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.
Answer:
The correct answer is "0.32 mL".
Explanation:
The given values are:
Density of gold bar,
d = 19.3 g/mL
Mass of gold bar,
m = 6.3 grams
Now,
The volume will be:
⇒ 
or,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
<span>An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.5 m above the ground. The railcar has a mass of 38,500 kg and is moving to the right at a constant speed of 8.7 m/s on a frictionless rail...
</span>
Answer:
A. 0.199 J
B. 0.0663 C
C = 0.0221 F
D. 12.68 ohms
Explanation:
From the question:
time duration, t = 0.28 seconds
Average power, P = 0.71 W
Average voltage, V = 3 V
A) Energy is given as:
E = P * t
=> E = 0.71 * 0.28 = 0.199 J
B) Electrical energy is also given as:
E = qV
where q = charge
=> q = E / V
∴ q = 0.199 / 3 = 0.0663 C
C) Capacitance is given as charge over voltage:
C = q / V
=> C = 0.0663 / 3 = 0.0221 F
D) Electrical power, P, can also be given as:
P = 
where R = resistance
=> R = 
R = 