The string of a bow and arrow is pulled tight.

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 you place a paper clip very close to a magnet, the paper clip

Alik [6]

The answer is “D. all of the above”!

Metal from the paper clip is attracted to the magnet, so it will naturally move toward and stick to the magnet. This will cause the paper clip to temporarily become a magnet for other metals. I hope this helped!

7 0

2 years ago

Determine the frequency of light whose wavelength is 4.257 x10-7 cm

Bas_tet [7]

<span>Frequency x Wavelength = Speed of light Now, speed of light = 3 x 10^5 km/s = 3 x 10^8 m/s = 3 x 10^10 cm/s Frequency = speed/Wavelength = (3 x 10^10)/(4.257 x 10^-7) =7 x 10^16 Hz</span>

6 0

3 years ago

HELP!!!!!! Need Help !!!

denis23 [38]

Answer:

its y

Explanation:

3 0

2 years ago

Question 4 - If Angelica starts out at 30m/s, and in 16 s speeds up to 84 m/s, what is her acceleration?

Triss [41]

Answer:

C. $3.375 m/s^2$

Explanation:

The acceleration of an object can be found using the equation:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time it takes for the velocity to change from u to v

In this problem:

u = 30 m/s is the initial velocity of Angelica

v = 84 m/s is the final velocity

t is the time

Substituting into the equation, we find the acceleration:

$a=\frac{84-30}{16}=3.375 m/s^2$

4 0

3 years ago