The average speed of the car is 93.33 km/hr and the average velocity of the car is 40 km/hr.
The total distance cover in east direction is=100*3=300 km
The total distance cover in the west direction=80*1.5=120 km
The total distance covered is =300+120=420 km
And Total displacement of the car is =300-120=180 km
As we know that the average speed is given as
Avg Speed =Total Distance / Total time
=420/4.5=93.33 km/hr
As we know that the average velocity is given as
Avg Speed =Total Displacement/ Total time
=180/4.5=40 km/hr
Therefore, The average speed of the car is 93.33 km/hr and the average velocity of the car is 40 km/hr.
(1) You must find the point of equilibrium between the two forces,
<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2
So,
x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.
</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and
R - x = 3.46*10^8 m
from the center of the Earth.
(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
The coefficient of static friction is 0.222
Explanation:
In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where
is the coefficient of static friction
m is the mass of the car
g is the acceleration of gravity
v is the speed of the car
r is the radius of the track
In this problem, we have:
r = 564 m
v = 35 m/s

And re-arranging the equation for
, we can find the coefficient of static friction:

Learn more about friction:
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Answer:
521 nm
Explanation:
Given the values and units we are given, I'm assuming 5.76*10^14 Hz is frequency.
The formula to use here is λ * υ = c, where λ is wavelength, υ is frequency, and c is the speed of light.
λ = 
Answer:
distance between object and image = 18.9 cm
Explanation:
given data
radius of curvature = 18 cm
focal length = 1/2 radius of curvature
magnification = 40%
to find out
distance between object and image
solution
we know lens formula that is
1/f = 1/v + 1/u ....................1
here f = 18 /2 and v and u is object and image distance
and we know m = 40% = 0.40
so 0.40 = -v / u
so here v = - 0.40 u
so from equation 1
1/f = 1/v + 1/u
2/18 = - 1/0.40u + 1/u
u = -13.5 cm ..................2
and
v = -0.40 (- 13.5)
v = 5.4 cm ......................3
so from equation 2 and 3
distance between object and image = 5.4 + 13.5
distance between object and image = 18.9 cm