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77julia77 [94]
3 years ago
5

At the point of fission, a nucleus of ^235 U that has 92 protons is divided into two smaller spheres, each of which has 46 proto

ns and a radius of 5.90 X 10 ^-15m. What is the magnitude of the repulsive force pushing these two spheres apart
Physics
1 answer:
IrinaK [193]3 years ago
3 0

Answer : F = 3.5\times10^{3}\ N

Explanation :

Given that

Radius of sphere r = 5.90\times 10^{-15}\ m

The distance between the centers of the two spheres is

r = 2\times 5.90\times 10^{-15}\ m

The charge of the sphere q = 46\times1.6\times10^{-19} C

The magnitude of the repulsive force between the charges pushing them a part is

Using coulomb law

F = \dfrac {kq_{1}q_{2}}{r^{2}}

F = \dfrac{9\times10^{9}\times (46\times1.6\times10^{-19})^{2}C^{2}}{2\times(5.90\times10^{-15})^{2}\ m^{2}}

F = 3501.3\ N

F = 3.5\times10^{3}\ N

Hence, this is the required solution.









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<span>So the nurse must exert 80.34 N of force</span>

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A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.
kondaur [170]

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

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F_k = \mu_k N = ma

where;

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N is the normal reaction = mg

μkmg = ma

μkg = a

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3 0
3 years ago
Help please, I need it
love history [14]

Answer:

1.97×10⁻²¹ J

Explanation:

Use ideal gas law to find temperature.

PV = nRT

(9 atm) (9 L) = (83.3 mol) (0.0821 L·atm/mol/K) T

T = 11.9 K

The average kinetic energy per atom is:

KE = 3/2 kT

KE = 3/2 (1.38×10⁻²³ J/K) (11.9 K)

KE = 2.46×10⁻²² J

For a mass of 5.34×10⁻²⁶ kg, the kinetic energy is:

KE = (5.34×10⁻²⁶ kg) (1 mol / 0.004 kg) (6.02×10²³ atom/mol) (2.46×10⁻²² J)

KE = 1.97×10⁻²¹ J

5 0
3 years ago
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