What force is required to accelerate a body with a mass of 15kilograms at a rate of

A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to

jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

b) Current at the center of the circular coil $=2.48\times 10^{5}\ A$

c) Current near the center of a solenoid $236.8\ A$

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

Conversion $1\times 10^6 A = 1\ MA$ ,and $2cm=\frac{2}{100}=0.02\ m$

$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

8 0

3 years ago

A certain refrigerator has a COP of 5.00. When the refrigerator is running, its power input is 500 W. A sample of water of mass

yulyashka [42]

Answer:

t=20s

Explanation:

To solve this problem we must apply the first law of thermodynamics, which indicates that the energy that enters a system is the same that must come out, resulting in the following equation

For this problem we will assume that the water is in a liquid state, since it is a domestic refrigerator

q=m.cp.(T2-T1)

q=heat

m=mass of water =600g=0.6Kg

cp=

specific heat of water=4186J/kgK

T2=temperature in state 2=20°C

T1=temperature in state 1=0°C

solving:

q=(0.6)(4186)(20-0)=50232J

A refrigerator is a device that allows heat to be removed to an enclosure (Qin), by means of the input of an electrical energy (W) and the heat output (Qout), the coefficient of performance COP, allows to know the ratio between the heat removed ( Qin) and the added electrical power (W), the equation for the COP is

$COP=\frac{Qin}{Win}$

To solve this exercise we must know the value of the heat removed to the water (Qin)

solving for Qin

Qin=(COP)(Win)

Qin=(5)(500W)=2500W

finally we remember that the definition of power is the ratio of work over time

w=work

p=power=500w

$P=\frac{W}{t} \\t=\frac{W}{P} \\t=\frac{50232}{2500} \\t=20.09s$

3 0

3 years ago

An athlete stretches a spring an extra 40.0 cm beyond its initial length. how much energy has he transferred to the spring, if t

marissa [1.9K]

The energy transferred to the spring is given by:
$U= \frac{1}{2}kx^2$
where
k is the spring constant
x is the elongation of the spring with respect its initial length

Let's convert the data into the SI units:
$k=52.9 N/cm = 5290 N/m$
$x=40.0 cm=0.4 m$

so now we can use these data inside the equation ,to find the energy transferred to the spring:
$U= \frac{1}{2}kx^2= \frac{1}{2}(5290 N/m)(0.4m)^2=423.2 J$

4 0

3 years ago

An organism tissues and organ are made or cells. T or F

marta [7]

The answer is true
Hope it helped

3 0

3 years ago

A rectangular plate has a length of (21.7 ± 0.2) cm and a width of (8.2 ± 0.1) cm. Calculate the area of the plate, including it

Grace [21]

Answer:

(177.94 ± 3.81) cm^2

Explanation: