Integrating both the sides from x = L to x = 2L

\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx

\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]

\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}

Similarly,

For the field in between the range 2L< x < 3L:

\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx

\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]

\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}

Now,

If at x = 2L,

\vec{E_{2L}} = 500 N/C

Then at x = 3L:

\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C

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