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3 years ago

9

If you place a magnet under a clear dish, and sprinkle iron fillings over it, above what part of the magnet will most of the fil

lings gather

2 answers:

choli [55]3 years ago

8 0

Mostly all around the center the highest magnetic force and also around it because of the magnets magnetic field

xxTIMURxx [149]3 years ago

5 0

round the corners of the magnet that where it is stronger

hope this helps :)

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A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len

weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}$

$\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}$

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}$

$\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}$

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

6 0

2 years ago

A car has two horns, one emitting a frequency of 199 hz and the other emitting a frequency of 203 hz. what beat frequency do the

andriy [413]

Beat frequency, fb = |f2-f1|

That is, beat frequency is the absolute difference between two frequencies. Is is as a results of destructive and constructive inferences.

Therefore, in this case:

fb = 203 - 199 = 4 Hz

3 0

2 years ago

Change of 5000 c flows through a circuit in one hour. what is the current intensity?

AleksandrR [38]

Answer:

I = 1.38 A

Explanation:

Given that,

Charge, q = 5000 C

Time, t = 1 hour = 3600 s

We need to find the current intensity. The current intensity is equal to the electric charge per unit time. It can be given by :

$I=\dfrac{q}{t}$

Substitute all the values in the above formula

$I=\dfrac{5000\ C}{3600\ s}\\\\=1.38\ A$

So, the current intensity is 1.38 A.

8 0

2 years ago

g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

7 0

2 years ago

By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.1 kg and

puteri [66]

Answer:

14.8 kg

Explanation:

We are given that

$m_1=43.7 kg$

$m_2=12.1 kg$

$g=9.8 m/s^2$

$a=\frac{1}{2}(9.8)=4.9 m/s^2$

We have to find the mass of the pulley.

According to question

$T_2-m_2 g=m_2 a$

$T_2=m_2a+m_2g=m_2(a+g)=12.1(9.8+4.9)=177.87 N$

$T_1=m_1(g-a)=43.7(9.8-4.9)=214.13 N$

Moment of inertia of pulley= $I=\frac{1}{2}Mr^2$

$(T_2-T_1)r=I(-\alpha)=\frac{1}{2}Mr^2(\frac{-a}{r})=\frac{1}{2}Mr(-4.9)$

Where $\alpha=\frac{a}{r}$

$(177.87-214.13)=-\frac{1}{2}(4.9)M$

$-36.26=-\frac{1}{2}(4.9)M$

$M=\frac{36.26\times 2}{4.9}=14.8 kg$

Hence, the mass of the pulley=14.8 kg

6 0

3 years ago