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lora16 [44]
2 years ago
7

An ideal spring with spring constant k is hung from the ceiling and a block of mass

Physics
1 answer:
xz_007 [3.2K]2 years ago
6 0
Using the idea that the total energy of the system can be thought to be zero at the starting point and remains zero.

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Which change would decrease the total current, I, flowing through this circuit?
iren [92.7K]

Answer:

A. Increasing the voltage of the battery

Explanation:

The relationship between voltage, V, current, I and resistance, R, is given as follows;

V = I × R

∴ I = V/R

From the above relationship, the current flowing in the circuit is directly proportional to the voltage of the battery, and inversely proportional to the resistance, 'R', of the circuit

Therefore, increasing the voltage, 'V', of the battery, increases the total current, 'I', flowing in the circuit.

7 0
3 years ago
Why is the sky blue?
Taya2010 [7]
A clear cloudless day-time sky is blue because molecules in the air scatter blue light from the sun more than they scatter red light. When we look towards the sun at sunset, we see red and orange colours because the blue light has been scattered out and away from the line of sight.
4 0
3 years ago
A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread non uniformly throu
Aloiza [94]
In other words a infinitesimal segment dV caries the charge 
<span>dQ = ρ dV </span>

<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>

<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>

<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
6 0
3 years ago
A 0.73-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar
Tasya [4]

Answer:

A)  B = 5.4 10⁻⁵ T, B) the positive side of the bar is to the West

Explanation:

A) For this exercise we must use the expression of Faraday's law for a moving body

            fem = -  \frac{d \phi }{dt}

            fem = - \frac{d (B l y}{dt}= - B l v- d (B l y) / dt = - B lv

            B = - \frac{fem}{l \ v}

we calculate

             B = - 7.9 10⁻⁴ /(0.73 20)

             B = 5.4 10⁻⁵ T

B) to determine which side of the bar is positive, we must use the right hand rule

the thumb points in the direction of the rod movement to the south, the magnetic field points in the horizontal direction and the rod is in the east-west direction.

Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West

4 0
3 years ago
In which of the following situations or places would you want to reduce the
sashaice [31]

Answer:

D

Explanation:

<em>The correct answer would be in the axle of the wheels while you ride your bicycle.</em>

Options A, B, and C requires that the forces of friction is increased in order to have more control.

However, option D requires that there is a minimal frictional force in the axle of the wheels of a bicycle while riding so that a little effort would be required to keep the bicycle moving.

<u>The lesser the friction, the lower the effort that would be needed to keep the bicycle moving and vice versa.</u>

4 0
3 years ago
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