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bulgar [2K]
3 years ago
10

A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed

of 2.5m/s , you start pushing on one edge of the box at a 45? angle (use degrees in your calculations throughout this problem) with a constant force of magnitude Fp = 23N , as shown in figure. (Figure 1) There is now a frictional force between the box and the wall of magnitude 13N . How fast is the box sliding 2.0s after you started pushing on it?
Using our simplified model, in which we know that the forces are constant (but we don't know what their magnitudes are), which of the following motion diagrams would be a reasonable representation of the motion of the box?
Physics
2 answers:
hoa [83]3 years ago
6 0

Answer:

The speed of the box after 2 s is 3 m/s.

Explanation:

The net force on the box in vertical direction:

{F_{net}} = {F_g} - f - {F_p}\sin {45^{\rm{o}}}F

Here, {F_g} is the gravitational force, f is the force of friction and {F_p} is the pushing force.

According to Newton’s second law, the net force is:

{F_{net}} = ma

Here, a is the acceleration in vertical direction

\begin{array}{c}\\ \Rightarrow ma = {F_g} - f - {F_p}\sin {45^{\rm{o}}}\\\\a = \frac{{{F_g} - f - {F_p}\sin {{45}^{\rm{o}}}}}{m}\\\end{array}

Substitute 30N for Fg, 13N for f and 23N for Fp

 \begin{array}{c}\\a = \frac{{\left( {30{\rm{ N}}} \right) - \left( {13{\rm{ N}}} \right) - \left( {23{\rm{ N}}} \right)\sin {{45}^{\rm{o}}}}}{{3.1{\rm{ kg}}}}\\\\ = \frac{{0.7365{\rm{ N}}}}{{3.1{\rm{ kg}}}}\\\\ = 0.24\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\\end{array}

Newton’s equation of motion:

v - u = at

Substitute 2.5m/s for u , 0.24m/s2 for a and 2s for t .

\begin{array}{c}\\v - 2.5{\rm{ m/s}} = \left( {0.24{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {2{\rm{ s}}} \right)\\\\v = 0.48{\rm{ m/s}} + 2.5{\rm{ m/s}}\\\\ = 2.98{\rm{ m/s}}\\\\ \approx {\bf{3 m}}{\rm{/}}{\bf{s}}\\\end{array}

The speed of the box after 2 s is 3 m/s.

Tems11 [23]3 years ago
5 0

Answer:

The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.

Explanation:

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2 years ago
Please answer the questions... I will surely mark you as the brainliest according to me :)
frutty [35]

Answer:

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A wave traveling in a string has a wavelength of 35 cm, an amplitude of 8. 4 cm, and a period of 1. 2 s. What is the speed of th
AlekseyPX

0.29 m/s (wave velocity = wavelength (lamda)/period (T) in metres)

35 / 1.2 = 29.16

29.16 ÷ 100 = 0.29

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3 0
1 year ago
When running a half marathon (13.1 miles), it took Kevin 8 minutes to run from mile marker 1 to mile marker 2, and 18 minutes to
Vsevolod [243]

Answer:

It took Kevin 26 minutes to run from markers 1 to 4

His average speed from mile markers 1 to 4 is 0.154 miles/minute

Kevin must run by average speed 0.1 miles/minute to finish the race

Explanation:

Lets explain how to solve the problem

A half marathon 13.1 miles

Kevin took 8 minutes to run from mile marker 1 to mile marker 2 and

18 minutes to run from mile marker 2 to mile marker 4

→ He took 8 minutes and 18 minutes to run from marker 1 to marker 4

→ The total time of the first 4 marker = 8 + 18 = 26 minutes

<em>It took Kevin 26 minutes to run from markers 1 to 4</em>

<em></em>

Average speed is total distance divided by total time

The average speed of Kevin as he ran from mile marker 1 to mile

marker 4 is the 4 miles divides by 26 minutes

→ Average speed = 4 ÷ 26 = \frac{2}{13} = 0.154 miles/minute

<em>His average speed from mile markers 1 to 4 is 0.154 miles/minute</em>

<em></em>

It took Kevin 71 minutes to pass mile marker 9

Kevin need to complete the race in 112 minutes, then what must

Kevin's average speed be as he travels from mile marker 9 to the

finish line?

The total distance of the race is 13.1 miles, he ran 9 miles

→ The remaining distance = 13.1 - 9 = 4.1 miles

He must run 4.1 miles to complete the race

The total time is 112 minutes, he used 71 minutes to run the first 9 miles

→ The remaining time = 112 - 71 = 41 minutes

He must finish the 4.1 miles in 41 minutes

→ His average speed for the last part of the race = 4.1 ÷ 41 = 0.1 mi/min

<em>Kevin must run by average speed 0.1 miles/minute to finish the race</em>

7 0
2 years ago
Read 2 more answers
A car travels 15 kilometers west in 10 minutes. After reaching the destination, the car travels back to the starting point, agai
jeka94

Speed = (distance traveled) / (time to travel the distance).
 
Strange as it may seem, 'velocity' is completely different. 

Velocity doesn't involve the total distance traveled at all. 
Instead, 'velocity' is based on 'displacement' ... the distance
between the start-point and end-point, regardless of the route
taken to get there.  So the displacement in driving once around
any closed path is zero, because you end up where you started. 

Velocity =

           (displacement during some time)
divided by
            (time for the displacement)

AND the direction from the start-point to the end-point.


For the guy who drove 15 km to his destination in 10 min, and then
back to his starting point in 5 min, (assuming he returned by way of
the same 15-km route):

         Speed = (15km + 15km) / (10min + 5min)  =  (30/15) (km/min)

                                                                                 =  2 km/min.

        Velocity = (end location - start position) / (15 min)  =  Zero .

5 0
3 years ago
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