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2 years ago

Please please help me, don't guess please my test is timed !!

Physics

1 answer:

dimulka [17.4K]2 years ago

4 0

Answer:

a) J = F t = 40 * .05 = 2 N-s

b) J = 2 N-s momentum changed by 2 N-s

c) Initial momentum appears to be zero

J = change in momentum = m v2 - m v1 = m v2 = 2 N-s

v2 = J / m = 2 / .057 = 35 m/s

d) if the impulse time was increased and the average force remained the same then the change in momentum would increase with a corresponding increase in velocity attained - note the increase in v2 in part c)

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I dont know what pfp to put on today so I'm asking u guys....

Mandarinka [93]

Answer:

i think number 2 should be your pfp

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2 years ago

If you walk 30 meters forwards, and then turn around and walk 25 meters backwards, what is the distance that you walked? What di

xeze [42]

Given :

Walk in forward direction is 30 m .

Walk in backward direction is 25 m .

To Find :

The distance and displacement .

Solution :

We know , distance is total distance covered and displacement is distance between final and initial position .

So , distance travelled is :

D = 30 + 25 m = 55 m .

Now , we first move 30 m in forward direction and then 25 m in backward direction .

So , displacement is :

D = 30 - 25 m = 5 m .

Therefore , distance and displacement covered is 55 m and 5 m respectively .

Hence , this is the required solution .

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3 years ago

Walking across a carpet is an example of charge being transferred by

Sati [7]

static electricity and friction

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3 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

3 years ago

Determine the number of atoms per unit cell in a (a) face-centered cubic, (b) body- centered cubic, and (c) diamond lattice.

fredd [130]

Answer:

a) 4

b) 2

c) 8

Explanation:

In a cubic lattice, each atom of the vertex is shared among other 8 unit cells, so the atoms on the vertex contribute to 1/8 to a given unit cell.

a) Ina face-centered cubic we have 8 atoms on the vertex and one in each face, which is shared with another unit cell, so it contribute to 1/2

Therefore, the total atoms are:

8*(1/8) + 6*(1/2) = 4

b) In the body centered cubic structure, the centered atom is not shared with another cell, therefore it contribute to 1 to the given cell:

The number of atoms per unit cel is:

8*(1/8) + 1 = 2

c) The diamond lattice is similar to the face-centered cubic lattice but it contains two identical atoms per lattice point.

Therefore it must contain twice atoms than the face-centered cubic lattice:

that is, it has 8 atoms per unit cell

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3 years ago