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Tanya [424]
2 years ago
9

A 0.24 kg mass with a speed of 0.60 m/s has a head-on collision with a 0.26 kg mass that is traveling in the opposite direction

at a speed of 0.20 m/s. Assuming that the collision is perfectly inelastic, what is the final speed of the combined masses?
Physics
1 answer:
8_murik_8 [283]2 years ago
4 0

The  final speed of the combined masses is 0.186m/s

According to the law of collision, the sum of the momentum of the bodies before the collision is equal to the momentum after the collision.

Mathematically;

m1u1 - m2u2 = (m1+m1)v

  • v is the final speed of the combined masses.

Substituting the given parameters;

0.24(0.6) - 0.26(0.2) = (0.24+0.26)v

0.144 - 0.052 = 0.5v

0.092 = 0.5v

v = 0.092/0.5

v = 0.184m/s

Hence the final speed of the combined masses is 0.186m/s

Learn more on collision here:  brainly.com/question/7538238

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Reil [10]
To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.

The formula for centripetal acceleration is:
Centripetal Acceleration = v squared divided by r

where:
v = velocity, m/s
r= radium, m

assuming the velocity does not change:

at the top of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 0.25 m
                                      = 81 m/s^2

at the bottom of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 1.25 m
                                      = 16.2 m/s^2

to find the change in centripetal acceleration, take the difference of the two.
change in centripetal acceleration = centripetal acceleration at the top of the hill - centripetal acceleration at the bottom of the hill

= 81 m/s^2 - 16.2 m/s^2
= 64.8 m/s^2 or 65 m/s^2
6 0
3 years ago
A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta
larisa [96]

Answer:

Part a)

x = 15.76 m

Part b)

y = 7.94 m

Part c)

x = 26.16 m

Part d)

y = 7.49 m

Part e)

x = 83.23 m

Part f)

y = -75.6 m

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

v_x = 19.9 cos39.9

v_x = 15.3 m/s

similarly we have

v_y = 19.9 sin39.9

v_y = 12.76 m/s

Part a)

Horizontal displacement in 1.03 s

x = v_x t

x = (15.3)(1.03)

x = 15.76 m

Part b)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.03) - 4.9(1.03)^2

y = 7.94 m

Part c)

Horizontal displacement in 1.71 s

x = v_x t

x = (15.3)(1.71)

x = 26.16 m

Part d)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.71) - 4.9(1.71)^2

y = 7.49 m

Part e)

Horizontal displacement in 5.44 s

x = v_x t

x = (15.3)(5.44)

x = 83.23 m

Part f)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(5.44) - 4.9(5.44)^2

y = -75.6 m

6 0
2 years ago
Two identical balls move directly toward each other with equal speeds. how will the balls move if they collide and stick togethe
Citrus2011 [14]
The momentum of both the identical balls would eventually be transferred to one another when it comes to a point wherein they will collide. In addition, the phenomenon is called an elastic collision wherein both the momentum and energy of the system would considered to be conserved.
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3 years ago
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Mekhanik [1.2K]

Answer:

c. probablistic view of nature.

Explanation:

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Then with the help of classical mechanics it can be prove that the particle can not cross the barrier but according to the quantum mechanics, there is a small but a finite probability to cross the barrier.

Therefore by the above discussion it can be concluded that quantum mechanics can be thought as a probablistic view of nature.

7 0
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Luda [366]

Answer:

mesa

Explanation:



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5 0
2 years ago
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