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3 years ago

The most dangerous thing for a motorcyclist passing parked cars is

Physics

1 answer:

jeka943 years ago

6 0

The main danger is vehicles making u-turns or pulling out without signalling.

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A mass of 10 g of oxygen fill a weighted pistonâ€“cylinder device at 20 kPa and 110Â°C. The device is now cooled until the tempe

mezya [45]

Answer:

The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

Explanation:

Given that,

Mass of oxygen = 10 g

Pressure = 20 kPa

Initial temperature = 110°C

Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

$\Delta V=V_{2}-V_{1}$

$\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})$

Put the value into the formula

$\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)$

$\Delta V=14.297\ L$

$\Delta V=14.3\times10^{-3}\ m^3$

Hence, The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

6 0

3 years ago

What will happen to the Sun after it loses so much mass that its gravity

Natali5045456 [20]

Answer:

The answer is C

Explanation:

7 0

2 years ago

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Do 10 grams of oxygen atoms and 10 grams of iron atoms contain the same number of particles

Sergio [31]

I’m lost at this question, sorry but I would’ve help !

4 0

3 years ago

A rollercoaster car has 2500 J of potential energy and 160 J of

lara31 [8.8K]

Answer:

E_{k2}=2660 [J] kinetic energy.

Explanation:

The energy in the initial state i.e. when the rollercoaster is at the top is equal to the energy in the final state i.e. when it is at the bottom of the hill.

These states can be represented by means of the second equation.

$E_{k1}+E_{p1}=E_{k2}\\160 + 2500 = E_{k2}\\E_{k2}=2660 [J]$

Since the rollercoaster is located in the bottom of the hill where the potential energy level is zero, therefore there is only kinetic energy in the second state.

8 0

3 years ago

One runner in a relay race has a 1.50 s lead and is running at a constant speed of 3.25 m/s. The runner has 30.0 m to run before

yarga [219]

The second runner must run 3.3m/s. If the leading runner is 1.5 seconds ahead and there are 30m left, the second runner would need to run slightly faster than the lead in order to finish at the same time. To calculate this I did 30/1.5 which gave me 0.05. I added this onto the speed of the lead runner to get 3.3m/s :)

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3 years ago

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