All categories

3 years ago

If you were standing at the center of curvature in front of a concave mirror, what image would be projected?

2 answers:

6 0

"<span>The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are" is the type of image among the choices given in the question that would be projected. The correct option among all the options that are given in the question is the first option. I hope it helps you.</span>

Bas_tet [7]3 years ago

5 0

Answer:

The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

Explanation:

As we know that the object position is at center of curvature of the mirror

So here the distance is given as

$d = R = 2f$

now by mirror formula of relation between distance of image and distance of object

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

$\frac{1}{d_i} + \frac{1}{2f} = \frac{1}{f}$

$d_i = 2f$

so image will form at the position of object itself

Now we will say that the magnification of the object will be

$M = -\frac{d_i}{d_o} = -1$

so image will be upside down and same as that the height of object

so correct answer would be

The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

You might be interested in

What is the soil fabric?

inessss [21]

Answer:

The term soil fabric refers to the geometrical arrangement of individual particles in a soil, including the geometrical distribution of pore spaces. Soil fabric may be used to describe the particle arrangement in both cohesive soils such as clays and granular soils such as silts, sands, and gravels.

5 0

3 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$