Physics help! I really need this homework done:

zloy xaker [14]

The Mercury's mass for the given acceleration due to gravity is 0.3152 x 10²⁴ kg.

The ratio of the calculated and accepted value of the Mercury's mass is 0.95.

Mass is the amount of matter present in the object.

The mass of the object is always constant, anywhere it is on the Earth or Moon or any other planet.

Given is the acceleration due to gravity of Mercury planet at North pole is g = 3.698 m/s² and the radius of Mercury planet is 2440 km.

The acceleration due to gravity is related with mass as

g = GM/R²

Substitute the values, we have

3.698 = 6.67 x 10⁻¹¹ x M/(2440 x1000)³

M = 2.2016 x 10¹³ / 6.67 x 10⁻¹¹

M = 0.3152 x 10²⁴ kg

Thus, the mercury's mass is 0.3152 x 10²⁴ kg.

(b) Accepted value of Mercury's mass is 3.301 x 10²³ kg

Ratio of the value of mass calculated and accepted is

Mcalc/M accep = 0.3152 x 10²⁴ kg / 3.301 x 10²³ kg

= 0.95

Thus, the ratio is 0.95

Learn more about mass.

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Which theory of language development and acquisition includes the concepts of reinforcement, stimuli control, deprivation, and a

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A skinner idea of language .-.

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4 years ago

How much work is required to accelerate a proton from rest up to a speed of 0.993 c? Express your answer with the appropriate un

Keith_Richards [23]

Answer:

(a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

Explanation:

Given that,

Speed = 0.993 c

We need to calculate the work done

Using work energy theorem

The work done is equal to the kinetic energy relative to the proton

$W=K.E$

$W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2$

Put the value into the formula

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=1.122\times10^{-9}\ J$

$W=7001\ MeV$

(b). We need to calculate the momentum of this proton

Using formula of momentum

$p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value into the formula

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}$

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}$

$p=1.404\times10^{-26}c$

$p=4.20\times10^{8}\ kg-m/s$

Hence, (a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

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3 years ago

particle moves over three quarters of a circle of radius R what is the magnitude of its displacement

fomenos

Answer:

is equal to threee idont know

Explanation:

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sanay akoy pag padendyahan niyo nay. hahahahhahahaha maawa kayo saakin mga aking mahal hahahhahahahahhaha hahah en then go char hahahhaa

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3 years ago

The force that pulls planets toward the Sun is called _____. nuclear fusion gravity magnetism differentiation

Alona [7]

The pair of equal forces that pull a planet and the Sun
toward each other are gravitational forces.

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3 years ago

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