Which type of atom has the same number of protons and electrons

Bambi the young dear was distracted Buy butterfly and jumped into the road in front of the two vehicles as shown in the diagram

bagirrra123 [75]

Speed of car A is given as

$v_a = 70 mph$

now we need to convert it into SI units

1 miles = 1609 m

1 hour = 3600 s

now we have

$v_a = 70 *\frac{1609}{3600} = 31.3 m/s$

now its distance from Bambi is given as

$d_a = 350 m$

time taken by it to hit the Bambi

$t = \frac{d}{v}$

$t = \frac{350}{31.3}$

$t = 11.2 s$

Now other car is moving at speed 50 mph

so its speed in SI unit will be

$v_b = 50* \frac{1609}{3600}$

$v_b = 22.35 m/s$

now its distance from Bambi is given as

$d_b = 590 feet$

as we know that 1 feet = 0.3048 m

$d_b = 590*0.3048 = 179.83 m$

now the time to hit the other car is

$t_2 = \frac{179.83}{22.35}$

$t_b = 8.05 s$

So Car B will hit the Bambi first

7 0

3 years ago

Determine the maximum r-value of the polar equation r =3+3 cos 0

AURORKA [14]

[r] =6

Solve for r by simplifying both sides of the equation, then isolating the variable.

<em> </em>I hope this makes sense

8 0

2 years ago

Read 2 more answers

Potassium ions (K+) move across a 7.0 -mm- thick cell membrane from the inside to the outside. The potential inside the cell is

Reil [10]

Explanation:

Relation between potential energy and charge is as follows.

U = qV

or, $\Delta U = q \times \Delta V$

= $1.6 \times 10^{-19} \times 70 \times 10^{-3}$

= $112 \times 10^{-22}$ J

or, = $1.12 \times 10^{20} J$

Therefore, we can conclude that change in the electrical potential energy $\Delta U$ is $1.12 \times 10^{20} J$ .

7 0

2 years ago

Plz help me and thx if you do!

Rina8888 [55]

I think it’s b... not sure tho sorry

3 0

2 years ago

One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel

Gekata [30.6K]

Answer:

a) $v_{U-235} = 2.68 \cdot 10^{5} m/s$

$v_{He-4} = -1.57 \cdot 10^{7} m/s$

b) $E_{He-4} = 8.23 \cdot 10^{-13} J$

$E_{U-235} = 1.41 \cdot 10^{-14} J$

Explanation:

Searching the missed information we have:

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

$p_{i} = p_{f}$

$m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

Since the plutonium nucleus is originally at rest, $v_{Pu-239} = 0$ :

$0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

$v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}$ (1)

Kinetic Energy:

$E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}$

$2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$ (2)

By entering equation (1) into (2) we have:

$1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}$

Solving the above equation for $v_{U-235}$ we have:

$v_{U-235} = 2.68 \cdot 10^{5} m/s$

And by entering that value into equation (1):

$v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s$

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

$E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J$

For U-235:

$E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J$

I hope it helps you!

3 0

3 years ago