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nevsk [136]
1 year 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]1 year 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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alexdok [17]

Answer:

20.25 m

Explanation:

  • <u>Centripetal acceleration </u>is given by; the square of the velocity, divided by the radius of the circular path.

That is;

         <em><u>ac = v²/r</u></em>

<em>         </em><em><u> Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the  radius, m</u></em>

Therefore;

r = v²/ac

  = 27²/36

  = 20.25 m

Hence the radius is 20.25 meters

5 0
3 years ago
Read 2 more answers
A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20 m/s by a 5620 N braking force actin
Dennis_Churaev [7]

Answer:

the distance traveled by the car is 42.98 m.

Explanation:

Given;

mass of the car, m = 2500 kg

initial velocity of the car, u = 20 m/s

the braking force applied to the car, f = 5620 N

time of motion of the car, t = 2.5 s

The decelaration of the car is calculated as follows;

-F = ma

a = -F/m

a = -5620 / 2500

a = -2.248 m/s²

The distance traveled by the car is calculated as follows;

s = ut + ¹/₂at²

s = (20 x 2.5) + 0.5(-2.248)(2.5²)

s = 50 - 7.025

s = 42.98 m

Therefore, the distance traveled by the car is 42.98 m.

5 0
2 years ago
A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0
3 years ago
What is the amount of heat required to raise the temperature of 200.0 g of aluminum by 10 c?
mash [69]
Q = 175.8J = 1.8• 10 2 J
5 0
3 years ago
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
grin007 [14]

Answer:

f_e = 1.51 cm

Explanation:

given.

magnification(m) = 400 x

focal length (f_0)= 0.6 cm

distance between eyepiece and lens (L)= 16 cm

Near point (N) = 25 cm

focal length of the eyepiece (f_e)= ?

using equation

m = -\dfrac{L-f_e}{f_o}.\dfrac{N}{f_e}

400 = \dfrac{16-f_e}{0.6}.\dfrac{25}{f_e}

9.6 = \dfrac{16-f_e}{f_e}

9.6f_e = 16-f_e

10.6 f_e = 16

f_e = 1.51 cm

3 0
3 years ago
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