<span>Even though the Sun has a greater mass than Earth, the Moon orbits Earth because it's closer to the Earth than to the Sun. Because of this proximity between the Earth and the Moon, the Earth has a stronger gravitational pull than the Sun does. Furthermore, the Earth's mass is 81 times that of the Moon, and so at this proximity, it is more than able to overpower what pull the Sun exerts on the Moon.</span>
Answer:
0.6
Explanation:
Angular acceleration is equal to Net Torque divided by rotational inertia, which is the rotational equivalent to Newton’s 2nd Law. Therefore, angular acceleration is equal to 3.6/6 which is 0.6. Hope this helped!
Answer:
A) 26V
Explanation:
(a) the potential difference between the plates
Initial capacitance can be calculated using below expresion
C1= A ε0/ d1
Where d1= distance between = 2.70 mm= 2.70× 10^-3 m
ε0= permittivity of space= 8.85× 10^-12 Fm^-1
A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2
If we substitute the values we
C1= A ε0/ d1
=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3
C1=2.589 ×10^-12 F= 2.59 pF
Initial charge can be determined using below expresion
q1= C1 × V1
V1=2.589 ×10^-12 F
V1= voltage=7.90 V
If we substitute we have
q1= 2.589 ×10^-12 × 7.90
q1= 20.45×10^-12C
20.45 pC
Final capacitance can be calculated as
C2= A ε0/ d2
d2=8.80 mm= /8.80× 10^-3
7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3
C1=0.794 ×10^-12 F= 0.794 pF
Final charge= initial charge
q2=q1 (since the battery is disconnected)
q2=q1= 20.45 pC
Final potential difference
V2= q/C2
= 20.45/0.794
= 26V
Answer:
A quantity that does not depend on the direction is called a scalar quantity. Vector quantities have two characteristics, a magnitude, and a direction. Scalar quantities have only a magnitude. When comparing two vector quantities of the same type, you have to compare both the magnitude and the direction.
Scalar quantities only have magnitude (size). Scalar quantities include distance...
A quantity that is specified by both size and direction is a vector. Displacement includes both size and direction and is an example of a vector. However, distance is a physical quantity that does not include a direction and isn't a vector.
Explanation:
hope this helps...
To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as
![R = 350ft](https://tex.z-dn.net/?f=R%20%3D%20350ft)
![a_t = 1.1ft/s^2](https://tex.z-dn.net/?f=a_t%20%3D%201.1ft%2Fs%5E2)
PART A )
![a_c = \frac{V^2}{R}](https://tex.z-dn.net/?f=a_c%20%3D%20%5Cfrac%7BV%5E2%7D%7BR%7D)
![a_c = \frac{V^2}{350}](https://tex.z-dn.net/?f=a_c%20%3D%20%5Cfrac%7BV%5E2%7D%7B350%7D)
Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is ![5.25ft/s^2](https://tex.z-dn.net/?f=5.25ft%2Fs%5E2)
![a = \sqrt{a_t^2+a_r^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Csqrt%7Ba_t%5E2%2Ba_r%5E2%7D)
![5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}](https://tex.z-dn.net/?f=5.25%20%3D%20%5Csqrt%7B%281.1%29%5E2%2B%28%5Cfrac%7Bv%5E2%7D%7B350%7D%29%5E2%7D)
![27.5625 = 1.21 + \frac{v^4}{122500}](https://tex.z-dn.net/?f=27.5625%20%3D%201.21%20%2B%20%5Cfrac%7Bv%5E4%7D%7B122500%7D)
![v=42.3877ft/s](https://tex.z-dn.net/?f=v%3D42.3877ft%2Fs)
Now calculate the angular velocity of the motorcycle
![v = r\omega](https://tex.z-dn.net/?f=v%20%3D%20r%5Comega)
![42.3877 = 350\omega](https://tex.z-dn.net/?f=42.3877%20%3D%20350%5Comega)
![\omega = 0.1211rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%200.1211rad%2Fs)
Calculate the angular acceleration of the motorcycle
![a_t = r\alpha](https://tex.z-dn.net/?f=a_t%20%3D%20r%5Calpha)
![1.1 = 350\alpha](https://tex.z-dn.net/?f=1.1%20%3D%20350%5Calpha)
![\alpha = 3.1428*10^{-3}rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%203.1428%2A10%5E%7B-3%7Drad%2Fs%5E2)
Calculate the time needed by the motorcycle to reach an acceleration of
![5.25ft/s^2](https://tex.z-dn.net/?f=5.25ft%2Fs%5E2)
![\omega = \alpha t](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Calpha%20t)
![0.1211 = 3.1428*10^{-3}t](https://tex.z-dn.net/?f=0.1211%20%3D%203.1428%2A10%5E%7B-3%7Dt)
![t = 38.53s](https://tex.z-dn.net/?f=t%20%3D%2038.53s)
PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is ![6.75ft/s^2](https://tex.z-dn.net/?f=6.75ft%2Fs%5E2)
![a = \sqrt{a_t^2+a_r^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Csqrt%7Ba_t%5E2%2Ba_r%5E2%7D)
![6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}](https://tex.z-dn.net/?f=6.75%20%3D%20%5Csqrt%7B%281.1%29%5E2%2B%28%5Cfrac%7Bv%5E2%7D%7B350%7D%29%5E2%7D)
![45.5625 = 1.21 + \frac{v^4}{122500}](https://tex.z-dn.net/?f=45.5625%20%3D%201.21%20%2B%20%5Cfrac%7Bv%5E4%7D%7B122500%7D)
![v=48.2796ft/s](https://tex.z-dn.net/?f=v%3D48.2796ft%2Fs)
PART C)
Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is ![21.5ft/s](https://tex.z-dn.net/?f=21.5ft%2Fs)
![a_r = \frac{v^2}{R}](https://tex.z-dn.net/?f=a_r%20%3D%20%5Cfrac%7Bv%5E2%7D%7BR%7D)
![a_r = \frac{21.5^2}{350}](https://tex.z-dn.net/?f=a_r%20%3D%20%5Cfrac%7B21.5%5E2%7D%7B350%7D)
![a_r =1.3207ft/s^2](https://tex.z-dn.net/?f=a_r%20%3D1.3207ft%2Fs%5E2)
Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is ![21.5ft/s](https://tex.z-dn.net/?f=21.5ft%2Fs)
![a = \sqrt{a_t^2+a_r^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Csqrt%7Ba_t%5E2%2Ba_r%5E2%7D)
![a = \sqrt{(1.1)^2+(1.3207)^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Csqrt%7B%281.1%29%5E2%2B%281.3207%29%5E2%7D)
![a = 1.7187ft/s^2](https://tex.z-dn.net/?f=a%20%3D%201.7187ft%2Fs%5E2)
PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is ![6.75ft/s^2](https://tex.z-dn.net/?f=6.75ft%2Fs%5E2)
![a = \sqrt{a_t^2+a_r^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Csqrt%7Ba_t%5E2%2Ba_r%5E2%7D)
![6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}](https://tex.z-dn.net/?f=6.75%20%3D%20%5Csqrt%7B%281.1%29%5E2%2B%28%5Cfrac%7Bv%5E2%7D%7B350%7D%29%5E2%7D)
![45.5625 = 1.21 + \frac{v^4}{122500}](https://tex.z-dn.net/?f=45.5625%20%3D%201.21%20%2B%20%5Cfrac%7Bv%5E4%7D%7B122500%7D)
![v=48.2796ft/s](https://tex.z-dn.net/?f=v%3D48.2796ft%2Fs)