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3 years ago

7

Assuming weightless pulleys and 100% efficiency, what is the minimum input force required to lift a 120 N weight using a single

fixed pulley?
A. 21 N
B. 61 N
C. 121 N
D. 241 N

1 answer:

Free_Kalibri [48]3 years ago

8 0

May be b I’m not for sure tho

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Yes, it does. Acceleration is defined as the rate of change of velocity. Velocity is the rate of change of position with a given direction i.e. speed with direction.So even though the speed of the turning car is constant the velocity is not. This means that the velocity is changing, and the rate of change of velocity is called acceleration. Hence, the car is accelerating.

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4 years ago

square root A 1400 kg car is coasting on a horizontal road with a speed of 18 m/s . After passing over an unpaved, sandy stretch

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Answer:

The net force on the car is 2560 N.

Explanation:

According to work energy theorem, the amount of work done is equal to the change of kinetic energy by an object. If ' $W$ ' be the work done on an object to change its kinetic energy from an initial value ' $K_{i}$ ' to the final value ' $K_{f}$ ', then mathematically,

$W = K_{f} - K_{i} = \dfrac{1}{2}~m~(v_{f}^{2} - v_{i}^{2})........................................(I)$

where ' $m$ ' is the mass of the object and ' $v_{i}$ ' and ' $v_{f}$ ' be the initial and final velocity of the object respectively. If ' $F_{net}$ ' be the net force applied on the car, as per given problem, and ' $s$ ' is the displacement occurs then we can write,

$W = F_{net}~.~s.......................................................(II)$

Given, $m = 1400~Kg,~v_{i} = 18~m~s^{-1}~v_{f} = 14~m~s^{-1}~and~s = 35~m$ .

Equating equations (I) and (II),

$&& - F_{net} \times 35~m = \dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})~m^{2}~s^{-2}\\&or,& F_{net} = \dfrac{\dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})}{35}~N\\&or,& F_{net} = 2560~N$

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3 years ago

The prismatic bar has a cross-sectional area

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$sum fx=0\\
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3 years ago

Please select the word from the list that best fits the definition I study French so I can talk to my cousin who lives in Paris

salantis [7]

Answer:

There is no list...

Explanation:

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3 years ago

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A spaceship travels at a constant speed from earth to a planet orbiting another star. When the spacecraft arrives, 12 years have

tigry1 [53]

Answer:

$7.2878\times10^{16} m$

Explanation:

Time elapsed on Earth $d_{t}$ = 12 years

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now r= $\frac{d_{t} }{d_t_o}$ = 12/9.2= 1.3043

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$v= 3\times10^8 \sqrt{1-\frac{1}{1.3043^2}}$

v= $1.9258\times10^8$

therefore distance L= $d_{t}\times v$

putting value we get

= $12\times\times1.9258\times10^8$

= $12\times\times365\times24\times3600\times1.9258\times10^8$

= $7.2878\times10^{16} m$

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3 years ago