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kenny6666 [7]
3 years ago
12

A sort of "projectile launcher" is shown below. A large current moves in a closed loop composed of fixed rails, a power supply,

and a very light, almost frictionless bar touching the rails. A magnetic field is perpendicular to the plane of the circuit. The bar has a length of 20 cm, a mass of 1.2 g, and is placed in a field of 1.7 T.
a) What constant current flow is needed in order for it to accelerate to 33 m/s in a distance of 1.0 m?

b) In what direction must the field point?

PLEASE ANSWER OR ATLEAST TRY!!! Please and thank you.
Physics
1 answer:
Zarrin [17]3 years ago
3 0

Answer:

wallah i don't understand anything with my stoopid brain

Explanation:

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The efficiency of a screw is low because of _____.
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The correct answer that would best complete the given statement above would be the second option. A screw is an inclined plane wrapped around a cylinder. <span>The efficiency of a screw is low because there is more input than output. In other words, it is because of friction. Hope this answer helps.</span>
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Answer:

Hope this help you!!

Explanation:

Crust : The crust is the thinnest layer of the Earth. It has an average thickness of about 18 miles below land, and around 6 miles below the oceans. The crust is the layer that makes up the Earth's surface and it lies on top of a harder layer, called the mantle.

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Which of the following shows the correct order in which organs take part in the process of digestion? Large intestine, small int
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A hockey ball accelerates from 0.m/s to 25m/s in 0.05 seconds what is the acceleration of the ball ?
S_A_V [24]

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3 0
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You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t
kati45 [8]

Answer:

The  coefficient of kinetic friction  \mu_k =  0.724

Explanation:

From the question we are told that

   The  length of the lane is  l =  36.0 \  m

    The speed of the truck is  v  =  22.6\  m/s

     

Generally from the work-energy theorem we have that  

    \Delta KE  =   N  *  \mu_k *  l

Here N  is the normal force acting on the truck which is mathematically represented as

     \Delta KE is the change in kinetic energy which is mathematically represented as

        \Delta KE =  \frac{1}{2} *  m *  v^2

=>     \Delta KE =  0.5  *  m *  22.6^2

=>      \Delta KE =  255.38m

        255.38m =    m *  9.8  *  \mu_k *   36.0

=>     255.38  =    352.8  *  \mu_k

=>   \mu_k =  0.724

 

6 0
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