Start by facing East. Your first displacement is the vector
<em>d</em>₁ = (225 m) <em>i</em>
Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,
<em>d</em>₂ = (350 m) <em>j</em>
Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,
<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )
<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>
The net displacement is
<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃
<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>
and its magnitude is
|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m
Kinetic energy of pieces A and B are 2724 Joule and 5176 Joule respectively.
<h3>What is the relation between the masses of A and B?</h3>
Mass of piece B = Mb
- Velocities of pieces A and B are Va and Vb respectively.
- As per conservation of momentum,
Ma×Va = Mb×Vb
So, 1.9Mb × Va = Mb×Vb
=> 1.9Va = Vb
<h3>What are the kinetic energy of piece A and B?</h3>
- Expression of kinetic energy of piece A = 1/2 × Ma × Va²
- Kinetic energy of piece B = 1/2 × Mb × Vb²
- Total kinetic energy= 7900J
=>1/2 × Ma × Va² + 1/2 × Mb × Vb² = 7900
=> 1/2 × Ma × Va² + 1/2 × (Ma/1.9) × (1.9Va)² = 7900
=> 1/2 × Ma × Va² ×(1+1.9) = 7900 j
=> 1/2 × Ma × Va² = 7900/2.9 = 2724 Joule
- Kinetic energy of piece B = 7900 - 2724 = 5176 Joule
Thus, we can conclude that the kinetic energy of piece A and B are 2724 Joule and 5176 Joule respectively.
Learn more about the kinetic energy here:
brainly.com/question/25959744
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Explanation:
The electric field of an isolated charged parallel-plate capacitor is given by :
........(1)
Where
q is the electric charge
A is the area of cross section of parallel plate
It is clear from equation (1) that the electric field of a parallel plate capacitor is directly proportional to the charge on the plate and inversely proportional to the area of cross section of a plate.
So, the correct option is (E) i.e. "none of the above".
Answer:
2.12/R mW
Explanation:
The electrical power, P generated by the rod is
P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?
So, the induced emf on the conductor is
E = BLv
= 0.575 T × 0.2 m × 0.4 m/s
= 0.046 V
= 46 mV
The electrical power, P generated by the rod is
P = B²L²v²/R
= B²L²v²/R
So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²
= 0.002116/R W
= 2.12/R mW
Hello!
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.
Data:
Hooke represented mathematically his theory with the equation:
F = K * Δx
On what:
F (elastic force) = 2 N
K (elastic constant) = 4 N/cm
Δx (deformation or elongation of the elastic medium or distance from a spring) = ?
Solving:




simplify by 2


Answer:
B.) 1/2 cm
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I Hope this helps, greetings ... Dexteright02! =)