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2 years ago

10. What is the acceleration of a 1000 kg car subject to a 500 N net force?

Physics

1 answer:

Simora [160]2 years ago

3 0

Answer:

0.5m/s^2

Explanation:

We can use the formula [ F = ma ] but solve for "a" since that is what we are looking for.

F = ma

F/m = a

We know the net force and mass so substitute those values and simplify.

500/1000 = 0.5m/s^2

Best of Luck!

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Mary and her younger brother Alex decide to ride the 26 ft diameter carousel at the State Fair. Mary sits on one of the horses i

hammer [34]

Answer:

$a_M=1.92a_A$

Explanation:

$\omega_M=\omega_A$ = Angular speed

$r_M$ = Distance of Mary = 11.5 ft

$r_A$ = Distance of Alex = 6 ft

Ratio of centripetal acceleration is given by

$\dfrac{a_M}{a_A}=\dfrac{\omega_M^2r_M}{\omega_A^2r_A}\\\Rightarrow \dfrac{a_M}{a_A}=\dfrac{r_M}{r_A}\\\Rightarrow a_M=a_A\dfrac{r_M}{r_A}\\\Rightarrow a_M=\dfrac{11.5}{6}a_A\\\Rightarrow a_M=1.92a_A$

Mary's centripetal acceleration is 1.92 times the centripetal acceleration of Alex

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3 years ago

Multiple-Concept Example 6 reveiws the principles that play a role in this problem. A nuclear power reactor generates 2.3 x 109

r-ruslan [8.4K]

Answer:

change in mass = 2.41*10^{8}kg

Explanation:

The change in the mass can be computed by using the relation

$E=\Delta mc^2\\\Delta m=\frac{E}{c^2}$ (1)

That is, the energy liberated comes from the mass of the nuclear fuel. The energy generated in one year is

$E=Pt=2.3*10^{9}\frac{J}{s}*1 year*\frac{365.25 day}{1 year}*\frac{24h}{1 day}*\frac{3600s}{1h}=7.25*10^{16}J$

Hence, by replacing in the equation (1) you have (c=3*10^{8}m/s)

$\Delta m=\frac{7.25*10^{16}J}{3*10^{8}\frac{m}{s}}=2.41*10^{8}kg$

HOPE THIS HELPS!!

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3 years ago

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Answer:

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garik1379 [7]

Answer:

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Explanation:

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2 years ago