Answer:
Points downward, and its magnitude is 9.8 m/s^2
Explanation:
The motion of a projectile consists of two independent motions:
- A uniform horizontal motion, with constant velocity and zero acceleration. In fact, there are no forces acting on the projectile along the horizontal direction (if we neglect air resistance), so the acceleration along this direction is zero.
- A vertical motion, with constant acceleration g = 9.8 m/s^2 towards the ground (downward), due to the presence of gravity wich "pulls" the projectile downward.
The total acceleration of the projectile is given by the resultant of the horizontal and vertical components of the acceleration. But we said that the horizontal component is zero, therefore the total acceleration corresponds just to its vertical component, therefore it is a vector with magnitude 9.8 m/s^2 which points downward.
The speed is constant since there is no opposite force facing the car
That’s my best guest hope it helps !! :)
Answer:
11:1
Explanation:
At constant acceleration, an object's position is:
y = y₀ + v₀ t + ½ at²
Given y₀ = 0, v₀ = u, and a = -g:
y = u t − ½g t²
After 6 seconds, the ball reaches the maximum height (v = 0).
v = at + v₀
0 = (-g)(6) + u
u = 6g
Substituting:
y = 6g t − ½g t²
The displacement between t=0 and t=1 is:
Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]
Δy = 6g − ½g
Δy = 5½g
The displacement between t=6 and t=7 is:
Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]
Δy = (42g − 24½g) − (36g − 18g)
Δy = 17½g − 18g
Δy = -½g
So the ratio of the distances traveled is:
(5½g) / (½g)
11 / 1
The ratio is 11:1.
Answer:
Electric current is defined as the rate of flow of electric charge in a circuit from point one point to another. This is carried by electrically charged particles within the circuit. Current is represented by the symbol I and its unit measured in Amperes. It is therefore related to the voltage and resistance of the circuit. If the current in the circuit reduces, the rate at which the charge and current on the capacitor reduces also proportionally in an exponential manner.
Explanation:
Since a decrease in the flow of current in the circuit is observed, the implication for the rate at which the charge and voltage on the capacitor is also an exponential decrease in the rate of flow with time. This is because the electric current is directly proportional to the electric charge and the time.
This problems a perfect application for this acceleration formula:
Distance = (1/2) (acceleration) (time)² .
During the speeding-up half: 1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half: 1,600 meters = (1/2) (1.3 m/s²) T²
Pick either half, and divide each side by 0.65 m/s²:
T² = (1600 m) / (0.65 m/s²)
T = square root of (1600 / 0.65) seconds
Time for the total trip between the stations is double that time.
T = 2 √(1600/0.65) = <em>99.2 seconds</em> (rounded)
