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4 years ago

Which lens can produce a virtual image and a real image?

Physics

2 answers:

Licemer1 [7]4 years ago

4 0

Answer:

Explanation:

When the two rays of light actually meets after refraction from a lens, the image is said to be real.

When the two rays of light appears to meet after refraction from a lens, the image is said to be virtual.

A concave lens always forms a virtual image, because it diverge the rays of light.

A convex lens forms a real image when the object placed between infinity and the focus of lens but t forms a virtual image when the object is placed between optical centre and the focus of the lens.

diamong [38]4 years ago

4 0

Answer:

Convex APEX Verified

Explanation:

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A car going 22 m/s increases its speed to pass a truck. Five seconds later the car is going 35 m/s. Calculate the acceleration o

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3 years ago

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During a race, a runner runs at a speed of 6 m/s. Four seconds later, she is running at a speed of 14 m/s. What is the runner's

DedPeter [7]

Answer:

D. 2 m/s²

Step-by-step explanation:

Initial speed of the runner (u) = 6 m/s

Final speed of the runner (v) = 14 m/s

Time taken (t) = 4 s

By using equation of motion, we get:

$\bf \longrightarrow v = u + at \\ \\ \rm \longrightarrow 14 = 6 + 4a \\ \\ \rm \longrightarrow 14 - 6 = 6 - 6 + 4a \\ \\ \rm \longrightarrow 8 = 4a \\ \\ \rm \longrightarrow 4a = 8 \\ \\ \rm \longrightarrow \dfrac{4a}{4} = \dfrac{8}{4} \\ \\ \rm \longrightarrow a = 2 \: m {s}^{ - 2}$

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3 years ago

What force must a rocket overcome to reach space

Nat2105 [25]

Hello! Gravity is lost when you reach the outside of the Earth's orbit! So you would have to overcome gravity in order to officially be in space. Gravity is the first thing a rocket must overcome to reach space.

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4 years ago

15 points!

VMariaS [17]

Answer:

$4.12\times 10^{-5}\ J$ .

Explanation:

Given that,

Capacitance, $C=1.4\times 10^{-7}\ F$

Charge stored in the capacitor, $Q=3.4\times 10^{-6}\ C$

We need to find the electric potential energy stored in the capacitor. The formula for the electric potential energy stored in the capacitor is given by :

$E=\dfrac{Q^2}{2C}$

Put all the values,

$E=\dfrac{(3.4\times 10^{-6})^2}{2\times 1.4\times 10^{-7}}\\\\=4.12\times 10^{-5}\ J$

So, the required electric potential eenergy is equal to $4.12\times 10^{-5}\ J$ .

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3 years ago