Which year was pluto no longer considered a planet?.

Find the right answer please

Ghella [55]

Answer:

3.78

Explanation:

bc ik it is :3

5 0

3 years ago

A amusement park moves riders in a circle at a rate of 6.0m/s if the radius is 9.0 meters what is the acceleration of the ride

oksano4ka [1.4K]

Centripetal acceleration is (speed-squared) / (radius)

CA = (6 m/s)² / (9 m)

CA = (36 m²/s²) / (9 m)

CA = (36/9) (m²/m·s²)

<em>Centripetal acceleration = 4 m/s²</em>

4 0

3 years ago

A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they

Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁ + m₂u₂

P₁ = (1000 x 25) + (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(1000 x 25) + (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

= 27.727(1000 + 1200)

= 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

8 0

2 years ago

A satellite is launched to orbit the Earth at an altitude of 3.25 107 m for use in the Global Positioning System (GPS). Take the

Korolek [52]

Answer:Orbital period =21.22hrs

Explanation:

given that

mass of earth M = 5.97 x 10^24 kg

radius of a satellite's orbit, R= earth's radius + height of the satellite

6.38X 10^6 + 3.25 X10^7 m =3.89 X 10^7m

Speed of satellite, v= $\sqrt GM/R$

where G = 6.673 x 10-11 N m2/kg2

V= \sqrt (6.673x10^-11 x 5.97x10^ 24)/(3.89 X 10^ 7m)

V =10,241082.2

v= 3,200.2m/s

a) Orbital period

$\sqrt GM/R$ = $\frac{2\pi r}{T}$

V= $\frac{2\pi r}{T}$

T= 2 $\pi$ r/ V

= 2 X 3.142 X 3.89 X 10^7m/ 3,200.2m/s

=76,385.1 s

60 sec= 1min

60mins = 1hr

76,385.1s =hr

76,385.1/3600=21.22hrs

3 0

3 years ago

Jeremy ran a race at a speed of 15 m/s in a time of 8 seconds. Calculate the distance he ran.

Masja [62]

Answer:

120 m/s if m/s means miles per second than

Explanation:

15 × 8

8 0

2 years ago

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