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klemol [59]
3 years ago
12

A man jogs at a velocity of 2.5 m/s west for 1,200 s. He then walks at a velocity 1.0 m/s east for 500 s and then stops to rest

. What is the displacement of man when he stops to rest?
Physics
1 answer:
olganol [36]3 years ago
5 0

Speed of man towards West =2.5 m/s

Time to travel in the given direction = 1200 s

Total displacement towards west

d_1 = v_1 t

d_1 = 2.5 \times 1200 = 3000 m

Now speed of man towards East = 1 m/s

time to travel in east direction = 500 s

total displacement towards East

d_2 = v_2 t

d_2 = 1 \times 500 = 500 m

Now total displacement will be

d = d_1 - d_2

d = 3000 - 500 = 2500 m

so it is 2500 m towards west

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Time taken by the rubber-soled shoes to reduce a person's potential from 3.00×10³ V to 100V is 3.91s

The initial potential difference \Delta V_o is related to the final potential difference \Delta V by

\Delta V=\Delta V_{o} e^{-t / \tau}

Our target is to find t, so we rearrange equation (1) for t to be

\Delta V=\Delta V_{o} e^{-t / RC}\\ \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) = \frac{-t}{RC}\\ t= -RC \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) \tag{2}

The body has a capacitance 150 pF while the foot has a capacitance 80 pF and both are in parallel connection. So, the equivalent capacitance is

C= 150 \mathrm{~pF} + 80 \mathrm{~pF} = 230 \mathrm{~pF}

The shoes of the rubber-soled has resistance R=5000 \mathrm{~M\Omega}. Plug the values for \Delta V, \Delta V_o and C into equation (2) to get the time t.

t&= -RC \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) \\ &= -( 5000 \times 10^{6} \mathrm{~\Omega})(230 \times 10^{-12} \mathrm{~F}) \ln \left ( \frac{100\mathrm{~V}}{3000 \mathrm{~V}} \right ) \\ &= {3.91 \mathrm{~s}}

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A car traveling 90km/hr is 100 m behind a truck traveling 50km/hr. How long will it take the car to reach the truck?
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The faster car behind is catching up/closing the gap/gaining on
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to reach the truck.

(1/400 hour) x (3,600 seconds/hour) = 3600/400 = <em>9 seconds</em>, exactly 


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What is the frequency of an event?
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"Frequency" just means "often-ness" ... how often something happens.
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the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during
Ivanshal [37]

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

Mv+mu=Mv_1+mv_2

(900 x 47) + (200 x -30)  = (900 x v_1) + (200 x v_2)

36300 =  (900 x v_1) + (200 x v_2)

9v_1 + 2v_2 = 363 ..............(i)

9v_1 = 363 - 2v_2

v_1=\frac{363 - 2v_2}{9}

The mathematical expression for the conservation of kinetic energy is

\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2

\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2    ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$  

21681 = 9v_1^2+2v_2^2

Substituting (i) in (ii)

21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2

(363-2v_2)^2+18v_2^2=195129

(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0

22v_2^2-145v_2-63360=0

Solving the equation, we get

v_2=96 \ m/s, -30 \ m/s

The negative velocity is neglected.

Therefore, substituting 96 m/s for v_2 in (i), we get

v_1=\frac{363-(2 \times 96)}{9}

     = 19

Thus, only impulse of importance is used to find final velocity.

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3 years ago
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