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3 years ago

7

Thirty-five percent of natural sunlight is made up of

2 answers:

dedylja [7]3 years ago

6 0

Thirty-Five percent of natural sunlight is made up of visible light

Elden [556K]3 years ago

5 0

Answer: Visible light makes up 35% of natural light.

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A 7.0-kg rock is subject to a variable force given by the equation f(x)=6.0n−(2.0n/m)x+(6.0n/m2)x2 if the rock initially is at r

stealth61 [152]

For Newton's second law, the force is equal to the product between the mass and the acceleration of the rocket:
$F=ma$
From which we can rewrite the acceleration as
$a(x)= \frac{F}{m} = \frac{1}{m} (6-2x+6x^2)$
where m=7.0 kg.

The velocity of the rocket is the derivative of the acceleration:
$v(x) = \frac{1}{m} (-2+12 x)$
and if we substitute x=9.0 m, we find the rocket velocity after 9.0 m:
$v(9)= \frac{1}{7}(-2+12\cdot 9)=15.1 m/s$

4 0

3 years ago

Do transistors amplfy a.c or d.c<br>

Gnoma [55]

Explanation:

thet amplify DC, because of the voltage ( small current input signal)

5 0

3 years ago

Read 2 more answers

A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed

Liono4ka [1.6K]

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

$F = \dfrac{mv^2}{r}$

m is the mass, v is the velocity and r is the radius.

It follows that $F \propto v^2$ , provided m and r are constant.

When v is doubled, the new force, $F_1$ , is

$F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F$

Hence, the tension in the string is quadrupled.

8 0

3 years ago

Show how the alternative definition of power, found in your book, can be derived by substituting the definitions of work and spe

Harman [31]

Let us consider body moves a distance S due to the force F.

Hence the work by the body W = FS

If the force is not along the direction of displacement,then the work by a body for travelling a distance S will be -

$W=[ Fcos\theta]*S$ where $Fcos\theta$ is the component of the force along the direction of displacement.

$Hence\ W= FScos\theta$

$= F.S$

As per the question the power P is given as -

$P=\frac{W}{\delta t}$

$=\frac{F.S}{\delta t}$

$= F.\frac{S}{\delta t}$

$= \ F.V$

Hence alternative definition of power P = F.V

8 0

3 years ago

Read 2 more answers

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

Stels [109]

Answer:

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

$E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta$

$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

$r^2=d^2+x^2$

So

$E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}$

Now by integrating above equation

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

4 0

3 years ago