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3 years ago

If you are givin force and time, you can determine power if you can know.

Physics

2 answers:

Anastaziya [24]3 years ago

8 0

If you are givin force and time, you can determine power if you can know.

C. Joules

pentagon [3]3 years ago

7 0

Answer:

D. Distance

Explanation:

Power is work divided by time. Work is force times distance, therefore you need distance in order to find power.

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A rocket is fired in deep space, where gravity is negligible. If the rocket has a mass of 6000 kg and ejects gas at a relative v

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25 miles per second worth of gas

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3 years ago

The water from a fire hose follows a path described by y equals 3.0 plus 0.8 x minus 0.40 x squared (units are in meters). If

ivanzaharov [21]

Explanation:

It is given that, the water from a fire hose follows a path described by equation :

$y=3+0.8x-0.4x^2$ ........(1)

The x component of constant velocity, $v_x=5\ m/s$

We need to find the resultant velocity at the point (2,3).

Let $\dfrac{dx}{dt}=v_x$ and $\dfrac{dy}{dt}=v_y$

Differentiating equation (1) wrt t as,

$\dfrac{dy}{dt}=0.8\times \dfrac{dx}{dt}-0.8x\times \dfrac{dx}{dt}$

$v_y=0.8\times v_x-0.8x\times v_x$

$v_y=0.8v_x(1-x)$

When x = 2 and $v_x=5\ m/s$

So,

$v_y=0.8\times 5\times (1-2)$

$v_y=-4\ m/s$

Resultant velocity, $v=\sqrt{v_x^2+v_y^2}$

$v=\sqrt{5^2+(-4)^2}$

v = 6.4 m/s

So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.

8 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS AND I NEED ALL CORRECT ANSWERS ONLY!!!

ryzh [129]

iron, cobalt and Nickel is the answer

6 0

3 years ago

You will begin with a relatively standard calculation.Consider a concave spherical mirror with a radius of curvature equal to 60

Strike441 [17]

a) 30.0 cm

For any mirror, the radius of curvature is twice the focal length:

r = 2f

where

r is the radius of curvature

f is the focal length

For the mirror in this problem, we have

r = 60.0 cm is the radius of curvature

Therefore, solving the equation above for f, we find its focal length:

$f=\frac{r}{2}=\frac{60.0 cm}{2}=30.0 cm$

b) 90 cm

The mirror equation is:

$\frac{1}{s'}=\frac{1}{f}-\frac{1}{s}$

where

s' is the distance of the image from the mirror

f is the focal length

s is the distance of the object from the mirror

For the situation in the problem, we have

f = +30.0 cm is the focal length (positive for a concave mirror)

s = 45.0 cm is the object distance from the mirror

Solving the formula for s', we find

$\frac{1}{s'}=\frac{1}{30.0 cm}-\frac{1}{45.0 cm}=0.011 cm^{-1}$

$s'=\frac{1}{0.011 cm^{-1}}=90 cm$

c) -2

The magnification of the mirror is given by

$M=-\frac{s'}{s}$

where in this problem we have

s' = 90 cm is the image distance

s = 45.0 cm is the object distance

Solving the equation, we find:

$M=-\frac{90 cm}{45 cm}=-2$

So, the magnification is -2.

d) -12.0 cm

The magnification can also be rewritten as

$M=\frac{y'}{y}$

where

y' is the height of the image

y is the heigth of the object

In this problem, we know

y = 6.0 cm is the height of the object

M = -2 is the magnification

Solving the equation for y', we find

$y'=My=(-2)(6.0 cm)=-12.0 cm$

and the negative sign means that the image is inverted.

Part e and f are exactly identical as part b) and c).

6 0

4 years ago

A wire of resistivity rhorho must be replaced in a circuit by a wire of the same material but 4 times as long. If, however, the

kap26 [50]

The resistivity of a wire is given by:

$R=\rho\frac{l}{\pi r^2}$

Where $\rho$ is the electrical resistivity of the wire, $l$ is the length of the wire and r is the radius os the wire, Recall that $r=\frac{d}{2}$ . So:

$R=\rho\frac{l}{\pi(\frac{d}{2})^2}\\R=4\rho\frac{l}{\pi d^2}$

We have $l'=4l, \rho'=\rho, R'=R$ . Thus:

$R'=4\rho'\frac{l'}{\pi d'^2}\\R=4\rho\frac{4l}{\pi d'^2}\\4\rho\frac{l}{\pi d^2}=4\rho\frac{4l}{\pi d'^2}\\\frac{1}{d^2}=\frac{4}{d'^2}\\d'^2=4d^2\\d'=2d$

The diameter of the new wire must be twice of the original wire.

5 0

3 years ago