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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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Suppose a rocket ship in deep space moves with constant acceleration equal to 9.80 m/s2, which gives the illusion of normal grav
DochEvi [55]

Answer:

a) 3673469.39 seconds

b) 6.61×10¹⁴ m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0.12×3×10⁸ m/s

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

Equation of motion

v=u+at\\\Rightarrow 0.12\times 3\times 10^8=0+9.8t\\\Rightarrow t=\frac{0.12\times 3\times 10^8}{9.8}=3673469.39\ s

Time taken to reach 12% of light speed is 3673469.39 seconds

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{(0.12\times 3\times 10^8)^2-0^2}{2\times 9.8}\\\Rightarrow s=6.61 \times 10^{14}\ m

The distance it would have to travel is 6.61×10¹⁴ m

7 0
3 years ago
a man hits a golf ball (0.2kg) which accelerates at a rate of 20 m/s what amount of force acted on the ball
MAVERICK [17]

The ball only accelerates during the brief time that the club is in contact
with it. After it leaves the club face, it takes off at a constant speed.

If it accelerates at 20 m/s² during the hit, then

   Force = (mass) x (acceleration) = (0.2kg) x (20 m/s²) = <em>4 newtons</em> .


8 0
3 years ago
Read 2 more answers
(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r
Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface U_1=\frac{GM_em}{R_e}

Gravitational Potential Energy at height h is U_2=\frac{GM_em}{R_e+h}

Energy required to lift the satellite E_1=U_1-U_2

E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}

Now Energy required to orbit around the earth

E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}

\Delta E=E_1-E_2

\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}

E_1=E_2  (given)

\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0

\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0

h=\frac{R_e}{2}

h=3.19\times 10^6\ m

(b)For greater height E_1  is greater than E_2

thus energy to lift the satellite is more than orbiting around earth

4 0
2 years ago
A rabbit is moving in the positive x-direction at 1.10 m/s when it spots a predator and accelerates to a velocity of 10.9 m/s al
anzhelika [568]

Answer:

aₓ = 0 ,       ay = -6.8125 m / s²

Explanation:

This is an exercise that we can solve with kinematics equations.

Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.

x axis

          vₓ = v₀ₓ = 1.10 m / s

          aₓ = 0

y axis

initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration

         v_{y}= v_{oy} -ay t

          ay = (v_{oy} -v_{y}) / t

          ay = (0 -10.9) / 1.6

          ay = -6.8125 m / s²

the sign indicates that the acceleration goes in the negative direction of the y axis

8 0
3 years ago
Time dilation means that
Anastaziya [24]
The answer is D. time really does pass more slowly in a rest frame of reference relative to a frame of reference that is moving
5 0
3 years ago
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