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Roman55 [17]
3 years ago
13

When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 n backward on the ground for 0.800 s. what is his f

inal speed?
Physics
1 answer:
VladimirAG [237]3 years ago
6 0
Given:
m = 70 kg, mass
F = 650 N, force
t = 0.8 s, time

Calculate the acceleration.
a = F/m = (650 N)/(70 kg)
   = 9.2857 m/s²

Because the initial velocity is zero, the final velocity (after time 0.8 s) is 
v = a*t = (9.2857 m/s²)*(0.8 s)
   = 7.43 m/s

Answer: 7.43 m/s  (nearest hundredth)
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Find the acceleration a body whose velocity increases from 11m/s to 33m/s in 10 seconds
solong [7]

Answer:

1.1 m/(s)^2

Explanation:

u=11 m/s

v=33 m/s

t=10s

v=u+at

=> 33=22+(a)(10)

=> 33-22=10a

=> 10a=11

=> a=11/10=1.1 m/(s)^2

7 0
2 years ago
What is the momentum of a two-particle system
Harlamova29_29 [7]

Answer:

-67,500 kgm/s

Explanation:

1300 * 20 + 1100 * (-85) = -67,500 kgm/s

8 0
2 years ago
What energy is directly dependent upon velocity and mass?
erastova [34]

Answer:

Kinetic energy

Explanation:

3 0
3 years ago
Even in the most advanced circuits, we cannot oscillate electrons back and forth at that rate through wires. But we can oscillat
den301095 [7]

Answer:

the oscillations of the electrons must be in the 10⁸ Hz = 100 MHz range

Explanation:

The speed of a wave of radio, television, light, heat, all are manifestations of electromagnetic waves that are oscillations of electric and magnetic fields that support each other, the speed of all these waves is the same and the vacuum is equal to c = 3 108 m / s

All waves have a relationship between the speed of the wave, its frequency and wavelength

          c = λ f

          f = c /λ

for this case lam = 1 m

          f = 3 10⁸/1

          f = 3 10⁸ Hz

the oscillations of the electrons must be in the MHz range

It should be clarified that the speed of light in air is a little lower

          n = c / v

          v = c / n

the refractive index of vacuum is n = 1 and the refractive index of air is n = 1.000002

5 0
3 years ago
With detailed explaniation
belka [17]
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

4 0
2 years ago
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