Answer:
The average recoil force on the gun during that 0.40 s burst is 45 N.
Explanation:
Mass of each bullet, m = 7.5 g = 0.0075 kg
Speed of the bullet, v = 300 m/s
Time, t = 0.4 s
The change in momentum of an object is equal to impulse delivered. So,

For 8 shot burst, average recoil force on the gun is :

So, the average recoil force on the gun during that 0.40 s burst is 45 N.
<h3>Hello There!!</h3>
<h3><u>Given</u>,</h3>
Force(F) = 150N
Mass(m) = 90kg
<h3><u>To </u><u>Find,</u></h3>
Acceleration(a) = ?
<h3><u>We know,</u></h3>
F= m×a


<h3>Hope this helps</h3>
Answer:
Explanation:
As the sum of the two right directed forces match exactly the left directed force, the only unbalanced force, and thus the net force, is the upward 25 N force.
Answer:
(a) W= 44N
(b)W= 31.65 N
Explanation:
Data
T=44 N : Maximum force that the rope can withstand without breaking
Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
(a) We apply the formula (1) at constant speed , then, a=0
W: heaviest fish that can be pulled up vertically
∑F = 0
T-W =0
W = T
W= 44N
(b) We apply the formula (1) , a= 1.26 m/s²
W: heaviest fish that can be pulled up vertically
W= m*g
m= W/g
g= 9.8 m/s² : acceleration due to gravity
∑F = 0
T-W = m*a
T= W+(W/g)*a
44=W*(1+1/9.8)* (1.26 )
44= W* 1.39
W= 44/1.39
W= 31.65 N
Answer:
The wavelength is 3500 nm.
Explanation:
d= 
n= 1
θ= 30°
λ= unknown
Solution:
d sinθ = nλ
λ = 
λ = 3500 nm