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Valentin [98]
2 years ago
15

If runner A is running at 7.50 m/s and runner B is running at 7.90 m/s, how long will it take runner B to catch runner A if runn

er A has a
55.0-m head start?
Physics
1 answer:
yaroslaw [1]2 years ago
3 0

Answer:

t= 137.5 s

Explanation:

So if we are wanting to figure out how long it takes runner B to catch runner A. we must first set the slope of each runner equal to one another

<u>Slopes:</u>

Runner A:    y = 7.50x + 55

Runner B:    y = 7.90 x

sooooo

  7.50 x + 55 = 7.90 x

- 7.50 x          - 7.50 x

 55 = .40 x

55/.40 = .40 x / .40

x = 137.5 s

t= 137.5 s

7.50 * 137.5 + 55 =1086.25 m

7.90 * 137.5 =  1086.25 m

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A string of mass 60.0 g and length 2.0 m is fixed at both ends and with 500 N in tension. a. If a wave is sent along this string
Darya [45]

Answer:

a

The  speed of  wave is   v_1  = 129.1 \ m/s

b

The new speed of the two waves is v =  129.1 \ m/s

Explanation:

From the question we are told that

    The mass of the string is  m  =  60 \ g  =  60 *10^{-3} \ kg

    The length is  l  =  2.0 \ m

    The tension is  T  = 500 \ N

Now the velocity of the first wave is mathematically represented as

     v_1  = \sqrt{ \frac{T}{\mu} }

Where  \mu is the linear density which is mathematically represented as

      \mu  =  \frac{m}{l}

substituting values    

     \mu  =  \frac{ 60 *10^{-3}}{2.0 }

     \mu  =  0.03\ kg/m

So

   v_1  = \sqrt{ \frac{500}{0.03} }

   v_1  = 129.1 \ m/s

Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )

     

8 0
4 years ago
13. A 50 kg cart is attached to a 75 kg donkey. Two men, each using 100 N of force, try to pull the
Arturiano [62]
  • F1=-100N
  • F2=500N

\\ \sf\longmapsto F_{net}=F_1+F_2

\\ \sf\longmapsto F_{net}=-100+500

\\ \sf\longmapsto F_{net}=400N

7 0
3 years ago
A balloon drifts 140 m toward the west in 45 s; then the wind suddenly changes and the balloon flies 90 m toward the east in the
rusak2 [61]

Answer:

Explanation:

Given

1 ) 140 m west in 45 s .

2 ) 90 m east in 25 s .

a )

distance travelled in first 45 s = 140 m

b ) distance travelled in next 25 s = 90 m

c ) Total distance travelled = 140 m + 90 m

= 230 m

d ) average speed in first 45 s

= distance in 45 s  45

= 140 / 45 = 3.11 m /s

e ) average speed in next 25 s

distance in 25 s / 25

= 90 / 25 = 3.6 m /s

f ) average in entire trip

=  total distance / total time

= (140 + 90) / ( 25 + 45 )

= 3.28 m /s

g )

displacement in first 45 s = 140 m towards west

h )

displacement in next 25 s = 90 m towards east

i )

total displacement = 140 - 90

= 50 m towards west .

3 0
3 years ago
Consider two planets in space that gravitationally attract each other. if the masses of both planets are doubled, and the distan
amid [387]
Yhuihoifjhh <span>F = Gm1m2 / r^2 

if the masses are doubled then the force is increased by a factor of 4 


if the distance is doubled the force is decreased by a factor of 1/ 2^2 
the net result is no change in force</span>
6 0
3 years ago
Read 2 more answers
The modern standard of length is 1 m and the speed of light is approximately 2.99792 x 10^8 m/s. Find the time change in t for l
ICE Princess25 [194]

Explanation:

Distance = rate × time

1 m = (2.99792×10⁸ m/s) t

t = 3.33565×10⁻⁹ s

6 0
3 years ago
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