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Elenna [48]
2 years ago
9

In the diagram below, point A is southwest of point B. Jaleh walks for 400 seconds along path 1, going from point A to point B.

She then runs for 100 seconds from point B back to point A, following path 3. What is Jaleh's velocity for the first part of her trip? (Remember that velocity is calculated using displacement and time.)
A. 0 m/s

B. 0.5 m/s northeast

C. 1.4 m/s southwest

D. 1 m/s northeast

Physics
2 answers:
Maslowich2 years ago
6 0
The answer is indeed A according to my calculations
timama [110]2 years ago
4 0

Answer:

C

Explanation:

velocity = displacement (m) / change of time (s)

velocity = (400 + 300) / (100 + 400)

velocity = 1.4 m/s

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Answer:

\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine (\sf \eta)

Explanation:

\boxed{ \bold{v =  \frac{2}{9}  \frac{( {r}^{2} ( \rho -  \sigma)g)}{ \eta} }}

\sf \implies \eta =  \frac{2}{9}  \frac{( {r}^{2}( \rho -  \sigma)g )}{v}

Substituting values of r, ρ, σ, v & g in the equation:

\sf \implies \eta =  \frac{2}{9}  \frac{( {(0.15)}^{2}  \times  (7.85 - 1.25) \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times  \frac{145.6191}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times 64.7196

\sf \implies \eta =  2 \times 7.191

\sf \implies \eta =  14.382 \: poise

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Answer

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T₁ and T₂ be the tension at the ends of the wire

equating all the vertical force

T₁ + T₂ = 600 + 844

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taking moment about T₂

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1181.58+W=1813.05

W= 631.47 kg

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