Answer: hello your question is incomplete below is the missing part
A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.
answer: 
- q
Explanation:
Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero 
given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q
 
        
             
        
        
        
<span>Most low-level radioactive waste (LLW) is typically sent to land-based disposal immediately following its packaging for long-term management. This means that for the majority (~90% by volume) of all of the waste types produced by nuclear technologies, a satisfactory disposal means has been developed and is being implemented around the world.
</span>
Radioactive wastes are stored so as to avoid any chance of radiation exposure to people, or any pollution.The radioactivity of the wastes decays with time, providing a strong incentive to store high-level waste for about 50 years before disposal.Disposal of low-level waste is straightforward and can be undertaken safely almost anywhere.Storage of used fuel is normally under water for at least five years and then often in dry storage.<span>Deep geological disposal is widely agreed to be the best solution for final disposal of the most radioactive waste produced.
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Answer:
Q = 12540  J
Explanation:
It is given that,
Mass of water, m = 50 mL = 50 g
It is heated from 0 degrees Celsius to 60 degrees Celsius.
We need to find the energy required to heat the water. The formula use to find it as follows :

Where c is the specific heat of water, c = 4.18 J/g°C
Put all the values, 

So, 12540 J of energy is used to heat the water.
 
        
             
        
        
        
Answer:
   r = 3.787 10¹¹ m
Explanation:
We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration
     F = ma
     G m M / r² = m a
The centripetal acceleration is given by
     a = v² / r
For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship
     v = d / t
The distance traveled Esla orbits, in a circle the distance is
     d = 2 π r
Time in time to complete the orbit, called period
      v = 2π r / T
Let's replace
     G m M / r² = m a
     G M / r² = (2π r / T)² / r
     G M / r² = 4π² r / T²
     G M T² = 4π² r3
      r = ∛ (G M T² / 4π²)
Let's reduce the magnitudes to the SI system
      T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)
      T = 1.03 10⁸ s
Let's calculate
       r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]
       r = ∛ (21.44 10³⁵ / 39.478)
       r = ∛(0.0543087 10 36)
       r = 0.3787 10¹² m
       r = 3.787 10¹¹ m
 
        
             
        
        
        
Answer:
Explanation:
net force on the skier = mg sin 39 - μ mg cos39 
mg ( sin39 - μ cos39 ) 
= 73 x 9.8 ( .629 - .116)
= 367 N 
impulse = net force x time = change in momentum .
= 367 x 5 = 1835 kg m /s 
velocity of the skier after 5 s = 1835 / 73 
= 25.13 m /s 
b ) 
net force becomes zero 
mg ( sin39 - μ cos39 ) = 0 
μ = tan39 
= .81 
c )
net force becomes zero , so he will continue to go ahead with constant speed of 25.13 m /s
so he will have speed of 25.13 m /s after 5 s .