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4 years ago

Which substance(s) have no fixed shape and no fixed volume?

Engineering

2 answers:

Black_prince [1.1K]4 years ago

5 0

Answer:

I think gas substances

stiks02 [169]4 years ago

4 0

Answer:

Gas substances have no fixed shape or volume.

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Remote?? maybe I’m not really sure

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3 years ago

Find the number of Btu conducted through a wall in 8 hours. The wall is 8 feet high by 24 feet long and has a total R-value of 1

dedylja [7]

Answer:

ΔQ = 4930.37 BTu

Explanation:

given data

height h = 8ft

Δt = 8 hours

length L = 24 feet

R value = 16.2 hr⋅°F⋅ft² /Btu

inside temperature t1 = 68°F

outside temperature t2 = 16°F

to find out

number of Btu conducted

solution

we get here number of Btu conducted by this expression that s

$\frac{\Delta Q}{\Delta t} =\frac{-A}{R} (t2 -t1)$ ......................1

here A is area that is = h × L = 8 × 24 = 1492 ft²

put here value we get

$\frac{\Delta Q}{8} =\frac{-192}{16.2} (16-68)$

solve it we get

ΔQ = 4930.37 BTu

7 0

3 years ago

Please help for our stem project :(

Darina [25.2K]

Answer:

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2 years ago

. Two rods, with masses MA and MB having a coefficient of restitution, e, move

GarryVolchara [31]

Answer:

a) $V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}$

$V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}$

b) $U_A = 3.66 m/s$

$V_B = 4.32 m/s$

c) Impulse = 0 kg m/s²

d) percent decrease in kinetic energy = 47.85%

Explanation:

Let $U_A$ be the initial velocity of rod A

Let $U_B$ be the initial velocity of rod B

Let $V_A$ be the final velocity of rod A

Let $V_B$ be the final velocity of rod B

Using the principle of conservation of momentum:

$M_AU_A + M_BU_B = M_AV_A + M_BV_B$ ............(1)

Coefficient of restitution, $e = \frac{V_B - V_A}{U_A - U_B}$

$V_A = V_B - e(U_A - U_B)$ ........................(2)

Substitute equation (2) into equation (1)

$M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B$ ..............(3)

Solving for $V_B$ in equation (3) above:

$V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}$ ....................(4)

From equation (2):

$V_B = V_A + e(U_A -U_B)$ ......(5)

Substitute equation (5) into (1)

$M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))$ ..........(6)

Solving for $V_A$ in equation (6) above:

$V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}$ .........(7)

$M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?$

Rod A is said to be at rest after the impact, $V_A = 0 m/s$

Substitute these parameters into equation (7)

$0 = \frac{(2 - 0.65*1)U_A - (1*3)(1+0.65)}{2+1}\\U_A = 3.66 m/s$

To calculate the final velocity, $V_B$ , substitute the given parameters into (4):

$V_B = \frac{(2*3.66)(1+0.65) - (1 - (0.65*2))*3}{2+1}\\V_B = 4.32 m/s$

c) Impulse, $I = M_AV_A + M_BV_B - (M_AU_A + M_BU_B)$

$I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))$

I = 0 $kg m/s^2$

d) % $\triangle KE = \frac{(0.5 M_A V_A^2 + 0.5 M_B V_B^2) - ( 0.5 M_A U_A^2 + 0.5 M_B U_B^2)}{0.5 M_A U_A^2 + 0.5 M_B U_B^2} * 100\%$

% $\triangle KE = \frac{((0.5*2*0) + (0.5 *1*4.32^2)) - ( (0.5 *2*3.66^2) + 0.5*1*(-3)^2))}{ (0.5 *2*3.66^2) + 0.5*1*(-3)^2)} * 100\%$

% $\triangle KE = -47.85 \%$

7 0

4 years ago

Pls paki answers naman Ito pls sa inyo

netineya [11]

:nanjan lang din yan sa binasa mo ne........

7 0

3 years ago