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3 years ago

Please help me, please i will mark brainliest

Physics

2 answers:

abruzzese [7]3 years ago

5 0

Answer:

conductor,neither,insulator

Explanation:

conductors allow electricity to flow through while insulators do not

HOPE IT HELPS

bixtya [17]3 years ago

3 0

Conductor, neither, insulator

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A company designed and sells an ultrasonic receiver, which detects sounds unable to be heard by the human ear. The receiver can

Kruka [31]

Answer:

the diameter of the outside edge of the receiver is $8\sqrt{18} \ \ \ inches$

Explanation:

From the schematic free body diagram illustrating what the question is all about below;

Let represent A to be the vertex where the receiver is being placed

S to be the focus

BP to be equal to r (i.e radius of the outer edge)

BC to be 2 r (i.e the diameter)

Given that AS = 4 in and AP is 18 in

Let AP be x- axis and AY be y -axis

A=(0,0)

S=(4,0) = (0,0)

So that the equation of the parabolic path of the receiver will be:

$y^2 =4 ax \\ \\ y^2 = 4*4*x \\ \\ y^2 = 16x$

B = (AP, BP)

B = (18, r)

B lies y² = 16 x

r² = 16 x

r² 16 × 18

$r = \sqrt{16*18 } \\ \\ r = 4\sqrt{18}$

Diameter BC = 2r

$2* 4\sqrt{18} \\ \\= 8\sqrt{18 } \ \ \ inches$

6 0

4 years ago

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 block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The sprin

Mariana [72]

Answer:

The velocity of block = 0.188 $\frac{m}{s}$

Explanation:

Mass m = 5.6 kg

k = 1040 $\frac{N}{m}$

$\mu$ = 0.26

$x_{1} =$ 0.035 m , $v_{1}$ = 0

$x_{2} =$ 0.02 m

From work energy theorem

$K_{1} + U_{1} + W_{other} = K_{2} + U_{2}$ --------- (1)

Kinetic energy

$K = \frac{1}{2} k x^{2}$ ------- (1)

Potential energy

$U = \frac{1}{2} k x^{2}$ ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

$R_{N}$ = mg

$R_{N}$ = 5.6 × 9.81 = 54.9 N

Friction force = 0.4 × 54.9 = 21.9 N

Now the work done by the friction

$W_{f}$ = - 21.9 × 0.015

$W_{f}$ = - 0.329 J

Now kinetic energy

At point 1

$K_{1} = \frac{1}{2} m v^{2} _{1}$

$K_{1} = 0$

$U_{1} = \frac{1}{2} k x^{2}$

$U_{1} = \frac{1}{2} (1040) 0.035^{2}$

$U_{1} =$ 0.637 J

At point 2

$K_{2} = \frac{1}{2} (5.6) v^{2} _{2}$

$K_{2} = 2.8 v_{2} ^{2}$

Potential energy

$U_{2} = \frac{1}{2} k x_2^{2}$

$U_{2} = \frac{1}{2} (1040) 0.02^{2}$

$U_{2} = 0.208$ J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8 $v_{2} ^{2}$ + 0.208

2.8 $v_{2} ^{2}$ = 0.1

$v_{2} =$ 0.188 $\frac{m}{s}$

This is the velocity of block.

6 0

4 years ago

Latitude where each of the following types of winds-

Dmitriy789 [7]

westerlies

polar easterlies and trade winds are in a different type

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3 years ago

In graphing enthalpy entropy and state changes which two variables are included ? A. volume and temperature B. amount of heat ad

Blababa [14]

In graphing enthalpy entropy and state changes the two variables that are included are amount of heat added and pressure. The answer is letter D. The rest of the choices do not answer the question above.

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4 years ago

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