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notka56 [123]
3 years ago
12

The coefficient of static friction between waxed skis and now snow is 0.14. What force will be needed to get a 700 N skier at re

st moving?
A: 28 N
B: 70 N
C: 98 N
D: 980 N
Physics
2 answers:
TiliK225 [7]3 years ago
8 0

Answer:

C. 98 N

Explanation:

The force needed to get the skier at rest moving must be at least equal to the maximum static frictional force acting on the skier, which is given by

F=\mu W

where

\mu = 0.14 is the coefficient of static friction

W = 700 N is the weight of the skier

Substituting into the equation, we find

F=(0.14)(700 N)=98 N

balu736 [363]3 years ago
3 0

Answer:

c

Explanation:

μ = f/N where μ is the coefficient of friction; f is the amount of force that resists motion, and N is the normal force. You must solve for f here so 700 N x .14 = 98 N.

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On what principle does a bottle opener work
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Answer:

Bottle opener works on a fulcrum.

Explanation:

A bottle opener is a second-class lever because the pivot point is at one end of the opener and the load is in the middle.

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3 years ago
Un cuerpo gira con movimiento circular uniforme un ángulo total de 1080° en 20
LekaFEV [45]

Velocidad angular = (angulo total) / (tiempo total)

Velocidad angular = (1080 grados) / (20 segundos)

Velocidad angular = (1080/20) g/s

Velocidad angular = 54 g/s

Pero 180 grados = π radianes

V.A. = (54 g/s) x (π rad / 180 g)

V.A. = (54π gr-rad / 180 seg-gr)

<em>V.A. = 0.3π rad/seg</em>

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7 0
3 years ago
A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ
Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3

(c) From any of the equations, solve for Length of the string (L);

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m

(a) the fundamental frequency is calculated as;

f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o =  65 \ Hz

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

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3 years ago
A capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d
yawa3891 [41]
I think its true i dont kno for sure
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State one use and one disadvantage of the expansion of materials when they are heated.
Reil [10]

An example of the use of expansion of materials when they are heated is in

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It contains mercury which expands as temperature increases. This is used

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The disadvantage of expansion of materials when they are heated can be

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8 0
2 years ago
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