Answer:![0.084 m/s^2](https://tex.z-dn.net/?f=0.084%20m%2Fs%5E2)
Explanation:
Given
Total time=27 min 43.6 s=1663.6 s
total distance=10 km
Initial distance ![d_1=8.13 km](https://tex.z-dn.net/?f=d_1%3D8.13%20km)
time taken=25 min =1500 s
initial speed ![v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s](https://tex.z-dn.net/?f=v_1%3D%5Cfrac%7B8.13%5Ctimes%201000%7D%7B25%5Ctimes%2060%7D%3D5.6%20m%2Fs)
after 8.13 km mark steve started to accelerate
speed after 60 s
![v_2=v_1+at](https://tex.z-dn.net/?f=v_2%3Dv_1%2Bat)
![v_2=5.6+a\times 60](https://tex.z-dn.net/?f=v_2%3D5.6%2Ba%5Ctimes%2060)
distance traveled in 60 sec
![d_2=v_1\times 60+\frac{a60^2}{2}](https://tex.z-dn.net/?f=d_2%3Dv_1%5Ctimes%2060%2B%5Cfrac%7Ba60%5E2%7D%7B2%7D)
![d_2=336+1800 a](https://tex.z-dn.net/?f=d_2%3D336%2B1800%20a)
time taken in last part of journey
![t_3=1663.6-1560=103.6 s](https://tex.z-dn.net/?f=t_3%3D1663.6-1560%3D103.6%20s)
distance traveled in this time
![d_3=v_2\times t_3](https://tex.z-dn.net/?f=d_3%3Dv_2%5Ctimes%20t_3)
![d_3=\left ( 5.6+a\times 60\right )103.6](https://tex.z-dn.net/?f=d_3%3D%5Cleft%20%28%205.6%2Ba%5Ctimes%2060%5Cright%20%29103.6)
and total distance![=d_1+d_2+d_3](https://tex.z-dn.net/?f=%3Dd_1%2Bd_2%2Bd_3)
![10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6](https://tex.z-dn.net/?f=10000%3D8.13%5Ctimes%201000%2B336%2B1800%20a%2B%5Cleft%20%28%205.6%2Ba%5Ctimes%2060%5Cright%20%29103.6)
![1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6](https://tex.z-dn.net/?f=1870%3D336%2B1800%20a%2B%5Cleft%20%28%205.6%2Ba%5Ctimes%2060%5Cright%20%29103.6)
![a=0.084 m/s^2](https://tex.z-dn.net/?f=a%3D0.084%20m%2Fs%5E2)
Answer:
Vertical velocity decreases.
Explanation:
The motion of the ball is a projectile ball, which consists of two independent motions:
- a horizontal motion, with constant velocity
- a vertical motion, with constant acceleration g=9.8 m/s^2 towards the ground
In the vertical motion, there is a constant acceleration directed downward: this means that the vertical velocity decreases as the ball goes higher. In fact, it decreases following the equation
![v(t)=v_0 -gt](https://tex.z-dn.net/?f=v%28t%29%3Dv_0%20-gt)
And it decreases until the ball reaches its maximum height, then it starts increasing again.
Answer:
r = 1.63×10^5 mi
Explanation:
Let r = distance of object from earth
Rs = distance between earth and sun
Ms = mass of the sun
= 3.24×10^5 Me (Me = mass of earth)
At a distance R from earth, the force Fs exerted by the sun on the object is equal to the force Fe exerted by the earth on the object. Using Newton's universal law of gravitation,
Fs = Fe
GmMs/(Rs - r)^2 = GmMe/r^2
This simplifies to
Ms/(Rs - r)^2 = Me/r^2
(3.24×10^5 Me)/(Rs - r)^2 = Me/r^2
Taking the reciprocal and then its square root, this simplifies further to
Rs - r = (569.2)r ----> Rs = 570.2r
or
r = Rs/570.2 = (9.3×10^7 mi)/570.2
= 1.63×10^5 mi