A basketball player is running at a constant speed of 2.5 m/s when he tosses a basketball upward with a speed of 6.0 m/s. How fa

Response:

65.14 g

Clarification:

The entropy change resulting from mixing two metals is

Here, R is the real gas constant, n_{cu} denotes the moles of copper, while n_{Ni} signifies the moles of nickel.

Therefore, determining the moles of Nickel yields

=1.7 moles

Then substituting n_{Ni} = 1.7 into the equation produces ΔS= 15,

with R= 8.314, and solving gives

Upon resolving, the quantity of moles of copper in the mixture equals

n_cu= 1.025

Thus, the copper mass is calculated as n_cu×M_cu = 1.025×63.55 = 65.14 g

Consequently, the required mass is found to be = 65.14 g

Answer:

  v = 54.2 m/s

Explanation:

We can utilize conservation of energy to solve this issue.

Initial condition Higher

         Em₀ = U = m g h

Final condition. Lower

        = K = ½ m v²

        Em₀ = Em_{f}

        m g h = ½ m v²

         v² = 2gh

         v = √ (2gh)

Now let's perform the calculation

         v = √ (2 * 9.8 * 150)

         v = 54.2 m/s

Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To calculate the kinetic energy variation, we can utilize the work-energy theorem.

W = ΔK

∫ F .dx = K - K₀

If the object starts from rest, then K₀ = 0.

So, ∫ F dx cos θ = K.

As the force and displacement directions align, the angle is zero, and hence the cosine is 1.

Now we can substitute and perform integration:

α ∫ x³ dx + β ∫ dx = K.

Thus, α x⁴ / 4 + β x = K.

Next, we evaluate from the limits F = 0 to F:

α (x⁴ / 4 - 0) + β (x - 0) = K.

Consequently, K = αX⁴ / 4 + β x.

This results in K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x.

To finalize the computation, we need to ascertain the displacement.

Answer:

Height (h) = 17 m

Velocity (v) = 18.6 m/s

Explanation: This problem can be solved using kinematic motion equations.

Given Data

Initial velocity (u) = 0

Acceleration (a) = g

Time (t) = 1.9 seconds

First, we calculate the height.

Then, we find the final velocity

The acceleration graph is a linear representation described by y=9.8, as it remains constant:

The velocity graph can be represented by y=9.8x (where y signifies velocity and x indicates time):

The displacement graph can be described as y=4.9x^2 (with x as time and y as displacement):

These graphs apply exclusively from x=0 to x=1.9, so disregard other sections of the graphs.

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

= Reservoir pressure = 10 atm

= Reservoir temperature = 300 K

= Exit pressure = 1 atm

= Exit temperature

= Specific gas constant = 287 J/kgK

= Specific heat ratio = 1.4 for air

Assuming isentropic flow

Flow temperature at exit is 155.38424 K

Density at exit can be derived using the ideal gas equation

Flow density at exit measures 2.2721 kg/m³