Answer:
They are separated by a distance of 18 m. Find the gravitational attraction between them. r= 1 m r= 1 m 18 m. Mass = 1.5 kg. Mass = 8.5 kg.
Answer:
Explanation:
a ) It is given that bomb was at rest initially , so , its momentum before the explosion was zero.
b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.
If v be the velocity of the third part along a direction making angle θ
with x axis ,
x component of v = vcosθ
So momentum along x axis after explosion of third part = mv cosθ
= 10 v cosθ
Momentum along x of first part = - 5 x 42 m/s
momentum of second part along x direction =0
total momentum along x direction before explosion = total momentum along x direction after explosion
0 = - 5 x 42 + 10 v cosθ
v cosθ = 21
Similarly
total momentum along y direction before explosion = total momentum along y direction after explosion
0 = - 5 x 38 + 10 v sinθ
v sinθ= 21
squaring and and then adding the above equation
v² cos²θ +v² sin²θ = 21² +19²
v² = 441 + 361
v = 28.31 m/s
Tanθ = 21 / 19
θ = 48°
Answer:
True
Explanation:
For any equation to be balanced, be it elemental equation or radioactive decay equation, the number of atoms on the left must balance the number of atoms of the right. The same is true for alpha decay as the number of protons on the left side must equal the number of protons on the right side?
Answer:
0.003333 s to 0.000125s or from 3.33ms to 0.125ms wher m is for milli
1.1m to 0.04125 m
Explanation:
T= 1/f=
if f= 300Hz then T = 1/300 =0.003333 s
if f= 8000 then T= 1/8000 = 0.000125s
now v=f×wave length
or wavelength = speed/ frequency
when f = 300 Hz
wavelength = 330/300=1.1 m
wavelength = 330/8000 = 0.04125m
note : i have taken speed of sound as 330 m/s you can take any value given in between 330m/s to 340m/s
