All categories

3 years ago

What is the nature of image of Sun obtained on a paper with the help of concave mirror

Physics

1 answer:

Natalka [10]3 years ago

5 0

Answer:

Because concave mirror focus the sunlight at one particular point . At that particular point , amount of sunlight is converged and heat is trapped at that point . This is because , it burn the paper on which the image is focussed .

You might be interested in

Ahmad pulls a sled across level snow by 75 N . If the sled moves 15 m , how much work does Ahmad do on the sled?

Nataly_w [17]

Answer:

1125 J

Explanation:

W=F x s

= 75 x 15

=1125

4 0

2 years ago

A 0.5 kg block slides down a frictionless semicircular track from a height, h, and collides with a second 1.5 kg block at the bo

Ede4ka [16]

Answer:

$h' = 0.062\cdot h$

Explanation:

The speed of the 0.5 kg block just before the collision is found by the Principle of Energy Conservation:

$U_{g} = K$

$m\cdot g \cdot h = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{2\cdot g \cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot h}$

$v \approx 4.429\cdot \sqrt{h}$

Knowing that collision is inelastic, the speed just after the collision is determined with the help of the Principle of Momentum Conservation:

$(0.5\,kg) \cdot (4.429\cdot \sqrt{h}) = (2\,kg)\cdot v$

$v = 1.107\cdot \sqrt{h}$

Lastly, the height reached by the two blocks is:

$K = U_{g}$

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot h'$

$h' = \frac{v^{2}}{2\cdot g}$

$h' = \frac{(1.107\cdot \sqrt{h})^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h' = 0.062\cdot h$

3 0

3 years ago

What is longitudinal wave?

nignag [31]

Answer: The definition is a longitudinal wave is a wave consisting of a periodic disturbance or vibration that takes place in the same direction as the advance of the wave

Explanation:

7 0

3 years ago

A hot-air balloon is rising straight up with a speed of 4.6 m/s. a ballast bag is released from rest relative to the balloon at

kherson [118]

The time taken by the ballast bag to reach the ground is 2.18 s

The ballast bag at rest with respect to the balloon has the upward velocity (u) of 4.6 m/s , which is the velocity of the balloon. When it is dropped from the balloon, its motion is similar to an object thrown upwards with an initial velocity u and it falls under the acceleration due to gravity g.

Taking the upward direction as positive and the downward direction as negative, the following equation of motion may be used.

$s=ut+\frac{1}{2}at^2$