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GREYUIT [131]
3 years ago
9

Please can someone answer these questions

Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
4 0

Answer:

Kkk

Explanation:

Syd hsf and I 4 want to be 16 and just want to

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6. How would the measurements for potential difference and current change if a 200 Ω resistor was used in Circuit 1 instead of t
Maksim231197 [3]

Answer:

Explanation:

Resistance is defined as the opposition to the flow of an electric current in a circuit. This means that a higher amount of resistance tends to reduce the amount of current flowing through the resistance. The lower the current, the greater the possibility for the resistor to allow current to pass through it.  if a 200 Ω resistor was used in Circuit 1 instead of the 100 Ω resistor, then the current in the circuit will tends to increase since we are replacing the load with a lesser resistor and a smaller resistance tends to allow more current to flow through it

For the potential difference, a decrease in the resistance value will onl decrease the potential difference flowing in the circuit according to ohm's law. According to the law the pd in a circuit is directly proportional to the current which means an increase in the resistance value will cause an increase in the corresponding pd and vice versa.

7 0
3 years ago
A blue and a green billiard ball, each with a mass of 0.15 kg, collide directly. Before the collision, the blue ball had a speed
cestrela7 [59]

Answer:

.

Explanation:

Can you specify what is the question asking for

4 0
2 years ago
What gas makes up 75 of earth's atmosphere?
dexar [7]
"Nitrogen gas" makes up 75% (approx. 78%) of the earth's atmosphere

Hope this helps!
6 0
3 years ago
The fiugure below shows a birds eye view of ta car going around a highway curve
luda_lava [24]
Okay whats the question though?
5 0
4 years ago
Point charges q1 and q2 are separated by a distance of 60 cm along a horizontal axis.
amm1812

Answer:

38 cm from q1(right)

Explanation:

Given, q1 = 3q2 , r = 60cm = 0.6 m

Let that point be situated at a distance of 'x' m from q1.

Electric field must be same from both sides to be in equilibrium(where EF is 0).

=> k q1/x² = k q2/(0.6 - x)²

=> q1(0.6 - x)² = q2(x)²

=> 3q2(0.6 - x)² = q2(x)²

=> 3(0.6 - x)² = x²

=> √3(0.6 - x) = ± x

=> 0.6√3 = x(1 + √3)

=> 1.03/2.73 = x

≈ 0.38 m = 38 cm = x

8 0
3 years ago
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