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elena-s
9 days ago
11

Four particles with masses 2 kg, 5 kg, 2 kg, and 2 kg are connected by rigid rods of negligible mass as shown. assume the system

rotates in the yz plane about the x axis with an angular speed of 5 rad/s. x y 5 rad/s o 5 rad/s 5 m 5 m 2 kg 5 kg 2 kg 2 kg find the moment of inertia of the system about the x axis. answer in units of kg · m2 .
Physics
1 answer:
Maru [2.9K]9 days ago
8 0
The answer is 10pi. I believe this will be helpful.
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A charge of 4.9 x 10-11 C is to be stored on each plate of a parallel-plate capacitor having an area of 150 mm2 and a plate sepa
Keith_Richards [2907]

Answer:2.53*10^-10F

Explanation:

C=£o£r*A/d

Where £ represents the permittivity constant

£o= 8.85*10^-12f/m

£r=6.3

A=150mm^2=0.015m^2

d=3.3mm= 0.0033m

C=8.85*10^-12*6.3*0.015/0.0033

C=8.85*6.3*10^-12*0.015/0.0033

C=55.755*0.015^-12/0.003

C=8.36/3.3*10^-13+3

C=2.53*10^-10F

7 0
1 month ago
A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or
Softa [2668]

Answer:

 Baseball mass:  m_b=3.992*10^{14}kg  

Explanation:

 Circumference of a baseball is calculated using 2πr = 23 cm

 Thus, the radius comes out to be 3.66 cm, which equals 3.66*10^{-2} m

 The mass density of the baseball matches that of a neutron or proton.

 Proton mass = 10^{-27} kg  

 Proton diameter = 10^{-15} m

 Proton radius =  5*10^{-16} m

 Volume of the baseball is \frac{4}{3} \pi r^3

 Now by substituting all values into the mass per unit volume equation for the baseball, we get:  

         \frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}

         \frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}  

         m_b=3.992*10^{14}kg

       Therefore, the baseball mass amounts to m_b=3.992*10^{14}kg              

5 0
26 days ago
Read 2 more answers
1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
Softa [2668]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle \theta to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

where y_o represents the initial height from ground level, v_{oy} is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

v_y=v_{oy}-gt

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of v_o.

The peak height is attained when v_y=0, which allows us to compute the time to reach that height.

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum height reached is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is equal to 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Continuing with the calculations for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Substituting known values yields

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

5 0
1 month ago
A person weighing 55 kg walks by applying 160 N of force on the ground, while pushing a 10-kg object. If the person accelerates
Sav [2826]

Answer:

I'm uncertain

Explanation:

since I didn't provide a correct answer, continue with my inquiries and you can use 'I'm uncertain' for 100 points.

6 0
29 days ago
A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea
serg [3228]

Answer:

v = 3369.2 m/s

Explanation:

The beacon is rotating at an angular speed of

f = 10 rev/min

so we have

\omega = 2\pi f

\omega = 2\pi(\frac{10}{60})

\omega = 1.047 rad/s

We know that

v = r \omega

At this point we have

r = 2 miles = 2(1609 m)

r = 3218 m

So we can conclude with

v = 3218(1.047)

v = 3369.2 m/s

6 0
1 month ago
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