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Aneli [31]
3 years ago
13

Pls help i have test

Physics
2 answers:
Ne4ueva [31]3 years ago
7 0
I think no is the correct ans for the question ✅
ludmilkaskok [199]3 years ago
5 0

Answer:

I think the answer is a no. I guess

Explanation:

don't mind if it is write or wrong

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Solar cells are used to trap sunlight energy (light energy) and convert it to electric energy for domestic and other purposes.

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What role does team collaboration play in successfully planning a mission to Mars?
mihalych1998 [28]

Answer:it helps get everyone in the same page

Explanation:

This is important so everyone knows what they should be doing

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3 years ago
I NEED THIS ASAP PLEASE
kolbaska11 [484]

Answer:

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Explanation:

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5 0
3 years ago
Determine e when I = 0.50 A and R = 12 W.
sammy [17]

Answer:

The correct answer is "24 V".

Explanation:

The given values are:

Current,

I = 0.50 A

Resistance,

R = 12 W

As we know,

⇒ I = 0.5\times (\frac{E}{2R})

On substituting the given values, we get

⇒ 0.5= (\frac{E}{4\times 12} )

⇒ 0.5= (\frac{E}{48} )

⇒   E=24 \ V  

7 0
3 years ago
An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.
CaHeK987 [17]

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm

B) Converging

The focal length is:

- Positive for a converging lens

- Negative for a diverging lens

In this case, the focal length is positive, so it is a converging lens.

C) 18.0 mm

The magnification equation states that:

\frac{h_i}{h_o}=-\frac{q}{p}

where

h_i is the heigth of the image

h_o is the height of the object

q=36.0 cm

p=16.0 cm

Solving the formula for h_i, we find

h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm

So the image is 18 mm high.

D) Inverted

From the magnification equation we have that:

- When the sign of h_i is positive, the image is erect

- When the sign of h_i is negative, the image is inverted

In this case, h_i is negative, so the image is inverted.

4 0
4 years ago
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