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svetlana [45]
3 years ago
8

A 2.3 kg particle-like object moves in a plane with velocity components vx = 40 m/s and vy = 75 m/s as it passes through the poi

nt with (x, y) coordinates of (3.0, −4.0) m. (Express your answers in vector form.)
(a) What is its angular momentum relative to the origin at this moment?
(b) the point located at (-2.0, -2.0) m?
Physics
1 answer:
Sonbull [250]3 years ago
6 0

Answer:

(a) \overrightarrow{L}=885.5\widehat{k}

(b) \overrightarrow{L}=1046.5\widehat{k}

Explanation:

mass, m = 2.3 kg

vx = 40 m/s

vy = 75 m/s

(a) Angular momentum is given by

\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}

Where, p is the linear momentum and r is the position vector about which the angular momentum is calculated.

Here, \overrightarrow{r}=3\widehat{i}-4\widehat{j}

\overrightarrow{p}=m\overrightarrow{v}

\overrightarrow{p}=2.3\left ( 40\widehat{i}+75\widehat{j} \right )

\overrightarrow{p}= 92\widehat{i}+172.5\widehat{j}

So, the angular momentum

\overrightarrow{L}=\left ( 3\widehat{i}-4\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )

\overrightarrow{L}=885.5\widehat{k}

(b) Here, \overrightarrow{r}=(3+2)\widehat{i}+(-4+2)\widehat{j}

\overrightarrow{r}=5\widehat{i}-2\widehat{j}

\overrightarrow{L}=\left ( 5\widehat{i}-2\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )

\overrightarrow{L}=1046.5\widehat{k}

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Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

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Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

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