To get the solution you must need to draw a force triangle. Attach the head of the 60N north force arrow with the tail of the 60N east force arrow. The subsequent is the arrow connecting he tail and head of the two arrows.
You get a right angled triangle, and the resultant is (60^2 + 60^2) ^0.5 = 84.85 N or 85 N northeast.
Answer:
v₁ = 4 [m/s].
Explanation:
This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.
where:
P = linear momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
where:
m₁ = mass of the tank = 500 [kg]
v₁ = velocity of the tank after firing the missile [m/s]
m₂ = mass of the missile = 20 [kg]
v₂ = velocity of the missile after firing = 100 [m/s]
![(500*v_{1})=(20*100)\\v_{1}=2000/500\\v_{1}=4[m/s]](https://tex.z-dn.net/?f=%28500%2Av_%7B1%7D%29%3D%2820%2A100%29%5C%5Cv_%7B1%7D%3D2000%2F500%5C%5Cv_%7B1%7D%3D4%5Bm%2Fs%5D)
Answer:
The flux will be nine times as great.
Explanation:
The electric flux due to a charge Q located in the center of a sphere can be obtained using Gauss's law. Considering a Gaussian surface in the form of a sphere of radius r:
The electric field (E) is parallel to the surface vector (dS), so 
Since the electric flux is proportional to the square of the sphere's radius, if radius of sphere were tripled, the flux will be nine times as great.
Answer:
Option D. 9.47 V
Explanation:
We'll begin by calculating the equivalent resistance of the circuit. This can be obtained as follow:
Resistor 1 (R₁) = 20 Ω
Resistor 2 (R₂) = 30 Ω
Resistor 3 (R₃) = 45 Ω
Equivalent Resistance (R) =?
R = R₁ + R₂ + R₃ (series connections)
R = 20 + 30 + 45
R = 95 Ω
Next, we shall determine the current in the circuit. This can be obtained as follow:
Voltage (V) = 45 V
Equivalent Resistance (R) = 95 Ω
Current (I) =?
V = IR
45 = I × 95
Divide both side by 95
I = 45 / 95
I = 0.4737 A
Finally, we shall determine, the voltage across R₁. This can be obtained as follow:
NOTE: Since the resistors are in series connection, the same current will pass through them.
Current (I) = 0.4737 A
Resistor 1 (R₁) = 20 Ω
Voltage 1 (V₁) =?
V₁ = IR₁
V₁ = 0.4737 × 20
V₁ = 9.47 V
Therefore, the voltage across R₁ is 9.47 V.
