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tekilochka [14]
3 years ago
6

How to do this question

Physics
1 answer:
Yuki888 [10]3 years ago
7 0

Answer:

64°

Explanation:

The triangle is an isosceles triangles (both legs are equal to the radius of the circle), so that means the base angles are the same.

Angles of a triangle add up to 180°, so:

128 + 2x = 180

2x = 52

x = 26

∠1 is complementary to the base angle, so:

∠1 = 90 − 26

∠1 = 64

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C.    amplitude....................


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it could kill the fish in water

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A fan that is switched on for 1 minute uses 500 W usefully but also wastes 300 W through the emission of sound and heat. What's
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Explanation:

Efficiency is defined as the ratio between the useful output over the total amount consumed. E = \frac{useful}{total}

The fan does 500W of useful work while wasting 300 W. The total power consumption is 800 W (500 + 300).

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Hydrogen gas is generated when acids come into contact with certain metals. When excess hydrochloric acid reacts with 2.2 g of (
CaHeK987 [17]

To solve the problem it is necessary to apply the concepts related to Byle's Law and Avogadro's Law.

The ideal gas equation would help us find the final solution to the problem, defined by

PV = nRT

Where,

T= Temperature of the gas

R = Universal as constant

n = number of moles

V = Volume

P = Pressure

For our case we have that the mass of Zn is 2.2g in moles would be

[tex]Zn = \frac{2.2}{65}[/tex]

Zn = 0.0338

We know that 1 mole of hydrogen gas is proceed by 1 mole of zinc and the result is Zn^{2+}, then Hydrogen can produce the same quantity,

H_2 = 0.0338

Applying the previous equation we have that

V= \frac{nRT}{P}

V = \frac{0.0338*0.08206*293.15}{0.98}

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6 0
3 years ago
A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image
Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}

=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}

Re-arrange to get i

\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

m=\frac{h'}{h} =-\frac{i}{o}

substitute the values to get the height of image h'

\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm

5 0
3 years ago
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