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3 years ago

9

Coiling wire around an iron core and applying an electric current through the wire creates a temporary magnet called an electrom

agnet. What would increase the strength of an electromagnet? A) Use a smaller nail. B) Use a battery with less voltage. C) Use a battery with more voltage. D) Use fewer coils of wire around the nail.

2 answers:

Serggg [28]3 years ago

5 0

Answer:

C) Use a battery with more voltage.

Explanation:

The equation for the magnetic field around a coil is given by,

B = μ₀NI

where,

B = Magnetic flux density

μ₀ = permeability

N = number of turns per meter

I = Current in the wire

So when using a higher voltage battery, more current passes through the battery as resistance of the wire remains the same.

nydimaria [60]3 years ago

4 0

Answer:

C) Use a battery with more voltage.

Explanation:

To increase the strength of the electromagnet use a battery with more voltage.

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When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0

lora16 [44]

Answer:

$d=0.165m$

Explanation:

Given

$m=0.25kg$ , $x_{1}=5cm*\frac{1m}{100cm}=0.05m$ , $x_{2}=4cm*\frac{1m}{100cm}=0.04m$ , $v=2\frac{rev}{s}$

The tension of the spring is

$F_{k}=K*x_{1}=m*g$

$K=\frac{m*g}{x_{1}}$

$K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m$

The force in the spring is equal to centripetal force so

$F_{c}=\frac{m*v^2}{r}$

$v=w*r=2\pi*r$

But Fc is also

Fc=KxΔr

$F_{c}=K*(r-x_{2})$

Replacing

$m*4\pi^2*r=K*(r-x_{2})$

$0.25kg*4\pi^2*r=49*(r-0.04m)$

$r=0.205m$

total distance is

$d=0.205-0.04=0.165m$

3 0

3 years ago

A mass is attached to a spring with an unknown spring constant. The spring gains 10 J of elastic potential energy if stretched b

Firlakuza [10]

From the given information in the question, the correct option is Option 1: 14 cm.

A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:

PE = $\frac{1}{2}$ k $x^{2}$

Where PE is the energy, k is the spring constant and x is extension.

i. Given that: PE = 10 J and x = 10 cm, then;

PE = $\frac{1}{2}$ k $x^{2}$

10 = $\frac{1}{2}$ k $10^{2}$

20 = 100k

k = 0.2 J/cm

ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.

PE = $\frac{1}{2}$ k $x^{2}$

20 = $\frac{1}{2}$ (0.2) $x^{2}$

40 = 0.2 $x^{2}$

$x^{2}$ = 200

x = $\sqrt{200}$

= 14.1421

x = 14.14 cm

So that;

x is approximately 14.00 cm.

Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.

For more information, check at: brainly.com/question/1352053.

8 0

3 years ago

Given that water at standard pressure freezes at 0∘c, which corresponds to 32∘f, and that it boils at 100∘c, which corresponds t

Zolol [24]

The temperature difference of 1 K is equivalent to the temperature difference of 1 °C. Therefore, we find the relationship between the change in °F and °C.
A change of 212 - 32 °F is the same as a change of 100 - 0 °C. Thus:
(212 - 32) °F = (100 - 0) °C
1 °C = 1.8 °F
1 K = 1.8 °F

6 0

3 years ago

6. A light ray strikes a reflective plane surface at an angle of 560 with the surface.

Zolol [24]

Answer:

deez nouts

Explanation:

5 0

3 years ago

A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the

Marrrta [24]

Our values can be defined like this,

$m = 65kg$

$v = 3.5m / s$

$d = 0.55m$

The problem can be solved for part A, through the Work Theorem that says the following,

$W = \Delta KE$

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

$\frac {1} {2} mv ^ 2$

So,

$Fd = \frac {1} {2} m ^ 2$

Clearing F,

$F = \frac {mv ^ 2} {2d}$

Replacing the values

$F = \frac {(65) (3.5)} {2 * 0.55}$

$F = 723.9N$

B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.

6 0

3 years ago