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4 years ago

Which of the following statement describes an actual orbit

2 answers:

7 0

I suppose right answer is d because staellite means an object that move around the larger object and Jupiter also moves around the Sun

Papessa [141]4 years ago

6 0

Answer:

The Moon is a satellite of Earth

Explanation:

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Connecting many devices in a single socket does not affect the flow of current in a

eimsori [14]

Answer:- False

Explanation:-

False because it can leads to overloading and further to short circut.

7 0

3 years ago

How do you add scientific notation if the problem is like this: 3.72 x 10^9 + 5.46 x 10^8?

Crazy boy [7]

You can only add or subtract numbers in scientific notation if they have
the same power of 10. If they're different, then you have to change one
to match the other one.

Here are both ways to do your example:

3.72x 10^9 = 37.2x10^8

(37.2 x 10^8)+(5.46 x 10^8) = (37.2+5.46) x 10^8 = 42.66x10^8 = 4.266x10^9

=======================

5.46 x 10^8 = 0.546 x 10^9

3.72 x 10^9 + 0.546 x 10^9 = (3.72 + 0.546) x 10^9 = 4.266 x 10^9

3 0

4 years ago

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a

Pie

Answer:

Part a)

$v = 7407.1 m/s$

Part b)

$v_{rel} = 1.05 \times 10^4 m/s$

Explanation:

Part a)

As we know that orbital velocity at certain height from the surface of Earth is given as

$v = \sqrt{\frac{GM}{R+h}}$

here we know that

$M = 5.98 \times 10^{24} kg$

$R = 6.37 \times 10^6 m$

$h = 900 km = 9.0 \times 10^5 m$

now we have

$v = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9.0 \times 10^5}}$

$v = 7407.1 m/s$

Part b)

When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as

$v_{rel} = \sqrt{2} v$

$v_{rel} = 1.05 \times 10^4 m/s$

6 0

3 years ago

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The coefficient of static friction between t

Wewaii [24]

Answer:

a) a = 6.1 m/s^2

b) a = 0.98m/s^2

Explanation:

Mass of slab = 40kg

Mass of block = 10kg

Coefficient of static friction (Us) = 0.60

Kinetic coefficient (UK) = 0.40

Horizontal force = 100N

The normal reaction from 40kg slab on 10 kg block = 10*9.81

= 98.1N

Static frictional force = Us*R

= 98.1*0.6

= 58.86N

This is less than the force applied

If 10 kg block will slide on the 40 kg slab, net force = 100 - kinetic force

Kinetic force (Uk*R) = 0.4*98.1

= 39.28N

= 39N

Net force = 100 -39

= 61N

Recall that F = ma

For 10 kg block

a = F/m

a = 61/10

a = 6.1m/s^2

b) Frictional force on 40 kg slab by 10 kg = 98.1*0.4

= 39.24

= 39N

F = ma

a = F/m

For 40kg slab

a = 39/40

a = 0.98m/s^2

3 0

3 years ago

023 (part 1 of 2) 10.0 points

Annette [7]

Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

$\Delta KE_{rotational} = \dfrac{1}{2} \cdot I \cdot \omega ^2 = \dfrac{1}{2} \cdot m \cdot v ^2$

I = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

$\Delta KE_{rotational}$ = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

3 0

3 years ago