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riadik2000 [5.3K]
3 years ago
14

A push of magnitude P acts on a box of weight W as shown in the figure. The push is directed at an angle θ below the horizontal,

and the box remains a rest. The box rests on a horizontal surface that has some friction with the box. The normal force on the box due to the floor is equal to
Physics
1 answer:
Kipish [7]3 years ago
3 0

Answer:

its on wheels and they are supposed to make it eas

Explanation:

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Because the snow suddenly gets too slushy, you decide to carry your 100-n sled the rest of the way home. How much work do you do
soldier1979 [14.2K]

 Work  =  (weight) x (distance lifted)

           =  (100 N)  x  (0.5 m)

           =       50 joules.


If you carry the sled level all the way to your house, you're not moving it
against gravity. 

Although you're huffing and puffing and sweating by the time you reach
your house, still, according to the Physics definition of work, you have
done NO work on the sled all the way home.
5 0
3 years ago
QUICK
gregori [183]

Answer:C

Explanation:

I did the test

7 0
3 years ago
What can hold a greater volume of water than a mug
Volgvan
A jug would hold more volume than a mug
7 0
3 years ago
Read 2 more answers
Consider the uniform electric field \vec{E} =(4.00~\hat{j}+3.00~\hat{k})\times 10^3~\text{N/C} ​E ​⃗ ​​ =(4.00 ​j ​^ ​​ +3.00 ​k
Tema [17]

Answer:

Electric flux, \phi=6.668\times 10^4\ Nm^2/C

Explanation:

It is given that,

Electric field, E=(4j+3k)\times 10^3\ N/C

We need to find the electric flux through a circular area of radius 2.66 m that lies in the xy-plane. A=Ak

The electric flux is given by :

\phi=E{\cdot}A

\phi=(4j+3k)\times 10^3{\cdot}Ak

Since, k.k=i.i=j.j = 1

So,

\phi=3\times 10^3\times \pi\times (2.66)^2\ k

\phi=6.668\times 10^4\ Nm^2/C

So, the electric flux through a circular area is \phi=6.668\times 10^4\ Nm^2/C.Hence, this is the required solution.

6 0
3 years ago
Water at 20oC flows through a long elliptical duct 30 cm wide and 22 cm high. What average velocity, in m/s, would cause the wei
mars1129 [50]

Explanation:

The given data is as follows.

   Fluid is water so, density \rho = 1000 kg/m^{3}

  Weight flow rate = 500 lbf/s = 2224.11 N/sec

  Cross-sectional area (A) = \pi \times \frac{30}{2} \times \frac{22}{2}

                                         = 0.05184 m^{2}

Hence, weight flow rate will be given as follows.

        w = \rho \times g \times A \times V

    2224.11 N/sec = \rho \times g \times A \times V

        V = \frac{2224.11}{1000 \times 9.81 \times 0.05184} m/s

            = 4.373 m/s

Thus, we can conclude that average velocity in the given case is 4.373 m/s.

8 0
3 years ago
Read 2 more answers
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