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Jet001 [13]
3 years ago
8

Pls,help ASAP .the question is in attachment​

Physics
1 answer:
Alex777 [14]3 years ago
8 0

Answer:c

Explanation:

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A thin rod rotates at a constant angular speed. Consider the tangential speed of each point on the rod for the case when the axi
ASHA 777 [7]

Answer:

    v = R w    

With this expression we see that for each point at different radius the tangential velocity is different

Explanation:

They indicate that the angular velocity is constant, that is

            w = dθ / dt

Where θ is the radius swept angle and t the time taken.

The tangential velocity is linear or

           v = dx / dt

Where x is the distance traveled in time (t)

 

In the definition of radians

          θ = s / R

Where s is the arc traveled and R the radius vector from the pivot point, if the angle is small the arc (s) and the length (x) are almost equal

         θ = x / R

We substitute in the speed equation

         v = d (θ R) / dt

The radius is a constant for each point

         v = R dθ / dt

         v = R w

With this expression we see that for each point at different radius the tangential velocity is different

6 0
3 years ago
A motorcyclist is traveling at 58.3 mph on a flat stretch of highway during a sudden rainstorm. The rain has reduced the coeffic
Novosadov [1.4K]

Explanation:

Below is an attachment containing the solution.

8 0
4 years ago
Number 5 and 6 i need just give answer
sashaice [31]

Explanation:

voltage = current × resistance

5.

12 V = 4.2 A × resistance

resistance = 12 V / 4.2 A = 2.857142857... Ohm

FYI :

4.2 A would be a lot for a small electronic device like a CD player. that would be 12×4.2 = 50.4 Watt, and the CD player would get really hot.

6.

120 V = current × 12 Ohm

current = 120 V / 12 Ohm = 10 A

8 0
2 years ago
Goals can be big or small? Question 1 options: True False
pav-90 [236]

Answer:

Probably big

Explanation:

lol

7 0
3 years ago
A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de
EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/πr^{2}

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / πr^{2}ex

E = \frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6}    } = 17.368 * 10^{9} Pa = 17.4 GPa

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = \frac{1}{E} (бy - v (бx + бz)) = -\frac{v}{E}бx

= \frac{vFx}{Epir^{2} } = \frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9}  } = -0.63 *10^{-6}

Finally

ey = Δr / r

Δr = ey * r = 10 * -0.63* 10^{-6} mm = -6.3 * 10^{-6} mm

Δd = 2Δr = -12.6 * 10^{-6} mm

Explanation:

5 0
3 years ago
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