The height is h = 17 10⁶ meters above the surface of Mars. To determine this, we apply Newton's second law according to the universal law of gravitation, represented by F = m a. The centripetal acceleration a is expressed as v² / r. Applying the gravitational force we have G m M / r² = m v² / r. Given that the speed of the object remains constant, we derive v from d / t, where d is the circumference and t is the orbital period. Substituting gives us d = 2π r and v = 2π r / T. Replacing these values leads to the equation G M / r² = (4π² r² / T) / r, so r³ = G M T² / 4π². Converting time into SI units, T = 24.66 h converts to 88776 seconds. Ultimately, the computed value of r is 2,045 10⁶ m, and after subtracting Mars’ radius of 3.39 10⁶ m, we find the height h to be 17 10⁶ m.
To counteract a 58 mph crosswind, the western component of the trajectory must be accounted for. Consequently, directing towards the northwest creates a 45-degree angle, aligning with the destination. This triangle's third vertex is located at the destination, with the right angle positioned there. The western aspect of their flight represents the triangle's base, while the vertical side reflects the resultant path, and the hypotenuse indicates the actual distance traveled. Since the 58 mph crosswind was countered by flying in a northwest direction, the distance from the starting point to the destination should equal the westward segment of their journey. The hypotenuse can be determined via the square root of twice the dimension of the identical sides.
c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02
An alternative method:
c = sqrt (2) * 58 = 1.414 * 58 = 82.02
Thus, they must fly at 82.02 mph.
A hiker proceeds 200 m west and subsequently another 100 m north, resulting in a displacement of 223 m. The direction can be determined using the trigonometric function where sin(angle) = opposite/hypotenuse, yielding an angle of 26.6 degrees. Therefore, the total displacement is 223 m at an angle of 26.6 degrees north of west.
Given the values:
m = 1160 kg
g = 9.81 m/s²
v = 2.5 m/s
Unknowns include:
k (spring constant)
x (spring compression)
1. To address this, use Newton's second law and Hook's law:
2. The total energy in the spring must correspond to the satellite's energy:
By combining these two equations
The overall force acting on the vehicle is zero
Explanation:
Let's evaluate the situation separately for the vertical direction and the horizontal direction along the slope.
Considering the direction perpendicular to the slope, two forces are in effect:
- The weight component acting perpendicular to the slope,
, directed into the slope - The normal force N, directed outward from the slope
Equilibrium exists here, indicating the net force in this direction is zero.
Now let’s examine the parallel direction to the slope. We have two forces present:
- The weight component aligned with the slope,
, directed down the slope - The frictional force
, acting up the slope
The car moves at a constant speed in this direction, indicating that its acceleration is zero.
Thus, according to Newton's second law,
implying the net force is zero:
Learn more about slopes and friction:
, which makes sense because it's as though the ball collided with a wall and simply reversed its velocity post-collision.