Explanation:
As we know that
time = distance/speed
The time used for firs half of the trip was
(1 mi)(12 mi/hour) = 1/12 hours = 5 minutes
The last half of the trip will took 10 minutes, 1/6 hour.
Speed = distance/ time
(1 mil) = (1/6h) = 6 mil/h
so the speed for last half of the trip was = 6mph
the average speed was
(2mil)(1/4 hour) = 8 mil/hour
So the ling's average speed was 8mph.
Answer:30 cm
Explanation:
Given
object distance ![u=30\ cm](https://tex.z-dn.net/?f=u%3D30%5C%20cm)
focal length of concave mirror ![f=15\ cm](https://tex.z-dn.net/?f=f%3D15%5C%20cm)
height of object ![h_o=1.8\ cm](https://tex.z-dn.net/?f=h_o%3D1.8%5C%20cm)
Using mirror formula
![\Rightarrow \frac{1}{v}+\frac{1}{u}=\frac{1}{f}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7B1%7D%7Bv%7D%2B%5Cfrac%7B1%7D%7Bu%7D%3D%5Cfrac%7B1%7D%7Bf%7D)
![\Rightarrow \frac{1}{v}=\frac{1}{15}-\frac{1}{30}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7B1%7D%7Bv%7D%3D%5Cfrac%7B1%7D%7B15%7D-%5Cfrac%7B1%7D%7B30%7D)
![\Rightarrow \frac{1}{v}=\frac{2-1}{30}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7B1%7D%7Bv%7D%3D%5Cfrac%7B2-1%7D%7B30%7D)
![\Rightarrow v=30\ cm](https://tex.z-dn.net/?f=%5CRightarrow%20v%3D30%5C%20cm)
and magnification is
![\Rightarrow m=\frac{-v}{u}=\frac{h_i}{h_o}](https://tex.z-dn.net/?f=%5CRightarrow%20m%3D%5Cfrac%7B-v%7D%7Bu%7D%3D%5Cfrac%7Bh_i%7D%7Bh_o%7D)
![\Rightarrow \frac{-30}{30}=\frac{h_i}{1.8}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7B-30%7D%7B30%7D%3D%5Cfrac%7Bh_i%7D%7B1.8%7D)
![\Rightarrow h_i=-1.8\ cm](https://tex.z-dn.net/?f=%5CRightarrow%20h_i%3D-1.8%5C%20cm)
So height of object is same as of object .
Position :image is formed at the spot where object is placed.
I think the vehicle relation to the hearer is : Past the hearer
Have you ever walk when a motor cycle run past you ? You will notice that you will hear less sound as the motor cycle goes further
Hope this helps
Answer:
distance = 6.1022 x10^16[m]
Explanation:
To solve this problem we must use the formula of the average speed which relates distance to time, so we have
v = distance / time
where:
v = velocity = 3 x 10^8 [m/s]
distance = x [meters]
time = 6.45 [light years]
Now we have to convert from light-years to seconds in order to get the distance in meters.
![t = 6.45 [light-years]*365[\frac{days}{1light-year}]*24[\frac{hr}{1day}] *60[\frac{min}{1hr}]*60[\frac{seg}{1min} ] =203407200 [s]](https://tex.z-dn.net/?f=t%20%3D%206.45%20%5Blight-years%5D%2A365%5B%5Cfrac%7Bdays%7D%7B1light-year%7D%5D%2A24%5B%5Cfrac%7Bhr%7D%7B1day%7D%5D%20%2A60%5B%5Cfrac%7Bmin%7D%7B1hr%7D%5D%2A60%5B%5Cfrac%7Bseg%7D%7B1min%7D%20%5D%20%3D203407200%20%5Bs%5D)
Now using the formula:
distance = v * time
distance = (3*10^8)*203407200
distance = 6.1022 x10^16[m]