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Sunny_sXe [5.5K]
4 years ago
15

Two automobiles of equal mass approach an intersection. One vehicle is traveling with speed 13.0 m/s toward the east, and the ot

her is traveling north with speed v2i. Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid marks at an angle of 55.08 north of east. The speed limit for both roads is 35 mi/h, and the driver of the northward-moving vehicle claims he was within the speed limit when the collision occurred. Is he telling the truth? Explain your reasoning.
Physics
1 answer:
puteri [66]4 years ago
7 0

Answer:

Explanation:

We shall apply law of conservation of momentum to know the Speed of northward moving vehicle before collision to check the veracity of driver's statement .

Let v be the velocity of composite mass after collision

Applying law of conservation of momentum in north direction

m v₂ = 2m v sin55.08

Applying law of conservation of momentum in east  direction

m x 13 = 2m v cos55.08

Dividing these two equations

v₂ / 13 = tan55.08

v₂ = 13  tan55.08

= 18.62 m/s

= (18.62 x60 x 60) / 1000

= 67 km/h

= 67 x 5/8 mi/h

= 42 mi/h

So he is lieing.

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For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with d
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Answer:

The maximum mass the bar can support without yielding = 32408.26 kg

Explanation:

Yield stress of the material (\sigma) = 200 M Pa

Diameter of the bar = 4.5 cm = 45 mm

We know that yield stress of the bar is given by the formula

                Yield Stress = \frac{Maximum load}{Area of the bar}

⇒                                \sigma = \frac{P_{max} }{A}  ---------------- (1)

⇒ Area of the bar (A) = \frac{\pi}{4} ×D^{2}

⇒                            A  = \frac{\pi}{4} × 45^{2}

⇒                            A = 1589.625 mm^{2}

Put all the values in equation (1) we get

⇒ P_{max} = 200 × 1589.625

⇒ P_{max} = 317925 N

In this bar the P_{max} is equal to the weight of the bar.

⇒ P_{max} = M_{max} × g

Where M_{max} is the maximum mass the bar can support.

⇒ M_{max} = \frac{P_{max} }{g}

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3 0
4 years ago
A vehicle starts from rest and accelerates uniformly for 12 seconds to a
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The value of the acceleration is 0.76 m/s² and the total time taken by the vehicle is 39 seconds.

<h3>Acceleration of the vehicle</h3>

The acceleration of the vehicle before coming to rest is calculated as follows;

v² = u² + 2as

where;

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  • u is the initial velocity
  • a is the acceleration
  • s is the distance traveled before stopping

the car came to rest with constant velocity attained after 12 seconds.

the initial velocity of the car before 12 seconds is zero.

v² = 0 + 2as

a = v²/2s

a = (10²)/(2 x 66)

a = 0.76 m/s²

<h3>Time of motion of the vehicle</h3>

d = ut + ¹/₂at²

where;

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  • t is the time of motion
  • a is acceleration
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580 = 0 + ¹/₂(0.76)t²

580 = 0.38t²

t² = 580/0.38

t² = 1,526.3

t = √1,526.3

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Thus, the value of the acceleration is 0.76 m/s² and the total time taken by the vehicle is 39 seconds.

Learn more about time of motion here: brainly.com/question/2364404

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Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.
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Answer:

10 J.

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We need to find the kinetic energy gained by the rake. We know that,

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K = 2 N × 5 m

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So, 10 J of kinetic energy is gained by the rake.

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Longer pendulums swing slower.

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