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DedPeter [7]
3 years ago
6

An elevator cab is pulled directly upward by a single cable. The elevator cab and its single occupant have a mass of 2300 kg. Wh

en that occupant drops a coin, its acceleration relative to the cab is 6.80 m/s² downward. What is the tension in the cable?
Physics
1 answer:
Murrr4er [49]3 years ago
3 0

Answer:

15.64 KN

Explanation:

mass of the elevator cab with a single occupant= 2300 kg

acceleration relative to the cab a_{ce}= 6.80 m/s^2

acceleration of the coin relative to the cab in the opposite direction of motion of cab so we can consider it as a= -6.80 m/s^2

The acceleration of elevator cab relative to the ground a_{cg}

now we can say that

a_{ce} +a_{eg} =a_{cg}

=-6.80+ a_{eg}= -9.8

[tex]a_{eg}=-9.8+6.80=-3.8

The forces that act on elevator cab are tension and gravitational, applying newtons second law

T- mg= ma_{eg}

Then the tension in the cable is

T= 2300(-3.8)+2300×9.8=  15640 N= 15.64 KN

therefore tension in the string will be 15.64 KN

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A 0.500-kg potato is fired at an angle of 80.0° above the horizontal from a PVC pipe used as a "potato gun" and reaches a height
lions [1.4K]

Answer:

(a) 47.15ms^{-1}

(b) 2470.13ms^{-2}

(c) 1235.06N and 252.05 as a ratio

Explanation:

From Newton's second law of motion

F=ma where m is mass, a is acceleration and F is net force

Also from kinetic equation of motion, velocity and displacement are related using equation

v^{2}=u^{2}+2sa

Where v is final velocity, u is initial velocity, a is acceleration and s is displacement

From the free body diagram attached, final velocity at maximum height is 0 and initial velocity is usin80^{o}

Also, the vertical component can be written as

v_{y}^{2} }=u_{y}^{2} } -2gs The negative sign before 2gs means displacement is opposite the gravitational force

Where g is acceleration due to gravity,u_{y} is vertical component of initial velocity and v_{y} is vertical component of final velocity

Since v_{y} is 0

u_{y}^{2} } =2gs

s=110 and g is taken as 9.8

u_{y}=\sqrt{2*9.8*110}=46.43275

u_{y}=46.43ms^{-1}

Also, it's evident that the vertical component of initial velocity is u_{y}=u_{i}sin \theta where \theta is angle of projection and u_{i} is resultant velocity

Making u_{i} the subject we obtain u_{i}=\frac {u_{y}}{sin \theta}

Since u_{y} and \theta are known as 46.43ms^{-1} and 80^{o} respectively, then u_{i}=\frac {46.43ms^{-1}}{sin 80^{o}}=47.15ms^{-1}

Therefore, the velocity of potato is 47.15ms^{-1}

(b)

Displacement depends on length of tube hence s=0.450m hence going back to kinetic equation v^{2}=u^{2}+2sa

The final velocity v is answer in part a which is 47.15ms^{-1}, initial velocity u is 0ms^{-1} hence the equation is re-written as

v^{2}=2sa and making a the subject we obtain

a=\frac {v^{2}}{2s}

a=\frac {47.15^{2}}{2*0.450}=2470.13ms^{-2}

Therefore, average acceleration is 2470.13ms^{-2}

(c)

From Newton's second law of motion, F=ma where m=0.500kg and a is 2470.13ms^{-2}

Therefore, the average force of potato is

F=0.5*2470.13=1235.06N

F=1235.06N

The weight, W of potato is given by W=mg

Taking R as ratio of average force and weight of potato

R=\frac {F}{W}=\frac {F}{mg} and since F=1235.06, m=0.500kg and g=9.8

R=\frac {1235.06}{0.500*9.8}=252.05

Therefore, ratio of average force to weight is 252.05

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Displacement = velocity * time

Just substitute the value, & solve the equation.

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B I think is the answer
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Answer:

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