If 2 different processes driven by heat transfer start at state 1 and end at state 2 BUT follow different paths, is the amount of energy transferred by heat the same for each path? Yes or no?

Answers

Answer 1

Answer: The answer yes I’m not sure

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A 70.0 kg man jumping from a window lands in an elevated fire rescue net 11.0 m below the window. He momentarily stops when he has stretched the net by 1.50 m. Assuming that mechanical energy is conserved during this process and that the net functions like an ideal spring, find the elastic potential energy of the net when it is stretched by 1.50 m. (10 pts)

Answers

Answer:

70.15 Joule

Explanation:

mass of man, m = 70 kg

intial length, l = 11 m

extension, Δl = 1.5 m

Let K is the spring constant.

In the equilibrium position

mg = K l

70 x 9.8 = K x 11

K = 62.36 N/m

Potential energy stored, U = 0.5 x K x Δl²

U = 0.5 x 62.36 x 1.5 x 1.5

U = 70.15 Joule

If 2.0 x 10^-4 C charge passes a point in 5.0 x 10^-5 s, what is the rate of current flow?4.0 x 10^-1 A
1.0 x 10^2 A
1.0 x 10^-10 A
4.0 x 10^0 A

Answers

Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:

I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
I = 4 A

Q= I×t

Where Q is charge in Coulombs C,

I is current in Amperes A,

t is in seconds s.

The rate of current flow is defined as the charge passing through a point in the circuit per second.

Rearranging formula gives I= $(Q)/(t)$

⇒I = $(2.0*10^-4)/(5.0*10^-5)$ = 4 Amperes

∴ current flow = 4.0×10^0 A

If the velocity of an object changes from 65 m/s to 98 m/s during a time interval of 12 s, what is the acceleration of the object?

Answers

a=acceleration
vf=final velocity
v₀=original velocity
t=time period

a=(vf-v₀)/t
a=(98 m/s-65 m/s)/12s=33 m/s / 12 s=2.75 m/s²

Answer: the acceleration of the object is 2.75 m/s²

   Size of acceleration = (change in speed) / (time for the change)

Change in speed = (end speed) - (start speed)

   =    (98 m/s) - (65 m/s) = 33 m/s .

Time for the change = 12 s

Size of acceleration = (33 m/s) / (12 s)

   = (33/12) (m/s²)

   = 2.75 m/s² .

P waves move through the crust at a speed of about 6.5 km/s. How far will the P wave move in five seconds?

Answers

Distance= speed x time
D=6.5x5
D=32.5km

a very large tank, closed from above and not open to the atmosphere, is partially filled with water. the pressure at the top surface of the water is kept at three times atmospheric pressure by constantly injecting pressurized air from above. water emerges from the tank through a circular nozzle that is located close to the base of the tank and is open to the atmosphere. at the instant when the water stands high above the nozzle, what is the water flow rate from the nozzle? the cross-sectional area of the nozzle is . g