Answer:
22.22m/s
Explanation:
The momentum before a collision = momentum after collision so...
work out the momentum of the first object (the bullet)
its p = mv
0.04 kg × 300 m/s = 0.54 kg × v
rearrange this to find v which is 0.04 x 300 = 12
so 12 = 0.54 x v
12/0.5 = v
v = 22.22m/s
hope this helps!
For how long? Or is it just speed ??
Answer:
Explanation:
The formula for this, the easy one, is
where No is the initial amount of the element, t is the time in years, and H is the half life. Filling in:
and simplifying a bit:
and
N = 48.0(.0625) so
N = 3 mg left after 12.3 years
Answer:
(A) 
Explanation:
Given:
Charge of one particle (q₁) = -0.0050 C
Charge of another particle (q₂) = 0.0050 C
Separation between them (d) = 0.025 m
We know that, from Coulomb's law, electric force acting between two charged particles is given as:
Plug in the given values and solve for electric force, 
. This gives,
Therefore, option (A) is correct. Negative sign implies that the nature of electric force is attraction.
Answer:
v₀ₓ = 63.5 m/s
v₀y = 54.2 m/s
Explanation:
First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:
K.E = (0.5)(mv₀²)
where,
K.E = initial kinetic energy of projectile = 1430 J
m = mass of projectile = 0.41 kg
v₀ = launch velocity of projectile = ?
Therefore,
1430 J = (0.5)(0.41)v₀²
v₀ = √(6975.6 m²/s²)
v₀ = 83.5 m/s
Now, we find the launching angle, by using formula for maximum height of projectile:
h = v₀² Sin²θ/2g
where,
h = height of projectile = 150 m
g = 9.8 m/s²
θ = launch angle
Therefore,
150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)
Sin θ = √(0.4216)
θ = Sin⁻¹ (0.6493)
θ = 40.5°
Now, we find the components of launch velocity:
x- component = v₀ₓ = v₀Cosθ = (83.5 m/s) Cos(40.5°)
<u>v₀ₓ = 63.5 m/s</u>
y- component = v₀y = v₀Sinθ = (83.5 m/s) Sin(40.5°)
<u>v₀y = 54.2 m/s</u>
