Gravity, acceleration, kinetic energy, the atmosphere
Answer:
![x=22.65m](https://tex.z-dn.net/?f=x%3D22.65m)
Explanation:
We have an uniformly accelerated motion, with a negative acceleration. Thus, we use the kinematic equations to calculate the distance will it take to bring the car to a stop:
![v_f^2=v_0^2-2ax\\\frac{0^2-v_0^2}{-2a}=x\\x=\frac{v_0^2}{2a}](https://tex.z-dn.net/?f=v_f%5E2%3Dv_0%5E2-2ax%5C%5C%5Cfrac%7B0%5E2-v_0%5E2%7D%7B-2a%7D%3Dx%5C%5Cx%3D%5Cfrac%7Bv_0%5E2%7D%7B2a%7D)
The acceleration can be calculated using Newton's second law:
![\sum F_x:F_f=ma\\\sum F_y:N=mg](https://tex.z-dn.net/?f=%5Csum%20F_x%3AF_f%3Dma%5C%5C%5Csum%20F_y%3AN%3Dmg)
Recall that the maximum force of friction is defined as
. So, replacing this:
![\mu N=ma\\\mu mg=ma\\a=\mu g\\a=0.901(9.8\frac{m}{s^2})\\a=8.83\frac{m}{s^2}](https://tex.z-dn.net/?f=%5Cmu%20N%3Dma%5C%5C%5Cmu%20mg%3Dma%5C%5Ca%3D%5Cmu%20g%5C%5Ca%3D0.901%289.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%5C%5Ca%3D8.83%5Cfrac%7Bm%7D%7Bs%5E2%7D)
Now, we calculate the distance:
![x=\frac{v_0^2}{2a}\\x=\frac{(20\frac{m}{s})^2}{2(8.83\frac{m}{s^2})}\\x=22.65m](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bv_0%5E2%7D%7B2a%7D%5C%5Cx%3D%5Cfrac%7B%2820%5Cfrac%7Bm%7D%7Bs%7D%29%5E2%7D%7B2%288.83%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%7D%5C%5Cx%3D22.65m)
Kinetic energy is a result of mass in motion at a certain velocity.
<span>1 Joule = 1 kg • (m/s)<span>2
</span></span>the force as a function of mass of the object.
Answer:
Explanation:
same idea as before Liam, first, find the parallel resistance in 35 || 20
(35*20) / (35+20) = 700 / 55 = 12.727272 ohms
now add the 12.727272 + 15 = 27.727272 ohms total resistance
V = IR
10 = I * 27.727272
10 / 27.727272 = I
0.360655 = I
V = IR (again, but across the 15 ohm resistor)
V = 0.360655 * 15
V = 5.4098
Answer:
5.77×10¹⁰ m
Explanation:
From the question,
Applying
Kepler's third law
P² = d³...................... Equation 1
Where P = Planet's period, d = distance between the center of the planet and the sun.
make d the subject of the formula n equation 1
d =
.................. Equation 2
Given: P = 0.24 sidereal.
Substitute the value of P into equation 2
d =
d = 0.386 Au
d = 0.386×1.496×10¹¹
d = 5.77×10¹⁰ m