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3 years ago

PLEASE HELP ME Color corresponds to the ______________ of light waves. wave speed cycles wavelength

1 answer:

5 0

Wavelength. Each wavelength is a certain color. For instance, shorter wavelengths (like 470nm) will be blue or violet, while longer wavelengths (like 650nm) will be red. Hope this helps! :)

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The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st

Aleonysh [2.5K]

Answer:

$a = 5.83 \times 10^{-4} m/s^2$

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = $1.50 m/s^2$

now we know that

$F = ma$

$F = 70(1.50) = 105 N$

now for the space station will be same as above force

$F = ma$

$105 = 1.8 \times 10^5 (a)$

$a = \frac{105}{1.8 \times 10^5}$

$a = 5.83 \times 10^{-4} m/s^2$

3 0

3 years ago

A 5 kg wooden block sitson a flat straight-away12 meters fromthe bottom of an infinitely long ramp, which has an angle of 20 deg

saveliy_v [14]

Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

(b) 74.56338m.

Explanation:

We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

Following formulas will be used.

$x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2$

$F =ma$

$V(t) = V_{o} +at$

$x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2$

(a) Calculating velocity right before going up the ramp.

Wooden block is going on a straightaway and has net for on it.

$F_{n} =F-F_{s} = F-uF_{n} = 100N-0.4*9.8m/s^2*5kg$ = $81N$

and this force produces acceleration of

$a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .$

With this acceleration, wooden block would reach at the foot of ramp in.

$x(t) = 12m = 16.2m/s^2*t^2$

$t = 1.217s$

and final velocity will be

$v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.$

this velocity of wooden box just before going up the ramp.

(b) How far up the ramp will the wooden block go before stopping.

Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.

$V= V_{h}cos(20) = 18.5288m/s$

and deceleration along the ramp is

$a = \frac{F_{s} }{m}$

Where $F_{s}$ force of friction along the inclined plane.

$F_s = uF_n = u*m*a$

$a = 9.8m/s^2*cos(20) = 9.2089m/s^2$

is a component of g along normal of the inclined plane.

$F_{s} = 0.25*5kg*9.2089m/s^2$

$= 11.5112N$

$a = \frac{11.5112N}{5kg} = 2.3022m/s^2$

And with this deceleration time needed to get wooded block to stop is.

$v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0$

$t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s$

and in that time wooden block would travel

$x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m$

This is how up wooden box will go before coming to stop.

3 0

3 years ago

If a 20 N object has been lifted 5 meters above the ground, how much gravitational potential energy does it have?

Andreyy89

The gravitational potential energy of the object is 100 J.

Gravitational potential energy stored in an object is the work done in raising the object to a height <em>h</em> against the gravitational force acting on it.

The gravitational force acting on a body is its weight mg, where m is its mass and g, the acceleration due to gravity.

Work done by a force is equal to the product of the force and the displacement made by the point of application of the force.

$GPE = mgh$