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olga_2 [115]
3 years ago
13

10. If one body is positively charged and another body is negatively charged, free electrons tend to

Physics
1 answer:
marta [7]3 years ago
7 0

Answer: B.move from the negatively charged body to the positively charged body.

Explanation: If one body is positively charged and another body is negatively charged, free electrons tend to move from the negatively charged body to the positively charged body. This happens because they have different charges and therefore electrons are more attracted to the positive or its opposite charge.

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HIIIIIIIIIIIIIIIIIII WHAT IS 50X550
jekas [21]
50 x 550 is 27500!!!!
3 0
4 years ago
Sharks are generally negatively buoyant; the upward buoyant force is less than the weight force. This is one reason sharks tend
Tresset [83]

Answer:

8.67807 N

34.7123 N

Explanation:

m = Mass of shark = 92 kg

\rho_{se} = Density of seawater = 1030 kg/m³

\rho_{f} = Density of freshwater = 1000 kg/m³

\rho_{sh} = Density of shark = 1040 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Net force on the fin is (seawater)

F_n=mg-V_s\rho_{se}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{se}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1030\times 9.81\\\Rightarrow F_n=8.67807\ N

The lift force required in seawater is 8.67807 N

Net force on the fin is (freshwater)

F_n=mg-V_s\rho_{f}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{f}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1000\times 9.81\\\Rightarrow F_n=34.7123\ N

The lift force required in a river is 34.7123 N

6 0
4 years ago
Carlos has eight marbles that are all the same size. he knows that one marble has a mass of 10 grams. to find out the total mass
AURORKA [14]
Carlos lost his marbles 
4 0
3 years ago
Exercice 1
Mariana [72]

Answer:

Injections of aqueous solution of fruct

levulose of formula C, H, O,,

For

prevent dehydration such solutions are obtained

dissolving a mass m = 25g of fructose for a volume of 50

final solution.

1.1 Calculate the molecular molar mass of fructose

1.2 Determine the amount of corresponding fructose material

1.3 Calculate the molar concentration of these fructose solutions

1.4

6 0
3 years ago
A circular hole in an aluminum plate is 2.739 cm in diameter at 0.000°C. What is the change in its diameter when the temperature
prisoha [69]

Answer:

L = 2.746 cm

Explanation:

As we know that thermal expansion coefficient of aluminium is given as

\alpha = 24 \times 10^{-6} per ^oC

now we also know that after thermal expansion the final length is given as

L = L_o(1 + \alpha \Delta T)

here we know that

L_o = 2.739 cm

\alpha = 24 \times 10^{-6}

\Delta T = 108 - 0= 108^oC

now we will have

L = 2.739(1 + 24 \times 10^{-6} (108))

L = 2.746 cm

3 0
4 years ago
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