Answer:
5558643.69 N
Explanation:
F = Force
v = Velocity = 31.5 knots
Converting to m/s


Power is given by

The forward force is exerted on the ship at this highest attainable speed is 5558643.69 N
Answer:
The acceleration of the car, a = -3.75 m/s²
Explanation:
Given data,
The initial velocity of the airplane, u = 75 m/s
The final velocity of the plane, v = 0 m/s
The time period of motion, t = 20 s
Using the I equations of motion
v = u + at
a = (v - u) / t
= (0 - 75) / 20
= -3.75 m/s²
The negative sign indicates that the plane is decelerating
Hence, the acceleration of the car, a = -3.75 m/s²
Answer:
554.27N
Explanation:
(a) The max frictional force exerted horizontally on the crate and the floor is,
Substitute the values,
μs=0.5
mass=113kg
g=9.81m/s
Ff=μsN
=μsmg
=(0.5 x 113 x 9.81)
Ff=554.27N
To solve this problem we will apply the definitions given in Newtonian theory about the Force of gravity, and the Force caused by weight. Both will be defined below, and in equal equilibrium condition to clear the variable concerning acceleration due to gravity. Finally, with the values provided in the statement, it will be replaced.
The equation for the gravitational force between the Earth and the object on the surface of the Earth is

Where,
G = Universal gravitational constant
= Mass of Earth
= Distance between object and center of earth
= Mass of Object
The equation for the gravitational pulling force on the object due to gravitational acceleration is

Equation the two expression we have


This the acceleration due to gravity which is composite constant.
Replacing with our values we have then


The value of composite constant is
. Here, the composite constant is nothing but the acceleration due to gravity which is constant always.
Answer:
0.195 m
Explanation:
Speed is distance moved per unit time, expressed as s=d/t and making d the subject of the formula then d=st
Where d is distance/depth moved, s is rhe speed of waves and t is time in seconds.
Substituting s with 1300 m/s and t with 0.00015 s then the depth of metal segment will be
D=1300*0.00015=0.195 m
Therefore, the depth is equivalent to 0.195 m