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2 years ago

7

Slow moving vehicles that travel at a speed less than ___ mph will have an orange reflective triangle with red around the edges

on the rear of the vehicle that should alert other drivers that their speed is not going to increase.

1 answer:

nika2105 [10]2 years ago

7 0

Slow moving vehicles that travel at a sped less than 25 MPH will have ...........
Slow moving vechicles [SMV] are those vehicles that travel on the road at a speed that is less than the recommended speed, example of a slow moving vehicle is tractor. These vehicles must have SMV signs on them to notify other drivers that they would not be travel fast.

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Answer <br> A<br> B<br> C<br> D and <br> E

harina [27]

Answer:

what is the question I cannot click the

Explanation:

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2 years ago

A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio

Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

$F_c = m\frac{v^2}{r}$

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

$v = r\omega$

with ω denoting the angular velocity, which we are given. With that, the above becomes:

$F_c = m\frac{v^2}{r}=m\omega^2 r$

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

$F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}$

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

$\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}$

and so 45 rev/min = 4.71 rad/s.

$\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27$

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.

3 0

2 years ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

So

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

So

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

So

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

4 0

2 years ago

What is the index of refraction of a material in which the speed of light is 7.50% slower than the speed of light in vacuum?

Inessa [10]

Answer:

The value is $n = 1.081$

Explanation:

From the question we are told that

The speed of in a vacuum is $c = 3.0 *10^{8} \ m/s$

The speed of light in the material is $v = c -0.075c = 0.925 c = 0.925 * 3.0*10^{8} = 2.775 *10^{8} \ m/s$

Generally the reflection of the material is mathematically represented as

$n = \frac{c}{v}$

=> $n = \frac{3.0*10^{8}}{2.775 *10^{8}}$

=> $n = 1.081$

8 0

2 years ago

What are some things that you do to make tasks like cleaning your room or doing the dishes more fun?

Anna007 [38]

Answer:

You can listen to music while doing either one, or you can get someone else to help you that way you have someone to talk to also you finish faster.

5 0

2 years ago

Read 2 more answers