Answer:
True
Explanation:
When a ray travelling parallel to the principle axis of a concave mirror then the light ray reflect out through the mirrors and passing through the focus.
When a light ray travelling through focus of a concave mirror then after reflection the light ray reflect out through the mirror and go parallel to principle axis.
Therefore, rays travelling parallel to the principle axis of a concave mirror will reflect out through the mirrors focus.
It is true.
Answer:
<em><u>Assuming that the vertical speed of the ball is 14 m/s</u></em> we found the given values:
a) V₀ = 23.4 m/s
b) h = 27.9 m
c) t = 0.96 s
d) t = 4.8 s
Explanation:
a) <u>Assuming that the vertical speed is 14 m/s</u> (founded in the book) the initial speed of the ball can be calculated as follows:
![V_{f}^{2} = V_{0}^{2} - 2gh](https://tex.z-dn.net/?f=%20V_%7Bf%7D%5E%7B2%7D%20%3D%20V_%7B0%7D%5E%7B2%7D%20-%202gh%20)
<u>Where:</u>
: is the final speed = 14 m/s
: is the initial speed =?
g: is the gravity = 9.81 m/s²
h: is the height = 18 m
b) The maximum height is:
![V_{f}^{2} = V_{0}^{2} - 2gh](https://tex.z-dn.net/?f=%20V_%7Bf%7D%5E%7B2%7D%20%3D%20V_%7B0%7D%5E%7B2%7D%20-%202gh%20)
![h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m](https://tex.z-dn.net/?f=%20h%20%3D%20%5Cfrac%7BV_%7B0%7D%5E%7B2%7D%7D%7B2g%7D%20%3D%20%5Cfrac%7B%2823.%204%20m%2Fs%29%5E%7B2%7D%7D%7B2%2A9.81%20m%2Fs%5E%7B2%7D%7D%20%3D%2027.9%20m%20)
c) The time can be found using the following equation:
![V_{f} = V_{0} - gt](https://tex.z-dn.net/?f=%20V_%7Bf%7D%20%3D%20V_%7B0%7D%20-%20gt%20)
![t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7BV_%7B0%7D%20-%20V_%7Bf%7D%7D%7Bg%7D%20%3D%20%5Cfrac%7B23.4%20m%2Fs%20-%2014%20m%2Fs%7D%7B9.81%20m%2Fs%5E%7B2%7D%7D%20%3D%200.96%20s%20)
d) The flight time is given by:
![t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s](https://tex.z-dn.net/?f=%20t_%7Bv%7D%20%3D%20%5Cfrac%7B2V_%7B0%7D%7D%7Bg%7D%20%3D%20%5Cfrac%7B2%2A23.4%20m%2Fs%7D%7B9.81%20m%2Fs%5E%7B2%7D%7D%20%3D%204.8%20s%20)
I hope it helps you!
Explanation:
B. All use generators to produce electrical current
<span>First draw a free-body diagram. Torque T = Force F x Distance d where force is the component of gravitational force g and d is the lever arm distance to the pivot point. Since the pivot point is at the back tire we subtract that from the length of the car resulting in d = 1.12 - 0.40 = 0.72 meters = d. We are interested in the perpendicular component of the force exerted on the car jack so use sin 8 degrees then T=1130 kg x 9.81 m/s^2 x sin(8 degrees) x0.72 m = 1,110.80 Newton-meters</span>
Answer:
Explanation:
We know that,
Neptune is 4.5×10^9 km from the sun
And given that,
Earth is 1.5×10^8km from sun
Then,
Let P be the orbital period and
Let a be the semi-major axis
Using Keplers third law
Then, the relation between the orbital period and the semi major axis is
P² ∝ a³
Then,
P² = ka³
P²/a³ = k
So,
P(earth)²/a(earth)³ = P(neptune)² / a(neptune)³
Period of earth P(earth) =1year
Semi major axis of earth is
a(earth) = 1.5×10^8km
The semi major axis of Neptune is
a (Neptune) = 4.5×10^9km
So,
P(E)²/a(E)³ = P(N)² / a(N)³
1² / (1.5×10^8)³ = P(N)² / (4.5×10^9)³
Cross multiply
P(N)² = (4.5×10^9)³ / (1.5×10^8)³
P(N)² = 27000
P(N) =√27000
P(N) = 164.32years
The period of Neptune is 164.32years