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Ghella [55]
3 years ago
7

A car's horn produces a constant frequency of 350 Hz as it passes by Suzy. What is the best estimate of the

Physics
2 answers:
Arada [10]3 years ago
8 0

Answer:

330 Hz

Explanation:

edg2020

Vladimir79 [104]3 years ago
7 0

Answer:

330Hz

Explanation:

on edge 2020

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A 70 kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m . What is the effective spring
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Answer:

29856.521 N/m

Explanation:

m=Mass\ of\ diver=70\ kg\\x=Length\ compressed\ by\ spring=2.3\times 10^{-2}\ m\\a=Acceleration\ due\ to\ gravity=9.81\ m/s^2\\F=Force\ exerted\ by\ diver=m\times a\\\Rightarrow F=70\times 9.81\\\Rightarrow F=686.7\ N\\k=spring\ constant\\F=k\times x\\\Rightarrow k=\frac {F}{x}\\\Rightarrow k=\frac {686.7}{2.3\times 10^{-2}}\\\therefore k=29856.521\ N/m

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4 years ago
if there are infinite universes is it possible that there is a universe where 1+1=3 or 4 or something
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3 years ago
Write the equation that links current, potential difference, and resistance
mrs_skeptik [129]

You can write the equation in 3 different ways, depending on which quantity you want to be the dependent variable.  Any one of the three forms can be derived from either of the other two with a simple algebra operation.  They're all the same relationship, described by "Ohm's Law".

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==> Potential difference = (current) x (resistance)

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Why are eight electrons (four pairs) surrounding each non-hydrogen atom the optimal electronic arrangement for covalent molecule
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3 0
2 years ago
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0
3 years ago
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