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Dima020 [189]
3 years ago
7

Two train tracks run parallel to each other, except for a short distance where they meet and become one track over a narrow brid

ge. One morning, a train speeds onto the bridge. Another train coming from the opposite direction, also speeds onto the bridge. Neither train can stop on the short bridge, yet there is no collision. How is this possible?
Physics
1 answer:
Darya [45]3 years ago
7 0

Answer:

This situation is possible.

Explanation:

It is said in the problem that one morning two trains are speeding up from opposite directions onto the bridge. It may seem from this statement that the trains are arriving at the same time. But this should be a wrong assumption.

Because it is only said in the problem that the trains are passing the bridge on same morning. No explicit timing is given.

So it may so happen that both the trains pass the narrow bridge on the same morning but at two difeerent times, making no collision with each other.

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A stone with a weight of 5.30 N is launched vertically from ground level with an initial speed of 23.0 m/s, and the air drag on
frozen [14]

Answer:

a) h=25.7\ m

b) v'=21.8733\ m.s^{-1}

Explanation:

Given:

  • weight of the stone, w=5.3\ N
  • initial velocity of vertical projection, u=23\ m.s^{-1}
  • air drag acting opposite to the motion of the stone, D=0.266\ N

The mass of the stone:

m=\frac{w}{g}

m=\frac{5.3}{9.8}

m=0.5408\ kg

Now the acceleration of the stone opposite of the motion:

D=m.d

where:

d = deceleration

0.266=0.5408\times d

d=0.4918\ m.s^{-2}

<u>In course of going up the net acceleration on the stone will be:</u>

g'=g+d

g'=9.8+0.4918

g'=10.2918\ m.s^{-2}

a)

Now using the equation of motion:

v^2=u^2-2 g'.h

where:

v= final velocity when the stone reaches at the top of the projectile = 0

h = height attained by the stone before starting to fall down

0^2=23^2-2\times 10.2918\times h

h=25.7\ m

b)

during the course of descend from the top height of the projectile:

initial velocity, v=0\ m.s^{-1}

The acceleration will be:

g"=g-d

g"=9.8-0.4918

g"=9.3082\ m.s^{-2}

here the gravity still acts downwards but the drag acceleration acts in the direction opposite to the motion of the stone, now the stone is falling down hence the drag  acts upwards.

Using equation of  motion:

v'^2=v^2+2g".h (+ve acceleration because it acts in the direction of motion)

v'^2=0^2+2\times 9.3082\times 25.7

v'=21.8733\ m.s^{-1}

8 0
3 years ago
Experiment to determine the refractive index of a glass prism FOCAL LENGTH = 9.5cm i° e° d° 30 43 69 40 41 61 45 39 56 50 37 48
yarga [219]
We are given the data for the angle of incidence and angle of refraction.
The first part of the problem is to plot the deviation and the angle of incidence. The minimum deviation is 37 with 65 degrees as the corresponding angle of incidence. The maximum deviation is 69 with a corresponding angle of incidence of 30 degrees. It is not advisable to use small values for the angle of incidence since it would result to a higher deviation from Snell's Law.
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3 years ago
For the ride to be comfortable, the magnitude of acceleration must not exceed 32 m/s2. What is the fastest constant speed that a
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Answer:

the maximum possible constant speed is  8 m/sec  

Explanation:

from the image, Given that

r(t) = (2t, t²,t²/3), -5 ≤ t ≤ 5  

Given that the curvature K(t)  = 2 / ( t² + 2)²  

note that t² + 2 ≥ 2  

(t² + 2)² ≥ 4

1 / (t² + 2)² ≤ 1/4

2 / (t² + 2)² ≤ 1/2

Also note that k(0) = 1/2

The normal component of acceleration satisfies aN = kv²

where v = ║v(t)║is the speed of the roller coaster.

The maximum possible normal component of acceleration is 32

so, aN ≤ 32 every where on the track

aN = kv² ≤  1/2v² ≤ 32

v² ≤ 64

Therefore, the maximum possible constant speed is  8 m/sec  

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The state of matter depends upon how close the individual particles are together
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The battery provides electrical energy to run an electric circuit.
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