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murzikaleks [220]
4 years ago
11

A person driving her car at 48 km/h approaches an intersection just as the traffic light turns yellow. She knows that the yellow

light lasts only 2.0 s before turning to red, and she is 30 m away from the near side of the intersection (Fig. 2-29). Should she try to stop, or should she make a run for it? The intersection is 15 m wide. Her car's maximum deceleration is -7.0 m/s2, whereas it can accelerate from 48 km/h to 70 km/h in 6.6 s. Ignore the length of her car and her reaction time. If she hits the brakes, how far will she travel before stopping?
Physics
1 answer:
trasher [3.6K]4 years ago
4 0

Answer:

She must stop the car  before interception, distance traveled 12.66 m

Explanation:

We will take all units to the SI system

Vo = 48Km / h (1000m / 1Km) (1h / 3600s) = 13.33 m / s

V2 = 70 Km / h = 19.44 m / s

We calculate the distance traveled before stopping

X = Vo t + ½ to t²

Time is what it takes traffic light to turn red  is t = 2.0 s

X = 13.33 2 + 1.2 (-7) 2²

X = 12.66 m

It stops car before reaching the traffic light turning to red

Let's analyze what happens if you accelerate, let's calculate the acceleration of the vehicle

     V2 = Vo + a t2

      a = (V2-Vo) / t2

      a = (19.44-13.33) /6.6

      a = 0.926 m / s2

This is the acceleration to try to pass the interception, now let's calculate the distance it travels in the time the traffic light changes from yellow to red (t = 2.0 s)

X = Vo t + ½ to t²

X = 13.33 2 + ½ 0.926 2²

X = 28.58 m

Since the vehicle was 30 m away, the interception does not happen

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You’re still in a redecorating mood and want to move a 35kg bookcase to a different place in the living room. You exert 58 N on
CaHeK987 [17]

Answer: 0.16

Explanation:

Newton's second law states that the resultant of the forces exerted on the bookcase is equal to the product between the mass (m) and the acceleration (a):

F-F_f = ma

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F = 58 N is the force applied

Ff is the frictional force

Substituting, we find the frictional force

F_f = F-ma=58 N-(35 kg)(0.12 m/s^2)=53.8 N

The frictional force has the form:

F_f = \mu mg

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\mu=\frac{F_f}{mg}=\frac{53.8 N}{(35 kg)(9.8 m/s^2)}=0.16

4 0
4 years ago
Two 60 cm parallel disks are separated by 40 cm and are aligned directly on top of each other. Both disks are black surfaces wit
Crazy boy [7]

Answer:

775.48 W

Explanation:

given,

diameter of disk = 0.6 cm

length of the disk = 0.4 m

T₁ = 450 K         T₂ = 450 K      T₃ = 300 K

\dfrac{d}{r_1}=\dfrac{0.4}{0.3} = 1.33

now,

the value of view factor (F₁₂)corresponding to 1.33

F₁₂ = 0.265

F₁₃ = 1 - 0.265 = 0.735

now,

net rate of radiation heat transfer from the disk to the environment:

=\dot{Q_{1-3}+Q_{2-3}} = 2 \dot{Q_{1-3}}

       = 2 F₁₃ A₁ σ (T₁⁴ - T₃⁴)

       = 2 x 0.735 x π x (0.3)² x (5.67 x 10⁻⁸ W/m²) (450⁴ - 300⁴)

       = 775.48 W

Net radiation heat transfer from the disks to the environment = 775.48 W

3 0
4 years ago
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