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3 years ago

When a potential difference of 13 V is placed across a resistor, the current in the resistor is 1.4

Physics

1 answer:

vitfil [10]3 years ago

8 0

Just divide the two numbers with each other.
I mean 13/1.4=9.2857...

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Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

$dI=\lambda x^2dx$

So total moment of inertia

$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

$x_c= \dfrac{5}{9}L$

So mass moment of inertia $I=\dfrac {7}{12}\lambda_ 0 L^3$ and location of center of mass $x_c= \dfrac{5}{9}L$

8 0

3 years ago

What kind of rock is this?

Katyanochek1 [597]

The picture is kinda blurry but it looks like granite which is metamorphic.

5 0

3 years ago

Read 2 more answers

The Grand Canyon is 1800 meters deep at its deepest point. A rock is dropped from the rim above this point. Write the height of

snow_tiger [21]

Answer:

Explanation:

Given

height of grand canyon is $h=1800\ m$

Rock is dropped i.e. initial velocity is zero

$u=0$

Using Equation of motion

$h=ut+\frac{1}{2}at^2$

h=height

u=initial velocity

a=acceleration

t=time

here a=acceleration due to gravity

$1800=0+\frac{1}{2}\times 9.8\times t^2$

$t^2=\frac{2\times 1800}{9.8}$

$t=19.166\ s$

3 0

3 years ago

Which statement is true for particles of the medium of an earth quake p wave

katen-ka-za [31]

I am pretty sure that the only statement which is true for particles of the medium of an earthquake P-wave is being shown in the option : b)vibrate parallel to the wave, forming compressions and rarefactions. As you know, it can be formed in two ways : from alternating compressions and rarefactions or primary wave. I bet you will agree with me.

8 0

3 years ago

190 kg of water is to be raised by a water pump to a height of 25 meters from the bottom of a well in 60 seconds. What should be

ladessa [460]

Answer:

776.6 w

1.04 hp

Explanation:

given:

Mass, m = 190kg

height change, h = 25m

time elapsed, t = 60 s

acceleration due to gravity, g = 9.81 m/s²

Potential energy required raising 190 kg of water to a height of 25m

= mgh

= 190 x 9.81 x 25

= 46,597.5 J

Power required in 60 s

= Energy required ÷ time elapsed

= 46,597.5 ÷ 60

= 776.6 Watts (Use conversion 1 W = 0.00134102 hp)

= 776.6 w x 0.00134102 hp/w

= 1.04 hp

6 0

3 years ago