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shepuryov [24]
3 years ago
10

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1500 kg, which is required to travel upward 57 m in

1.8 min, starting and ending at rest. The elevator's counterweight has a mass of only 840 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable?
Physics
1 answer:
Kazeer [188]3 years ago
8 0

Answer:

Explanation:

it equals16 2211

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Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) .
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Explanation:

Given that,

Force, F=((-8i)+6j)\ N

Position of the particle, r=(3i+4j)\ m

(a) The toque on a particle about the origin is given by :

\tau=F\times r

\tau=((-8i)+6j) \times (3i+4j)

Taking the cross product of above two vectors, we get the value of torque as :

\tau=(0+0-50k)\ N-m

(b) Let \theta is the angle between r and F. The angle between two vectors is given by :

cos\theta=\dfrac{r.F}{|r|.|F|}

cos\theta=\dfrac{(3i+4j).((-8i)+6j)}{(\sqrt{3^2+4^2} ).(\sqrt{8^2+6^2}) }

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A 65 kg student sits 3 m away from a 70 kg student. What is the magnitude of the gravitational force between the two students?
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Answer:

F=3.37\times 10^{-8}\ N

Explanation:

Given that,

Mass of student 1, m₁ = 65 kg

Mass of student 2, m₂ = 70 kg

The distance between the students, d = 3 m

We need to find the magnitude of the gravitational force between the two students. The formula for the magnitude of the gravitational force between two masses is given by :

F=G\dfrac{m_1m_2}{d^2}\\\\F=6.67\times 10^{-11}\times \dfrac{65\times 70}{(3)^2}\\F=3.37\times 10^{-8}\ N

So, the gravitational force between the two students is 3.37\times 10^{-8}\ N.

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