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1 month ago

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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 Â· m2 .

1 answer:

Maru [3.3K]1 month ago

8 0

The answer is 10pi. I believe this will be helpful.

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An object experiences an acceleration of -6.8 m/s2. As a result, it accelerates from 54 m/s to a complete stop. How much dista

Maru [3345]

The distance covered during the acceleration phase is d = 214.38 m. Given parameters include the acceleration of the object, a = -6.8 m/s², initial speed, u = 54 m/s, and final speed, v = 0. The equation for acceleration is defined as a = (v - u) / t. Therefore, rearranging gives t = (v - u) / a, resulting in t = (0 - 54) / (-6.8) = 7.94 s. The average speed of the object is V = (54 + 0)/2 = 27 m/s. The displacement is calculated as d = V x t = 27 x 7.94 = 214.38 m. Thus, the total distance traveled during that period of acceleration is 214.38 m.

3 0

1 month ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

kicyunya [3294]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

The law of conservation of mass states that the rate of fluid mass ( $m_{1}$ ) entering a system equals the rate at which the fluid mass ( $m_{2}$ ) exits the system.

The mass flow rate can be expressed as follows:

$m = \rho A v$

where $\rho$ denotes the fluid density, $A$ signifies the cross-sectional area through which fluid flows, and $v$ represents the fluid's velocity.

Based on the problem conditions, as the fluid's density remains constant, we can write:

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas for the fluid flow, while $v_{1}$ and $v_{2}$ are the corresponding velocities across those areas.

Given the conditions in the problem, $A_{2} > A_{1}$ , leading from the formula to $v_{2} < v_{1}$ .

Furthermore, fluid pressure arises from the fluid's movement through any specific area. When the fluid accelerates, part of its energy increases its speed in the direction of flow, resulting in lower pressure.

Thus, in this instance, $v_{2} < v_{1}$ the pressure in the larger cross-sectional area $P_{2}$ will exceed the pressure $P_{1}$ in the smaller cross-sectional area, implying

$P_{2} > P_{1}$ .

6 0

2 months ago

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

serg [3582]

Response:

a) The mug makes contact with the ground 0.7425m from the bar's end. b) |V|=5.08m/s θ= -72.82°

Clarification:

To address this issue, we begin with a diagram depicting the situation. (refer to the attached illustration).

a)

The illustration shows that the problem involves motion in two dimensions. To determine how far from the bar the mug lands, we need to find the time the mug remains airborne by examining its vertical motion.

To compute the time, we utilize the following formula with the known values:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

We have $y_{f}=0$ and $v_{y0}=0$ , allowing us to simplify the equation to:

$0=y_{0}+\frac{1}{2}at^{2}$

Now, we can calculate for t:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

We know $y_{0}=1.20m$ and $a=g=-9.8m/s^{2}$

The negative gravity indicates the downward motion of the mug. Hence, we substitute these values into the provided formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

Which results in:

t=0.495s

This time helps us evaluate the horizontal distance the mug traverses. Since:

$V_{x}=\frac{x}{t}$

Solving for x, we have:

$x=V_{x}t$

Substituting the known values yields:

x=(1.5m/s)(0.495s)

This calculates to:

x=0.7425m

b) With the time determining when the mug strikes the ground established, we can find the final velocity in the vertical direction using the formula:

$a=\frac{v_{f}-v_{0}}{t}$

The initial vertical velocity being zero simplifies our calculations:

$a=\frac{v_{f}}{t}$

Thus, we can determine the final velocity:

$V_{yf}=at$

Given that the acceleration equates to gravity (showing a downward effect), we substitute that alongside the previously found time:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

This leads to:

$V_{yf}=-4.851m/s$

Now, we ascertain the velocity components:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

Next, we find the speed by calculating the vector's magnitude:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

<pThus:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

Yielding:

|V|=5.08m/s

Lastly, to ascertain the impact direction, we apply the equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

<pFulfilling this provides:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

<pLeading to:

$\theta = -72.82^{o}$

4 0

2 months ago

The Hall effect can be used to calculate the charge-carrier number density in a conductor. If a conductor carrying a current of

kicyunya [3294]

Answer:

$6.6*10^{27}e/m^3$

Explanation:

When calculating Hall voltage, it is crucial to have the current, magnetic field strength, length, area, and number of charge carriers available. The Hall voltage can be expressed using the equation:

$V_h = \frac{iB}{neL}$

Where:

i= the current

B= the magnetic field strength

L = the length

n = the number of charge carriers

e= charge of an electron

We need to replace values and solve for n:

$n= \frac{iB}{V_h e L}$

$n= \frac{2*1.2}{4.5*10^{-6}*5^10^{-3}*1.6*10^{-19}}$

$n= 6.6*10^{27}electron.m^{-3}$

As a result, the charge carrier density is $6.6*10^{27}e/m^3$

5 0

2 months ago