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Scorpion4ik [409]
3 years ago
15

The stoplights on a street are designed to keep traffic moving at 24 mi/h. The average length of a street block between traffic

lights is about 80 m. What must be the time delay between green lights on successive blocks to keep the traffic moving continuously? There are 1.609 × 103 m in a mile. Answer in units of s.
Physics
1 answer:
Crazy boy [7]3 years ago
5 0

Answer:

t = 7.5 s

Explanation:

As we know that the average speed between two lights is given as

v = 24 mph

here we know that

v = 24 \times \frac{1602 m}{3600 s}

v = 10.68 m/s

now the distance between two lights is 80 m

so here if all the cars are moving continuously between two lights then the time delay between two lights is given as

t = \frac{L}{v}

t = \frac{80}{10.68}

t = 7.5 s

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What two factors affect the strength of a magnetic field
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Answer:

Factors Affecting the Strength of the Magnetic Field of an Electromagnet: Factors that affect the strength of electromagnets are the nature of the core material, strength of the current passing through the core, the number of turns of wire on the core and the shape and size of the core.

Explanation:

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7 0
3 years ago
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How old are spiral galaxies?<br> I WILL MAKE YOU BRAINLIEST, 5 STARS, AND THANKS! PLEASE ASAP
kondor19780726 [428]

Answer:

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7 0
3 years ago
Read 2 more answers
A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway
Andru [333]

Answer:

<em>63.44 rad/s</em>

<em></em>

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet v_{1} = 250 m/s

final velocity of bullet v_{2} = 140 m/s

loss of kinetic energy of the bullet = \frac{1}{2}m(v^{2} _{1} - v^{2} _{2})

==> \frac{1}{2}*0.0033*(250^{2}  - 140^{2} ) = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

KE = \frac{1}{2} mv^{2}

70.785 = \frac{1}{2}*0.25*v^{2}

566.28 = v^{2}

v= \sqrt{566.28} = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 =<em> 63.44 rad/s</em>

5 0
3 years ago
An excited 92 kg football player celebrates a touchdown by carelessly running straight into the goalpost at 9.4 m/s. He bounces
yKpoI14uk [10]

Answer:

i. 15.6 m/s

ii. I = 1.44 KNs

Explanation:

The impulse, I, on a body is the product of force applied on it and the time it acts.

i.e I = F x t

Impulse is sometimes expressed as the change in momentum of a body. It is measured in Ns.

i. mass, m, of the player = 92 kg

initial velocity of the player, u = 9.4 m/s

final velocity of the player, v = 6.2 m/s

Since he bounces back on hitting the pole, then the sign of initial and final velocities are of opposite sign.

So that,

change in velocity of the player = final velocity - initial velocity

                                          = 6.2 - (-9.4)

                                         = 6.2 + 9.4

                                         = 15.6 m/s

change in velocity of the player is 15.6 m/s

ii. Impulse, I = m(v - u)

                    = 92 x 15.6

                    = 1435.2

Impulse on the player is 1.44 KNs.

7 0
3 years ago
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