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Leto [7]
2 years ago
10

Mazie stands on her kitchen floor. The coefficient of kinetic friction between her socks and the floor is 0.35, and the coeffici

ent of static friction is 0.42. She has a mass of 58 kg.
a. If you push Mazie with a force of 200 N, will she start to slide? Explain your answer. (2 points)


b. Mazie slides across the floor at a speed of 1.3 m/s. What is the force of kinetic friction acting on her? (2 points)




c. Mazie climbs up on the roof of her house, which has an angle of 25°. The coefficient of static friction between her and the roof is 0.35. Will she slip off the roof? Explain your answer. (3 points)



d. Given that μs = 0.35, what is the maximum angle the roof could have without her slipping? (3 points)
Physics
1 answer:
Natasha2012 [34]2 years ago
4 0

Static friction is a force that keeps an object at rest. Static friction definition can be written as: The friction experienced when individuals try to move a stationary object on a surface, without actually triggering any relative motion between the body and the surface on which it is on

<h3>How can calculate the Fs by finding your normal force and multiplying it by your static friction force?</h3>

N = mg = (9.8)×(58) = 568.4N

Fs = NKs = (568.4)×(0.42) =238.7N

b) So this is an interesting question, she's going at a speed of 0.6 m/s. Kinetic friction is pushing her back so theres a negative force which is the force of kinetic friction right. So its NKf = (238.7)×(0.35) = 83.5N

To learn more about Static friction, refer

brainly.com/question/13680415

#SPJ9

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An unbalanced force of 40. newtons keeps a
Pachacha [2.7K]
(4) 4.0 m/s is the correct answer





4 0
3 years ago
Read 2 more answers
Particle A of charge 2.79 10-4 C is at the origin, particle B of charge -5.64 10-4 C is at (4.00 m, 0), and particle C of charge
kirill [66]

Answer:

a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N  g) 16.9 N

h) 24.3 N θ = 44.2º

Explanation:

a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.

So, Fcax = 0

b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.

Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²

Fyca = 29. 9 N

c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:

rbc² = (3.00 m)² + (4.00m)² = 25.0 m²

⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²

⇒ Fbc = 21.7 N

d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:

Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:

cos θ = x/r = 4.00 / 5.00 m =

Fcbx = 21.7*(-0.8) = -17.4 N

e) The  y component can be calculated in the same way, projecting the force over the y-axis, as follows:

Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N

f) The sum of both x components gives :

Fcx = 0 + (-17.4 N) = -17.4 N

g) The sum of both y components gives :

Fcy = 29.9 N + (-13.0 N) = 16.9 N

h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:

Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)²\sqrt{((17.4)^{2} +(16.9)^{2}} = 24.3 N

The angle from the horizontal can be found as follows:

Ф = arc tg (16.9 / 17.4) = 44.2º

4 0
3 years ago
A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle res
zhannawk [14.2K]

Answer:

a) 250 N/m

b) 22.4 rad/s , 3.6 Hz , 0.28 sec

c) 0.3125 J

Explanation:

a)

F = force applied on the spring = 7.50 N

x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m

k = spring constant of the spring

Since the force applied causes the spring to stretch

F = k x

7.50 = k (0.03)

k = 250 N/m

b)

m = mass of the particle attached to the spring = 0.500 kg

Angular frequency of motion is given as

w = \sqrt{\frac{k}{m}}

w = \sqrt{\frac{250}{0.5}}

w = 22.4 rad/s

f = frequency

Angular frequency is also given as

w = 2 π f

22.4 = 2 (3.14) f

f  = 3.6 Hz

T = Time period

Time period is given as

T = \frac{1}{f}

T = \frac{1}{3.6}

T = 0.28 sec

c)

A = amplitude of motion = 5 cm = 0.05 m

Total energy of the spring-block system is given as

U = (0.5) k A²

U = (0.5) (250) (0.05)²

U = 0.3125 J

5 0
4 years ago
If an object is moving, then the magnitude of its ____ cannot be zero.
Bad White [126]
If an object is moving, then its speed and the magnitude
of its velocity aren't zero. 

If its velocity is constant (speed doesn't change and it's
moving in a straight line), then its acceleration is zero.

Looks like choice-D sums it up nicely.
7 0
3 years ago
The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?
BigorU [14]

1. Frequency: 7.06\cdot 10^{14} Hz

The frequency of a light wave is given by:

f=\frac{c}{\lambda}

where

c=3\cdot 10^{-8} m/s is the speed of light

\lambda is the wavelength of the wave

In this problem, we have light with wavelength

\lambda=425 nm=425\cdot 10^{-9} m

Substituting into the equation, we find the frequency:

f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz


2. Period: 1.42 \cdot 10^{-15}s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

The frequency of this light wave is 7.06\cdot 10^{14} Hz (found in the previous exercise), so the period is:

T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s


4 0
3 years ago
Read 2 more answers
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