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3 years ago

What kind of lens is found in both a magnifying glass and a microscope

Physics

1 answer:

mina [271]3 years ago

5 0

I found this on google hope it’s helps and can I have brainless question

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Please help find the total resistance of these circuits!! I will mark brainliest if right!!

Elan Coil [88]

Answer:

If you know the total current and the voltage across the whole circuit, you can find the total resistance using Ohm's Law: R = V / I. For example, a parallel circuit has a voltage of 9 volts and total current of 3 amps. The total resistance RT = 9 volts / 3 amps = 3 Ω.

Explanation:

4 0

3 years ago

Which of the following is the type of rock

viva [34]

Answer:

d. intrusion

Explanation:

An intrusion is molten rock from the Earth's interior that squeezes into existing rock and cools. Folding Folding occurs when rock layers bend and buckle from Earth's internal forces.

6 0

3 years ago

Imagina que compras una placa rectangular de metal de 2mm de alto, 10mm de ancho x 50mm de largo, y una masa de 0.02kg. El vende

shusha [124]

Answer:

Densidad de la placa = 20 g/cm³.

La placa no es de oro.

Explanation:

Para encontrar la densidad de la placa rectangular primero debemos hallar su volumen:

$V = 2 mm*10 mm*50 mm = 1000 mm^{3}*\frac{1 cm^{3}}{(10 mm)^{3}} = 1 cm^{3}$

Ahora, encontremos al densidad de la placa:

$d = \frac{m}{V} = \frac{20 g}{1 cm^{3}} = 20 g/cm^{3}$

Dado que la densidad del oro es 19.32 g/cm³ y que la densidad de la placa rectangular calculada es 20 g/cm³, podemos decir que dicha placa no es de oro.

Espero que te sea de utilidad!

6 0

3 years ago

The planet should move around the elliptical orbit, and two segments of the orbit should become shaded in green. What aspect(s)

strojnjashka [21]

Answer: not sure

Explanation:

4 0

3 years ago

Use the definition of scalar product, a overscript right-arrow endscripts times b overscript right-arrow endscripts = ab cos θ,

makkiz [27]

Answer: $\theta=cos^{-1}0.991=7.69^o$

The following vectors have been given: $\vec{a}=3.0\widehat{i}+3.0\widehat{j}+3.0\widehat{k}\\ \vec{b}=5.0\widehat{i}+7.0\widehat{j}+6.0\widehat{k}$

The angle between these two vectors can be found by:

$cos\theta=\frac{\vec{a}.\vec{b}}{||\vec{a}|| ||\vec{b}||}\\
||\vec{a}=\sqrt{a_x^2+a_y^2+a_z^2}$

$\vec{a}.\vec{b}=a_xb_x+a_yb_y+a_zb_z\\ \vec{a}.\vec{b}=3\times5+3\times7+3\times6=15+21+18=54$

$||\vec{a}||=\sqrt{3^2+3^2+3^2}=\sqrt{27}\\ ||\vec{b}||=\sqrt{5^2+7^2+6^2}=\sqrt{110}$

$cos\theta=\frac{54}{\sqrt{27}\times\sqrt{110}}\\=0.991\\ \Rightarrow \theta=cos^{-1}0.991=7.69^o$

7 0

3 years ago