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Kay [80]
2 years ago
10

A small sandbag is dropped from rest from a hovering hot-air balloon. (assume the positive direction is upward.) (a) after 1.5 s

, what is the velocity of the sandbag?
Physics
1 answer:
marishachu [46]2 years ago
3 0

solution:

We know v0 = 0, a = 9.8, t = 4.0. We need to solve for v

so,

we use the equation:

v = v0 + at

v = 0 + 9.8*4.0

v = 39.2 m/s

Now we just need to solve for d, so we use the equation:

d = v0t + 1/2*a*t^2

d = 0*4.0 + 1/2*9.8*4.0^2

d = 78.4 m

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A 4.0 cm × 4.2 cm rectangle lies in the xy-plane. You may want to review (Pages 664 - 668) . Part A What is the electric flux th
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(A). The flux is 0.336 N.m²/C

(B). The flux is zero.

Explanation:

Given that,

Length = 4.2 cm

Width = 4.0 cm

Electric field E=(150 i-200 k)\ N/C

Area vector is perpendicular to xy plane

(A). We need to calculate the flux

Using formula of flux

\phi=E\cdot A

Where, E = electric field

A = area

Put the value into the formula

\phi=(150 i-200 k)\times(4.2\times10^{-2}\times4.0\times10^{-2})k

\phi=-200\times4.2\times10^{-2}\times4.0\times10^{-2}

\phi=-0.336\ N.m^2/C

(B). Given electric field

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We need to calculate the flux

Using formula of flux

\phi=E\cdot A

Put the value into the formula

\phi=(150 i-200 j)\times(4.2\times10^{-2}\times4.0\times10^{-2})k

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So, the flux is

\phi=0

Hence, (A). The flux is -0.336 N.m²/C

(B). The flux is zero.

7 0
3 years ago
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