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saw5 [17]
3 years ago
6

Three people pull simultaneously on a stubborn donkey. jack pulls eastward with a force of 92.5 n, jill pulls with 89.9 n in the

northeast direction, and jane pulls to the southeast with 163 n. (since the donkey is involved with such uncoordinated people, who can blame it for being stubborn?) find the magnitude of the net force the people exert on the donkey.
Physics
1 answer:
alekssr [168]3 years ago
7 0
Jack------------ force of 92.5 n   eastward-------Fjack(X)=92.5 n   Fjack(Y)=0

<span>jill ------------------------------- force of 89.9 n   northeast
Fjill(X)=cos45*89.9=63.57 n
</span>Fjill(Y)=sin45*89.9=63.57 n<span>

</span>jane -----------------------------force of 163 n   southeast
Fjane(X)=cos45*163=115.26 n
Fjane(X)=-sin45*163=-115.26 n

Ftotal (X)=92.5+63.57+115.26=271.33 n
Ftotal (Y)=0+63.57-115.26=-51.69 n

Fotal=((271.33)^2+(-115.26)^2) ^0.5=294.80 n southeast

the magnitude of the net force the people exert on the donkey. is 294.80 n southeast
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Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

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Let the volume of bubble at the surface of the lake is V2.

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P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

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\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

By substituting the values, we get

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6 0
3 years ago
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It is given that,

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Let \theta is the direction of the truck’s displacement from the warehouse from south of east.

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So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.

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To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.

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Define discrimination and write a sentence using the word.
seraphim [82]

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Answer:

Explanation:

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ε = 40×0.35×9.503×10^-3/0.1

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