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kari74 [83]
3 years ago
13

Magnetic field strength decrease when what is increases

Physics
1 answer:
Elis [28]3 years ago
6 0
Ans: One way to increase or decrease the strength of the magnetic field is to change the number of loops in the coil. The more loops you add, the stronger the field will become. The more loops you remove, the weaker the field will become.
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A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.4 ft/s,
Art [367]

Answer:

\dfrac{d\theta}{dt} =-0.233\ rad/s

Explanation:

given,

length of ladder = 10 ft

let x be the distance of the bottom and y be the distance of the top of ladder.

x² + y² = 100

differentiating with respect to time we get

2 x\dfrac{dx}{dt}+2y\dfrac{dy}{dt} = 0..............(1)

when x = 8 and y = 6 and when \dfrac{dx}{dt} = 1.4ft/s

from equation (1)

now,

16\times 1.4 + 12\dfrac{dy}{dt} = 0

\dfrac{dy}{dt} = -\dfrac{5.6}{3}

let the angle between the ladders be θ

tan\theta = \dfrac{y}{x}

y = xtan θ

\dfrac{dy}{dt} =\dfrac{dy}{dt} tan\theta + x sec^2\theta\dfrac{d\theta}{dt}

-\dfrac{5.6}{3} =1.4\times \dfrac{6}{8} + 8 (1+\dfrac{9}{16})\dfrac{d\theta}{dt}

\dfrac{25}{2} \dfrac{d\theta}{dt} =\dfrac{-17.5}{6}

\dfrac{d\theta}{dt} =-0.233\ rad/s

6 0
3 years ago
A subducting oceanic plate
beks73 [17]

Maybe this would help understand it better.

<span>Tectonic plates can transport both continental crust and oceanic crust, or they may be made of only one kind of crust. Oceanic crust is denser than continental crust. At a subduction zone, the oceanic crust usually sinks into the mantle beneath lighter continental crust</span>

4 0
2 years ago
An object starts from rest, and accelerates at 2m/s2 for 10s. How far has it gone in that time
blagie [28]

Answer:

100m

Explanation:

s = ut +  \frac{1}{2} a {t}^{2}

u=0;t=10sec;a=2m/s²

s = 0(10) +  \frac{1}{2}(2 \times  {10}^{2} )

s=10²;100m

7 0
3 years ago
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f
aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

    gm = 1/6 ge

    gm = 1/6  9.8 m/s² = 1.63 m/s²

We calculate the range

    R = Vo² sin 2θ  / g

    R = 25² sin (2 30) / 1.63

    R= 332 m

We will calculate the time of flight,

   Y = Voy t – ½ g t2  

   Voy = Vo sin θ

When the ball reaches the end point has the same initial  height Y=0

0 = Vo sin  t – ½  g t2

0 = 25 sin (30)  t – ½ 1.63 t2

0= 12.5 t –  0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

 t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

Let's calculate the reach on earth

R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

5 0
3 years ago
In Hooke’s law, what does the x represent?
Arte-miy333 [17]
X Represents the distance the spring is stretched or compressed away from its equilibrium or rest position.
7 0
3 years ago
Read 2 more answers
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