The transfer of heat is through direct contact of particles is called

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

Vladimir79 [104]

Answer:

The answer is " $1.01 \times 10^{-13}$ "

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

$KE=U=\frac{kqq'}{r}\\\\$

Calculating the closest distance:

$\to r=\frac{kqq'}{KE}\\\\$

$=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\$

$=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}$

5 0

2 years ago

Pls how to solve this problem. help would be appreciated

Bingel [31]

Answer:

80 m/s

Explanation:

Given:

a = -5 m/s²

v = 0 m/s

Δx = 640 m

Find: v₀

v² = v₀² + 2a(x − x₀)

(0 m/s)² = v₀² + 2(-5 m/s²) (640 m)

v₀ = 80 m/s

7 0

3 years ago

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,

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

$v_2 = 0.074m/s$

<em />

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

7 0

3 years ago

Why does a lone pair of electrons occupy more space around a central atom than a bonding pair of electrons?

Inessa [10]

Answer:

The lone pair of electrons occupy more space because the electrostatic force becomes weaker.

Explanation:

When there is a bond pair of electrons in the 2 positively charged the atomic nuclei draw the electron density towards them, thereby reducing the bond diameter.

In the case of the lone pair, only 1 nucleus is present, and the enticing electrostatic force becomes weaker and the intensity of the electrons will be increases. Therefore, the lone pair occupies more space than the pair of bonds.

5 0

2 years ago

Cliff is a postal worker, who placed a 0.60 kg package on a scale, compressing the scale by 0.03 m. What is the force constant o

lesya692 [45]

Hopefully this helps :)

7 0

3 years ago