Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is
, the final velocity is
, and the total time of the motion is
, so the acceleration is given by
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
So, the bullet penetrates the sandbag 1.8 meters.
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;
where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²
Therefore, the mutual force between the two point charges is 319.64 N
the unit for volt is v
The volt (symbol: V) is the derived unit for electric potential, electric potential difference (voltage), and electromotive force.
Answer:
a) The velocity is 2.94m/s
b) 0.441
Explanation:
a) Assume gravity is 9.8m/s^2
Use the equation below to solve for the velocity at 0.30 seconds
,
vf =unknown velocity vi= initial velocity vi=0m/s a= 9.8m/s^2 t=0.30seconds
Step 1: Substitute the variables with the knowns
Step 2: Solve
b)
Use the equation below to solve for the displacement at 0.30 seconds
Step 1: Substitute the same variables with the knowns
Note that vi*t=0 as vi=0m/s
Step 2: Solve
x=0.441m
Answer:
0.012-m
Explanation:
∆L = α × Lo × (T-To)
α is the coefficient of linear expansion = 12 × 10-6 K-1
Lo = Initial length = 25-m
∆L = Change in length
(T-To) = 40 K
∆L = 12 × 10-6 × 25 × 40
∆L = 0.012-m
