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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 small bolt with a mass of 33.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical dire
densk [106]

Answer:

0.027m

Explanation:

the bolt loses contact with the piston only when acceleration due to gravity equals acceleration of piston

ω² * A = g where ω is angular velocity, A amplitude, g acceleration due to gravity

ω is given by 2πf, ω² is 4π²f²

A= g/4π²f² depending on the value of g used either 10m/s² or 9.8m/s²,

i used 10m/s² in this answer

5 0
4 years ago
WHICH SHOULD NOT BE PART OF SCIENTIFIC INQUIRY
umka21 [38]

Answer:

Not collecting the info

Explanation:

5 0
3 years ago
A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi
stiks02 [169]

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

v=\sqrt{\frac{TL}{M}} (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

v= \lambda f (2)

We can equate expression (1) and (2):

\sqrt{\frac{TL}{M}}=\lambda f

Solving for T

T= \frac{M(\lambda f)^2}{L} (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic N_{harmonic} is:

\lambda_{harmonic}=\frac{2l}{N_{harmonic}}

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m

We can now find T on (3) using all the values we have:

T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}

T=75.22 N

3 0
3 years ago
Numeria
Vinil7 [7]

Answer:

hope it helps

plz mark as brainliest

5 0
3 years ago
Read 2 more answers
Water ice has a density of 0.91 g/cm2, so it will float in liquid water. Imagine you have a cube of ice, 10 cm on a side. a. Wha
Reptile [31]

Answer:

(i) W = 8.918 N

(ii) V = 9.1 \times 10^{-4} m^3

(iii) d = 9.1 cm

Explanation:

Part a)

As we know that weight of cube is given as

W = mg

W = \rho V g

here we know that

\rho = 0.91 g/cm^3

Volume = L^3

Volume = 10^3 = 1000 cm^3

now the mass of the ice cube is given as

m = 0.91 \times 1000 = 910 g

now weight is given as

W = 0.910 \times 9.8 = 8.918 N

Part b)

Weight of the liquid displaced must be equal to weight of the ice cube

Because as we know that force of buoyancy = weight of the of the liquid displaced

W_{displaced} = 8.918 N

So here volume displaced is given as

\rho_{water}Vg = 8.918

1000(V)9.8 = 8.918

V = 9.1 \times 10^{-4} m^3

Part c)

Let the cube is submerged by distance "d" inside water

So here displaced water weight is given as

W = \rho_{water} (L^2 d) g

8.918 = 1000(0.10^2 \times d) 9.8

d = 0.091 m

so it is submerged by d = 9.1 cm inside water

4 0
3 years ago
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