Answer:
C.
Measure the wear on his treads before and after riding a certain number of laps.
Answer:
Measure the wear on his treads before and after riding a certain number of laps.
Explanation:
By riding in a circular motion the inside of the tire will be in contact with the road more than the outside of the tire. Thus, to see if the constant circular motion had any effect on his tires David should measure the tread depth on both the inside and the outside of the tires before the experiment and measure the inside and the outside of the tires (at the same location on the tires) after the experiment. Then he can compare the tread loss on the inside of the tire to the tread loss on the outside of the tire.
Answer:
Energy dissipated = 13.453 Joules
Explanation:
In order to solve this problem, we first compute the gravitational potential energy the child has, and then find the kinetic energy at the lowest position.
The gravitational potential energy (relative to lowest position) is found as follows:
G.P.E = mass * gravity * height
Where, Height = 2 - 2 * Cos(34°)
Height = 0.3193 meters
G.P.E = 30 * 9.8 * 0.3193
G.P.E = 93.874 J
Kinetic energy:
K.E = 0.5 * mass * velocity^2
K.E = 0.5 * 30 * 2.31547^2
K.E = 80.421 J
Energy dissipated = G.P.E - K.E
Energy dissipated = 93.874 - 80.421
Energy dissipated = 13.453 J
To find the critical angle, we need to consider the forces acting on the system. The weight and frictional force must be taken into account. By equating the forces and solving for the critical angle, we can determine at what angle the system just begins to move.
To determine the critical angle for the system shown, we need to consider the forces acting on the objects. The force pulling m1 downwards is its weight, which is equal to its mass multiplied by the acceleration due to gravity. The force preventing m1 from moving is the frictional force, which is equal to the coefficient of friction multiplied by the normal force. The normal force is the force exerted by the surface perpendicular to it, which is equal to the weight of m2 minus the weight of the hanging part of the rope.
At the critical angle, the force of friction is at its maximum value, which is equal to the coefficient of friction multiplied by the normal force. The force pulling m1 downwards is equal to the force of friction. By equating these forces and solving for the critical angle, we can find the answer.
#SPJ2
If you do this on Earth, then the acceleration of the falling object is 9.8 m/s^2 ... NO MATTER what it's mass is.
If its mass is 10 kg, then the force pulling it down is 98.1 Newtons. Most people call that the object's "weight".
A racing car is travelling at 70 m/s and accelerates at -14 m/s^2. What would the car’s speed be after 3 s?
A racing car is travelling at 70 m/s and accelerates at -14 m/s^2.
v = u + at, we have
The car's speed after 3 s is 28 m/s.
Hope it helps
Here ΔI/Δt characterizes the rate at which the current I through the inductor is changing with time t.
Based on the equation given in the introduction, what are the units of inductance L in terms of the units of E, t, and I (respectively volts V, seconds s, and amperes A)?
What EMF is produced if a waffle iron that draws 2.5 amperes and has an inductance of 560 millihenries is suddenly unplugged, so the current drops to essentially zero in 0.015 seconds?
Answer:
Explanation:
E= −L ΔI / Δt.
L = E Δt / ΔI
Hence the unit of inductance may be V s A⁻¹
or volt s per ampere .
In the given case
change in current ΔI = - 2.5 A
change in time = .015 s
L = .56 H
E = − L ΔI / Δt.
= .56 x 2.5 / .015
= 93.33 V .
Answer:
The position of the arrows will not be on the target i.e. outside the bull's eye, neither will they be close to one another (widely scattered).
Explanation:
Accuracy refers to the closeness of a measurement to an actual or accepted value while precision refers to the closeness of measurements to one another.
Using archery as an illustration of precision and accuracy, measurements (arrows) that are neither accurate not precise are those arrows that will be far away or outside the bull's eye region (target) of the board and also far apart from one another.
In a nutshell, the arrows will be distant from the bull's eye or target (not accurate) and also distant from one another (not precise).