Answer:
In the previous section, we defined circular motion. The simplest case of circular motion is uniform circular motion, where an object travels a circular path at a constant speed. Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity, either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant.
You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force. The sharper the curve and the greater your speed, the more noticeable this effect becomes.
Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine Δs becoming smaller and smaller, then the acceleration would point exactly toward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration ac because centripetal means center seeking.
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By calculation, the diameter of the wire is 2.8 * 10^-3 m.
<h3>How do we obtain the length?</h3>
The following data are given in the question;
Mass of the wire = 1.0 g or 1 * 10^-3 Kg
Resistance = 0.5 ohm
Resistivity of copper = 1.7 * 10^-8 ohm meter
Density of copper = 8.92 * 10^3 Kg/m^3
V = m/d
But v = Al
Al = m/d
A = m/ld
Resistance = ρl/A
= ρl/m/ld =
l^2 = Rm/ρd
l = √ Rm/ρd
l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3
l = 1.82 m
A = πr^2
Also;
A = m/ld
A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3
Area of the wire = 6.2 * 10^-5 m^2
r^2 = A/ π
r = √A/ π
r = √6.2 * 10^-5 m^2/3.142
r = 1.4 * 10^-3 m
Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m
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Missing parts;
Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?
Answer:
Option D) 4A
Explanation:
As the cycle of the wave passes by, the amplitude gives the longest journey when the spot travels from the undistributed position. During each cycle the spot travels "Four times" .
Considering one of this cycle, if it begins to travel from it's undistributed position , there would be four movements i.e
* Upward movement through distance A
*Downward movement through distance A
*Downward again through distance A
*Upward through distance A.
Then it would travel back to its undistributed position held
Answer:
The wagon will move to the right.
Explanation:
From the question given above, the following data were obtained:
Force applied to the left (Fₗ) = 10 N
Force applied to the right (Fᵣ) = 30 N
Direction of the wagon =.?
To determine the direction in which the wagon will move, we shall determine the net force acting on the wagon. This can be obtained as follow:
Force applied to the left (Fₗ) = 10 N
Force applied to the right (Fᵣ) = 30 N
Net force (Fₙ) =?
Fₙ = Fᵣ – Fₗ
Fₙ = 30 – 10
Fₙ = 20 N to the right
From the calculations made above, the net force acting on the wagon is 20 N to the right. Hence the wagon will move to the right.
I would look this one up on Google
