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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 block of mass 10 kg slides down an inclined plane that has an angle of 30. If the inclined plane has no friction and the block
Helga [31]
No friction present means: Ek = Ep

So Ek = mgh = 10 * 9.8 * 2 = 196 J
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Noble Gases do not readily form compounds because they ___ chemically stable with 8 valence electrons
azamat

Answer : Noble Gases do not readily form compounds because they are chemically stable with 8 valence electrons.

Explanation :

Noble gases are the chemical elements that are present in group 18 in the periodic table.

The elements are helium, neon, argon, krypton, xenon and radon.

They are chemically most stable except helium due to having the maximum number of 8 valence electrons can hold their outermost shell that means they have a complete octet.

They are rarely reacts with other elements to form compounds by gaining or losing electrons since they are already chemically stable.

Hence, the noble Gases do not readily form compounds because they are chemically stable with 8 valence electrons.

8 0
3 years ago
What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?
viktelen [127]

Answer: 27.21 V

Explanation:

The <u>electric potential</u> V_{E} due to a point charge is expressed as:

V_{E}=k\frac{q}{r}

Where:

k=9(10)^{9}\frac{Nm^{2}}{C^{2}} is the <u>electric constant</u>

q=1.6(10)^{-19}C is the <u>electric charge of the hydrogen nucleus</u>, which is positive

r=5.292(10)^{-11}m is the <u>distance</u>

Rewritting the equation with the known values:

V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}

Finally:

V_{E}=27.21 V

5 0
3 years ago
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi
OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

8 0
3 years ago
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