Answer:
λ = 596 nm.
Explanation:
Fringe width = λ D / d
λ is wave length , D is screen distance and d is slit separation.
Putting the values
1.62 x 10⁻² =( λ x 5.3 ) / .195 x 10⁻³

λ = 596 nm.
2500 centimeters is 25 meters. 25 decimeters is 2.5 meters. 2500 centimeters is 10 times longer than 25 decimeters.
Answer:
The coefficient of friction and acceleration are 0.37 and 2.2 m/s²
Explanation:
Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.
Given that,
Mass of block = 60 kg
Acceleration = 2.0 m/s²
Mass = 100 kg
Horizontal force = 340 N
Let the frictional force be f.
We need to calculate the frictional force
Using balance equation

Put the value into the formula



We need to calculate the coefficient of friction
Using formula of friction force




We need to calculate the acceleration of the 100 kg block
Using formula of newton's law




Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²
Answer:
The nest must be about 4.15 meters above ground
Explanation:
Use the velocity equation under accelerated motion (acceleration of gravity ):

which for this case has initial velocity = 0 (falls from the nest), final velocity = 9 m/s, and a = 9.8 m/s^2, then we can find the time needed in air while falling to reach the required speed:

We now use this time value to find the distance covered in free fall during 0.92 seconds:

Answer:
(a) -1.18 m/s
(b) 0.84 m/s
Explanation:
(a)
The total linear momentum before the lumberjack begins to move is zero because all parts of the system are at res
From the law of conservation of momentum
m1v1+m2v2=0 hence m1v1=-m2v2 where m1 is mass of lumberjack, v1 is velocity of lumberjeck, m2 is mass of floating log, v2 is velocity of the floating log.
Substituting M1 for 103 Kg, V1 for 2.93 m/s, M2 for 255 Kg into the above equation we obtain
103Kg*2.93 m/s=-255Kg*V2
V2=-(103 kg*2.93 m/s)/255=-1.183490196 m/s
Hence V2=-1.18 m/s
(b)
For the second log
V(M1+M2)=m1v1 where V is the common velocity
V(103 Kg+255 Kg)=103 Kg*2.93 m/s
V=(103 Kg*2.93 m/s)/(103 Kg+255 Kg)=0.842988827 m/s
V=0.84 m/s