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2 years ago

What is the acceleration of a car that goes from 0 to 20m/s in 7 seconds?

Physics

1 answer:

nadezda [96]2 years ago

6 0

Answer:

2m/s^2

Explanation:

Clculate the acceleration:

V = u +at

20m/s = 0 + a*10s

a = 20m//10s

a = 2m/s²

From the data given , it is not possible to calculate the displacement , because no direction of motion is given

But it is possible to calculate the distance travelled

Distance = ut + ½ *a*t²

distance = 0 + ½ * 2m/s * 10²s

distance = 100m

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What is the atomic mass of an element?

Alina [70]

Atomic mass is the mass of an atom, particle, or molecule. The atomic mass is determined by the number of protons and neutrons in the atom. For example, Oxygen has 8 protons (as seen by the atomic number) and 8 neutrons which gives oxygen an atomic mass of 16.

5 0

2 years ago

A cylinder which is in a horizontal position contains an unknown noble gas at 4.63 × 104 Pa and is sealed with a massless piston

AleksandrR [38]

Answer:

The change in internal energy of the system is -17746.78 J

Explanation:

Given that,

Pressure $P=4.63\times10^{4}\ Pa$

Remove heat $\Delta U= -1.95\times10^{4}\ J$

Radius = 0.272 m

Distance d = 0.163 m

We need to calculate the internal energy

Using thermodynamics first equation

$dU=Q-W$ ...(I)

Where, dU = internal energy

Q = heat

W = work done

Put the value of W in equation (I)

$dU=Q-PdV$

Where, W = PdV

Put the value in the equation

$dU=-1.95\times10^{4}-(4.63\times10^{4}\times3.14\times(0.272)^2\times(-0.163))$

$dU=-17746.78\ J$

Hence, The change in internal energy of the system is -17746.78 J

3 0

3 years ago

Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$