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stiks02 [169]
3 years ago
10

A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t

o the bar. The linear speed of the end points of the bar is vvv. What is the magnitude of the angular momentum LLL of the bar?
Physics
1 answer:
Pie3 years ago
7 0

Answer:

I = \frac{mvb}{6}

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

       L = Iω ....                        (1)

moment of inertia of rod rotating about its center of length b

 

      I = \frac{ mb^2}{12}  ........               .(2)  

using         v = ωr  

where w is angular velocity

and r is radius of  rod which is equal to b

        so we get  2v =  ωb  

                            ω  = 2v/b  .................            (3)    

here velocity is two time because two opposite ends  are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

                 L   =   \frac{mb^2}{12}×\frac{2v}{b}

              so final answer will be      L  =   \frac{mvb}{6}

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3 years ago
You have a spring-loaded air rifle. When it is loaded, the spring is compressed 0.3 m and has a spring constant of 150 N/m. In j
Feliz [49]

The potential energy of the spring is 6.75 J

The elastic potential energy stored in the spring is given by the equation:

E= \frac{1}{2} kx^2

where;

k is the spring constant

x is the compression/stretching of the string

In this problem, we have the spring as follows:

k = 150 N/m is the spring constant

x = 0.3 m is the compression

Substituting in the equation, we get

E=\frac{1}{2} (150) (0.3)^2

E=6.75J

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6 0
1 year ago
A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult
Dvinal [7]

Answer:

Explanation:

Given

time taken t=2\ s

Speed acquired in 2 sec v=42\ m/s

Here initial velocity is zero u=0

acceleration is the rate of change of velocity in a given time

a=\frac{v-u}{t}

a=\frac{42-0}{2}=21\ m/s^2

Distance travel in this time

s=ut+0.5at^2

where  

s=displacement

u=initial velocity

a=acceleration

t=time

s=0+\0.5\times 21\times (2)^2

s=42\ m

so Jet Plane travels a distance of 42 m in 2 s                                

5 0
3 years ago
A spring has a natural length of 28.0 cm. If a 23.0-N force is required to keep it stretched to a length of 36.0 cm, how much wo
krek1111 [17]

Answer:

0.23 J

Explanation:

k*(36 - 28) = 23

so k = 23/8 N/cm

W = k(32 - 28)²/2 = 23/8 * 4²/2 = 23 N-cm = 0.23 J

4 0
2 years ago
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