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cupoosta [38]
3 years ago
12

When three people with a total mass of 2.00 x 102 kg step into their 1.200 x 103 kg car, the car’s

Physics
1 answer:
Aleksandr-060686 [28]3 years ago
3 0

Answer:

a

 k =  457333.3 N/m

b

x_a  =0.09\ m      

Explanation:

From the question we are told that

    The total mass of  three people is  M  = 2.00*10^{2} \ kg

     The mass of the car is  m_c  =  1.200 *10^{3} \ kg

     The compression of the car spring is  x = 3 \ cm  = 0.03 \ m

     

Generally the spring constant is mathematically represented as

          k =  \frac{F}{x}

Here F is the force exerted by the mass of three people and that of the car , this is mathematically represented as        

=>       F = (M +m_c) *g

=>       F = ([2.0*10^{2} ]+[ 1.200*10^{3}]) * 9.8

=>       F = 13720 \  N

So

        k =  \frac{13720}{0.03}

=>     k =  457333.3 N/m

Generally if the mass which the car is loaded with is  m  =  3.00*10^{2} \ kg

Then the force experienced by the spring is  

         =>       F_a = (m +m_c) *g

         =>       F_a = (3.00*10^{3} + 1.200 *10^{3}) * 9.8

         =>       F_a = 41160 \  N

Generally from the above formula the compression is  

       x_a  = \frac{F_a}{k}

=>    x_a  = \frac{41160}{457333.3}

=>    x_a  =0.09\ m      

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A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

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

Compute U if the bulb remains on for 5h

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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A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm. What is the image height?
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Answer:

<h2>6.36 cm</h2>

Explanation:

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v = image distance

Given f = 16.0 cm, u = 24.8 cm

1/v = 1/16 - 1/24.8

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v = 1/0.02218

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To get the image height, we will us the magnification formula.

Mag = v/u = Hi/H

Hi = image height = ?

H = object height = 3.50 cm

45.09/24.8 = Hi/3.50

Hi = (45.09*3.50)/24.8

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