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Arte-miy333 [17]
2 years ago
7

To understand the vector nature of momentum in the case in which two objects collide and stick together. In this problem we will

consider a collision of two moving objects such that after the collision, the objects stick together and travel off as a single unit. The collision is therefore completely inelastic. You have probably learned that "momentum is conserved" in an inelastic collision. But how does this fact help you to solve collision problems?
Physics
2 answers:
ludmilkaskok [199]2 years ago
5 0

Answer:

Answer : The momentum is equal to the momentum of object 1 plus the momentum of object 2.

When masses collide with each other and stick together, the collision is said to be inelastic. The kinetic energy before collision and after collision are not equal. For the objects to stick together, they move in the same direction. Therefore the momentum is equal to the momentum of object 1 plus the momentum of object 2.

Explanation:

Plz give me brainliest

vlada-n [284]2 years ago
4 0

Answer:

Applying the law's theory and utilizing the equation of momentum ie. p=mv

Explanation:

The law of conservation of linear momentum states that the momentum in a <em>closed</em> system remains constant. Because a collision is inelastic, this proves that the system is closed. So the equation of momentum is p=mv, p is momentum, m is mass and v is velocity.

Because the momentum is conserved, the momentum (p) before the collision should be equal to the p after the collision, so we can equate them and solve for the unknown:

p=m.v

p(before) = p(after)

m(before) x v(before) = m(after) x v(after)

using this equation, you solve it and this helps you solve collision problems.

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The eiffel tower has a mass of 7.3 million kilograms and a height of 324 meters. its base is square with a side length of 125 me
uranmaximum [27]

Since the tower base is square with a side length of  125 m,

Therefore,

(125\ m)^2+ (125\ m)^2=31250 m^2

Square root of 31250 = 176.776953 (Diameter) , so this is the diameter of the cylinder to enclose it, and radius, r = 88.38834765 m and height, h = 324 m.

The volume of cylinder,

=\pi r^2h=3.14(88.38834765 m)^2\times 324 m =7948168.803\ m^3

Thus, the mass of the air in the cylinder,

=1.225\ kg/m^3 \times 7948168.803\ m^3=9736506.78\ kg

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4 0
1 year ago
Given that the CFL bulb is rated at 10% efficiency, if the LED bulb consumes half the amount of electrical power, for the same a
Butoxors [25]

Answer and explanation:

The efficiency of the LED bulb is 20% while The efficiency of the Inc bulb is 1%.

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\epsilon_{CFL}=\frac{L_0}{P_0}=0.1

Therefore:

\epsilon_{led}=\frac{L_0}{0.5P_0}=2 \cdot0.1=0.2\\\epsilon_{INC}=\frac{L_0}{10P_0}=0.1 \cdot 0.1=0.01

4 0
1 year ago
8. Rubbing a plastic bag and a balloon with a cloth gives both objects a net negative charge. The balloon's
Dafna1 [17]

Answer:

0.214 m

Explanation:

In order for the bag to levitate and not fall down, the electrostatic force between the bag and the balloon must balance the weight of the bag.

Therefore, we can write:

k\frac{q_1 q_2}{r^2}=mg

where

k is the Coulomb constant

q_1=-1\cdot 10^{-10}C is the charge on the balloon

q_2=-1\cdot 10^{-5} C is the charge on the bag

r is the separation betwen the bag and the balloon

m=0.02 g=2\cdot 10^{-5} kg is the mass of the bag

g=9.8 m/s^2 is the acceleration due to gravity

Solving for r, we find the distance at which the bag must be held:

r=\sqrt{\frac{kq_1 q_2}{mg}}=\sqrt{\frac{(9\cdot 10^9)(-1\cdot 10^{-10})(-1\cdot 10^{-5})}{(2\cdot 10^{-5})(9.8)}}=0.214 m

5 0
1 year ago
Solve A and B using energy considerations.
Alisiya [41]
 <span>Use the kinematic equation vf^2 = vi^2 + 2ad where; 
vf = ? 
vi = 0 m/s 
a = 9.8 m/s^2 
d1 = 10 m 
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final velocity at the ground (d1): vf = sqrt(2)(9.8)(10) = 14 m/s 
final velocity to the bottom of the cliff (d2): vf = sqrt(2)(9.8)(25) = 22.14 m/s 
</span>
7 0
2 years ago
A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it
makkiz [27]

Answer:D

Explanation:

Given

mass of object m=5 kg

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velocity acquired v=12 m/s

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

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Final Energy=\frac{1}{2}mv^2+W_{f}

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where W_{f} is friction work if any

490=360+W_{f}

W_{f}=130 J

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

4 0
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