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3 years ago

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The fundamental rule for the attraction and repulsion of magnets is magnetic forces follow the inverse-square law. opposite char

ges attract each other and like charges repel. like poles repel each other; opposite poles attract. magnetic poles attract other magnetic poles.

1 answer:

Hitman42 [59]3 years ago

7 0

Answer:

Explanation:

It is true: the fundamental rule for the attraction and repulsion of magnets is magnetic forces follow the inverse-square law.

The force is given by

$F\alpha \frac{1}{r^{2}}$

It is true: opposite charges attract each other and like charges repel.

It is true: like poles repel each other; opposite poles attract.

It is not true: magnetic poles attract other magnetic poles.

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How will creat thunderstrom

emmasim [6.3K]

Answer:

the air has to be unstable as well as it needs to be moved upwards.

Explanation:

it needs to be moved upwards and also needs to have unstable air.

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1 year ago

natulia [17]

Answer:

$\% Error = 2.6\%$

Explanation:

Given

$x: 1.54, 1.53, 1.44, 1.54, 1.56, 1.45$

Required

Determine the percentage error

First, we calculate the mean

$\bar x = \frac{\sum x}{n}$

This gives:

$\bar x = \frac{1.54+ 1.53+ 1.44+ 1.54+ 1.56+ 1.45}{6}$

$\bar x = \frac{9.06}{6}$

$\bar x = 1.51$

Next, calculate the mean absolute error (E)

$|E| = \sqrt{\frac{1}{6}\sum(x - \bar x)^2}$

This gives:

$|E| = \sqrt{\frac{1}{6}*[(1.54 - 1.51)^2 +(1.53- 1.51)^2 +.... +(1.45- 1.51)^2]}$

$|E| = \sqrt{\frac{1}{6}*0.0132}$

$|E| = \sqrt{0.0022}$

$|E| = 0.04$

Next, calculate the relative error (R)

$R = \frac{|E|}{\bar x}$

$R = \frac{0.04}{1.51}$

$R = 0.026$

Lastly, the percentage error is calculated as:

$\% Error = R * 100\%$

$\% Error = 0.026 * 100\%$

$\% Error = 2.6\%$

4 0

3 years ago

A 17-mm-wide diffraction grating has rulings of 530 lines per millimeter. White light is incident normally on the grating. What

Sedbober [7]

Answer:

377 nm

Explanation:

Number of lines per meter is, $N &=530 \times 1000 \\ &=530000 \text { lines } / \mathrm{m} \end{aligned}$

Grating element is, $d=\frac{1}{N}$

$=1.8868 \times 10^{-6} \mathrm{~m}[tex]
Order is, n=5
Condition for maximum intensity is, [tex]d \sin \theta=n \lambda$

$\lambda &=\frac{1.8868 \times 10^{-6}}{5(\sin 90)} \\
&=0.377 \times 10^{-6} \mathrm{~m} \\
&=377 \mathrm{~nm}$

7 0

2 years ago

A block of mass m = 2.0 kg lies on a rough ramp that is inclined at an angle θ = 20oto the horizontal. A force F of magnitude 5.

Marina86 [1]

Answer:

a) 0.64 b) 2.17m/s^2 c) 8.668joules

Explanation:

The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,

Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move

Frictional force = mgsin20o + 5N = 6.71+5N = 11.71

The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44

Coefficient of static friction = 11.71/18.44= 0.64

Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)

b) coefficient of kinetic friction = frictional force/ normal force

Fr = 0.4* mgcos 20o = 7.375N

F due to motion = ma = total force - frictional force

Ma = 11.71 - 7.375 = 4.335

a= 4.335/2(mass of the block) = 2.17m/s^2

C) work done = net force *distance = 4.335*2= 8.67Joules

8 0

2 years ago

A cheetah can accelerate at 4.5 m/s from rest to a speed of 30.0 m/s. Calculatethe distance

anzhelika [568]

Answer:

d= 100m

Explanation:

Cheetah kinematic

The cheetah moves with uniformly accelerated movement, and the formulas that describe this movement are:

d= v₀*t + (1/2)*a*t² Formula (1)

vf²=v₀²+2*a*d Formula (2)

vf=v₀+a*t Formula (3)

Where:

d:distance in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

t: time in seconds (s)

Known Data

v₀ =0

a = 4.5 m/s²

vf= 30 m/s.

Problem development

We apply the formula (2) that has known data to calculate the distance :

vf²=v₀²+2*a*d

(30)²= 0 + 2* 4.5* d

$d= \frac{900}{9}$

d= 100m

8 0

3 years ago