If your wanting to know how to talk to them, just act like you know them really well be very causal and be confident in what you say, they also like it when you make the first move to talk to them
Answer:
400Pa
Explanation:
use pressure = force/area
Answer:
i) 21 cm
ii) At infinity behind the lens.
iii) A virtual, upright, enlarged image behind the object
Explanation:
First identify,
object distance (u) = 42 cm (distance between object and lens, 50 cm - 8 cm)
image distance (v) = 42 cm (distance between image and lens, 92 cm - 50 cm)
The lens formula,
![\frac{1}{v} -\frac{1}{u} =\frac{1}{f}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv%7D%20-%5Cfrac%7B1%7D%7Bu%7D%20%3D%5Cfrac%7B1%7D%7Bf%7D)
Then applying the new Cartesian sign convention to it,
![\frac{1}{v} +\frac{1}{u} =\frac{1}{f}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv%7D%20%2B%5Cfrac%7B1%7D%7Bu%7D%20%3D%5Cfrac%7B1%7D%7Bf%7D)
Where f is (-), u is (+) and v is (-) in all 3 cases. (If not values with signs have to considered, this method that need will not arise)
Substituting values you get,
i) ![\frac{1}{42} +\frac{1}{42} =\frac{1}{f}\\\frac{2}{42} =\frac{1}{f}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B42%7D%20%2B%5Cfrac%7B1%7D%7B42%7D%20%3D%5Cfrac%7B1%7D%7Bf%7D%5C%5C%5Cfrac%7B2%7D%7B42%7D%20%3D%5Cfrac%7B1%7D%7Bf%7D)
f = 21 cm
ii) u =21 cm, f = 21 cm v = ?
Substituting in same equation![\frac{1}{v} =\frac{1}{21} =\frac{1}{21} \\\\\frac{1}{v} = 0\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv%7D%20%3D%5Cfrac%7B1%7D%7B21%7D%20%3D%5Cfrac%7B1%7D%7B21%7D%20%5C%5C%5C%5C%5Cfrac%7B1%7D%7Bv%7D%20%3D%200%5C%5C)
v ⇒ ∞ and image will form behind the lens
iii) Now the object will be within the focal length of the lens. So like in the attachment, a virtual, upright, enlarged image behind the object.
Answer:
600 J
Explanation:
The formula to find the kinetic energy of an object is
- m = mass in kg
- V = velocity in m/s
- KE is measured in Joules just as all other forms of energy.
Now, let's plug in the variables we're given and simplify.
Thus, the answer is 600 Joules.
Answer:
Explanation:
mass of first marble ![m_1=m=0.015\ kg](https://tex.z-dn.net/?f=m_1%3Dm%3D0.015%5C%20kg)
Initial velocity of the first marble
considering right side as positive
Mass of second marble
After collision first marble moves to the left with a velocity of 18 cm/s
i.e. ![v_1=-18\ cm/s](https://tex.z-dn.net/?f=v_1%3D-18%5C%20cm%2Fs)
considering
be the velocity of second marble after collision
The Coefficient of restitution is 1 for an elastic collision
![e=\frac{v_2-v_1}{u_1-u_2}](https://tex.z-dn.net/?f=e%3D%5Cfrac%7Bv_2-v_1%7D%7Bu_1-u_2%7D)
Putting values
![1=\frac{v_2-(-18)}{22.5-(-18)}\\22.5+18=v_2+18\\v_2=22.5\ m/s](https://tex.z-dn.net/?f=1%3D%5Cfrac%7Bv_2-%28-18%29%7D%7B22.5-%28-18%29%7D%5C%5C22.5%2B18%3Dv_2%2B18%5C%5Cv_2%3D22.5%5C%20m%2Fs)
So, the velocity of the second marble is 22.5 m/s to the right after the collision
(b)Initial kinetic energy =![0.5\times 0.015\times (22.5\times 10^{-2})^2+0.5\times 0.015\times (18\times 10^{-2})^2=6.22\times 10^{-4}\ J](https://tex.z-dn.net/?f=0.5%5Ctimes%200.015%5Ctimes%20%2822.5%5Ctimes%2010%5E%7B-2%7D%29%5E2%2B0.5%5Ctimes%200.015%5Ctimes%20%2818%5Ctimes%2010%5E%7B-2%7D%29%5E2%3D6.22%5Ctimes%2010%5E%7B-4%7D%5C%20J)
Final kinetic energy=
![0.5\times 0.015\times (18\times 10^{-2})^2+0.5\times 0.015\times (22.5\times 10^{-2})^2=6.22\times 10^{-4}\ J](https://tex.z-dn.net/?f=0.5%5Ctimes%200.015%5Ctimes%20%2818%5Ctimes%2010%5E%7B-2%7D%29%5E2%2B0.5%5Ctimes%200.015%5Ctimes%20%2822.5%5Ctimes%2010%5E%7B-2%7D%29%5E2%3D6.22%5Ctimes%2010%5E%7B-4%7D%5C%20J)