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3 years ago

6

Apakah soalan tingkatan 4 bm negeri selangor

1 answer:

noname [10]3 years ago

8 0

What does this mean? What do you need help with?

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The layer of earth that has the lightest elements is the

Sonja [21]

If I remember correctly (from my studies long time ago) the layers are from the outer to the center:
SiAl : Silicon-Aluminum
SiMa : Silicon-Magnesium (although should be Mg)
NiFe : Nickel-Iron

The SiMa layer should have the lightest elements (Magnesium is lighter than Aluminum)

8 0

3 years ago

What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL

lozanna [386]

Answer:

$\displaystyle \Delta x=1.74\ m$

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

$\displaystyle \Delta x=\swrt{\frac{2PE}{k}}$

Substituting:

$\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}$

Calculating:

$\displaystyle \Delta x=\sqrt{3.0175}$

$\boxed{\displaystyle \Delta x=1.74\ m}$

6 0

3 years ago

A diffraction grating with 750 slits per mm is illuminated by light which gives a first-order diffraction angle of 34.0°. What i

lesya [120]

<h2>Answer: 745.59 nm</h2>

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d$ is the width of the slit

$\lambda$ is the wavelength of the light

$n$ is an integer different from zero

Now, the first-order diffraction angle is given when $n=1$ , hence equation (1) becomes:

$dsin\theta_{1}=\lambda$ (2)

We know:

$\theta_{1}=34\°$

In addition we are told the diffraction grating has 750 slits per mm, this means:

$d=\frac{1mm}{750}$

Solving (2) with the known values we will find $\lambda$ :

$\lambda=(\frac{1mm}{750})sin(34\°)$ (3)

$\lambda=0.00074559mm$ (4)

Knowing $1mm=10^{6}nm$ :

$\lambda=745.59nm$ >>>This is the wavelength of the light, wich corresponds to red.

6 0

3 years ago

A motorist wishes to travel 40 kilometers at an average speed of 40 km/h. During the first 20 kilometers, an average speed of 40

allochka39001 [22]

poste en français s’il vous plaît

4 0

3 years ago

The intensity I of light from a light bulb, measured in watts per square meter (w/m squared), varies inversely as the square

DerKrebs [107]

Answer:

$157.5W/m^2$

Explanation:

We are given that

$I=70w/m^2$

When d=3 m

We have to find the intensity when the distance from the light bulb is 2 m away.

According to question

Intensity, $I\propto\frac{1}{d^2}$

$I=\frac{kd^2}$

Where k=Proportionality constant.

Substitute the values

$70=\frac{k}{3^2}$

$k=70\times 3^2=630$

Intensity when d=2 m

$I=\frac{630}{2^2}=157.5W/m^2$

6 0

3 years ago