A car has a mass of 1600 kg. It is stuck in the snow and is being pulled out by a cable that applies a force of 7560 N due north

3.258 m/s Explanation: The spring constant is assumed to be 263 N/m and the displacement of the spring is also assumed to be 0.7 m; the coefficient of friction between blocks is 0.4. The energy stored in the spring is described by . Given the conservation of energy in the system, the speed of the 8 kg block just prior to collision is 3.258 m/s.

Answer:

A) and B) are valid.

Explanation:

When an object remains at rest, it is indicative that no net force acts upon it.

The downward gravitational force from Earth must be counterbalanced by an upward force of equal magnitude in order to maintain rest.

This upward force is provided by the normal force, which adjusts to satisfy Newton’s 2nd Law and is always perpendicular to the surface supporting the object (in this instance, the ground).

At the molecular level, this normal force comes from the ground's bonded molecules acting like tiny springs, compressed by the object’s molecules, providing an upward restorative force.

Thus, statements A) and B) are true.

Answer:

a) factor

b) factor

c) factor

d) factor

Explanation:

For an oscillating spring-mass system, the time period is expressed as:

where:

represents the frequency of oscillation

signifies the mass linked to the spring

is the spring's stiffness constant

a) If the mass is doubled:

• New mass,

Thus, the new time period:

this leads to factor as per the question.

b) When the stiffness constant is quadrupled, holding other factors constant:

New stiffness constant,

Thus, the new time period:

this results in factor as required.

c) When both mass and stiffness constant are quadrupled:

New stiffness,

New mass,

Thus, the new time period:

which leads to factor as stated in the question.

d) If amplitude is quadrupled, the time period remains unaffected because T does not depend on amplitude as demonstrated by the equation.

Thus, factor

The complete question is;

A ski jumper descends a ramp and exits the ski track at a horizontal speed of 24 m/s. The slope at the landing site angles downwards at θ = 59◦. The acceleration due to gravity is 9.8 m/s².

What is the value of the relative angle φ at which the ski jumper makes contact with the slope? Provide the answer in degrees.

Answer:

14.08°

Explanation:

The time taken can be calculated using the formula;

t = (2V_x•tan θ)/g

t = (2 × 24 × tan 59)/9.8

t = 8.152 s

Next, the slope of the trajectory at the impact point is determined by;

tan α = V_y/V_x

Given V_x = 24 m/s

We will find V_y using;

v = gt

Therefore;

V_y = gt

V_y = 9.8 × (8.152) = 78.89 m/s

Thus;

tan α = 78.89/24

tan α = 3.2871

α = tan^(-1) 3.2871

α = 73.08°

Consequently;

Relative angle φ = α - θ = 73.08 - 59 = 14.08°