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Serhud [2]
3 years ago
14

Two men decide to use their cars to pull a

Physics
1 answer:
Ulleksa [173]3 years ago
7 0
The resultant force in the direction the truck is headed is:

734N*cos(31) + 1084N*cos(23)
629N+998N = 1627N
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Explain how the message is transmitted from the receiving neuron’s dendrites down the axon all the way to the axon terminal (ele
Vikki [24]

The vesicles release neurotransmitters. These cross the synapse and are accepted by the receptors in the dendrites of the next neuron.

Explanation:

An axon, or nerve fiber, is a long slender projection of a nerve cell, or neuron, that conducts electrical impulses away from the neuron's cell body. Axons are in effect the primary transmission lines of the nervous system, and as bundles they help make up nerves.

When an action potential reaches the axon terminal, it depolarizes the membrane and opens voltage-gated Na+ channels. Na+ ions enter the cell, further depolarizing the presynaptic membrane.

6 0
2 years ago
Hello, how to do qn 5?:)l
Hitman42 [59]

Answer:

A

Explanation:

Please see the attached picture for the full solution.

Since we are only concerned about the decrease in gravitational potential energy of the car, we look at the decrease in height of the car as it moves from point X to point Y, instead of the distance travelled by the car.

7 0
2 years ago
Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea
Alexxx [7]

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

         

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   \frac{m}{m+ \frac{I}{r^2} }

          v² = 2gh    \frac{1}{1 + \frac{I}{m r^2} }

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = g \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

let's use trigonometry

         sin θ = h / L

         

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    \frac{1}{1 + \frac{mr^2 }{m r^2 } }1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  \frac{1}{1 + \frac{1}{2}  \frac{mr^2}{mr^2} } = g sin θ   \frac{1}{1+ \frac{1}{2} }

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   \frac{1}{1 + \frac{2}{3} }

     a₃ = g sin θ \frac{3}{5}

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  \frac{1 }{1 + \frac{2}{5} }

     a₄ = g sin θ  \frac{5}{7}

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      \frac{105}{210}

b) a₂ = g sinθ ⅔ = g sin θ     \frac{140}{210}

c) a₃ = g sin θ \frac{3}{5}= g sin θ       \frac{126}{210}

d) a₄ = g sin θ \frac{5}{7} = g sin θ      \frac{150}{210}

the order of acceleration from lower to higher is

   

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

8 0
3 years ago
Use the diagram to answer each question.
valentina_108 [34]

Figure A shows cross section of a land form or rock. In Figure B, compression stress is applied on it. When compression stresses are applied on a rock, it squeezes the rock cause fold or fracture. The fault formed by compression stress is called thrust fault. If the compression stresses/ force continue to act on a rock it will converge and form thrust fault. In Figure C, tension stresses is applied on the rock. When a tension stress applied on a rock it deforms/ lengthen. There are three type of deformations occur due to tension stresses. One is elastic deformation, in which, rock retains it original shape when force/stresses are removed. Second is plastic deformation, in which rock lengthen and change occur permanently. Third type of deformation is result into fracture or breaking of rock. In Figure C, shear stresses are applied on rock. Shear stresses are applied with equal magnitude but in opposite direction. It cause breaking of rock.

3 0
3 years ago
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natali 33 [55]
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6 0
3 years ago
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