The magnitude of the resultant is
√ (22² + 2.2²) = √ (484 + 4.84) = √488.84 = 22.11 m/s .
The direction of the resultant is
tan⁻¹(22N / 2.2E) = tan⁻¹(10) = 5.71° east of north .
Answer:
Mass of the oil drop, 
Explanation:
Potential difference between the plates, V = 400 V
Separation between plates, d = 1.3 cm = 0.013 m
If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :


, e is the charge on electron
E is the electric field, 


So, the mass of the oil drop is
. Hence, this is the required solution.
The frequency of the wave is 6800 Hz
<u>Explanation:</u>
Given:
Wave number, n = 20
Speed of light, v = 340 m/s
Frequency, f = ?
we know:
wave number = 

Therefore, the frequency of the wave is 6800 Hz
The final velocity of the bullet+block is 0.799 m/s
Explanation:
We can solve this problem by applying the principle of conservation of momentum: in fact, the total momentum of the bullet-block system must be conserved before and after the collision.
Mathematically, we can write:

where
m = 0.001 kg is the mass of the bullet
u = 800 m/s is the initial velocity of the bullet
M = 1 kg is the mass of the block
U = 0 is the initial velocity of the block (initially at rest)
v is the final combined velocity of the bullet and the block
Solving the equation for v, we find the final velocity:

Learn more about conservation of momentum:
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Answer:
a)
= 0.25 m / s b) u = 0.25 m / s
Explanation:
a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved
We will write the data
m₁ = 0.40 kg
v₁₀ = 9.0 m / s
m₂ = 14 kg
v₂₀ = 0
Initial
po = m₁ v₁₀
Final
= (m₁ + m₂) vf
po = pf
m₁ v₁₀ = (m₁ + m₂) 
= v₁₀ m₁ / (m₁ + m₂)
= 9.0 (0.40 / (0.40 +14)
= 0.25 m / s
b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass
u = 0.25 m / s
In the direction of movement of the ball
c) Let's calculate the kinetic energy in both moments
Initial
K₀ = ½ m₁ v₁₀² +0
K₀ = ½ 0.40 9 2
K₀ = 16.2 J
Final
= ½ (m₁ + m₂)
2
= ½ (0.4 +14) 0.25 2
= 0.45 J
ΔK = K₀ - 
ΔK = 16.2-0.445
ΔK = 1575 J
These will transform internal system energy
d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.
v₁₀’= v₁₀ -u
v₁₀’= 9 -.025
v₁₀‘= 8.75 m / s
v₂₀ ‘= v₂₀ -u
v₂₀‘= - 0.25 m / s
‘=
- u
= 0
Initial
K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²
Ko = ½ 0.4 8.75² + ½ 14.0 0.25²
Ko = 15.31 + 0.4375
K o = 15.75 J
Final
= ½ (m₁ + m₂) vf’²
= 0
All initial kinetic energy is transformed into internal energy in this reference system