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Kazeer [188]
1 year ago
14

When starting a foot race, a 64 kilogram sprinter exerts an average force of 693 newtons backward on the ground for 0.59 seconds

. how far does he travel in meters during this time?
Physics
1 answer:
cluponka [151]1 year ago
6 0

The distance traveled by the sprinter in meters is determined as 1.88 m.

<h3>Acceleration of the sprinter</h3>

The acceleration of the sprinter is the rate of change of velocity of the sprinter with time.

The acceleration of the sprinter is calculated as follows;

Apply Newton's second law of motion as follows;

F = ma

a = F/m

where;

  • F is the applied force by the sprinter
  • m is mass of the sprinter
  • a is acceleration of the sprinter

a = 693 N / 64 kg

a = 10.83 m/s²

<h3>Distance traveled by the sprinter</h3>

The distance traveled by the sprinter is calculated as follows;

s = ut + ¹/₂at²

where;

  • u is initial velocity = 0

s = ¹/₂at²

where;

  • t is time of motion
  • a is acceleration

s = (0.5)(10.83)(0.59²)

s = 1.88 m

Thus, the distance traveled by the sprinter in meters is determined as 1.88 m.

Learn more about distance here: brainly.com/question/2854969

#SPJ1

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lutik1710 [3]

Answer:

The temperature of the metal is  T_m  =  376.8 ^o C

Explanation:

From the question we are told that

     The mass of the metal is  M =  60 \ kg

     The specific heat of the metal is  c_p  =  0.1027 kcal/(kg \cdot ^oC)

       The mass of the oil is M_o  =  810 \ kg

       The temperature of the oil is  T_o  =  35^oC

       The specific heat of oil is  c_o  =  0.7167 kcal/(kg \cdot ^oC )

       The equilibrium temperature is T_e  =  39 ^oC

According to the law of energy conservation

     Heat lost by metal  =  heat gained by the oil

So  

   The quantity  of heat lost by the metal is mathematically represented as

               Q =  - Mc_p \Delta T

=>            Q =  -Mc_p (T_m  -  T_c)

Where T_ m  the temperature of metal before immersion

The negative sign show heat lost

The quantity  of gained t by the metal is mathematically represented as      

           Q =  M_o c_o \Delta T

=>        Q =  M_o c_o (T_c - T_o)

So  

         Mc_p (T_m  -  T_c)   =   M_o c_o (T_c - T_o)

substituting values

          - 60 * 0.1027 (T_m  - 39)   =   810 * 0.7167 *  (39 - 35)

=>       T_m  =  376.8 ^o C

         

6 0
3 years ago
How far can a sound wave travel in 90 seconds when the ambient air temperature is 10 C?
Ksju [112]

Answer:

s = 30330.7 m = 30.33 km

Explanation:

First we need to calculate the speed of sound at the given temperature. For this purpose we use the following formula:

v = v₀√[T/273 k]

where,

v = speed of sound at given temperature = ?

v₀ = speed of sound at 0°C = 331 m/s

T = Given Temperature = 10°C + 273 = 283 k

Therefore,

v = (331 m/s)√[283 k/273 k]

v = 337 m/s

Now, we use the following formula to calculate the distance traveled  by sound:

s = vt

where,

s = distance traveled = ?

t = time taken = 90 s

Therefore,

s = (337 m/s)(90 s)

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If he happens to be walking north, then it takes him

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