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3 years ago

How have moral and ethical issues influenced science

1 answer:

6 0

Often times, moral and ethical issues could hindrance the development of science.
For example, the cloning technology has been hindranced by these type of issues since the beginning of its creation.
The moral and ethical aspect of society somehow oppose the technology because they feel that the sicentists are 'playing God' and doing something that humans are not suppose to be doing, even though there might be overwhelming benefits from the technology.

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Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the

sdas [7]

Answer:

The new distance is d = 0.447 d₀

Explanation:

The electric out is given by Coulomb's Law

F = k q₁ q₂ / r²

This electric force is in balance with tension.

We reduce the charge of sphere B to 1/5 of its initial value ( $q_{B}$ =q₂ = q₂ / 5) than new distance (d = n d₀)

dat

q₁ = $q_{A}$

q₂ = $q_{B}$

r = d₀

In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

F = k q₁ q₂ / d₀²

F = k q₁ (q₂ / 5) / (n d₀)²

.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)

5 n² = 1

n = √ 1/5

n = 0.447

The new distance is

d = 0.447 d₀

6 0

3 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

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3 years ago

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Anton [14]

The answer is C conduction

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Volumetric flasks are most accurate

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A total electric charge of 4.00 nC is distributed uniformly over the surface of a metal sphere with a radius of 26.0 cm. The pot

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