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fiasKO [112]
3 years ago
12

Calculate the maximal friction force for a parked car between the rubber tires and a wet street. Assume the car’s mass is 1600 k

g with a weight of 15680 N. The static friction coefficient for rubber and wet asphalt is μs=0.8.
Physics
1 answer:
NikAS [45]3 years ago
5 0

Answer:

Fr=12544N

Explanation:

1. Find the equation of eht maximal friction force:

The maximal friction force is given by the equation Fr=usmg, where μs is the static friction coefficient, m is the car´s mass and g is the gravitational force.

2. Replace values in the equation to find the answer:

Fr=0.8*1600kg*9.8\frac{m}{s^{2}}

Fr=12544N

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3 0
3 years ago
A scientist directs monochromatic light toward a single slit in an opaque barrier. The light has a wavelength of 580 nm and the
vredina [299]

Answer:

a) 9.72 mm

b) 4.86 mm

Explanation:

wave length of light  λ is  580 nm = 580 \times 10⁻⁹ m

Width of slit d = 0.215\times 10⁻³ m

Distance of screen D  = 1.8 m.

Width of one fringe = \frac{\lambda\times D}{d}

Putting the values we get fringe width

= \frac{580\times10^{-9}\times1.8}{.000315}

=4.86 mm.

a) Width of central maxima = 2 times width of one fringe

= 2 times 4.86

=9.72 mm

b) width of each fringe except central fringe  is same , no matter what the order is.Only brightness changes .

So width of either of the two first order bright fringe will be same and it will be  

= 4.86 mm.

3 0
3 years ago
As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

F1->

7 0
3 years ago
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