Answer: 10.29 sec.
Explanation:
Neglecting drag and friction, and at road level , the energy developed during the time the car is accelerating, is equal to the change in kinetic energy.
If the car starts from rest, this means the following:
ΔK = 1/2 m*vf ²
As Power (by definition) is equal to Energy/Time= 75000 W= 75000 N.m/seg, in order to get time in seconds, we need to convert 100 km/h to m/sec first:
100 (Km/h)*( 1000m /1 Km)*(3600 sec/1 h)= 27,78 m/sec
Now, we calculate the change in energy:
ΔK= 1/2*2000 Kg. (27,78)² m²/sec²= 771,728 J
<h2>If P= ΔK/Δt, </h2><h2>Δt= ΔK/P= 771,728 J / 75,000 J/sec= 10.29 sec.</h2>
Explanation:
Step1
Absolute pressure is the pressure above zero level of the pressure. Absolute pressure is considering atmospheric pressure in it. Absolute pressure is always positive. There is no negative absolute pressure.
The expression for absolute pressure is given as follows:
Here, 
is absolute pressure, 
is gauge pressure and
is atmospheric pressure.
Step2
Gauge pressure is the pressure that measure above atmospheric pressure. It is not considering atmospheric pressure. It can be negative called vacuum or negative gauge pressure. Gauge pressure used to simplify the pressure equation for fluid analysis.
Explanation:
thermal expansion ∝L = (δL/δT)÷L ----(1)
δL = L∝L + δT ----(2)
we have δL = 12.5x10⁻⁶
length l = 200mm
δT = 115°c - 15°c = 100°c
putting these values into equation 1, we have
δL = 200*12.5X10⁻⁶x100
= 0.25 MM
L₂ = L + δ L
= 200 + 0.25
L₂ = 200.25mm
12.5X10⁻⁶ *115-15 * 20
= 0.025
20 +0.025
D₂ = 20.025
as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0
The pressure of water is 7.3851 kPa
<u>Explanation:</u>
Given data,
V = 150×
m = 1 Kg
= 2 MPa
= 40°C
The waters specific volume is calculated:
= V/m
Here, the waters specific volume at initial condition is , the containers volume is V, waters mass is m.
= 150× /1
= 0.15 / Kg
The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 / Kg and 0.13 / Kg.
= 350+(400-350) 
= 395.17°C
Hence, the initial temperature is 395.17°C.
The volume is constant in the rigid container.
= = 0.15 / Kg
In saturated water labels for = 40°C.
= 0.001008 / Kg
= 19.515 / Kg
The final state is two phase region < < .
In saturated water labels for = 40°C.
= 
= 7.3851 kPa
= 7.3851 kPa
Water can freeze in cold weather and cause brake failure.
