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Alona [7]
3 years ago
10

An electromagnet is strong enough to lift 200 kg of mass. If you double the number of turns of wire around the core and triple t

he current, how much will it be able to lift?
Physics
1 answer:
VLD [36.1K]3 years ago
4 0

Answer:1200 kg

Explanation:

Given

Electromagnet can lift a mass of  200\ kg

Magnetic force necessary to hold mass is proportional to Magnetic field

and magnetic field (B) is proportional to NIA

where N=no of turns

I=current in the magnet

A=area of cross-section

When we double the no of turns and triple the current then

N becomes 2 N

I becomes 3 I

therefore magnetic field becomes 6 times

Thus magnetic force becomes 6 times and we can say that electromagnet can hold the 6 times of given weight

i.e. 6\times 200=1200\ kg

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I NEED HELP PLEASE, THANKS! :)
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Answer:

1. Largest force: C;  smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

F=K\dfrac{ q_{1}q_{2}}{r^{2}}

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write  

F=\dfrac{k}{r^{2}}

1. Net force on each particle

Let's

  • Call the distance between adjacent charges d.
  • Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}}  - \dfrac{k}{(2d)^{2}}  +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(b) Force on B

\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}}  - \dfrac{k}{d^{2}}  + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(C) Force on C

\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}}  + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(d) Force on D

\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}}  - \dfrac{k}{(2d)^{2}}  - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

F_{A} : F_{B} : F_{C} : F_{D}  =  \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\

2. Ratio of largest force to smallest

\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}

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kaheart [24]

Answer:

500; 300 feet.

Explanation:

A turn signal lamp can be defined as an amber blinking lamp which indicates a driver's intent to change direction. It is extremely important that drivers gives a turn signal (flashing light) on the side toward which he or she is turning either left or right.

Simply stated, the turn signal lamp indicate the driver's intent to turn either leftward or rightward by displaying flashing lights to the front and rear of his or her vehicle.

A turn signal lamp must be visible in normal sunlight at a distance of at least 500 feet from the front and rear of the vehicle if the vehicle is at least 80 inches wide, and at least 300 feet from front and rear of the vehicle if the vehicle is less than 80 inches wide according to the transportation traffics code.

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3 years ago
_________ is an umbrella term that applies to location-aware technologies that allow devices to communicate securely with each o
ratelena [41]

Answer:Near-field communication

Explanation:

Near Field Communication is an umbrella term that applies to location-aware technologies that allow devices to communicate securely with each other over short distances. Near-field communication is a short-range wirelessconnectivity standard that uses magnetic field induction to enable communication between devices when they're touched together, or brought within a few centimeters to each other.It also specifies a way for the devices to establish a peer-to-peer network to exchange data. After the peer to peer network has been configured, another wireless communication technology, such as Bluetooth or Wi-Fi, can be used for longer-range communication or for transferring larger amounts of data.

3 0
3 years ago
Read 2 more answers
A quantity of a gas is an absolute pressure of 400 K PA in the absolute temperature of 110° Kelvin when the temperature of the g
aliina [53]

Given:

P1 = 400 kPa
T1 = 110 K

T2 = 235K

Required:

P2

Solution:

Apply Gay-Lussac’s law where P/T = constant

P1/T1 = P2/T2

P2 = T2P1/T1

P2 = (235K)(400kPa) / (110K)

P2 = 855 kPa

4 0
3 years ago
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