Answer the following question about the attached diagram:1) Will the image form at A or B?
2) Will the image be erect or inverted?
3) Will the image be enlarged, diminished, or the same size?
4) What kind of mirror is this?
Answers
Answer:
1) Image is formed at B
2) The image is an inverted real image
3) The image is a diminished image
4) The mirror is a concave mirror.
Explanation:
According to the diagram, object placed at this point in front of the curved mirror (concave) is placed beyond the center of curvature of the mirror.
To form images by an object placed in front of a curved mirror, ray diagrams are used.
According to the diagram, the first ray from the object incidented on the mirror and parallel to the principal axis converges at the focus (f) after reflection. The second incident ray pass through the focus and then reflect back with reflected ray parallel to the principal axis.
The two reflected ray intersects back at a point B according to the diagram and the image will be formed at that point B with the image head pointing towards the point of intersection.
According to the diagram above, the image will be formed at B and the nature of the image will be inverted, real, diminished (smaller than the object) and formed between the center of curvature and the focus of the mirror.
The mirror in the diagram is a concave mirror since the plane surface of the mirror is bent inwards.
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The statement “In power plants, generators convert mechanical energy into chemical energy” is false. It should be that the generator will convert chemical energy from the raw material such as fuel oil that it receives into mechanical energy to generate energy.
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Answers
Answer:
Given :-
Initial Velocity (u) = 5 m/s
Final Velocity (v) = 15 m/s
Time (t) = 7.5 seconds
To Find :-
Distance (s) = ?
Solution :-
First of all we will find the acceleration of car :
→ Acceleration = Final Velocity - Initial Velocity ÷ Time
→ Acceleration = 15 - 5 ÷ 7.5
→ Acceleration = 10 ÷ 7.5
→ Acceleration = 1.3 m/s²
Now, by using Second equation of motion we can find distance travelled by the car :
→ s = ut + ½ at²
→ s = 5 × 7.5 + ½ × 1.3 × (7.5)²
→ s = 37.5 + ½ × 1.3 × 56.25
→ s = 37.5 + 1.3 × 28.125
→ s = 37.5 + 29.425
→ s = 66.925 m
Therefore,the distance travelled by the car is 66.925 m.
Answers
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Answers
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Which statement best compares the potential
difference across each resistor?
(1) The potential difference across the 6-ohm
resistor is the same as the potential
difference across the 3-ohm resistor.
(2) The potential difference across the 6-ohm
resistor is twice as great as the potential
difference across the 3-ohm resistor.
(3) The potential difference across the 6-ohm
resistor is half as great as the potential
difference across the 3-ohm resistor.
(4) The potential difference across the 6-ohm
resistor is four times as great as the potential
difference across the 3-ohm resistor.
Answers
Answer: The potential difference across both the resistors is the same.
Explanation :
Given that, two resistors 3 -ohm and 6 ohms are connected in parallel across a 9-volt battery. The equivalent resistance, in this case, is given by :
In parallel combination, the current across each resistor divides and the potential difference remains the same.
So, the correct statement is (1) "The potential difference across the 6-ohm resistor is the same as the potential difference across the 3-ohm resistor."
The statement best compares the potential difference across each resistor will be 1.
What is resistance?
Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.
The equivalent resistance is given by;
When two resistors are connected in parallel, the current across each resistor splits while the potential difference stays constant.
The potential difference across the 6-ohmresistor is the same as the potential difference across the 3-ohm resistor.
Hence the statement best compares the potential difference across each resistor will be 1.
To learn more about the resistance refer to the link;