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xeze [42]
3 years ago
14

Object A has of mass 7.20 kilograms, and object B has a mass of 5.75 kilograms. The two objects move along a straight line towar

d each other with velocities +2.00 meters/second and -1.30 meters/second respectively. What is the total kinetic energy of the objects after the collision, if the collision is perfectly elastic?
Physics
2 answers:
cestrela7 [59]3 years ago
8 0
Two things:
1) All collisions conserve momentum
2) Elastic collisions also conserve kinetic energy.

So basically the kinetic energy before the collision is the same as the kinetic energy after the collision.

KE = ½mv²

For object A:
KE = ½(7.20 kg)(2.00 m/s)² = 14.4 J

For object B:
KE = ½(5.75 kg)(-1.30 m/s)² = 4.86 J

total kinetic energy = 14.4 J + 4.86 J = 19.3 J
Genrish500 [490]3 years ago
7 0
The equation for Kinetic Energy in an Elastic Collision is this:
1/2(m₁v₁) + 1/2(m₂v₂) = 1/2(7.20kg)(2.00m/s) + 1/2(5.75kg)(-1.30m/s) =
3.4625m/s

I think....technically the equation above equates to the same thing except measuring the final velocity of the given masses, however you were not given that in the equation, making the first half of the full equation the only viable solution.

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If a small number of tree frogs migrate to a mat of vegetation that is already home to an established population of tree frogs a
Bingel [31]

Answer:

D

Explanation:

Gene flow is the transfer of genetic variation from one population to another

5 0
3 years ago
an object of mass 8 kg is whriled round in a vertical circle of radius 2m with a constant speed of 6m/s .Then the maximum and mi
algol13

Answer:

Maximum Tension=224N

Minimum tension= 64N

Explanation:

Given

mass =8 kg

constant speed = 6m/s .

g=10m/s^2

Maximum Tension= [(mv^2/ r) + (mg)]

Minimum tension= [(mv^2/ r) - (mg)]

Then substitute the values,

Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N

Minimum tension= [8 × 6^2)/2 -(8×9.8)]

=64N

Hence, Minimum tension and maximum Tension are =64N and 2224N respectively

5 0
2 years ago
Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center
kiruha [24]

Answer:

The angular  velocity is w_f =  1.531 \ rad/ s

Explanation:

From the question we are told that

     The mass of each astronauts is  m =  50 \ kg

      The initial  distance between the two  astronauts  d_i  =  7 \  m

Generally the radius is mathematically represented as r_i  =  \frac{d_i}{2} = \frac{7}{2}  =  3.5 \  m

      The initial  angular velocity is  w_1 = 0.5 \  rad /s

       The  distance between the two astronauts after the rope is pulled is d_f =  4 \  m

Generally the radius is mathematically represented as r_f  =  \frac{d_f}{2} = \frac{4}{2}  =  2\  m

Generally from the law of angular momentum conservation we have that

           I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}

Here I_{k_1 } is the initial moment of inertia of the first astronauts which is equal to I_{p_1} the initial moment of inertia of the second astronauts  So

      I_{k_1} = I_{p_1 } =  m *  r_i^2

Also   w_{k_1 } is the initial angular velocity of the first astronauts which is equal to w_{p_1} the initial angular velocity of the second astronauts  So

      w_{k_1} =w_{p_1 } = w_1

Here I_{k_2 } is the final moment of inertia of the first astronauts which is equal to I_{p_2} the final moment of inertia of the second astronauts  So

      I_{k_2} = I_{p_2} =  m *  r_f^2

Also   w_{k_2 } is the final angular velocity of the first astronauts which is equal to w_{p_2} the  final angular velocity of the second astronauts  So

      w_{k_2} =w_{p_2 } = w_2

So

      mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2

=>   2 mr_i^2 w_1 = 2 mr_f^2 w_2

=>   w_f =  \frac{2 * m * r_i^2 w_1}{2 * m *  r_f^2 }

=>    w_f =  \frac{3.5^2 *  0.5}{  2^2 }

=>   w_f =  1.531 \ rad/ s

       

3 0
2 years ago
The eye is actually a multiple-lens system, but we can approximate it with a single-lens system for most of our purposes. When t
hram777 [196]

Explanation:

Given that,

The optical power of the equivalent single lens is 45.4 diopters.

(a) The relationship between the focal length and the focal length is given by:

f=\dfrac{1}{P}

f=\dfrac{1}{45.4}

f = 0.022 m

or

f = 2.2 cm

(b) We need to find how far in front of the retina is this "equivalent lens" located. It is given by using lens formula as :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Here, u = infinity

\dfrac{1}{v}=\dfrac{1}{2.2}

v = 2.2 cm

So, at 2.2 cm in front of the retina is this "equivalent lens" located.

Hence, this is the required solution.

5 0
3 years ago
A chipmunk with a mass of 2 kg sits on a branch that is 10 m off the ground.
vlada-n [284]

I hope I'm not too late.

GPE = mass * gravity * height

GPE = 2 kg * 9.8 m/s * 10

GPE = 196 Joules

3 0
2 years ago
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