Answer:
= 4.38 × 10³⁴kgm²/s
Explanation:
Given that,
mass of moon m = 9.5 × 10²²kg
Orbital radius r = 4.28 × 10⁵km
Orbital period T = 28.9days
T = 28.9 × 24 × 60 × 60
= 2,496,960s
Angular momentum of the moon about the planet
L = mvr
L = mr²w
![L = mr^2\frac{2\pi }{T} \\\\L = \frac{9.5 \times 10^2^2 \times(4.28\times10^8)^2\times2\times3.14}{2496960} \\\\L = 4.389.5 \times 10^3^4kgm^2/s](https://tex.z-dn.net/?f=L%20%3D%20mr%5E2%5Cfrac%7B2%5Cpi%20%7D%7BT%7D%20%5C%5C%5C%5CL%20%3D%20%5Cfrac%7B9.5%20%5Ctimes%2010%5E2%5E2%20%5Ctimes%284.28%5Ctimes10%5E8%29%5E2%5Ctimes2%5Ctimes3.14%7D%7B2496960%7D%20%5C%5C%5C%5CL%20%3D%204.389.5%20%5Ctimes%2010%5E3%5E4kgm%5E2%2Fs)
Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:
![V_{Ax}=V_{A}cos\theta](https://tex.z-dn.net/?f=V_%7BAx%7D%3DV_%7BA%7Dcos%5Ctheta)
![V_{Ax}=(21m/s)cos(-14^{o})](https://tex.z-dn.net/?f=V_%7BAx%7D%3D%2821m%2Fs%29cos%28-14%5E%7Bo%7D%29)
![V_{Ax}=20.38 m/s](https://tex.z-dn.net/?f=V_%7BAx%7D%3D20.38%20m%2Fs)
![V_{Ay}=V_{A}sin\theta](https://tex.z-dn.net/?f=V_%7BAy%7D%3DV_%7BA%7Dsin%5Ctheta)
![V_{Ay}=(21m/s)sin(-14^{o})](https://tex.z-dn.net/?f=V_%7BAy%7D%3D%2821m%2Fs%29sin%28-14%5E%7Bo%7D%29)
![V_{Ay}=-5.08 m/s](https://tex.z-dn.net/?f=V_%7BAy%7D%3D-5.08%20m%2Fs)
we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:
![x_{A}=V_{Ax}t](https://tex.z-dn.net/?f=x_%7BA%7D%3DV_%7BAx%7Dt)
![x_{A}=(20.38m/s)(0.317s)](https://tex.z-dn.net/?f=x_%7BA%7D%3D%2820.38m%2Fs%29%280.317s%29)
![x_{A}=6.46m](https://tex.z-dn.net/?f=x_%7BA%7D%3D6.46m%20)
Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:
![V_{Bx}=20.88 m/s](https://tex.z-dn.net/?f=V_%7BBx%7D%3D20.88%20m%2Fs)
![V_{By}=-2.195 m/s](https://tex.z-dn.net/?f=V_%7BBy%7D%3D-2.195%20m%2Fs)
![y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}](https://tex.z-dn.net/?f=y_%7BBf%7D%3Dy_%7BB0%7D%2BV_%7B0%7Dt-%5Cfrac%7B1%7D%7B2%7Dat%5E%7B2%7D)
![0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}](https://tex.z-dn.net/?f=0%3D2.1m%2B%28-2.195m%2Fs%29t-%5Cfrac%7B1%7D%7B2%7D%28-9.8m%2Fs%5E%7B2%7D%29t%5E%7B2%7D)
![-4.9t^{2}-2.195t+2.1=0](https://tex.z-dn.net/?f=-4.9t%5E%7B2%7D-2.195t%2B2.1%3D0)
![t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E%7B2%7D-4ac%7D%7D%7B2a%7D)
![t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-%28-2.195%29%5Cpm%5Csqrt%7B%28-2.195%29%5E%7B2%7D-4%28-4.9%29%282.1%29%7D%7D%7B2%28-4.9%29%7D)
t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:
![x_{B}=V_{Bx}t](https://tex.z-dn.net/?f=x_%7BB%7D%3DV_%7BBx%7Dt)
![x_{B}=(20.88m/s)(0.468s)](https://tex.z-dn.net/?f=x_%7BB%7D%3D%2820.88m%2Fs%29%280.468s%29)
![x_{B}=9.77m](https://tex.z-dn.net/?f=x_%7BB%7D%3D9.77m)
So once we got the two distances we can now find the difference between them:
![x_{B}-x_{A}=9.77m-6.46m=3.31m](https://tex.z-dn.net/?f=x_%7BB%7D-x_%7BA%7D%3D9.77m-6.46m%3D3.31m)
so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Answer:
because of the raindrop velocity relative of the car has a vertical and horizontal component
Explanation:
- The car moves in a <em>horizontal direction </em>relative to the ground. The raindrops fall in the <em>vertical direction</em> relative to the ground.
- Their velocity relative to the moving car has both vertical and horizontal components and this is the reason for the diagonal streaks on the side window.
-
The diagonal streaks on the windshield arise from a different reason.
-
The drops are pushed off to one side of the windshield because of air resistance.
Answer:
Net force: 20 N to the right
mass of the bag: 20.489 kg
acceleration: 0.976 m/s^2
Explanation:
Since the normal force and the weight are equal in magnitude but opposite in direction, they add up to zero in the vertical direction. In the horizontal direction, the 195 N tension to the right minus the 175 force of friction to the left render a net force towards the right of magnitude:
195 N - 175 N = 20 N
So net force on the bag is 20 N to the right.
The mass of the bag can be found using the value of the weight force: 201 N:
mass = Weight/g = 201 / 9.81 = 20.489 kg
and the acceleration of the bag can be found as the net force divided by the mass we just found:
acceleration = 20 N / 20.489 kg = 0.976 m/s^2
Answer: Mabye like an ocean with dolphins swiming/jumping? Or even use the blue as a sky and then put green grass and do foxes or and a phoenix flying with a fox under it?
Explanation:
Just some ideas!