Can anyone help me out with this

Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl

tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

$f_{o}=603\ Hz$

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

$v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s$

For Train 2 Velocity of the Source,

$v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s$

Frequency of Source,

$f_{s}=500\ Hz$

To Find:

Frequency of Observer,

$f_{o}=?$ (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source $v_{s}$ and observer $v_{o}$ , the original frequency of the sound waves $f_{s}$ and the observed frequency of the sound waves $f_{o}$ ,

The Equation is

$f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})$

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

$f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz$

Therefore,

The frequency heard by the engineer on train 1

$f_{o}=603\ Hz$

7 0

3 years ago

If a wave were to hit a flexible, moveable surface what will it do?

Vika [28.1K]

what ever force the wave hit it at (on the x-axis) would create another wave from the force of the first wave. though the wave would be smaller because of other forces

4 0

4 years ago

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When the blood flows through the capillary bed, most of the plasma will return to the heart through the venules. What happens to

Bas_tet [7]

The answer is the 2nd sentence.

8 0

3 years ago

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Tarzan is in the path of a pack of stampeding elephants when Jane swings in to the rescue on a rope vine, hauling him off to saf

timurjin [86]

Answer:

h = 3.56 m

Explanation:

Assuming no friction on the rope or air resistance, we can apply the conservation of mechanical energy principle to the first part of the trajectory, from when Jane starts her swing till she catches Tarzan at the ground level.

⇒ $\Delta K + \Delta U = 0 (1)$

Rearranging terms, we get:

$U_{ij} + K_{ij} = U_{fj} + K_{f} (2)$

Now, as Jane starts from rest, Kij =0.
if we choose the ground level as our zero reference level, it will be also Ufj = 0.
Replacing in (2) by the expressions of Uij and Kfj, we have:

$m_{j} * g* h_{ij} = \frac{1}{2} * m_{j} * v_{fj} ^{2} (3)$

Replacing in (3) by the givens, and rearranging terms, we can solve for vfj, as follows:

$v_{fj} =\sqrt{2*9.8 m/s2*27 m} = 23 m/s (4)$

Now, as once jane catches Tarzan, both continue swinging together, we can take the catching moment as a completely inelastic collision.
Assuming no external forces act during the collision, total momentum must be conserved.

⇒ $p_{o} = p_{f} (5)$

Assuming that Tarzan is at rest when Jane catches him, the initial momentum will be simply as follows:

$p_{o} = m_{j} * v_{j} = 49 kg * 23 m/s = 1127 kg*m/s (6)$

The final momentum, will be just the product of the combined mass of Jane and Tarzan times the common speed for them after the collision:

$p_{f} = (m_{j} + m_{t} ) * v_{jt} = 135 kg* v_{jt} (7)$

As (6) and (7) are equal each other, we can solve for vjt, as follows:

$v_{jt} = \frac{1127 kg*m/s}{135 kg} = 8.35 m/s (8)$

Finally we can apply the same energy conservation principle to the last part of the trajectory, as follows:

$U_{ijt} + K_{ijt} = U_{fjt} + K_{fjt} (9)$

We know that Uijt = 0 and also that Kfjt = 0, due to both starts from the ground level and reach to the highest point before starting to fall down, so at this point, the kinetic energy will be zero.
Replacing by the givens and the result from (8), we can solve for the h as follows:
$h_{f} =\frac{v_{ijt} ^{2} }{2*g} = \frac{69.7m2/s2}{2*9.8m/s2} = 3.56 m$

8 0

3 years ago

Why screw drivers with long arm are better in use?

Molodets [167]

Answer:

Explanation:

Long screwdrivers are easier to turn because the grip is in the palm of the hand, thus allowing all fingers to contribute. A shorter screwdriver is turned with the thumb and first two or three fingers, providing much less torque.

7 0

3 years ago

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