Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. The heat capacity of Object A is

Answer:

The runner's deceleration is -23.33

Given:

Initial speed = 3.5

Final speed = 0

Time taken = 0.15 s

To determine:

Deceleration of the runner =?

Used Formula:

Using the first equation of motion,

v = u + at

Where, v = final speed

u = initial speed

a = deceleration

t = duration

Solution:

v = u + at

Where, v = final speed

u = initial speed

a = deceleration

t = duration

0 = 3.5 + a (0.15)

-3.5 = 0.15 (a)

a =

a = -23.33

The negative sign indicates that this represents deceleration.

Hence, the deceleration of the runner is -23.33

Answer:

W = -510.98 J

Explanation:

Force = 43 N, 61° SW

Displacement = 12 m, 22° NE

The work done is calculated using:

W = F*d*cos(A)

where A is the angle between the applied force and displacement.

The angle A between the force and displacement is determined as A = 61 + 90 + 22 = 172°

Hence, W = 43 * 12 * cos(172)

This results in W = -510.98 J

The negative result indicates that the work is done contrary to the direction of the force applied.

Answer:

The amount of heat that enters the gas throughout this two-step process totals 120 cal.

Explanation:

Given that,

Moles present = 3

Heat capacity at volume held constant = 4.9 cal/mol.K

Heat capacity at pressure held constant = 6.9 cal/mol.K

Starting temperature = 300 K

Ending temperature = 320 K

We are tasked with determining the heat absorbed by the gas at constant pressure

Employing the heat formula

Substituting the values into the equation

Next, we calculate the heat absorbed by the gas at constant volume

Using the corresponding heat formula

Insert the values into the formula

Now, it's necessary to evaluate the total heat flow into the gas during both steps

Using the total heat formula

Thus, the heat that transfers into the gas throughout this two-step process amounts to 120 cal.