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Varvara68 [4.7K]
3 years ago
6

The diagram above shows four light bulbs wired in a circuit. What would happen if bulb 3 burned out

Physics
1 answer:
andre [41]3 years ago
6 0
The lights will no longer work because the circuit will be cut
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For a maximum superelevation of 0.08 ft/ft and a degree of curve of 4o, calculate the maximum safe speed for the curve assuming
Mamont248 [21]

Answer:

Explanation:

Given that

Superelation= 0.08ft/ft

Given curve= u•

Curve junction factor= 0.13

DR= 5729.57795

R = 5729.57795/D

R = 5729.57795/4

R = 1432.4ft

c + f = V^2/gG

0.08 + 0.13 = V^2 / (32*1432.4)

V^2 = 9625.728 or V = 98 ft/sec

The designed speed for a project considered is a minimum value which means the highway design elements will meet or exceed the standards for the design speed. The maximum safe speed under normal condition is significantly greater than design speed

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3 years ago
Four football players are running down the field at the same speed. Player 1 weighs 180 lbs and is running toward the south goal
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Player 2 because moment is mass times acceleration and since they are all going the same speed. Speed doesn't matter so the only thing that is left is mass/ weight and he has the most
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3 years ago
A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull
sdas [7]

In spring mass system we know that angular frequency is given as

\omega = 2\pi f

f = 8.38 Hz

\omega = 2\pi(8.38)

\omega = 52.65  rad/s

now we know that speed of SHM at its extreme position is given by

v = A\omega

here we know that

A = 17.5 cm

v = 0.175 (52.65)

v = 9.21 m/s

so maximum speed is 9.21 m/s

7 0
3 years ago
Period T= time of one revolution measured in seconds
ser-zykov [4K]
60 RPM equals one hertz (i.e., one revolution per second, or a period of one second). The SI unit for period is the second.
3 0
3 years ago
Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in secon
dedylja [7]

Explanation:

It is given that,

Semi major axis of the Jupiter, a=7.78\times 10^{11}\ m

Mass of the sun, M=1.99\times 10^{30}\ kg

(a) Let T is the period of Jupiter's orbit. It is given by :

T^2\propto a^3

T^2=\dfrac{4\pi^2}{GM}a^3

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.99\times 10^{30}}\times (7.78\times 10^{11})^3

T=3.74\times 10^8\ s

(b) We know that,

1\ year=3.154\times 10^7\ s

or

1\ s=3.171\times 10^{-8}\ year

3.74\times 10^8\ s={3.171\times 10^{-8}}\times {3.74\times 10^8}

T = 11.859 earth years

Hence, this is the required solution.

7 0
3 years ago
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