The diagram shows an electromagnet made with copper wire, a steel nail,
and a 1.5 V battery. Which action could cause this electromagnet to be
stronger?

A. Replace the steel nail with a plastic straw.

B. Replace the battery with a 6 V battery.

C. Reduce the number of coils of wire wrapped around the nail.

D. Reverse the direction of the battery in the circuit.

Correct answer is B!

Answer:

The fundamental frequency is  [tex]f_1 =128 \ Hz[/tex]

Explanation:

From the question we are told that

   The frequency of one harmonics is  [tex]f_x= 448 \ Hz[/tex]

    The next higher harmonic is  [tex]f_z = 576 \ Hz[/tex]

Generally the frequency of an air column open at both ends is mathematically represented as

              [tex]f_n = \frac{nv }{ 2 L }[/tex]

Here n  is the order of the harmonics (frequency)

        v is the velocity of the sound

        L  is the length of the column

So for one harmonics we have that

        [tex]f_k = \frac{n v }{2L}[/tex]

Then for the next higher harmonics

       [tex]f_x = \frac{n+1 ) v}{2 L }[/tex]

Generally the difference between these frequencies is mathematically represented as  

       [tex]f_z- f_x = \frac{(n+1 )v}{ 2L} - \frac{(n )v}{ 2L}[/tex]

=>    [tex]576 - 448 = \frac{vn + v - nv }{2L}[/tex]

=>    [tex]\frac{ v }{2L} = 128[/tex]

Generally for fundamental  frequency n =  1

So  

       [tex]f_1 = n * \frac{v}{2L}[/tex]

So

       [tex]f_1 =1 * 128[/tex]

=>    [tex]f_1 =128 \ Hz[/tex]

Answer:

in a spontaneous process, the entropy of the universe increases.

Explanation:

Entropy is a measure of of the degree of  randomness or disorderliness in a system.

The second law of thermodynamics can be stated as follows; "in any spontaneous process, the entropy of the universe increases."  

The universe here refers to the system's disorder and the disorder of the surroundings.  Therefore, a spontaneous process can occur, in which the entropy of the system decreases, only if the entropy increases in the surroundings.

For instance, when ice freezes, the entropy of liquid water decreases, that is, the entropy of the system decreases. However, heat is given off to the surroundings and the entropy of the surroundings increases. This is an obvious expression of this law.

Answer:

The answer is "Cerebral Cortex"

Explanation:

The neurotransmitter diffuses across the short distance of the synapse and ties to a receptor protein of the objective neuron. At the point when the sub-atomic sign ties to the receptor, the cell film of the objective neuron changes its electrical state and another evaluated expected starts. On the off chance that that evaluated potential is sufficiently able to arrive at limit, the subsequent neuron produces an activity potential at its axon hillock. The objective of this neuron is another neuron in the thalamus of the mind, the piece of the CNS that goes about as a transfer for tactile data.

At another neurotransmitter, synapse is delivered and ties to its receptor. The thalamus at that point sends the sensory information to the cerebral cortex, the furthest layer of dark issue in the brain, where cognizant view of that water temperature starts.

A region of the cortex is particular for imparting signs down to the spinal cord for development. The upper engine neuron is in this area, called the precentral gyrus of the frontal cortex, which has an axon that broadens right down the spinal cord. At the degree of the spinal cord at which this axon makes a neurotransmitter, a reviewed potential happens in the cell membrane of a lower engine neuron.

Answer:

word for word answers!

Explanation:

1) Part A: By pedaling a bicycle lightly, the rider can go a long way

2) Part B: The two examples mentioned in the video are the handlebars and the brakes

3) Yes, it’s a type of lever because the two pedals rotate around a fixed point

Answer:

a) [tex]H_{in}=39.34 kJ[/tex]

b) Efficiency=76.77%

Explanation:

a)

In order to solve this problem, we can use the following formula:

[tex]H_{in}=H_{out}+W[/tex]

the problem provides us with all the necessary information so we can directly use the formula:

[tex]H_{in}=9.14kJ+30.2kJ[/tex]

[tex]H_{in}=39.34 kJ[/tex]

b) In order to find the efficiency, we can use the following formula:

[tex]Efficiency=\frac{W}{H_{in}}*100\%[/tex]

so we get:

[tex]Efficiency=\frac{30.2kJ}{39.34kJ}*100\%[/tex]

Efficiency=76.77%

Answer:

D i think

Explanation:

Heavier objects (objects with more mass) are more difficult to move and stop.

Answer:

Increasing the mass of the object (option D in the list of answers)

Explanation:

Recall that F = m x a

therefore, if the mass increases, the force increases

Answer:

Explanation:

Step one:

given data

speed = 123m/s

radius of circle= 4330m

Step two:

We need to find the circumference of the circle, it represents the distance traveled

C=2πr

C= 2*3.142*4330

C= 27209.72m

Step three:

We know that velocity= distance/time

time= distance/velocity

time= 27209.72/123

time=221.2 seconds

in minute = 221.2/60

time= 3.7 min

Answer:

B. Photosynthesis occurs in all organisms

Hope this helps!! :)

Answer:

a) h(g) = 358,53 m

b) t = 8,16 s

c) t(t) = 16,71 s

Explanation:

Equations for vertical shooting are:

Vf = V₀ -  g * t   ;     h  =  V₀*t -  (1/2)*g*t²  ;    Vf² = V₀² - 2*g*h

And at maximum heigt Vf = 0 then

0 = V₀ - g * t

t = V₀/g                     V₀  = 80 m/s      and   g = 9,8 m/s²

t  =  80 / 9,8   (s)

t = 8,16 s

Then 8,16 s is the time to get maximum height

If we plug t = 8,16 (s) in equation  h  =  V₀*t -  (1/2)*g*t²

we get:     h (max)  =  (80)*8,16 - 0,5*9,8*(8,16)² (m)

h (max) = 652,8  -  326,27 m

h (max) = 326,53 m

Then relative to ground that height becomes

h(g) = 326,53 + 32

h(g) = 358,53 m

In order to get the time the arrow is in the air we proceed as follows:

a) for the arrow to be at the launched point will take the same time that from the launched point to the maximum height, and after that we have to find out the time the arrow takes from 32 m down to the ground level

Then  

t(t) = 8,16 + 8,16 + tₓ     (2)

Where tₓ  is the time from 32 m height to ground

h  =  V₀*tₓ -  (1/2)*g*tₓ²  but since the arrow now is going down then we change the sign of the second term on the right side of the equation

32 = (80)*tₓ  +  0,5 * 9,8 * tₓ²     Note that when the arrow is at 32 m height the speed is again V₀ = 80 m/s

32 = 80*tₓ  + 4,9*tₓ²

A second-degree equation for tₓ, solving it

4,9*tₓ² + 80*tₓ - 32 = 0

t₁,₂ = -80 ± √ 6400 + 627,2 / 9,8

t₁,₂ =( - 80 ± 83,8 ) / 9,8

there is not a negative time therefore we dismiss such solution and

t₁ = 3,8 / 9,8

t₁ = 0,39 s

And

t(t) = 8,16 + 8,16 + 0,39  s

t(t) = 16,71 s