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3 years ago

10

Fertilization occurs in the fallopian tubes.

1 answer:

Tamiku [17]3 years ago

7 0

What’s the question? Is it true or false?

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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

Two motorcycles are traveling in opposite directions at the same speed when one of the cyclists blasts her horn, which has frequ

Galina-37 [17]

Answer:

6ms^-1

Explanation:

Given that the frequency difference is

( 563- 544) = 19

So alsoThe wavelength of each wave is = v/f = 344 /544

and there are 19 of this waves

So it is assumed that each motorcycle has moved 0.5 of this distance

in one second thus the speed of the motorcycles will be

=> 19/2 x 344/544 = 6.0 m/s

5 0

2 years ago

A tuning fork has a frequency of 280 Hz and the wavelength of the sound produced is 1.5 meters. Calculate the wave's frequency a

s2008m [1.1K]

Answer:

The answer would be 420 m/s

Explanation:

Look in attachment ⬇

I Hope this Helps!!!

3 0

2 years ago

The -is the process by which gases in the atmosphere absorb and reradiate heat?

Nikolay [14]

Water vapor and carbon dioxide

3 0

2 years ago

using hooke's law, f spring = k triangle x, find the elastic constant of a spring that stretches 2 cm when a 4 newton force is a

Ksivusya [100]

As we know that spring force is given as

$F = kx$

here we know that

F = 4 N

x = 2 cm = 0.02 m

now from the above equation we will have

$4 = k(0.02)$

$k = 200 N/m$

so the elastic constant of the spring will be 200 N/m

8 0

3 years ago