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leva [86]
3 years ago
8

30. A box with mass 20kg is on a cement floor. The coefficient of static friction between the box and floor is 0.25. A man is pu

shing the box with a horizontal force of 35N. What is the magnitude of the force of static friction between the box and floor
Physics
1 answer:
Mnenie [13.5K]3 years ago
7 0

Answer:

84.05

Explanation:

  • F=mg×0.25
  • F=20x9.81×0.25
  • f=49.05N
  • F=35N

F=f+F

F=49.05+35

=84.05

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the goat weighs 900 N and is 1 meter from the fulcrum. The strongman pulls down on the lever 3 meters from the fulcrum. What is
garik1379 [7]

Answer: The smallest effort = 300N

Explanation:

Using one of the condition for the attainment of equilibrium:

Clockwise moment = anticlockwise moments

900 × 1 = 3 × M

Where M = the weight of the strong man

3M = 900

M = 900/3 = 300N

Therefore, 300N is the smallest effort that the strongman can use to lift the goat

6 0
3 years ago
Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp
slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

Q=Av

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

Q=1.20 m^3/s

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2

So we can re-arrange the equation to find the speed of the water:

v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

Q_1 = Q_2\\A_1 v_1 = A_2 v_2

where we have

A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s

and where A_2 is the cross-sectional area of the pipe at the second point.

Solving for A2,

A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2

And finally we can find the radius of the pipe at that point:

A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m

6 0
3 years ago
At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force t
Salsk061 [2.6K]

Answer:

The work done on the athlete is approximately 2.09 J

Explanation:

From the definition of the work done by a variable force:

\displaystyle{\int_{x_i}^{x_j}F(x)dx}

and substituting with the function of our problem:

\displaystyle{\int_{0}^{0.19}(140x-190x^2)dx\approx2.09\mathrm{J}}

5 0
4 years ago
With certain exceptions, Class E airspace extends upward from either 700 feet or 1,200 feet AGL to, but does not include,A) 14,5
DedPeter [7]

Answer:

B) 18,000 feet MSL

Explanation:

There are three-dimensional parts in the navigation airspace in the world. The class E airspace is mostly used in the regions with coastal areas that are relatively populated. If we consider certain forms of exceptions, the class E airspace can move in the upward direction to few feet (i.e. 1200 ft). However, this doesn't include 18,000 feet MSL.

6 0
3 years ago
Nvm i got the answer but <br> free points
faust18 [17]

Answer:

thank you 谢谢

Explanation:

8 0
3 years ago
Read 2 more answers
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