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3 years ago

Distinguish between pressure and force

1 answer:

3 0

Force is (mass × acceleration) measured in Newton

Pressure is the 'force' per unit area measured in Newton/m^2 (pascal)

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What remains constant during the trajectory of an object

S_A_V [24]

Answer:

projectile

Explanation:

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3 years ago

Huck Finn walks at a speed of 0.70 m/sm/s across his raft (that is, he walks perpendicular to the raft's motion relative to the

exis [7]

Answer:

Explanation:

Given

Velocity of Huck w.r.t to raft $v_{H,raft}=0.7\ m/s$

Perpendicular to the motion of raft

Velocity of Raft in the river $v_{raft,river}=1.6\ m/s$

As Huck is traveling Perpendicular to the raft so he possess two velocities i.e. vertical velocity and horizontal velocity of River when observed from bank

$v_{Huck,river\ bank}=0.7\hat{j}+1.6\hat{i}$

So magnitude of velocity is given by

$|v|=\sqrt{0.7^2+1.6^2}$

$|v|=\sqrt{0.49+2.56}$

$|v|=\sqrt{3.05}$

$|v|=1.74\ m/s$

For direction $\tan =\frac{0.7}{1.6}=0.4375$

$\theta =23.63^{\circ}$ w.r.t river bank

4 0

3 years ago

None hsbisneienbjsnsidns ueheijejeie idhdidnis idndidjd idjdi

lys-0071 [83]

Answer:

‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍

7 0

2 years ago

A 4500 kg car is pushed by 3 boys who exert a force totaling 1500 N. When the boys begin to push the car it is traveling 4.5 m/s

Brilliant_brown [7]

Answer:

The velocity of the car after 3 seconds is 5.5 m/s

Explanation:

Mathematically, we can get the acceleration of the car first;

F = m•a

where m is the mass, F is the applied force and a is the acceleration

from the question;

F = 1500 N , m = 4500 kg and a = ?

By rewriting the formula;

a = F/m

a = 1500/4500 = 1/3 m/s^2

Now to get the velocity of the car after 3 seconds;

Mathematically;

V = U + at

where V is the final

velocity, U is the initial velocity,

From the question, we are trying to get V;

where U = 4.5 m/s

a = 1/3 m/s^2 and t = 3 seconds

Thus;

V = 4.5 + (1/3)(3)

V = 4.5 + 1

V = 5.5 m/s

7 0

2 years ago

3 Water within a piston–cylinder assembly, initially at 10 lbf/in.2 , 500°F, undergoes an internally reversible process to 80 lb

Goryan [66]

Answer:

109.779BTU/lbm and 220.179BTU/lbm

Explanation:

We have the values of Pressure and Temperature,

$P_1 = 10psi\\T_1 = 500\°F\\P_2 = 80psi\\T_2 = 800\°F$

At the tables, this conditions are for enthalpy and specific energy,

$s_1=1.9693BTU/lbm.R\\s_2=1.8704BTU/lbm.R\\u_1=1182.2BTU/lbm\\u_2=1292.6BTU/lbm$

We calculate now the heat transfer for the system,

$q_{1-2}=\frac{T_1+T_2}{2}(s_1-s_2)$

$q_{1-2}=\frac{(500+460)(800+460)}{2}(1.9693-1.8704)$

$q_{1-2}=109.779BTU/lbm$

We can calculate now the work transfer given by,

$w=q-\Delta u$

$w= 109.779-(1182.2-1292.6)$

$w=220.179BTU/lbm$

6 0

3 years ago