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3 years ago

5

What is the acceleration of a 3 kg rock that is thrown with a force of 18 N?

1 answer:

Ket [755]3 years ago

7 0

Answer:6m/s^2

Explanation:

Mass =3kg

Force=18N

Acceleration =force ➗ mass

Acceleration =18 ➗ 3

Acceleration =6

Acceleration =6m/s^2

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asambeis [7]

Answer:

$\lambda = 2.57 \times 10^{-11} m$

Explanation:

Average velocity of oxygen molecule at given temperature is

$v_{rms} = \sqrt{\frac{3RT}{M}}$

now we have

M = 32 g/mol = 0.032 kg/mol

T = 27 degree C = 300 K

now we have

$v_{rms} = \sqrt{\frac{3(8.31)(300)}{0.032}$

$v_{rms} = 483.4 m/s$

now for de Broglie wavelength we know that

$\lambda = \frac{h}{mv}$

$\lambda = \frac{6.6 \times 10^{-34}}{(5.31\times 10^{-26})(483.4)}$

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3 years ago

A 30 ky child running at 7 m/s jumps onto a 10 kg sled which was initially at rest. What will be the

Ede4ka [16]

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3 years ago

A 35.4 kg girl is riding on the back of a 15.23 kg cart. the cart and the kid are both moving eastward at 4.25 m/s when she step

Grace [21]

Answer:

The final velocity of the cart is $v_c = 7.02 \ m/s$

Explanation:

From the question we are told that

The mass of the girl is $m_g = 35.4 \ kg$

The mass of the cart is $m_c = 15.23 \ kg$

The speed of the cart and kid(girl) is $v = 4.25 \ m/s$

The final velocity of the girl is $v_g = 3.06 \ m/s$

Let assume that velocity eastward is positive and velocity westward is negative (Note that if we assume vise versa it wouldn't affect the answer )

The total momentum of the system before she steps off the back of the cart

is mathematically evaluated as

$p__{T1}} = (m_g + m_c) * v$

substituting values

$p__{T1}} = (35.4 + 15.23) * 4.25$

$p__{T1}} =215.17 \ kg m /s$

The total momentum after she steps off the back of the cart is mathematically evaluated as

$p__{T2}} = (m_g * v_g ) +( m_c * v_c )$

Where $v_c$ is the final velocity of the cart

substituting values

$p__{T2}} = (35.4 * 3.06 ) +( 15.23 * v_c )$

$p__{T2}} = 108. 324 + 15.23 v_c$

Now according to the law of conservation of momentum

$p__{T1}} =p__{T2}}$

So

$215.17 \ kg m /s = 108. 324 + 15.23 v_c$

=> $v_c = 7.02 \ m/s$

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3 years ago

Sports car goes from a velocity of zero to a velocity of 12 m/s East in two seconds. What is the cars acceleration?

bogdanovich [222]

Answer:

6 m/sec

Explanation:

12/2=6

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3 years ago

Kinetic energy varies jointly as the mass and the square of the velocity. A mass of 1515 grams and velocity of 77 centimeters pe

Alexus [3.1K]

Answer:

The second kinetic energy is 162 J.

Explanation:

Given that,

Mass, $m_1=15\ g$

Velocity, $v_1=7\ cm/s$

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Mass, $m_2=10\ g$

Velocity, $v_2=9\ cm/s$

We need to find kinetic energy $K_2$ . Kinetic energy is given by :

$K=\dfrac{1}{2}mv^2$

So,

$\dfrac{K_1}{K_2}=\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}\\\\K_2=\dfrac{K_1}{\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}}\\\\K_2=\dfrac{147}{\dfrac{15}{10}\times \dfrac{7^2}{9^2}}\\\\K_2=162\ J$

So, the second kinetic energy is 162 J.

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3 years ago