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inn [45]
3 years ago
7

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

s of rotation is perpendicular to the rod at its center and at one end. Explain for each case whether there are any points on the rod that have the same tangential speeds.
Physics
1 answer:
ASHA 777 [7]3 years ago
6 0

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

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3 years ago
n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend
Kitty [74]

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

p_f=p_i\\mv_1+Mv_2=(m+M)v

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m

hence, the heigth is 68cm

4 0
3 years ago
What condition is required for cloud formation in the atmosphere?
Vedmedyk [2.9K]

Answer: To form a cloud, the air that rises must cool to the point where some of the water vapor molecules "clump together" at a faster pace than they are pulled apart by their thermal energy. These molecules then condense to form the clouds and water droplets.

7 0
4 years ago
An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper
bixtya [17]

Answer:

a. P=18.61\ W

b. \eta_c=0.6976=69.76\%

c. \eta=0.4718=47.18\%

Explanation:

Given:

  • temperature of the hotter reservoir, T_H=1250\ K
  • temperature of the colder reservoir, T_C=378\ K
  • heat absorbed by the engine, Q_H=1.42\times 10^{5}\ J
  • heat rejected to the cold reservoir, Q_L=7.5\times 10^4\ J
  • time duration of the energy transfer, t=1\ hr=3600\ s

<u>Now the work done by the engine:</u>

Using energy conservation,

W=Q_H-Q_L

W=14.2\times 10^4-7.5\times 10^4

W=6.7\times 10^4\ J

a.

<u>Hence the power output:</u>

P=\frac{W}{t}

P=\frac{6.7\times 10^4}{3600}

P=18.61\ W

b.

\eta_c=1-\frac{T_L}{T_H}

\eta_c=1-\frac{378}{1250}

\eta_c=0.6976=69.76\%

c.

now actual efficiency:

\eta=\frac{W}{Q_H}

\eta=\frac{6.7\times 10^4}{1.42\times 10^{5}}

\eta=0.4718=47.18\%

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4 years ago
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natali 33 [55]

Answer:

25 degrees

Explanation:

The angle of incidence equals the angle of reflection

7 0
3 years ago
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