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schepotkina [342]

2 years ago

6

Lifting a rope A mountain climber is about to haul up a 50-m length of hanging rope. How much work will it take if the rope weig

hs 0.624 N>m

1 answer:

timama [110]2 years ago

4 0

Answer:

780 J

Explanation:

W=\int _{\:0}^{50}0.624xdx

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An upward moving object must be experiencing (or at least usually does experience) an upward force?

Hoochie [10]

That's false. Think about a stone or a baseball, during the first
several seconds after you tossed it straight up, before it reaches
its maximum height and starts to come down again.
There's no upward force on it during that time.

Also, after a roller coaster reaches the top of the FIRST hill, there's
no upward force on it for the whole rest of the ride, even though it
coasts up many more hills.

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4 years ago

There isnt enough sanitidzer(alcohol)what solution to be replaced

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For me: WASH OUR HANDS REGULARLY

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A car has a distance between axles of 6.2 m and a center of mass located 2.2 m from the front axle. What is the

NARA [144]

Answer:

The correct option is C

Explanation:

From the question we are told that

The distance between axles of the car is $d_a = 6.2$

The position of the car's center of mass is $D = 2.2 \ m \ in \ front \ of \ the \ axel$

Now we can evaluate the distance of the center of mass to the rear axle as follows

$k = d_a - D$

substituting values

$k = 6.2 - 2.2$

$k = 4 \ m$

assuming the car is at equilibrium, taking moment about the center of mass

$W_{front \ axle} * D - W_{rear \ axle} * k = 0$

=> $W_{front \ axle} * D = W_{rear \ axle} * k$

=> $\frac{W_{front \ axle}}{W_{rear \ axle} } = \frac{k}{D}$

substituting values

$\frac{W_{front \ axle}}{W_{rear \ axle} } = \frac{4}{2.2}$

$\frac{W_{front \ axle}}{W_{rear \ axle} } = 1.81$ Note [ $W_{front \ axle}$ is the front axle weight and

$W_{rear \ axle}$ is the rear axle weight ]

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3 years ago

The triceps muscle in the back of the upper arm extends the forearm.

iren [92.7K]

To solve this problem it is necessary to apply the concepts related to Torque as a function of Force and distance. Basically the torque is located in the forearm and would be determined by the effective perpendicular lever arm and force, that is

$\tau = F \times r$

Where,

F = Force

r = Distance

Replacing,

$\tau = 2*10^3*0.03$

$\tau = 60N\cdot m$

The moment of inertia of the boxer's forearm can be calculated from the relation between torque and moment of inertia and angular acceleration

$\tau = I \alpha$

I = Moment of inertia

$\alpha$ = Angular acceleration

Replacing with our values we have that

$I = \frac{\tau}{\alpha}$

$I = \frac{60}{120}$

$I = 0.5kg\cdot m^2$

Therefore the value of moment of inertia is $0.5kg\cdot m^2$

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3 years ago

1. What are the three ways an object can accelerate?

solong [7]

Answer:

increase speed, decrease speed, and change direction

Explanation:

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3 years ago