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4 years ago

Two trains of length 90m and120m

Physics

1 answer:

Alenkasestr [34]4 years ago

6 0

Answer:Train A is 90 m long and running at 72 km/h.

Train B is 120 m long and running at 36 km/h.

When the trains are running in opposite directions, their relative speed gets added: 72+36 = 108 km/h.

The total length of A and B = 90+120 = 210 m

So the time taken to cross each other = 210 m*60*60/108*1000m

= 7 seconds

Explanation:

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Baseball homerun hitters like to play in Denver, but curveballpitchers do not. Why?

pishuonlain [190]

Answer:

Because of height and lower atmospheric pressure.

Explanation:

Atmospheric pressure affects aerodynamic drag, lower pressure means less drag. At the altitude of Denver the air has lower pressure, this allows baseball players to hit balls further away.

Another aerodynamic effect is the Magnus effect. This effect causes spinning objects to curve their flightpath, which is what curveball pitchers do. A lower atmospheric pressure decreases the curving of the ball's trajectory.

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3 years ago

Identical objects, Object X and Object Y, are tied together by a string and placed at rest on an incline, as shown in the figure

goblinko [34]

The slope of the velocity time graph of an object moving with constant acceleration is constant

It will take approximately 3 seconds for the center of mass of Object X to reach point J near the bottom of the incline

The reason why the above time value is correct is given as follows:

Known parameters:

Initial velocity of the objects, u = 0

The graph in the question is a straight line graph with data points

(0, 0), (0.5, 1.0), (1.0, 2), (3.0, 6), and (3.5, 7)

Given that the slope of the velocity-time graph is constant, we have that the acceleration is constant and is given as follows;

$a = \dfrac{\Delta v}{\Delta t } = \dfrac{v_2 - v_1}{t_2 - t_1}$

Therefore;

$a = \dfrac{6 - 2}{3.0 - 1.0} = 2$

The acceleration, a ≈ 2 m/s²

The distance from the center of mass of the Object X to the point J near the bottom = 9 m

The equation for distance travelled is given as follows;

$s = u\cdot t + \dfrac{1}{2} \cdot a \cdot t^2$

Which gives;

$9 = 0\times t + \dfrac{1}{2} \times 2 \times t^2 = t^2$

t = √9 = 3

The time it will take the center of mass of Object X to reach point J near the bottom of the incline is t = 3 seconds

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3 years ago

Tectonic plates are generally smaller than land masses on maps.

Sloan [31]

False because they can be really big

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4 years ago

Read 2 more answers

Famed stunt pilot, Cleonvia Thread, pulls out rapidly from a dive. He is traveling at 222 mi/h at the bottom of his trajectory,

Simora [160]

Answer:

0.39 m/s²

Explanation:

From the question,

a = v²/r.................... Equation 1

Where v = velocity, r = radius.

Given: v = 222 mi/h = (222×0.44704) m/s = 9.83 m/s, r = 820 ft = (820×0.3048) m = 249.94 m.

Substitute thses values into equation 1

a = 9.83²/249.94

a = 96.63/249.94

a = 0.39 m/s²

Hence the acceleration is 0.39 m/s²

8 0

3 years ago

Consider a 10 gram sample of a liquid with specific heat 2 J/gK. By the addition of 400 J, the liquid increases its temperature

stich3 [128]

40 J/g is the heat of vaporization of the liquid.

Answer: Option D

Explanation:

Given that mass of liquid sample: m = 10 g

And, Specific heat of the liquid: S = 2 J/g K

Also, the increase in the temperature of the liquid, $\Delta T = T_{2}-T_{1} = 10 K$

Therefore, the total amount of heat energy required is given by:

$q_{1} = m \times S \times\left(T_{2}-T_{1}\right) = 10 \times 2 \times 10 = 200 J$

According to the given data in the question,

Total heat energy supplied, q = 400 J

Rest of heat would be $q_{2}=q-q_{1}=400-200=200 \mathrm{J}$

Now, 200 J vaporizes the mass, half of the liquid from full portion boiled away. So,

$m^{\prime} = \frac{10}{2} = 5 \mathrm{g}$

Latent heat of vaporization of the liquid is $L_{v}$ . It can be calculated as below,

$q_{2} = m^{\prime} L_{v}$

$L_{v} = \frac{q_{2}}{m^{\prime}} = \frac{200}{5} = 40 \mathrm{J} / \mathrm{g}$

5 0

3 years ago

Two trains of length 90m and120m​

Two trains of length 90m and120m