I'm pretty sure that it's 815.
Answer:
Vertical distance= 3.3803ft
Explanation:
First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:
Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h
Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h
time= 0.00012731h × (3600s/h)= 0.458316s
With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:
Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m
Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft
This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.
R = U : I. U is in Voltage and I is in Ampère. That gives you R = 36 : 8 = 4,5 Ohm
Does this help?
When an object is
immersed in a fluid (in this case water, but may include both liquids and
gases) the fluid exerts an upward force on the object which is called buoyancy
force or <span>up-thrust. Archimedes’ Principle states that the buoyant
force (upward push or force) applied to an object is equal to the weight of the fluid that the object takes the space of by
that object. Thus when an object is
placed in water the rise in the water level is dictated by the mass of that
object.</span>
<span>
</span>
<span>So for example if you fill a bucket with water and you drop a stone in that bucket, if you measure the weight of the water that overflows from the bucket due to the stone being dropped into the bucket is equivalent to the pushing force that the water has on the stone (as the stone drops to the bottom of the bucket the water is pushing it to stay afloat but the rock is more dense than water and as such its downthrust exceeds water's upthrust).</span>
