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2 years ago

6

Meher is riding a bicycle on a slope. explain the different motions taking place during this time

1 answer:

gtnhenbr [62]2 years ago

7 0

Answer:

Mehar cant ride down the slope

Explanation:

She does not has a bicycle

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What is first class lever ? can you pease say where is does the fulcrum ,load and effort

valkas [14]

Answer:

In first class lever the fulcrum lies between the effort and the load the effort lies on the left side and load lies on the right side .

Explanation:

hope this helps

7 0

3 years ago

Which best describes the relationship between humidity and air pressure?

spayn [35]

Answer:<span>Humid air is lighter, so it has lower pressure.

The reason is the molecules of water are H2O, whose molar mass is 18 g/mol.

These molecules displaces molecules of N2 and O2, whose molar masses are:

N2: 2*14g/mol = 28 g/mol, and
O2: 2*16g/mol = 32 g/mol.

Then molecules of 28g/mol and 32 g/mol are being replaced with molecules of 18g/mol, leading to a lower weight of the same volume of air, which results in lower pressure.
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5 0

3 years ago

Read 2 more answers

A 1090 kg car moving +11.0 m/s

stiks02 [169]

The mass of the second car is 1434.21 kg

<u>Explanation:</u>

Using law of conservation of momentum,

$m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v$

Given:

$m_{1}$ = 1090 kg

$u_{1}$ = 11 m/s

$u_{2}$ = 0

v = 4.75 m/s

We need to find $m_{2}$

When substituting the given values in the above equation, we get

$(1090 \times 11)+\left(m_{2} \times 0\right)=\left(1090+m_{2}\right) 4.75$

$11990=5177.5+4.75 m_{2}$

$4.75 m_{2}=11990-5177.5$

$4.75 m_{2}=6812.5$

$m_{2}=\frac{6812.5}{4.75}=1434.21 \mathrm{kg}$

6 0

2 years ago

The velocity selector in in a mass spectrometer consists of a uniform magnetic field oriented at 90 degrees to a uniform electri

NeTakaya

Answer:

50k/h is the answer to iy

3 0

3 years ago

A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres

Lemur [1.5K]

Answer:

$V_2 = 0.125 m^3$

Work done = = 5 kJ

Explanation:

Given data:

volume of nitrogen $v_1 = 0.08 m^3$

$P_1 = 150 kPa$

$T_1 = 200 degree celcius = 473 Kelvin$

$P_2 = 80 kPa$

Polytropic exponent n = 1.4

$\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}$

putting all value

$\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}$

$\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS$

polytropic process is given as

$P_1 V_1^n = P_2 V_2^n$

$150\times 0.08^{1.4} = 80 \times V_2^{1.4}$

$V_2 = 0.125 m^3$

work done $= \frac{P_1 V_1 -P_2 V_2}{n-1}$

$= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}$

= 5 kJ

4 0

3 years ago