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kodGreya [7K]
4 years ago
10

A box is being pulled to the right. The free body diagram is shown.

Physics
2 answers:
Vinvika [58]4 years ago
7 0

Answer:

25N

Explanation:

algol [13]4 years ago
4 0

Answer:

the answer is A

A is 25N

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7 0
3 years ago
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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
Infant car seats are made to face the rear of the car. This is safer in a front end collision because of Newton's First law. New
kicyunya [14]
The answer is D hope it helps:)
5 0
3 years ago
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A spring is compressed 0.035 m inside a dart gun. (K=500 N/m). The spring has elastic energy. Calculate it.
NARA [144]

The elastic potential energy of the spring is 0.31 J

Explanation:

The elastic potential energy of a spring is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the compression/stretching of the spring

For the spring in this problem, we have:

k = 500 N/m (spring constant)

x = 0.035 m (compression)

Substituting, we find the elastic potential energy:

E=\frac{1}{2}(500)(0.035)^2=0.31 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

6 0
3 years ago
From the circuit above, predict which bulb (or bulbs) will be the brightest. Why do you think that?
inysia [295]

Answer:

the middle

Explanation:

the left one bulb gets power from the outher bulb

the one on right has more bulbs

7 0
3 years ago
Read 2 more answers
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