An object is placed at a distance 30cm from a thin lenses of power 4 diaptor. Discuss the nature of image

A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball

Akimi4 [234]

Answer:

Approximately $122.625\; {\rm m}$ (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ , the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let $v_{0}$ denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly $(-v_{0})$ . The velocity of the ball would be changed from $v$ to $(-v_{0})\!$ (such that $\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})$ ) within $t = 10\; {\rm s}$ .

Also because the drag on the ball is negligible, the acceleration of the ball would be $a = -g = -9.81\; {\rm m\cdot s^{-2}}$ . Thus:

$\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}$ .

Since $\Delta v = (-2\, v_{0})$ :

$-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}$ .

$\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}$ .

The ball reaches maximum height when its velocity is $v_{1} = 0\; {\rm m\cdot s^{-1}}$ . Apply the SUVAT equation $x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)$ to find the displacement $x$ between the original position (ground level, where $v_{0} = 49.05\; {\rm m\cdot s^{-1}}$ ) and the max-height position of the ball (where $v_{1} = 0\; {\rm m\cdot s^{-1}}$ .)

$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

7 0

2 years ago

Which of the following correctly describe electric field lines?

Xelga [282]

-- Electric field lines DO never cross. (A)

-- Electric field lines that are close together DO indicate a stronger electric field. (B)

-- Electric field lines DO not affect the charge that created them. (C)

-- Electric field lines DON'T begin on north poles and end on south poles. North and South "poles" are the way we talk about magnets, not electric charges.

6 0

3 years ago

Read 2 more answers

A metal wire has a resistance of 13.00 at a temperature of 25.0 degree celsius

Nady [450]

Explanation:

what exactly are you asking for?

4 0

2 years ago

Mars, which has a radius of 3.4 × 106 m and a mass of 6.4 × 1023 kg, orbits the Sun, which has a mass of 2.0 × 1030 kg at a dist

Arisa [49]

The correct answer is: revolution

6 0

3 years ago

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An RL series circuit is connected to an ac generator with a maximum emf of 20 V. If the maximum potential difference across the

Rama09 [41]

If the maximum emf of the ac generator is 20 V and the maximum potential difference across the resistor is 16 V Then the maximum potential difference across the inductor is 4 V.

Calculation:

Step-1:

It is given that the RL circuit is connected to a 20 V ac generator. The maximum potential difference across the resistor is 16 V. It is required to find the maximum potential drop across the inductor.

Step-2:

The maximum emf of the generator is equal to the sum of the maximum potential difference across the resistor and the maximum potential difference across the inductor.

Therefore,

The maximum potential difference across the inductor + Maximum maximum potential difference across the resistor = Maximum emf of the generator

Thus,

Maximum maximum potential difference across the inductor + 16 V = 20 V

Therefore,

Maximum maximum potential difference across the inductor = 20 V - 16 V = 4 V

Learn more about potential differences across resistor and inductor here,

brainly.com/question/15715072

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5 0

2 years ago