1) Focal length
We can find the focal length of the mirror by using the mirror equation:

(1)
where
f is the focal length

is the distance of the object from the mirror

is the distance of the image from the mirror
In this case,

, while

(the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:

from which we find

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.
3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.
4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
Answer:R=1607556m
θ=180degrees
Explanation:
d1=74.8m
d2=160.7km=160.7km*1000
d2=160700m
d3=80m
d4=198.1m
Using analytical method :
Rx=-(160700+75*cos(41.8))= -160755.9m
Ry= -(74.8+75sin(41.8))-198.1=73m
Magnitude, R:
R=√Rx+Ry
R=√160755.9^2+20^2=160755.916
R=160756m
Direction,θ:
θ=arctan(Rx/Ry)
θ=arctan(-73/160755.9)
θ=-7.9256*10^-6
Note that θ is in the second quadrant, so add 180
θ=180-7.9256*10^6=180degrees
Answer:
The amount of gas that is to be released in the first second in other to attain an acceleration of 27.0 m/s2 is

Explanation:
From the question we are told that
The mass of the rocket is m = 6300 kg
The velocity at gas is being ejected is u = 2000 m/s
The initial acceleration desired is 
The time taken for the gas to be ejected is t = 1 s
Generally this desired acceleration is mathematically represented as

Here
is the rate at which gas is being ejected with respect to time
Substituting values

=> 
=> 
=> 
=> 
Answer: the answer is d
Explanation: there are not more than 10 violations within a twelve month period hope this helps