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2 years ago

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The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope. The sl

eds move on an icy surface. The dogs exert a 240-N force on the rope attached to the front sled. The front rope pulls horizontally. Find the acceleration of the sleds.

1 answer:

KonstantinChe [14]2 years ago

6 0

Answer:

Acceleration, $a=1.2\ m/s^2$

Explanation:

Given that,

The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope, m = 100 kg

Force exerted by the doges on the rope attached to the front sled, F = 240 N

To find,

The acceleration of the sleds.

Solution,

Let a is the acceleration of the sleds. The product of mass and acceleration is called force. Its expression is given by :

F = ma

$a=\dfrac{F}{m}$

$a=\dfrac{240\ N}{2\times 100\ kg}$ (m = 2m)

$a=1.2\ m/s^2$

So, the acceleration of the sleds is $1.2\ m/s^2$ .

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otez555 [7]

Answer:

When reviewing the results, the correct one is C

Explanation:

The right hand rule is widely useful in knowing the direction of force in a maganto field,

The ruler sets the thumb in the direction of the positive particle, the fingers extended in the direction of the magnetic field, and the palm in the direction of the force.

Let's apply this to our exercise.

The thumb that is the speed goes in the negative direction of the axis,

The two extended that the magnetic field look negative x,

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When reviewing the results, the correct one is C

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2 years ago

A packing crate with mass 80.0 kg is at rest on a horizontal, frictionless surface. At t = 0 a net horizontal force in the +x-di

Nataly [62]

Answer:

Final speed of the crate is 15 m/s

Explanation:

As we know that constant force F = 80 N is applied on the object for t = 12 s

Now we can use definition of force to find the speed after t = 12 s

$F . t = m(v_f - v_i)$

so here we know that object is at rest initially so we have

$80 (12) = 80( v_f - 0)$

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Now for next 6 s the force decreases to ZERO linearly

so we can write the force equation as

$F = 80 - \frac{40}{3} t$

now again by same equation we have

$\int F .dt = m(v_f - v_i)$

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$80 t - \frac{40t^2}{6} = 80(v_f - 12)$

put t = 6 s

$480 - 240 = 80(v_f - 12)$

$v_f = 12 + 3$

$v_f = 15 m/s$

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1 year ago

Kimonoski takes a 9-minute shower every day. The shower uses about 1.8 gal per minute of water. He also uses 23 gallons of hot w

ioda

Answer:

$Q_{week} = 458884.6\, BTU$

Explanation:

The weekly water consumption of Kimonoski is:

$m_{bath,week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (1.8\,\frac{gal}{min} )\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (\frac{1\,min}{60\,s} )\cdot (9\,min)\cdot (\frac{60\,s}{1\,min} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)$

$m_{bath.week} = 948.205\,lbm$

$m_{others, week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (23\,gal)\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)$

$m_{others, week} = 1346.218\,lbm$

$m_{week} = m_{bath,week} + m_{others, week}$

$m_{week} = 2294.423\,lbm$

The total energy required per week for hot water is:

$Q_{week} = m_{week}\cdot c_{p,water}\cdot \Delta T$

$Q_{week} =(2294.423\,lbm)\cdot (1\,\frac{BTU}{lbm\cdot ^{\textdegree}F} )\cdot (50^{\textdegree}F)$

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1 year ago

An electric clock is hanging on a wall. As you are watching the second hand rotate, the clock's battery stops functioning, and t

Setler [38]

Answer:

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Explanation:

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initial angular velocity is termed as positive angular velocity

$\omega = positive$

now it comes to stop so angular acceleration is taken in opposite to the direction of angular speed

so we will have

$\alpha = negative$

so here correct answer is

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1 year ago

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Klio2033 [76]

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2 years ago