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lapo4ka [179]
3 years ago
14

If I push a box at a constant rate is there friction force acting on it?

Physics
1 answer:
yanalaym [24]3 years ago
3 0
Yes, the friction is acting in the opposite direction you are pushing.
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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
How many joules of work are done on an object when a force of 10 N pushes it 5 m?
zhenek [66]

Answer:

option C

Explanation:

given,                            

Force on the object = 10 N

distance of push = 5 m

Work done = ?              

we know,              

work done is equal to Force into displacement.

W = F . s            

W = 10 x 5              

W = 50 J                

Work done by the object when 10 N force is applied is equal to 50 J

Hence, the correct answer is option C

5 0
3 years ago
What combination of things happen on earth that helps make wind, and the weather on earth? (Please list at least two factors, an
Westkost [7]

Answer:

There are six main components, or parts, of weather. They are <u>temperature, atmospheric pressure, wind, humidity, precipitation, and cloudiness</u>. Together, these components describe the weather at any given time. These changing components, along with the knowledge of atmospheric processes, help meteorologists—scientists who study weather—forecast what the weather will be in the near future.

8 0
2 years ago
A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar
m_a_m_a [10]

Answer:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

#Where u is in meters and t in seconds.

Explanation:

Given that :m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s

From \omega=kL we have:

k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2

From F_d(t)=-\gamma u\prime(t) we have that:

\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m

Now,given that the initial value problem is given by:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

Hence,the position of u at time t is given by

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\, u in meters,t in seconds.

4 0
3 years ago
If f(x) = -x2+x-1 then value of f(f(2)) is
qaws [65]

Answer:

<u>-13</u>

Explanation:

<u>Function</u>

  • f(x) = -x² + x - 1

<u>Solving</u>

  • Substitute 2 in place of x
  • f(2) = -(2)² + 2 - 1
  • f(2) = -4 + 2 - 1
  • f(2) = -3
  • f(f(2)) = f(-3)
  • f(-3) = -(-3)² - 3 - 1
  • f(-3) = -9 - 3 - 1
  • f(f(2)) = <u>-13</u>
6 0
1 year ago
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