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The stoplights on a street are designed to keep traﬃc moving at 26 mi/h. the average length of a street block between traﬃc ligh

bonufazy [111]

We use the formula,

$v = \frac{d}{t}$ .

Here, v is velocity and its value given 26 mi/h ( in m/s, $\frac{26 \times 1.609 \times 10^{3} m}{60 \times 60 s} =11.62 \ m/s$ ) and d is distance and its value is given 80 m.

Substituting these values in above formula we get,

$t = \frac{80 m}{11.62m/s } = 6.88 \ s$

Thus, the time delay between green lights on successive blocks to keep the traﬃc moving continuously is 6.88 s

3 0

3 years ago

vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses

dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

$v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}$

(This is correct because the horizontal motion has acceleration zero). Then:

$v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s$

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

$E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}$

Then, plugging in the given values, we obtain:

$k=\frac{(61.2kg)(15.0m/s)^2}{(2.76m)^2}\\\\k=1808N/m$

Finally, the effective spring constant of the firing mechanism is 1808N/m.

3 0

3 years ago

A highway curves to the left with radius of curvature of 36 m and is banked at 28 ◦ so that cars can take this curve at higher s

iVinArrow [24]

Answer: 30.34m/s

Explanation:

The sum of forces in the y direction 0 = N cos 28 - μN sin28 - mg

Sum of forces in the x direction

mv²/r = N sin 28 + μN cos 28

mv²/r = N(sin 28 + μcos 28)

Thus,

mv²/r = mg [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]

v²/r = g [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]

v²/36 = 9.8 [(0.4695 + 0.87*0.8829) - (0.8829 - 0.87*0.4695)]

v²/36 = 9.8 [(0.4695 + 0.7681) / (0.8829 - 0.4085)]

v²/36 = 9.8 (1.2376/0.4744)

v²/36 = 9.8 * 2.6088

v²/36 = 25.57

v² = 920.52

v = 30.34m/s

5 0

3 years ago

A 6.89-nC charge is located 1.76 m from a 4.10-nC point charge. (a) Find the magnitude of the electrostatic force that one charg

iogann1982 [59]

Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive

Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:

$F=\frac{k*q1*q2}{d^{2}}$

Replacing the dat we obtain F=82 nN.

The force is repulsive because the points charged have the same sign.

5 0

3 years ago

What is measured using micrometer?<br><br>

Fantom [35]

Explanation:

A micrometer, sometimes known as a micrometer screw gauge, is a device incorporating a calibrated screw widely used for accurate measurement of components in mechanical engineering and machining as well as most mechanical trades, along with other metrological instruments such as dial, vernier, and digital calipers

6 0

2 years ago