This is an electrical device that changes the voltage of electricity through the use of two set of coils

A mug rests on an inclined surface, as shown in (Figure 1) , θ=17∘.

Anna [14]

Refer to the figure shown below.

g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.

Let μ = the coefficient of static friction.

The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N

For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058

The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306

Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306

7 0

3 years ago

Read 2 more answers

Why do people cover their flowers on a cool night. Think about radiation absorption conduction and temperature

Blababa [14]

When the temperature lowers the plants can freeze and die. One way to prevent that is to cover the plants so they dont freeze

3 0

3 years ago

1.Convert 340 cm into m *(answer=0.34m)

Nataly [62]

Answer:

1.for the first one 100centimeters make 1 meter therefore

100cm-1m

340cm-x

340/100

=3.4

the answer is supposed to be 3.4, maybe there's a mistake with the question or the answer

2.the weight of a body is given by the formula

mass×gravity,in this case the mass is 75kg and the gravity is 9.8

weight=75×9.8

=735N

3.for this one the mass of a body is given by the formula

mass=weight/gravity

=420/9.8

=42.8kg

I hope this helps

4 0

3 years ago

When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero fil

Anika [276]

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

$2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}$