Answer:
I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.
Explanation:
In order to increase the horizontal distance covered by the ball, we need to examine the variables involved in the formula of range of projectile. The formula for the range of projectile is given as follows:
R = V₀² Sin 2θ/g
where, g is a constant on earth (acceleration due to gravity) and θ is the angle of ball with ground at the time of launching. The value of θ should be 45° for maximum range. In this case we do not know the angle so, we can not tell if we should change it or not.
The only parameter here which we can increase to increase the range is launch velocity (V₀). The formula for V₀ in terms of horizontal and vertical components is as follows:
V₀ = √(V₀ₓ² + V₀y²)
where,
V₀ₓ = Horizontal Velocity
V₀y = Vertical Velocity
Hence, it is clear from the formula that we can increase both the horizontal and vertical velocity to increase the initial speed which in turn increases the horizontal distance covered by the ball.
<u>Therefore, I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.</u>
F=-ks
F=-(390)(.45)
F=-175.5 N
Work=force x displacement
Work= 175.5(0.45)
Work= 78.98 J
Work = ∆E =78.98 J
Answer=79J (first option)
It's a projectile near the earth under the influence of gravity only.
B and C because an earthquake does not directly affect gravity, and precipitation doesn’t fall, it is water surfacing
Answer:
The current in the motor in this case is 13.2 A
Explanation:
Given:
Resistance 
5 Ω
Emf 
V
Induced emf 
V
When motor run at half speed due to load increased then induced emf is also reduced to half of its value
So new induced emf in our case is given by,
V
Where 
A
Therefore, the current in the motor in this case is 13.2 A
