Answer:
Explanation:
the solution is given in the pictures attached below
Answer:
The velocity of the players will be <u>2.88 m/s</u> in the <u>east</u> direction.
Explanation:
Let 'v' be the velocity of the players after collision.
Consider the east direction as positive direction.
Given:
Mass of the first player is, 
kg
Initial velocity of the first player is, 
m/s
Mass of the second player is, 
kg
Initial velocity of the second player is, 
m/s
In order to solve this problem we use the law of conservation of momentum which says that the total momentum must be conserved before and after the collision. So we can write:
Solving for v, we get:
Therefore, their velocity after the collision is 2.88 m/s.
The sign of the velocity after collision is positive. So, the players will move in the east direction only after collision.
Answer:
The forces are the weight, drag, and lift. Lift and drag are actually two components of a single aerodynamic force acting on the ball. Drag acts in a direction opposite to the motion, and lift acts perpendicular to the motion.
Explanation:
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Answer:
A 10 N force pointing up
Explanation:
If the net acceleration of the object is horizontal pointing to the right, that means that all vertical forces must have canceled out, and the only ones "unbalanced" are the horizontal ones (10 N to the right minus 5 N to the left giving a net force of 5 N to the right).
Since they mentioned only one vertical force pointing down (10 N), there must be another one of same magnitude but pointing in opposite direction (up).
Then there must also be a 10 N force pointing up acting on the object.
Answer:
a)
two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign. both charges are positive(+) or Negative (-)
b)
both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C
Explanation:
Given that;
L = 0.26 m
k = 180 N/m
x = 0.039 m
a)
we know that two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign.
b)
Spring force F = kx
F = 180 × 0.039
F = 7.02 N
Now, Electrostatic force F = Keq²/r²
where r = L + x = ( 0.26 + 0.039 )
we know that proportionality constant in electrostatics equations Ke = 9×10⁹ kg⋅m3⋅s−2⋅C−2
so from the equation; F = Keq²/r²
Fr² = Keq²
q = √ ( Fr² / Ke )
we substitute
q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )
q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )
q = √ (0.627595 / 9×10⁹)
q = √(6.97 × 10⁻¹¹)
q = 8.35 × 10⁻⁶ C
Therefore both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C
