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aev [14]
3 years ago
10

Brandon buys a new seadoo he goes 12 km north from the beach he jumps wakes for 6 km to the east the chases a boat 10 km north w

hat distance did he cover what was his displacement
Physics
1 answer:
d1i1m1o1n [39]3 years ago
3 0

Answer:

Distance covered 28 km

displacement is 22.8 km North-East

Explanation:

Distance shows how far apart objects or points are from each other. The distance he covered is the sum of all the distance travelled. Therefore:

Distance covered = 12 km + 6 km + 10 km = 28 km

Displacement is a vector quantity (has direction). It is the overall change in position.

The total distance traveled north = 12 km + 10 km = 22 km

The distance traveled east = 6 km

The displacement (d) is:

d² = 22² + 6² = 484 + 36

d² = 520

d = √520 = 22.8 km

Therefore the displacement is 22.8 km North-East

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Answer:

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Explanation:

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Where given:

Atmospheric pressure (P1) = 1.013*10^5 Pa

T1 =  20+273.15 = 293.15 K

P2 = ?

T2 = 120+273.15 = 393.15 K

Using the gas equation: P1/T1 = P2/T2

Therefore, P2 = P1*T2/T1 = 1.013*10^5 *393.15/293.15 = 13.6*10^4 Pa

The net pressure = P2 - P1 = 13.6*10^4 - 1.013*10^5 = 34.6 kPa

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Area = 0.11 m^2

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The net force F_{120} = 34555.7*0.11 = 3801.13 N

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You have a 1.8m long copper wire. You want to make an N-turn current loop that generates a 2.0mT magnetic field at the center wh
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Answer:

The diameter is  0.022m.

Explanation:

The magnetic field B at the center of the coil is given by

(1). B = \dfrac{\mu_0 NI}{d}

where \mu_0 = 1.26*10^{-6} m\:kg\:s^{-2} A^{-2} is the magnetic constant, I is the current, N number of coils, and d is the diameter of the coil.

Now, if we call L the length of the wire, then it must be true that

\pi dN = L <em>(this says </em>N<em> coil circumferences (</em>c=\pi d<em>) fit into </em>L<em> )</em>

\therefore N = \dfrac{L}{\pi d }

putting this into equation (1) we get:

B = \dfrac{\mu_0 IL }{\pi d^2}

solve for d:

\boxed{d = \sqrt{\dfrac{\mu_0I L }{\pi B} }}

putting in the numerical values

\mu_0 = 1.26*10^{-6} m\:kg\:s^{-2} A^{-2}

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we get:

d = \sqrt{\dfrac{(1.26*10^{-6})(1.3) *1.8 }{\pi (2.0*10^{-3})} }

\boxed{d = 0.022m}

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4 years ago
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