The kilogram is the Standard International System of Units unit of mass. It is defined as the mass of a particular international prototype made of platinum-iridium and kept at the International Bureau of Weights and Measures.

8 0

3 years ago

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You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between

Anni [7]

Answer:

0.074m/s

Explanation:

We need the formula for conservation of momentum in a collision, this equation is given by,

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

Where,

$m_1$ = mass of ball

$m_2$ = mass of the person

$u_1$ = Velocity of ball before collision

$u_2$ = Velocity of the person before collision

$v_1$ = velocity of ball afer collision

$v_2$ = velocity of the person after collision

We know that after the collision, as the person as the ball have both the same velocity, then,

$v_1 = v_2$

$m_1u_1 + m_2u_2 = (m_1+m_2)v_2$

Re-arrenge to find $v_2$ ,

$v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}$

Our values are,

$m_1$ = 0.425kg

$u_1$ = 12m/s

$m_2$ = 68.5kg

$u_2$ = 0m/s

Substituting,

$v_2 = \frac{(0.425)(12)+(68.5)(0)}{0.425+68.5}$

$v_2 = 0.074m/s$

The speed of the person would be 0.074m/s after the collision between him/her and the ball

7 0

3 years ago

A point charge q1q1 is held stationary at the origin. A second charge q2q2 is placed at point aa, and the electric potential ene

Yuri [45]

Explanation:

The given data is as follows.

$U_{a} = 5.4 \times 10^{-8} J$

$W_{/text{a to b}} = -1.9 \times 10^{-8} J$

Electric potential energy ( $U_{b}$ ) = ?

Formula to calculate electric potential energy is as follows.

$U_{b} = U_{a} - W_{/text{a to b}}$

= $5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)$

= $7.3 \times 10^{-8} J$

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is $7.3 \times 10^{-8} J$ .

6 0

4 years ago