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max2010maxim [7]
10 months ago
8

For each question, select the right answer from the choices below:

Physics
1 answer:
Leni [432]10 months ago
3 0

Option (ii) B is the correct option. The object on the moon has greater mass.

To resolve this, utilize the formulas Force = Mass * Acceleration.

The equation can be used to find the mass given the force in Newtons, using 9.8 m/s² for the acceleration of gravity of the earth and 1.6 m/s² for the moon.

Calculating the mass on earth:

30 N = 9.8 m/s² * mass

This results in a mass of 3.0 kg for the object on Earth.

Calculating the mass of the moon:

30 N = 1.6 m/s²2 * mass

Thus, the moon's object has a mass of 19. kg.

This can be explained by the fact that the earth has a stronger gravitational pull than the moon, producing more force per kilogram of mass. As a result, the moon's mass must be bigger to produce the same amount of force at a lower acceleration from gravity (1.6 m/s² vs. 9.8 m/s²).

To know more about Mass, refer to this link :

brainly.com/question/13386792

#SPJ9

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Two students, Student X and Student Y, stand on a long skateboard that is at rest on a flat, horizontal surface, as shown. In or
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Answer:

the answer is B.

Explanation:

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3 years ago
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A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica
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Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed \omega =40rad/s

mass of m_2=0.8 kg dropped at r=0.3 m from center

let \omega _2 be the final angular velocity of cylinder

Conserving Angular momentum

L_1=L_2

\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2

\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2

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3 0
3 years ago
How much gas is there in 100cm3 of air?
sweet [91]
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3 years ago
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You have two vectors, which are 2.59 m at 30.0° north of east and 4.18 m at 60.0° north of west. What is the magnitude in meters
andrew11 [14]

Answer:\ec{r}=0.153\hat{i}+4.914\hat{j}v

Explanation:

Given

Vector 1

\vec{a}=2.59\left ( cos30\hat{i}+sin30\hat{j}\right )

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Resultant \vec{r}=\vec{a}+\vec{b}

\vec{r}=2.59\left ( cos30\hat{i}+sin30\hat{j}\right )+4.18\left ( -cos60\hat{i}+sin60\hat{j}\right )

\vec{r}=0.153\hat{i}+4.914\hat{j}

|r|=4.916

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along the lead, at right angles to the lead at the middle, and at right angles to the lead at one end.

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