A test car carrying a crash test dummy accelerates from 0 to 30 m/s and then crashes into a brick wall. Describe the direction o

Answer:

x₂=2×1

Explanation:

According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;

mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.

mgx=(kx)²/2

x=2mg/k----------------compression when the object is at rest

However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy

Thus, if kx²=mv² then

v=x *√(k/m) ----------------where v=0

<pDoubling v results in multiplying x *√(k/m) by 2, leading to x₂ being double x₁

<span>A centripetal force maintains an object's circular motion. When the ball is at the highest point, we can assume that the ball's speed v is such that the weight of the ball matches the required centripetal force to keep it moving in a circle. Hence, the string will not become slack. centripetal force = weight of the ball m v^2 / r = m g v^2 / r = g v^2 = g r v = sqrt { g r } v = sqrt { (9.80~m/s^2) (0.7 m) } v = 2.62 m/s Thus, the minimum speed for the ball at the top position is 2.62 m/s.</span>

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.

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:

Answer:

a. Angle= 28.82°

b. Approved. Although he might feel cold, he should be able to cross.

Explanation:

Velocity Vector

Velocity is a measure of how quickly something is moving in a specific direction. It is represented as a vector that has both magnitude and direction. If an object can only move in one direction, then speed can serve as the scalar equivalent of that velocity (only focusing on magnitude).

a.

The explorer aims to swim across a river to reach his campsite, as depicted in the image below. The river's velocity is vr and the explorer's swimming speed in still water is ve. If he were to swim straight towards the campsite, he would end up downstream due to the river's current. Therefore, he must swim at an angle that allows him to overcome the current while still moving towards his goal. This angle relative to the shore is what we need to determine. The explorer's speed can be broken down into its horizontal (vx) and vertical (vy) components. In order to counteract the river's flow:

We can calculate the vertical component of the explorer's swimming speed as

Thus

Finding the value of

Then the angle is given by

b.

The component of the explorer's velocity that goes horizontally is

This represents the actual velocity directed towards the campsite

Considering that

To find t

Calculating the duration for the explorer to cross the river

As this time is under the hypothermia threshold (300 seconds), the conclusion is

Approved. Although he will feel cold, he should manage to cross successfully.