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ruslelena [56]
3 years ago
8

If the force of static friction on a crate is 67 N and the weight of the create is 289 N, what is the coefficient of static fric

tion?
a
19363
b
0.23
c
4.31
d
222
Physics
1 answer:
RUDIKE [14]3 years ago
8 0

Answer:

B) μ = 0.23

Explanation:

The coefficient of friction is equal to μ = F / N where μ (mu) is the coefficient of friction, F is the friction force, and N is the normal force (the force of an object being applied onto the earth by gravity).

F, the friction force, is given as 67 N

Since the weight of the crate is 289 N, that means the normal force is 289 N.

This means that the coefficient of friction is  μ = F / N = 67 N / 289 N = 0.23183391 = 0.23

Therefore, B is the correct answer

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For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point

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Therefore 4.1 cm is λ/4

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Predict which molecule will show a similar relationship between its heat of fusion and its heat vaporization that water does
BaLLatris [955]

Question:

A) C6H6

B) CH3CH2CH2CH2CH2COH6

C) NaCl

D) NH3

Answer:

The correct option is;

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Explanation:

Heat of fusion = 6.02 kJ/mol

Heat of vaporization =40.8 kJ/mol

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Heat of fusion = 9.92 kJ/mol

Heat of vaporization =30.8kJ/mol

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The above compound is more likely solid

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8 0
3 years ago
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With detailed explaniation
belka [17]
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

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