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polet [3.4K]
2 years ago
11

12–139. Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is posted at 60 km>h, d

etermine the minimum acceleration experienced by the passengers.
(60)2



y

2

0)2 = 1


*12–140. Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is posted at 60 km>h, determine the maximum acceleration experienced by the passengers.

12–142. The race car has an initial speed vA = 15 m> s at
a. If it increases its speed along the circular track at the rate at = (0.4s) m> s2, where s is in meters, determine the time needed for the car to travel 20 m. Take r = 150 m.




x


Probs. 12–139/140

Physics
1 answer:
Sergio039 [100]2 years ago
5 0
Check the attached file for the answer.

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A 5.0-g marble is released from rest in the deep end of a swimming pool. An underwater video reveals that its terminal speed in
Temka [501]

Answer:

a) a = -g = 9.8 m/s² , b) a = 0 m/s² and c)   t1 = 0.0213 s

Explanation:

a) At the moment the marble is released its velocity is zero, so it has no resistance force, the only force acting is its weight, so the acceleration is the acceleration of gravity

       a = -g = 9.8 m / s²

b) When the marble goes its terminal velocity all forces have been equalized, therefore, the sum of them is zero and consequently if acceleration is also zero

      a = 0 m / s²

c) We have to assume a specific type of resistive force, for liquid in general the resistive force is proportional to the speed of the body.

The expression of this situation is

         v = mg / b (1 -e^{-bt/m} )

For a very long time the exponential is zero, so the terminal velocity is

        v_{T} = mg / b

        b = mg /  v_{T}

        b = 5 10-3 9.8 / 0.3

        b = 0.163

We already have all the data to calculate the time for v = ½ v_{T}

        ½ v_{T} = v_{T} (1 -e^{-bt/m})

        ½ = 1- e (- 0.163 t1 / 5 10-3)

        e (-32.6 t1) = 1-0.5              (by  ln())

       -32.6 t1 = ln 0.5

       t1 = -1 / 32.6 (-0.693)

       t1 = 0.0213 s

3 0
2 years ago
A physicist is constructing a solenoid. She has a roll of insulated copper wire and a power supply. She winds a single layer of
Leni [432]

Answer:

P =105.44 W

Explanation:

Given that

D= 10 cm ,L= 60 cm

d= 0.1 cm ,B= 6.4 mT

ρ= 1.7 x 10⁻⁸ Ω · m

The number of turns N

N= L/d

N= 60/0.1 = 600 turns

Length of wire

Lc= πDN

Lc= 3.14 x 0.1 x 600

Lc=188.4 m

The magnetic filed given as

B=\dfrac{\mu_oNI}{L}

I=\dfrac{BL}{\mu_oN}

Now by putting the values

I=\dfrac{0.0064\times 0.6}{4\pi \times 10^{-7}\times 600}

I=5.09 A

The resistance R given as

R=\dfrac{\rho L_c}{A}

A=\dfrac{\pi}{4}\times d^2

A=\dfrac{\pi}{4}\times (0.1\times 10^{-2})^2

R=\dfrac{1.7\times 10^{-8} \times 188.4}{\dfrac{\pi}{4}\times (0.1\times 10^{-2})^2}

R=4.07 Ω

Power P

p =I²R

P= 5.09² x 4.07 W

P =105.44 W

5 0
2 years ago
forces F1 (east) and F2 are simultaneously applied to a 3.0kg mass. when F2 is east, a=5.0m/s² east and when F2 is west a=1.0m/s
zhenek [66]

Answer:

345453-5676

its the right answer

5 0
2 years ago
Ryan and Becca are moving a folding table out of the sunlight. A cup of lemonade, with the message 0.44 kg is on the table. Becc
jarptica [38.1K]
Calculate the weight of the table through the equation,

   W = mg

where W is the weight, m is the mass, and g is the acceleration due to gravity. Substituting the known values,
   W = (0.44 kg)(9.8 m/s²) 
   <em>W = 4.312 N</em>

The components of this weight can be calculated through the equation,

   Wx = W(sin θ) 

and Wy = W(cos θ)

x - component:
   Wx = W(sin θ)
Substituting,
  Wx = (4.312 N)(sin 150°) = <em>2.156 N</em>

  Wy = (4.312 N)(cos 150°) =<em> -3.734 N</em>
6 0
2 years ago
There are two forces on the 2 kg box in the overhead view of the following figure, but only one is shown. For F1=20N, a= 12 m/s2
maw [93]

Answer:

second force = 32.784

Magnitude =\sqrt{32.784

θ = -90°

Explanation:

a)

Fnet = ma

F1 + F2 = ma

20N + F2 = 2(12 × cos30° + 12 ×sin30°)

F2 = 2 × 12 ( sin 30° + cos 30°)

    = 24 × ( 1 + √3 )÷ 2

  =12 (1 +√3 )

  = 32.784

b) \sqrt{12(1 +\sqrt{3}}

= \sqrt{12 ( 1+ 1.732)}

= \sqrt{12 (2.732)}

= \sqrt{32.784}  

c)

θ = 30° + 180°

θ = 210°

210° - 300°

θ = -90°

8 0
2 years ago
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