Answer:
The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.
(3) and (4) is correct option.
Explanation:
Given that,
Beat frequency f = 5.0 Hz
Frequency f'= 462 Hz
We need to calculate the possible frequencies for the A string of the other violinist
Using formula of frequency
...(I)
...(II)
Where, f= beat frequency
f₁ = frequency
Put the value in both equations
Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.
Answer:
A. 3.4 m
Explanation:
Given the following data;
Force = 56.7N
Workdone = 195J
To find the distance
Workdone is given by the formula;
Making "distance" the subject of formula, we have;
Substituting into the equation, we have;
Distance = 3.4 meters.
Answer:
speed of the bullet before it hit the block is 200 m/s
Explanation:
given data
mass of block m1 = 1.2 kg
mass of bullet m2 = 50 gram = 0.05 kg
combine speed V= 8.0 m/s
to find out
speed of the bullet before it hit the block
solution
we will apply here conservation of momentum that is
m1 × v1 + m2 × v2 = M × V .............1
here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet
put all value in equation 1
m1 × v1 + m2 × v2 = M × V
1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8
solve it we get
v2 = 200 m/s
so speed of the bullet before it hit the block is 200 m/s
No. The moon always keeps the same side facing us. Its rotation and revolution periods are equal.
Answer:
V=4.7m/s
Explanations:
Let Ma mass of cat A=7kg
Va velocity of cat A=7m/s
Mb mass of cat b=6.1kg
VB velocity of cat b=2m/s
From conservation of linear momentum
MaVa+MbVb=(Ma+Mb)V
7*7+6.1*2=(7+6.1)V
61.2=13.1V
V=4.7m/s
