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Dvinal [7]
3 years ago
14

A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second2. What is the net force on the car

Physics
1 answer:
Novay_Z [31]3 years ago
0 0

Answer:

4.5 *10^{3}\ kgm/s^{2}

Explanation:

Given:

Mass= 1.00 * 10^{3} \ kilograms,\\Acceleration =4.5\  meters/second^{2}

As we know that

Force=Mass * Acceleration

Putting the value of mass and Acceleration  we get ,

Force\  = \ 1.00 * 10^{3} * 4.5\ m/s^{2}

Force\  =4.5 * \ 10^{3} \  kgm/s^{2}

Therefore Net force is :4.5 *10^{3}\ kgm/s^{2}

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The work function (energy needed to remove an electron) of gold is 5.1 eV. Two pieces of gold (at the same potential) are separa
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Answer:

The value of L is 0.3 nm.

Explanation:

Given that,

Energy \phi= 5.1 eV

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We need to calculate the value of L

We know that,

Formula of tunneling probability

T=Ge^{-2kL}....(I)

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\phi = E-U

Put the value in equation (II)

k=\dfrac{\sqrt{2\times9.1\times10^{-31}\times5.1\times1.6\times10^{-19}}}{1.055\times10^{-34}}

k=1.155\times10^{10} m^{-1}

From equation (I)

ln T=-2kLG

L=\dfrac{ln T}{-2kG}

L=\dfrac{ln 10^{-3}}{-2\times1.155\times10^{10}\times1}

L=2.99\times10^{-10}\ m

L=0.299\times10^{-9}\ m

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7 0
3 years ago
Please help... <br>What forms of energy are present in lighting a table lamp?​
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The acceleration due to gravity is lower on the Moon than on Earth. Which one of the following statements is true about the mass
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Mass is the same, weight is less.

Explanation:

We know the following equation

W = mg

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You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.
Alja [10]

Answer:

Part a)

M = 7.25 \times 10^{25} kg

Part b)

M = 2\times 10^{30} kg

Explanation:

Part a)

As we know that the diameter of the planet is given as

d = 1.8 \times 10^7 m

now the radius of the planet is given as

r = 9 \times 10^6 m

now we know that the acceleration due to gravity of the planet is given by the equation

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here we know that

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now from above equation we have

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Part b)

Now by kepler's law we know that

time period of revolution of planet about the star is given as

T = 2\pi \sqrt{\frac{r^3}{GM}}

so we have

\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}

now we have

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mula of time period

402\times (365\times 24 \times 3600) = 2\pi \sqrt{\frac{(8.17\times 10^12)^3}{(6.67\times 10^{-11})M}}

Now we have

M = 2\times 10^{30} kg

7 0
3 years ago
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