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maks197457 [2]
2 years ago
5

Which of the following would increase the strength of an electromagnet ?

Physics
2 answers:
AleksAgata [21]2 years ago
6 0
<span>More coils of wire around the magnetic material. More current through the wire</span>
GalinKa [24]2 years ago
5 0

Increasing the number of coils and Increasing the power of the Voltage supply.

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Suppose Earth's mass increased but Earth's diame-
navik [9.2K]

Answer: It would increase.

Explanation:

The equation for determining the force of the gravitational pull between any two objects is:

F = G \frac{m1m2}{r^2}

Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.

Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.

Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.

7 0
2 years ago
Light hits a clear plastic bottle and a granite rock. Which choice most accurately describes the effect of visible light on the
balu736 [363]

Answer:

#4 is the accurate answer.

8 0
2 years ago
Read 2 more answers
A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm . It carries a current of 1.50 A . Part A How
tensa zangetsu [6.8K]

Answer:

The number of turns is  N  = 1750 \ turns

Explanation:

From the question we are told that

  The inner radius is r_i =  12.0 \  cm  =  0.12 \  m

   The outer radius is  r_o =  15.0 \  cm  =  0.15 \  m

   The current it carries is I =  1.50 \  A

    The magnetic field is  B  =   3.75 mT = 3.75 *10^{-3} \  T

   The distance from the center is d =  14.0 \ cm  =  0.14 \  m

Generally the number of turns is mathematically represented as

    N  =  \frac{2 *  \pi  * d  *  B}{ \mu_o *  r_o }

Generally  \mu_o is the permeability of free space with value  

    \mu_o  =  4\pi * 10^{-7} \ N/A^2

So

  N  =  \frac{2 *  3.142   * 0.14 *  3.75 *10^{-3} }{ 4\pi * 10^{-7}  * 0.15  }

  N  = 1750 \ turns

5 0
2 years ago
Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2
GarryVolchara [31]

Answer:

Therefore the required solution is

U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

Explanation:

Given vibrating system is

u''+\frac{1}{4}u'+2u= 2cos \omega t

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) =  - A ω² cosωt -B ω² sin ωt

Putting this in given equation

-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t

Equating the coefficient of sinωt and cos ωt

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2

\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0.........(1)

and

\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0

\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0........(2)

Solving equation (1) and (2) by cross multiplication method

\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}

\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}

\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}   and        B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}

Therefore the required solution is

U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

5 0
2 years ago
Some hydrogen gas is enclosed within a chamber being held at 200∘c with a volume of 0.0250 m3. the chamber is fitted with a mova
Mrac [35]

Answer: The final volume V₂ of the container is  0.039 m³.

Explanation:

Since the temperature is constant, the gas would expand isothermally.

For isothermal expansion,

P₁V₁=P₂V₂

Where, P₁ and P₂ are the initial and final pressure and V₁ and V₂ are initial and final volume.

It is given that:

V₁ = 0.0250 m³

P₁ = 1.5 × 10⁶ Pa

P₂ = 0.950 × 10⁶ Pa

V₂ = ?

⇒ 1.5 × 10⁶ Pa × 0.0250 m³ = 0.950 × 10⁶ Pa × V₂

⇒V₂ = 0.039 m³

Hence, the final volume V₂ of the container is  0.039 m³.

4 0
1 year ago
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