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3 years ago

In a concave mirror parallel rays falling on it convergs at

Physics

1 answer:

ella [17]3 years ago

5 0

Answer:

1) In a concave mirror parallel rays falling on it converges at F and 2F.

Explanation:

Spherical mirrors can be used for magnification of images. There are basically two types of spherical mirrors and they are converging mirror and diverging mirrors. The converging mirrors are also termed as concave mirrors and its basic work is to converge or combine light rays coming from a larger distance to a single point. Mostly the light beams falling parallel to the principle axis of the concave mirror will be acting as parallel rays. And when these parallel rays fall on the mirror, the converging point can be the focal point of the mirror.

Thus the location of converging point in concave mirrors will be based on the position or distance of object from the mirror. If the object distance is very far from the twice the focal length distance of mirror, then the converging point will be the focal point or F. And if the object is placed slightly greater than twice the distance of focal point, then the image will be obtained at 2F. But the parallel beams will be converging at F and 2F.

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An upright object 2.80 cm tall is placed 16.0 cm away from the vertex of a concave mirror with a center of curvature of 24.0 cm.

horrorfan [7]

Answer:

f = 12 cm

Explanation:

Center of Curvature:

The center of that hollow sphere, whose part is the spherical mirror, is known as the ‘Center of Curvature’ of mirror.

The Radius of Curvature:

The radius of that hollow sphere, whose part is the spherical mirror, is known as the ‘Radius of Curvature’ of mirror. It is the distance from pole to the center of curvature.

Focal Length:

The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’.

The focal length of the spherical (concave) mirror is approximately equal to half of the radius of curvature:

$f = \frac{R}{2}$

where,

f = focal length = ?

R = Radius of curvature = 24 cm

Therefore,

$f = \frac{24\ cm}{2}$

f = 12 cm

8 0

2 years ago

A vehicle moving with a uniform a acceleration of 2m/s and has a velocity of 4m/s at a certain time .

Evgesh-ka [11]

Answer:

1 second later the vehicle's velocity will be:

$v(1)= 6\,\,\frac{m}{s} \\$

5 seconds later the vehicle's velocity will be:

$v(5)=14\,\,\frac{m}{s}$

Explanation:

Recall the formula for the velocity of an object under constant accelerated motion (with acceleration " $a$ "):

$v(t)=v_0+a\,t$

Therefore, in this case $v_0=4\,\,\frac{m}{s}$ and $a=2\,\,\frac{m}{s^2}$

so we can estimate the velocity of the vehicle at different times just by replacing the requested "t" in the expression:

$v(t)=v_0+a\,t\\v(t)=4+2\,\,t\\v(1)=4+2\,(1) = 6\,\,\frac{m}{s} \\v(5)=4+2\,(5)=14\,\,\frac{m}{s}$

3 0

3 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0

3 years ago

Someone please help its a simple power problem.

SOVA2 [1]

Well 200 doubled or (x2)=400 if that’s what it means

7 0

2 years ago

a moving ball rolls into a stationary ball. the total momentum of both balls after the collision will be

rosijanka [135]

The momentum of the rolling ball will have less momentum than before the collision and the stationary ball will have more momentum after the collusion.

5 0

3 years ago

Read 2 more answers