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3 years ago

What are two ways you can increase the magnetic field of an electromagnetic?

1 answer:

3 0

1). Wind more turns of wire in the coil around the core.

2). Increase the electrical current flowing in the wire.
(You can do that by using a battery or power supply with a little more voltage.)

3). If the core of the electromagnet is anything else but pure iron,
take it out, throw it away, and replace it with a core of pure iron.

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A ranger in a national park is driving at 52 km/h when a deer jumps onto the road 87 m ahead of the vehicle. After a reaction ti

lys-0071 [83]

Answer:

Time, t = 0.23 seconds

Explanation:

It is given that,

Initial speed of the ranger, u = 52 km/h = 14.44 m/s

Final speed of the ranger, v = 0 (as brakes are applied)

Acceleration of the ranger, $a=-4\ m/s^2$

Distance between deer and the vehicle, d = 87 m

Let d' is the distance covered by the deer so that it comes top rest. So,

$d'=\dfrac{v^2-u^2}{2a}$

$d'=\dfrac{-(14.44)^2}{2\times -4}$

d' = 26.06 m

Distance between the point where the deer stops and the vehicle is :

D=d-d'

D=87 - 26.06 = 60.94 m

Let t is the maximum reaction time allowed if the ranger is to avoid hitting the deer. It can be calculated as :

$t=\dfrac{v}{D}$

$t=\dfrac{14.44}{60.94}$

t = 0.23 seconds

Hence, this is the required solution.

4 0

3 years ago

The de Broglie wavelength of an electron traveling at 7.0 x 107m/s is 1.0 x 10-11m. (Remember that me = 9.1 x 10-31kg and h = 6.

ivolga24 [154]

The De Broglie's wavelength of a particle is given by:

$\lambda=\frac{h}{p}$

where

$h=6.6 \cdot 10^{-34} Js$ is the Planck constant

p is the momentum of the particle

In this problem, the momentum of the electron is equal to the product between its mass and its speed:

$p=m_e v=(9.1 \cdot 10^{-31} kg)(7.0 \cdot 10^7 m/s)=6.4 \cdot 10^{-23} kg m/s$

and if we substitute this into the previous equation, we find the De Broglie wavelength of the electron:

$\lambda=\frac{h}{p}=\frac{6.6 \cdot 10^{-34} Js}{6.4 \cdot 10^{-23} kg m/s}=1.0 \cdot 10^{-11} m$

So, the answer is True.

7 0

3 years ago

An electron has an initial velocity of (19.0 j + 18.0 k) km/s and a constant acceleration of (3.00 ✕ 1012 m/s2)i in a region in

asambeis [7]

Answer:

$E=(-17.08 i +7.2 j -7.6 k )N/C$

Explanation:

$v= (19.0j+18.k)km/s$

$a=3.0x10^{13}m/s^2$

$\beta= 400x10{-6}T$

Electron information needed to solve the question:

$m_e=9.11x10^{-31}kg$

$q=-1.6x10{-19}C$

$F=F_E+F_B=q*(E+Vx \beta)$

$F=m*a$

$m*a=q*(E+Vx\beta)$

$E=\frac{m*a}{q}-(Vx\beta )$

$E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]$

$E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C$

$E=(-17.08 i +7.2 j -7.6 k )N/C$

6 0

2 years ago

The relationship that exists between gravity and distance and mass respectively.

ruslelena [56]

Answer:

D...............................

7 0

3 years ago

Assume your mass is 60 kg. The acceleration due to gravity is 9.8 m/s 2 . How much work against gravity do you do when you climb

Andre45 [30]

Answer:

W=1705.2 J

Explanation:

Given that

mass ,m= 60 kg

Acceleration due to gravity ,g= 9.8 m/s²

Height ,h= 2.9 m

As we know that work done by a force given as

W = F . d

F=force

d=Displacement

W=work done by force

Now by putting the values

F= m g (Acting downward )

d= h (Upward)

W= m g h ( work done against the force)

W= 60 x 9.8 x 2.9 J

W=1705.2 J

Therefore the answer will be 1705.2 J.

8 0

3 years ago