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Lynna [10]
3 years ago
13

How to find the coefficient of friction without the force of friction or vice versa

Physics
1 answer:
son4ous [18]3 years ago
7 0

Answer:

Finding the Force of Friction Without an Experiment

1.The following equation tells you the strength of the frictional force (with the static friction coefficient): F = μ s t a t i c N F=\mu_{static} N F=μstaticN.

2.If your surface is flat and parallel to the ground, you can use: ...

3.If it isn't, the normal force is weaker.

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Gold, which has a density of 19.32 g/cm3, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long
NNADVOKAT [17]

Answer:

a) 578.0 cm²

b) 25.18 km

Explanation:

We're given the density and mass, so first calculate the volume.

D = M / V

V = M / D

V = (6.740 g) / (19.32 g/cm³)

V = 0.3489 cm³

a) The volume of any uniform flat shape (prism) is the area of the base times the thickness.

V = Ah

A = V / h

A = (0.3489 cm³) / (6.036×10⁻⁴ cm)

A = 578.0 cm²

b) The volume of a cylinder is pi times the square of the radius times the length.

V = πr²h

h = V / (πr²)

h = (0.3489 cm³) / (π (2.100×10⁻⁴ cm)²)

h = 2.518×10⁶ cm

h = 25.18 km

3 0
3 years ago
Just asking how everyone is doing cause of the virus
Bess [88]

Answer:

We are all fiine

Explanation:

How are you doing also

3 0
3 years ago
Read 2 more answers
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

6 0
3 years ago
Vivian went on a bicycle trip through Germany with her family. One afternoon, she rode her bicycle along a long flat road at a c
natima [27]

d = distance traveled by her on her bicycle on a long flat road = 24 kilometer

t = time taken by her to travel distance "d" on her bicycle on a long flat road = 1.2 hours

v = average speed of vivian = ?

we know that average speed is given as

v = d/t

inserting the values in the above formula

v = 24 kilometer / 1.2 hour

v = 20 kilometer/hour


hence the correct choice is

C) 20 km/h



8 0
3 years ago
A generator converts mechanical energy into _____________________ energy.
erica [24]
Electrical energy is your answer.
6 0
3 years ago
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