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2 years ago

15

an object on a planet has a mass of 243 Kg. what is the acceleration of the object, if the radius of the planet is 2.32 x 10^7m

and the mass of the planet is 6.35 x 10^30 Kg estimate g as 6.67 x 10^-11 N (m/kg)

1 answer:

Ksenya-84 [330]2 years ago

3 0

Answer:

7.87x10^5m/s^2

Explanation:

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Write the examples of unit

OLga [1]

Answer:

The definition of a unit is a fixed standard amount or a single person, group, thing or number. An example of a unit is a single apartment in an apartment building

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2 years ago

A disk with mass 1.64 kg and radius 0.61 meters is spinning counter-clockwise with an angular velocity of 17.6 rad/s. A rod of m

Masja [62]

Answer:

The loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

Given;

mass of the disk, m₁ = 1.64 kg

radius of the disk, r = 0. 61 m

angular velocity of the disk, ω₁ = 17.6 rad/s

mass of the rod, m₂ = 1.51 kg

length of the rod, L = 1.79 m

angular velocity of the rod, ω₂ = 5.12 rad/s (clock-wise)

let the counter-clockwise be the positive direction

let the clock-wise be the negative direction

The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;

m₁ω₁ + m₂ω₂ = ωf(m₁ + m₂)

where;

ωf is the common final angular velocity

1.64 x 17.6 + 1.51(-5.12) = ωf(1.64 + 1.51)

21.1328 = ωf(3.15)

ωf = 21.1328 / 3.15

ωf = 6.709 rad/s

The moment of inertia of the disk is calculated as follows;

$I_{disk} = \frac{1}{2} mr^2\\\\I_{disk} = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk} = 0.305 \ kgm^2$

The moment of inertia of the rod about its center is calculated as follows;

$I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2$

The initial rotational kinetic energy of the disk and rod;

$K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i= \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \ \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J$

The final rotational kinetic energy of the disk-rod system is calculated as follows;

$K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J$

The loss in rotational kinetic energy due to the collision is calculated as follows;

$\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J$

Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.

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2 years ago

5. Name one career in the health and fitness field that interests you.

BartSMP [9]

Answer:

I've always wanted to be a nurse. growing up my auntie used to show me a lot of stuff about becoming a nurse since she us one.

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1 year ago

19,792,000,000 in scientific notation will have how many significant figures

xz_007 [3.2K]

Answer:

= 1.9792 × 10^10

Significant Figures= 5

Explanation:

Look at the attachment below

Hope this helps (:

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3 years ago

video for A door 1 m wide, of mass 15 kg, is hinged at one side so that it can rotate without friction about a vertical axis. It

natka813 [3]

Answer:

$\omega_f = 0.4\ rad/s$

Explanation:

given,

width of door dimension = 1 m

mass of door = 15 Kg

mass of bullet = 10 g = 0.001 Kg

speed of bullet = 400 m/s

$I_{total} =I_{door} + I_{bullet}$

$I_{total} =\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2$

a) from conservation of angular momentum

$L_i = L_f$

$mv\dfrac{W}{2} = I_{total}\omega_f$

$mv\dfrac{W}{2}= (\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f$

$\omega_f= \dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}$

$\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}$

$\omega_f= \dfrac{\dfrac{0.01\times 400}{2}}{\dfrac{15\times 1 }{3}+(\dfrac{0.01\times 1}{4})}$

$\omega_f = 0.4\ rad/s$

8 0

3 years ago