Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e., the length and radius have twice th

1) The projectile's motion follows
,
In order to determine the velocity, we must compute the derivative of h(t): Next, we will compute the speed at t=2 s and t=4 s: The negative value of the second speed suggests that the projectile has already attained its highest point and is now descending. 2) The maximum height of the projectile occurs when its speed equals zero: Thus, we have And solving yields


3) To determine the maximum height, we substitute the time at which the projectile reaches this peak into h(t), specifically t=2.30 s: 4) The time at which the projectile lands is when the height reaches zero; h(t)=0, which leads to This results in a second-degree equation, producing two answers: the negative root can be disregarded as it lacks physical significance; the second root is

, which indicates the landing time of the projectile. 5) The moment the projectile impacts the ground corresponds to the velocity at time t=4.68 s:

, carrying a negative sign to denote a downward direction.

A tennis player recovering from an ankle injury, restricted from pivoting, can keep her cardiovascular fitness by using the rowing machine, pedaling on a stationary bike with one leg, or swimming. These exercises do not necessitate directional changes and are safe for her injury.

Answer:

Please refer to the explanation

Explanation:

Race distance is 5km

Top speed = 45 yards

Converting yards to kilometers:

1km equals 1093.613 yards

x = 45 yards

(1093.613 * x) = 45

x = 45 / 1093.613

x = 0.0411480 km

Where x indicates the maximum distance he can sustain his highest speed in kilometers.

Thus, from the data available, we can determine that Lamar will not be able to maintain his maximum speed for the full 5km race, as he can only sustain it for 0.0411 kilometers.

Answer:

The distance before stopping is 1.52 m,

velocity is 4.0 m/s on the y-axis

Explanation:

The particle’s motion is two-dimensional due to acceleration along both the x and y axes; each axis can be addressed independently for calculations.

a) At the moment the particle starts to reverse, its velocity should be zero (Vfx = 0)

     Vfₓ = V₀ₓ + aₓ t  

     t = -  V₀ₓ/aₓ

     t = - 2.4/(-1.9)

     t=  1.26 s

When the particle stops, we calculate its position

     X1 = V₀ₓ t + ½ aₓ t²

     X1= 2.4 1.26 + ½ (-1.9) 1.26²

     X1= 1.52 m

At this point, the particle begins returning.

b) The velocity comprises both x and y components.

   For the x section, Vₓ = 0 m/s indicates a halt, but the y component retains a velocity

    Vfy= Voy + ay t

    Vfy= 0 + 3.2 1.26

    Vfy = 4.0 m/s

Thus, the velocity reads as

    V = (0 x^ + 4.0 y^) m/s

c) To graph the motion, we create a table listing position x and y at given time intervals; let's begin the calculations for equations

      X = V₀ₓ t+ ½  aₓ t²

      Y = Voy t + ½  ay t²

      X= 2.4 t + ½ (-1.9) t²

      Y= 0 + ½ 3.2 t²

      X= 2.4 t – 0.95 t²

      Y=   1.6 t²

With these equations, we construct two graphs for position against time, one for the x-axis and another for the y-axis

                       Chart for graphing

              Time (s)     x(m)            y(m)

                 0                0               0

                 0.5             0.960       0.4

          1       1.45          1.6

                 1.50      1.46      3.6

                2.00      1.00      6.4

Answer:

4.05 m/s

Explanation:

We will express the varying velocities as vectors.

Newton moves northward at 3.90 m/s from Daniel's stationary position.

V_n = 3.9 j

Assuming Pauli runs relative to Daniel at velocity X.

The relative velocity of Newton as seen by Pauli will be

3.9 j - X

Given that

the relative velocity of Newton with respect to moving Pauli = 1.1 i (1.1 towards the east).

Thus,

3.9 j - X = 1.1 i

X = -1.1 i + 3.9 j.

Magnitude of X

X² = 1.1² + 3.9²

X = 4.05 m/s

Therefore, Pauli runs relative to Daniel at 4.05 m/s.

The direction will be west of north at an angle θ,

Tan θ = 1.1 / 3.9