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Vera_Pavlovna [14]

3 years ago

7

What two forms of energy does the sun supply?

1 answer:

Nadya [2.5K]3 years ago

8 0

Nuclear fusion and heat (thermal) energy

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A 10.0 cm object is 5.0 cm from a concave mirror that has a focal length of 12 cm. What is the distance between the image and th

fiasKO [112]

Let's use the mirror equation to solve the problem:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$
where f is the focal length of the mirror, $d_o$ the distance of the object from the mirror, and $d_i$ the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for $d_i$ by using the numbers given in the text of the problem:
$\frac{1}{12 cm}= \frac{1}{5 cm}+ \frac{1}{d_i}$
$\frac{1}{d_i}= -\frac{7}{60 cm}$
$d_i = -8.6 cm$
Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.

6 0

3 years ago

Read 2 more answers

A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 2.50 × 1012 m/s2 in a machine. If the proton h

Veronika [31]

Answer:

(A) Speed will be $44.18\times 10^4m/sec$

(b) Change in kinetic energy = $1560\times 10^{-19}$

Explanation:

We have given mass of proton $m=1.67\times 10^{-27}kg$

Acceleration of the proton $a=2.50\times 10^{12}m/sec^2$

Initial velocity u = $1.60\times 10^4$ m/sec

Distance traveled by proton s = 3.90 cm = 0.039 m

(a) From third equation of motion we know that

$v^2=u^2+2as$

$v^2=(1.60\times 10^4)^2+2\times 2.5\times 10^{12}\times 0.039$

$v=44.18\times 10^4m/sec$

(b) Initial kinetic energy $KE_I=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (1.6\times 10^4)^2$

Final kinetic energy $KE_F=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (44.18\times 10^4)^2$

So change in kinetic energy $\Delta KE=KE_F-KE_I=\frac{1}{2}\times 1.6\times 10^{-27}\times 10^8\times (44.18^2-1.6^2)=1560\times 10^{-19}J$

6 0

3 years ago

a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f

harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for $\Delta$ DAC:

$tan\theta = \frac{d}{x}$

⇒ $\theta = tan^{-1} \frac{d}{x}$

Now for the $\Delta$ BAC:

$tan\theta = \frac{d + h}{x}$

⇒ $\theta = tan^{-1} \frac{d + h}{x}$

Now, differentiating w.r.t x:

$\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]$

For maximum angle, $\frac{d\theta }{dx}$ = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]

0 = $\frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}$

$\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}$

After solving the above eqn, we get

x = $\sqrt{\frac{d}{d + h}}$

The observer should stand at a distance equal to x = $\sqrt{\frac{d}{d + h}}$

4 0

3 years ago

How dose the circulatory systems interact with the digestive systems?

scoray [572]

The digestive system digests and makes nutrients out of food while the circulatory system distributes and circulates the nutrients i believe?

8 0

2 years ago

NewtonsXmeters / time is a unit of which of the following:

GuDViN [60]

Answer:

energy I think I'm not sure of the answer.

7 0

3 years ago