Answer:
the higher the ramp the less distance it will travel
Answer:

Explanation:
Recall the formula for acceleration:
, where
is final velocity,
is initial velocity, and
is elapsed time (change in velocity over this amount of time).
Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.
We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).
We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.
Substituting values in our formula, we have:

Alternative:
Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!
Answer:
Its inductance L = 166 mH
Explanation:
Since a current, I = 0.698 A is obtained when a voltage , V = 5.62 V is applied, the resistance of the coil is gotten from V = IR
R = V/I = 5.62/0.698 = 8.052 Ω
Since we have a current of I' = 0.36 A (rms) when a voltage of V' = 35.1 V (rms) is applied, the impedance Z of the coil is gotten from
V₀' = I₀'Z where V₀ = maximum voltage = √2V' and I₀ = maximum current = √2I'
Z = V'/I' = √2 × 35.1 V/√2 × 0.36 V = 97.5 Ω
WE now find the reactance X of the coil from
Z² = X² + R²
X = √(Z² - R²)
= √(97.5² - 8.05²)
= √(9506.25 - 64.8025)
= √9441.4475
= 97.17 Ω
Now, the reactance X = 2πfL where f = frequency of generator = 93.1 Hz and L = inductance of coil.
L = X/2πf
= 97.17/2π(93.1 Hz)
= 97.17 Ω/584.965 rad/s
= 0.166 H
= 166 mH
Its inductance L = 166 mH
Answer:
The Magnifying power of a telescope is 
Explanation:
Radius of curvature R = 5.9 m = 590 cm
focal length of objective
= 
⇒
= 
⇒
= 295 cm
Focal length of eyepiece
= 2.7 cm
Magnifying power of a telescope is given by,



therefore the Magnifying power of a telescope is 