Answer: 7.3seconds
Explanation:
Using the echo concept, before echo can occur, we must have an object making the sound (the teacher), the diatance between the object and the reflector (x)(i.e the wall), the time taken for the shadow to disappear (t) and the velocity of sound /movement of the object (v). Using the relationship 2x = vt
According to the question, x = 12m (initial distance), v = speed of object t = time of disappearance.
We have time before disappearance calculated as 2 × 12/1.1 = 21.8s
When he is 4m from the building, it will there take 4 × 21.8/12 = 7.3seconds for his shadow to decrease on the way.
<h2>
Answer:50</h2>
Explanation:
Let be the airspeed.
Let be the cross wind speed.
We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.
If and are two perpendicular vectors,the resultant vector has the magnitude
Given,
So,the ground speed is
Before we start thinking about the snowball, we need to remind
each other that energy is "conserved". That means that if you
ever see energy decrease in one place, then the missing amount
must have gone somewhere, and if you ever see energy increase
in one place, then the energy that appeared must have come from
somewhere. Energy does not magically appear or disappear.
So you toss a snowball out of your hand. As you let it go, you give
it some kinetic energy, and it starts rolling along the ground.
Once you let go of it, it can't get any more energy (unless it has
some kind of little tiny engine inside it).
Kinetic Energy = (1/2) (mass) (speed)² .
and that amount can't change.
So if extra snow sticks to it as it merrily rolls along, and its mass
increases, then it must slow down.
Radius of curvature<span> of a spherical mirror = 2 × </span>Focal length<span>
and the ray of light is parallel to the principle axis if the incident </span><span>Passes through focus
therefore, focal length =30/2=<span>15cm
</span></span><span>The mirror should be placed at a distance of 15cm from the light bulb to produce this parallel light beam.</span>