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2 years ago

Does the voltage of a battery affect the strength of an electromagnet?

1 answer:

swat322 years ago

7 0

I'm trying to make an electromagnet that's strength is constantly getting incremented by small amounts every second. I need to know, which would have a greater effect on the electromagnet's strength, amps or volts? (I know increasing the turns and/or density of the magnet wire will increase the strength, but I am looking for answers other than that particular one.)

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Please select the word from the list that best fits the definition Visualizes where items are located

Amanda [17]

Answer:I’m pretty sure it’s spatial

Explanation:

3 0

2 years ago

Read 2 more answers

An observer in frame S sees lightning simultaneously strike two points 100 m apart. The first strike occurs at xx1 = yy1 = zz1 =

Veseljchak [2.6K]

Answer:

a) 0, = -0.33 us

b) 140m

c) No, The event are not simultaneous i.e they did not occur at the same time, the second even (-0.33 usec) occurs 0.33 usec earlier than the first event.

Explanation:

the lorentz factor expression is written as;

y = 1₀ / √(1 - (v²/c²))

where v is the relative speed of an observer and c is the speed of light

so we were given that relative speed to be o.7c

therefore

y = 1 / √(1 - ((0.7c)² / c²))

y = 1 / √(1 - (0.49c² / c²))

y = 1 / √(1 - 0.49)

y = 1 / 0.7141

y = 1.4

1 - the coordinates of the first event, the s' frame of reference is,

x1 ' = y(x1 - vt1) = 0

y1 ' = y1, z1' = z1 and

t1 ' = y [t1 - v/c²x1]

= 0

2 - the coordinates of the second event, the s ' frame of reference is'

x2 ' = y(x2-vt2)

= 1.4(100m - 0)

= 140m

y2 ' = y2, z2 ' = z2

t2 ' = y [ t2 - v/c²x2 ]

= 1.4 [ 0 - 0.7c/c²(100) ]

using speed of light c as 3*10^8

1.4 [ 0 - (0.7*3*10^8) / (3*10^8)²(100) ]

= -0.33 us

distance between

delltaX' = X2' - X1'

= 140m - 0

= 140m

No, The event are not simultaneous i.e they did not occur at the same time.

the second even (-0.33 us) occurs 0.33 us earlier than the first event.

4 0

2 years ago

An object has a mass of 5 kg. What force is needed to accelerate it at 6 m/s2? (Formula: F=ma)

Studentka2010 [4]

Force needed is 5×6=30

5 0

3 years ago

Read 2 more answers

An electron and a proton are each placed at rest in an electric field of 500 N/C. Calculate the speed (and indicate the directio

Bumek [7]

Answer:

For proton: 2592 m/s In the same direction of electric field.

For electron: 4752000 m/s In the opposite direction of electric field.

Explanation:

E = 500 N/C, t = 54 ns = 54 x 10^-9 s,

Acceleration = Force /mass

Acceleration of proton, ap = q E / mp

ap = (1.6 x 10^-19 x 500) / (1.67 x 10^-27) = 4.8 x 10^10 m/s^2

Acceleration of electron, ae = q E / me

ae = (1.6 x 10^-19 x 500) / (9.1 x 10^-31) = 8.8 x 10^13 m/s^2

For proton:

u = 0, ap = 4.8 x 10^10 m/s^2, t = 54 x 10^-9 s

use first equation of motion

v = u + at

vp = 0 + 4.8 x 10^10 x 54 x 10^-9 = 2592 m/s In the same direction of electric field.

For electron:

u = 0, ae = 8.8 x 10^13 m/s^2, t = 54 x 10^-9 s

use first equation of motion

v = u + at

vp = 0 + 8.8 x 10^13 x 54 x 10^-9 = 4752000 m/s In the opposite direction of electric field.

6 0

2 years ago

Brayden and Riku now use their skills to work a problem. Find the equivalent resistance, the current supplied by the battery and

Liono4ka [1.6K]

a) $5 \Omega$ , 1.6 A

b) $6 \Omega$ , 1.33 A

Explanation:

In this situation, we have two resistors connected in series.

The equivalent resistance of resistors in series is equal to the sum of the individual resistances, so in this circuit:

$R=R_1+R_2$

where

$R_1=4\Omega$

$R_2=1 \Omega$

Therefore, the equivalent resistance is

$R=4+1=5 \Omega$

Now we can use Ohm's Law to find the current flowing through the circuit:

$I=\frac{V}{R}$

where

V = 8 V is the voltage supplied by the battery

$R=5\Omega$ is the equivalent resistance of the circuit

Substituting,

$I=\frac{8}{5}=1.6 A$

The two resistors are connected in series, therefore the current flowing through each resistor is the same, 1.6 A.

In this part, a third resistor is added in series to the circuit; so the new equivalent resistance of the circuit is

$R=R_1+R_2+R_3$

where:

$R_1=4\Omega\\R_2=1\Omega\\R_3=1\Omega$

Substituting, we find the equivalent resistance:

$R=4+1+1=6 \Omega$

Now we can find the current through the circuit by using again Ohm's Law:

$I=\frac{V}{R}$

where

V = 8 V is the voltage supplied by the battery

$R=6\Omega$ is the equivalent resistance

Substituting,

$I=\frac{8}{6}=1.33 A$

And the three resistors are connected in series, therefore the current flowing through each resistor is the same, 1.33 A.

3 0

2 years ago