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stealth61 [152]

3 years ago

10

An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8

seconds? If the net tangential force applied to the racer is constant, what is its value?

1 answer:

marissa [1.9K]3 years ago

6 0

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

P= 800 x 108.3

P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

$a=18.66\ m/s^2$

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

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Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 240°C. The water is slowly heated at constan

Digiron [165]

Answer:

The final temperature of water is 381.39 °C.

Explanation:

Given that

Mass of water = 5 kg

Heat transfer at constant pressure Q = 2960 KJ

Initial temperature = 240 °C

We know that heat transfer at constant pressure given as follows

$Q=mC_p\Delta T$

We know that for water

$C_p=4.187\ \frac{KJ}{kg.K}$

Lets take final temperature of water is T

So

$Q=mC_p\Delta T$

$2960=5\times 4.187(T-240)$

T=381.39 °C

So the final temperature of water is 381.39 °C.

7 0

3 years ago

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure

and we know that is also in term of change in density also

so

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

here ρ is density

and we know ρ for 20°C = 998 kg/m³

and ρ for 80°C = 972 kg/m³

so from equation 2 put all value

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

so now % change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

put here all value

0.0130 = $\frac{\pi }{4} *2^2*dl$

dl = 0.004138 m

so water level rise is 4.138 mm

8 0

3 years ago

A steel bar 110 mm long and having a square cross section 22 mm on an edge is pulled in tension with a load of 89,000 N, and exp

iragen [17]

Answer:

Elastic modulus of steel = 202.27 GPa

Explanation:

given data

long = 110 mm = 0.11 m

cross section 22 mm = 0.022 m

load = 89,000 N

elongation = 0.10 mm = 1 × $10^{-4}$ m

solution

we know that Elastic modulus is express as

Elastic modulus = $\frac{stress}{strain}$ ................1

here stress is

Stress = $\frac{Force}{area}$ .................2

Area = (0.022)²

and

Strain = $\frac{extension}{length}$ .............3

so here put value in equation 1 we get

Elastic modulus = $\frac{89,000\times 0.11}{0.022^2 \times 1 \times 10^{-4} }$

Elastic modulus of steel = 202.27 × $10^9$ Pa

Elastic modulus of steel = 202.27 GPa

3 0

3 years ago

2.) A fluid moves in a steady manner between two sections in a flow

Talja [164]

Answer:

$250\ \text{lbm/min}$

$625\ \text{ft/min}$

Explanation:

$A_1$ = Area of section 1 = $10\ \text{ft}^2$

$V_1$ = Velocity of water at section 1 = 100 ft/min

$v_1$ = Specific volume at section 1 = $4\ \text{ft}^3/\text{lbm}$

$\rho$ = Density of fluid = $0.2\ \text{lb/ft}^3$

$A_2$ = Area of section 2 = $2\ \text{ft}^2$

Mass flow rate is given by

$m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}$

The mass flow rate through the pipe is $250\ \text{lbm/min}$

As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

$m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}$

The speed at section 2 is $625\ \text{ft/min}$ .

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3 years ago

Which option identifies the type of device the engineer will develop in the following scenario?

Stells [14]

It would be actuator

4 0

2 years ago

Read 2 more answers