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3 years ago

What is The distance traveled by a vehicle in 12 minutes,if it’s speed is 35km/h

Physics

1 answer:

Delvig [45]3 years ago

3 0

Answer: 7 km

Explanation: 12 minutes is 1/5 of an hour. 35/5=7

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A mercury barometer reads 745.0 mm on the roof of a building and 760.0 mm on the ground. Assuming a constant value of 1.29 kg/m3

Ipatiy [6.2K]

Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

$\Delta P \propto \rho \cdot \Delta h$

$\Delta P = k \cdot \rho \cdot \Delta h$

Where:

$\Delta P$ - Manometric pressure difference, measured in kilopascals.

$\rho$ - Fluid density, measured in kilograms per cubic meter.

$\Delta h$ - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

$\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}$

$\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}$

Where:

$\Delta h_{air}$ - Height difference of the air column, measured in meters.

$\Delta h_{Hg}$ - Height difference of the mercury column, measured in meters.

$\rho_{air}$ - Density of air, measured in kilograms per cubic meter.

$\rho_{Hg}$ - Density of mercury, measured in kilograms per cubic meter.

If $\Delta h_{Hg} = 0.015\,m$ , $\rho_{air} = 1.29\,\frac{kg}{m^{3}}$ and $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$ , the height difference of the air column is:

$\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)$

$\Delta h_{air} = 158.140\,m$

The height of the building is 158.140 meters.

5 0

3 years ago

Read 2 more answers

A pilot can withstand an acceleration of up to 9g, which is about 88 m/s2, before blacking out. What is the acceleration experie

sweet-ann [11.9K]

Answer:

Acceleration, $a=97.67\ m/s^2$

Explanation:

Given that,

Speed of pilot, v = 475 m/s

Radius of the circle, r = 2310 m

If the pilot is moving in a circular path, it will experience a centripetal acceleration. It is given by the formula as :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(475\ m/s)^2}{2310\ m}$

$a=97.67\ m/s^2$

So, the acceleration experienced by a pilot flying in a circle is $97.67\ m/s^2$ . Hence, this is the required solution.

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3 years ago

Which of the circuit diagrams shown in Figure 21-1A is a parallel circuit?

Thepotemich [5.8K]

I and II only it’s has multiple paths for the electricity to flow

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2 years ago

A ball is thrown straight up from a point 2 m above the ground. The ball reaches a maximum height of 3 m above its starting poin

qwelly [4]

Answer:

2 m

Explanation:

The displacement of any body is the shortest distance in an object's path between its initial and final point.

The ball would travel 3 m from the point of throwing then fall down 5 m to the ground. The total distance traveled is 7 m.

The displacement of the ball will be the distance from the point of throwing to the ground i.e., 2 m as it is the shortest distance between the initial and final point of the ball's journey.

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He used truffula trees to make thneeds

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