Answer:
1. The nucleus is about 1/2 the size of the atom
Explanation:
Alpha particles are positive charge particles and they are bounced back by the nucleus because nucleus is also same size
Now in present experiment Rutherford found that very few alpha particles are bounced back along same path which shows that very small region inside the nucleus is having positive charge and rest part of the atom is empty.
Now if we found that half of the alpha particles are bounced back then it shows that size of the nucleus is very large now as compare to previous one because only nucleus can bounce back the alpha particles
so correct answer will be
1. The nucleus is about 1/2 the size of the atom
Answer:
The focal length of the appropriate corrective lens is 35.71 cm.
The power of the appropriate corrective lens is 0.028 D.
Explanation:
The expression for the lens formula is as follows;

Here, f is the focal length, u is the object distance and v is the image distance.
It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.
Put v= -71.4 cm and u= 24.0 cm in the above expression.


f= 35.71 cm
Therefore, the focal length of the corrective lens is 35.71 cm.
The expression for the power of the lens is as follows;

Here, p is the power of the lens.
Put f= 35.71 cm.

p=0.028 D
Therefore, the power of the corrective lens is 0.028 D.
Time taken by the package to reach the sea level= 13.7 s
height=h=925 m
initial velocity along vertical= vi=0
acceleration due to gravity=g=9.8 m/s^2
using the kinematic equation h= Vi*t + 1/2 gt^2
925=0(t)+1/2 (9.8)t^2
4.9 t^2=925
t= 13.7 s
Answer:
The velocity will be "76.8 m/s".
Explanation:
The given values are:
Acceleration,
a = 2.4 m/s²
Time,
t = 32 seconds
By equation of motion,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
<span>a change in allele frequencies in a population over time also known as evolution</span>