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andreev551 [17]

3 years ago

14

a rock is tied to the end of the string and swung in a circle with a radius of 1/2 meter. if the rock goes around once in 1/10 o

f a second what is the centripetal acceleration of the rock?

1 answer:

vaieri [72.5K]3 years ago

3 0

Answer:

1974.4328 m/s²

Explanation:

r = Radius = 0.5 m

t = Time taken = 0.1 second

Rotational speed

$v=\dfrac{2\pi r}{t}\\\Rightarrow v=\dfrac{2\pi 0.5}{0.1}\\\Rightarrow v=31.42\ m/s$

The centripetal acceleration is given by

$a_c=\dfrac{v^2}{r}\\\Rightarrow a_c=\dfrac{31.42^2}{0.5}\\\Rightarrow a_c=1974.4328\ m/s^2$

The centripetal acceleration of the rock is 1974.4328 m/s²

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76.8 kPa to mm of Hg

Bumek [7]

Answer:

The answer is 576.0473

Explanation:

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3 0

3 years ago

If the AMA of the inclined plane below is 2, calculate the IMA and efficiency. IMA = Efficiency =

wlad13 [49]

Answer:

IMA = 2.5 metres

EFFICIENCY = 80%

Explanation:

The AMA of a machine is referred to as the Actual Mechanical Advantage of a machine, calculated as the ratio of the output to the input force.

The Ideal Mechanical Advantage is the ratio of the input distance to the output distance.

From the diagram, the input distance which is also the distance moved by effort = 5metres

The load distance (output distance) = 2 metres

IMA = INPUT DISTANCE / OUTPUT DISTANCE

IMA = 5metres / 2 metres = 2.5 meters

Efficiency is the ratio of AMA TO IMA

AMA = 2, IMA = 2.5

EFFICIENCY = AMA / IMA

EFFICIENCY = (2 / 2.5) × 100%= 0.8 × 100%

EFFICIENCY = 80%

5 0

3 years ago

Is this statement true or false? The next generation of nuclear power plants being built in California and South Africa are even

alexira [117]

Answer:

True

Explanation:

Modern safer and cheaper nuclear reactors can not only meet the range of our long term energy demands, they can also fight global warming.

Modern techniques provide ways to reduce radioactive waste amount. "A closed fuel cycle may be switched on for new kinds of nuclear plants. Alternatively, the waste is chemically dissuaded to transform the reusable element into fuel. This implies that nuclear waste would not be buried.

8 0

3 years ago

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

3 years ago

a man hits a golf ball (0.2kg) which accelerates at a rate of 20 m/s what amount of force acted on the ball

MAVERICK [17]

The ball only accelerates during the brief time that the club is in contact
with it. After it leaves the club face, it takes off at a constant speed.

If it accelerates at 20 m/s² during the hit, then

Force = (mass) x (acceleration) = (0.2kg) x (20 m/s²) = <em>4 newtons</em> .

8 0

3 years ago

Read 2 more answers