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2 years ago

7

A solar charging system produces 14.5 volts on a sunny day the currents flows through the circuit at 43.5 ramps what is the resi

stance of the circuit

1 answer:

brilliants [131]2 years ago

8 0

Eeeeeeeeeeeeeeeeeeeee

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A man pushes a lawn mower on a level lawn with a force of 195 N. If 37% of this force is directed downward, how much work is don

densk [106]

Answer:

$W = 1032.6 J$

Explanation:

Net force by which man push the lawn mower is given as

$F = 195 N$

now it is given that 37% of this force is vertically downwards

so we will have

$F_y = 0.37 \times 195$

$F_y = 72.15 N$

now we also know that

$F_x^2 + F_y^2 = F^2$

here we have

$F_x^2 + 72.15^2 = 195^2$

$F_x = 181.2 N$

Now work done by this force to move the lawn mower is given as

$W = F_x . d$

$W = (181.2)(5.7 m)$

$W = 1032.6 J$

7 0

2 years ago

What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 43

kiruha [24]

Answer:

The magnification is $m = 0.3674$

Explanation:

From the question we are told that

The power of the lens is $P = -4.00 D(dioptre)$

Generally $1 dioptre = 1 \ meter$

The object distance is $u = -43 \ cm$ the negative sign is because the distance is measured in the opposite direction of incident light (i.e away )

Generally the focal length is mathematically represented as

$f = \frac{1}{P}$

=> $f = \frac{1}{4.00 }$

=> $f = 0.25 \ m$

converting to cm

=> $f = 0.25 \ m = 0.25 * 100 = 25 \ cm$

Generally from lens equation we have that

$\frac{1}{f} +\frac{1}{v} -\frac{1}{u}$

=> $\frac{1}{25} +\frac{1}{v} -\frac{1}{-43}$

=> $v = -15.8 \ cm$

Generally the magnification is mathematically represented as

$m = \frac{v}{u}$

=> $m = \frac{- 15.8}{-43}$

=> $m = 0.3674$

6 0

2 years ago

hichkok12 [17]

Gravitational potential energy i think

8 0

2 years ago

A 5. 3 ft -ft-tall girl stands on level ground. The sun is 30 ∘ above the horizon. How long is her shadow?

slamgirl [31]

The trigonometric ratios are sine, cosine and tangent. We can use cosine to find the length of the shadow from the calculation as 1.7 ft.

<h3>What is trigonometric ratio?</h3>

The term trigonometric ratio has to do with sine, cosine and tangent which are used to solve problems that involve the right angled triangle.

Using the geometry of the right angle triangle, we can find the tangent of the angle 30 in order to obtain the length of the shadow.

Tan 30° = opp/3

opp = 3Tan 30°

opp = 1.7 ft

The length of the girl's shadow from the calculation is 1.7 ft.

Learn more about tangent: brainly.com/question/14022348

8 0

1 year ago

Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an

8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 $m/s^2$ where as radial the component of acceleration is 8.77 $m/s^2$

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

$r=e^u$

So the transverse component of acceleration are given as

$a_{\theta}=(ru''+2r'u')\\$

Here

$r=e^u\\r=e^{\pi/4}\\r=2.1932 m$

$r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m$

So

$a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\$

The transverse component of acceleration is 26.32 $m/s^2$

The radial component is given as

$a_r=r''-r\theta'^2$

Here

$r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m$

So

$a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2$

The radial component of acceleration is 8.77 $m/s^2$

6 0

3 years ago