We can solve this problem by using the law of conservation of momentum. In fact, for an isolated system (= no external force), the total momentum must be conserved before and after the collision. So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$

where: in this case:

$m_1 = 91.5 kg$ is the mass of the first player

$u_1 = 2.73 m/s$ is the initial velocity of the first player (choosing east as positive direction)

$m_2 = 63.5 kg$ is the mass of the second player

$u_2 = 3.09 m/s$ is the initial velocity of the second player

$v$ is their combined velocity afterwards

Solving for v, we find:

$v = \frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{(91.5)(2.73)+(63.5)(3.09)}{91.5+63.5}=2.88 m/s$

And the sign is positive, so the direction is east.

Learn more about momentum here:

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7 0

2 years ago

. A laser beam shines along the surface of a block of transparent material (see Fig. E33.8). Half of the beam goes straight to a

Neporo4naja [7]

Answer:

n = 1,875

Explanation:

The speed of light in vacuum is constant (c) and in a material medium it is

v = d / t

The refractive index of a material is defined by

n = c / v

Let's look for the speed of light in the material, in general the length that light travels is known, this value is high, x = 1, when we place a block on the road, a small amount is lengthened by the length of the block, which in general is despised

These measurements are made on a digital oscilloscope that allows to stop the signals and measure their differences, that is, the zero is taken when the first ray arrives and the time for the second ray is measured,

v = d / t

v = 1 / 6.25 10⁻⁹

v = 1.6 10⁸ m / s

we calculate the refractive index

n = 3 10⁸ / 1.6 10⁸

n = 1,875

6 0

2 years ago

Dr. Kirwan is preparing a slide show that he will present to the executive board at tonight's committee meeting. He places a 3.5

lisov135 [29]

Answer:

A) d_o = 20.7 cm

B) h_i = 1.014 m

Explanation:

A) To solve this, we will use the lens equation formula;

1/f = 1/d_o + 1/d_i

Where;

f is focal Length = 20 cm = 0.2

d_o is object distance

d_i is image distance = 6m

1/0.2 = 1/d_o + 1/6

1/d_o = 1/0.2 - 1/6

1/d_o = 4.8333

d_o = 1/4.8333

d_o = 0.207 m

d_o = 20.7 cm

B) to solve this, we will use the magnification equation;

M = h_i/h_o = d_i/d_o

Where;

h_o = 3.5 cm = 0.035 m

d_i = 6 m

d_o = 20.7 cm = 0.207 m

Thus;

h_i = (6/0.207) × 0.035

h_i = 1.014 m

8 0

2 years ago

What part of the gun position equipment supports all of the elevating parts of the gun?

BARSIC [14]

The gunstock it’s also called stock or shoulder stock

5 0

3 years ago