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nevsk [136]
2 years ago
10

A glider with a mass of 2 kg is moving rightward at 1 m/s. A second glider, with a mass of 3 kg, is also moving rightward, at 5

m/s. After the two gliders collide, the first glider is moving rightward at 2 m/s. What is the velocity of the second glider after the collision
Physics
1 answer:
Sophie [7]2 years ago
7 0

The velocity of the second glider after the collision is 4.33 m/s rightward.

<h3>Velocity of the second glider after the collision</h3>

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

  • m₁ is mass of first glider
  • m₂ is mass of second glider
  • u₁ is initial velocity of first glider
  • u₂ is initial velocity of second glider
  • v is the final velocity of the gliders

(2)(1) + (3)(5) = (2)(2) + 3v₂

17 = 4 + 3v₂

3v₂ = 17 - 4

3v₂ = 13

v₂ = 13/3

v₂ = 4.33 m/s

Thus, the velocity of the second glider after the collision is 4.33 m/s rightward.

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

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An astronaut is a short distance away from her space station without a tether rope. She has a large wrench. What should she do w
Hitman42 [59]

Answer: b. Throw it directly away from the space station.

Explanation:

According to <u>Newton's third law of motion</u>, <em>when two bodies interact between them, appear equal forces and opposite senses in each of them.</em>  

To understand it better:  

Each time a body or object exerts a force on a second body or object, it (the second body) will exert a force of equal magnitude but in the opposite direction on the first.  

So, if the astronaut throws the wrench away from the space station (in the opposite direction of the space station), according to Newton's third law, she will be automatically moving towards the station and be safe.

3 0
3 years ago
Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
3 years ago
Kinetic energy depends on the height of an object above the ground?
s344n2d4d5 [400]
Annnnd? Ummmmmmmmmmmmmmm
5 0
3 years ago
What is the new volume of the gas if the pressure on 350 L of oxygen
Iteru [2.4K]

Answer:

420 L

Explanation:

Applying Boyle's Law,

PV = P'V'.................... Equation 1

Where P = Initial pressure, P' = Final pressure, V = Initial volume, V' = Final volume.

make V' the subject of the equation

V' = PV/P'.................... Equation 2

From the question,

Given: P = 720 mmHg, V = 350 L, P' = 600 mmHg

Substitute these values into equation 2

V' = (720×350)/600

V' = 252000/600

V' = 420 L

7 0
3 years ago
What current flows when a 35 v potential difference is imposed across a 1.4 kω resistor?
svetlana [45]

Current  =  (voltage) / (resistance)

              =    (35 volts) / (1,400 ohms)

              =         25 milliamperes 
5 0
3 years ago
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