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LenKa [72]
3 years ago
6

Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant

speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine:
Physics
1 answer:
levacccp [35]3 years ago
3 0

This question is incomplete, the complete question is;

Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).

Answer:

the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

Explanation:

Given the data in the question and image below and as illustrated in the second image;

distance S = 40 m

V_B = 54 km/hr

V_A = 72 km/hr

α = 100 m

now, angular velocity of Bxy will be;

ω_B = V_B / α

so, we substitute

ω_B = ( 54 × 1000/3600) / 100

ω_B = 15 / 100

ω_B = 0.15 rad/s

Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

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Zepler [3.9K]

Answer:

Decreases to half.

Explanation:

From the question given above, the following data were obtained:

Initial mass (m₁) = m

Initial force (F₁) = F

Initial acceleration (a₁) =?

Final mass (m₂) = ½m

Final force (F₂) = ¼F

Final acceleration (a₂) =?

Next, we shall determine a₁. This can be obtained as follow:

F₁ = m₁a₁

F = ma₁

Divide both side by m

a₁ = F / m

Next, we shall determine a₂.

F₂ = m₂a₂

¼F = ½ma₂

2F = 4ma₂

Divide both side by 4m

a₂ = 2F / 4m

a₂ = F / 2m

Finally, we shall determine the ratio of a₂ to a₁. This can be obtained as follow:

a₁ = F / m

a₂ = F / 2m

a₂ : a₁ = a₂ / a₁

a₂ / a₁ = F/2m ÷ F/m

a₂ / a₁ = F/2m × m/F

a₂ / a₁ = ½

Cross multiply

a₂ = ½a₁

From the illustrations made above, the acceleration of the car will decrease to half the original acceleration

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3 years ago
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Alec says the force of gravity is stronger on a piece of paper after it’s crumpled. His classmate, Jordan, disagrees. Alec “prov
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Explanation:

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You stand on a frictionless platform that is rotating with an angular speed of 5.1 rev/s. Your arms are outstretched, and you ho
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Answer:

\omega'=19.419\ rev.s^{-1}

Explanation:

Given:

angular speed of rotation of friction-less platform, \omega=5.1\ rev.s^{-1}

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moment of inertia with contracted weight, I'=2.6\ kg.m^2

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