Answer:
Velocity of truck will be 20.287 m /sec
Explanation:
We have given mass of the truck m = 4000 kg
Radius of the turn r = 70 m
Coefficient of friction 
Centripetal force is given 
And frictional force is equal to 
For body to be move these two forces must be equal
So 

Answer:
Speed will be equal to 1.40 m/sec
Explanation:
Mass of the rubber ball m = 5.24 kg = 0.00524 kg
Spring is compressed by 5.01 cm
So x = 5.01 cm = 0.0501 m
Spring constant k = 8.08 N/m
Frictional force f = 0.031 N
Distance moved by ball d = 15.8 cm = 0.158 m
Energy gained by spring

Energy lost due to friction

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J
This energy will be kinetic energy


v = 1.40 m/sec
Answer:
15 cm
Explanation:
= Diameter of the coin = 15 mm
= Diameter of the image of coin = 5 mm
= distance of the coin from mirror = 15 cm
= distance of the image of coin from mirror = ?
Using the equation


= - 5 cm
= radius of curvature
Using the mirror equation


= - 15 cm
<span>The Appalachian Mountains were formed when colliding tectonic plates folded and upthrust, mainly during the Permian Period and again in the Cretaceous Period. The folds and thrusts were then eroded and carved by wind, streams and glaciers. These erosive processes are ongoing, and the topography of the Appalachian Mountains continue to change. They have changed with the miles of land that are cleared of all vegetation and topsoil. In the 1970's coal miners literally blow away the top of a mountain to get to the coal underneath.</span>
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.