Two objects with different masses collide and bounce back after an elastic collision. Before the collision, thetwo objects were moving at velocities equal in magnitude but opposite in direction. After the collision,
A. the less massive object had gained momentum.
B. the more massive object had gained momentum.
C. both objects had the same momentum.
D. both objects lost momentum.

Answers

Answer 1

Answer:

The more massive object will lose momentum after collision while the less massive object will gain momentum after collision.

Let the mass of the first object = m₁
Let the mass of the second object = m₂
let the initial velocities of the two objects = u
let the final velocity of the first object after collision = v₁
Let the final velocity of the second object after collision = v₂

Apply theprinciple of conservation of linear momentum for elastic collision;

let the heavier object = m₁

m₁u + m₂(-u) = m₁(-v₁) + m₂v₂

m₁u - m₂u = -m₁v₁ + m₂v₂

where;

m₁u and m₂u are initial momentum of both objects before collision

m₁v₁ and m₂v₂ are final momentum of both objects after collision

Thus, from the equation above we can conclude the following, the more massive object will lose momentum after collision while the less massive object will gain momentum after collision.

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Answer 2

Answer:

Answer:A

Explanation:

Given

mass of two objects are $m_1$ and $m_2$

taking $m_1>m_2$

Suppose $u_1$ and $u_2$ are the velocities of $m_1$ and

$u_1=u$ and $u_2=-u$

therefore after elastic collision velocity of $m_1$ and $m_2/[tex] are [tex]v_1$ and $v_2$

$v_1=(m_1-m_2)/(m_1+m_2)\cdot u_1+(2m_2)/(m_1+m_2)\cdot u_2$

$v_2=(2m_1)/(m_1+m_2)\cdot u_1-(m_1-m_2)/(m_1+m_2)\cdot u_2$

for $u_1$ and $u_2$

$v_1=(m-m)/(m+m)\cdot u+(2m)/(m+m)\cdot (-u)$

$v_1=(m_1+m_2)/(m_1+m_2)\cdot u=u$

$v_2=(2m)/(m+m)\cdot u-(m-m)/(m+m)\cdot (-u)$

$v_2=(3m_1-m_2)/(m_1+m_2)\cdot u$

$v_2=u+2\cdot (m_1-m_2)/(m_1+m_2)\cdot u$

so velocity of mass $m_2$ is more as compared to $m_1$

so less massive object gained some momentum

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Dua buah vektor, yaitu P= 6i-5j+2k dan Q = 2i+2j+8k. Tentukan vektor R agar P-Q+R=0. Hitung pula besar vektor R?

Answers

As it is given to us

$P = 6i - 5j + 2k$

$Q = 2i + 2j + 8k$

also it is given that

P - Q + R = 0

so here we can rearrange it to find the value of R

$R = Q - P$

so here we have

$R = (2i + 2j + 8k) - (6i - 5j + 2k)$

$R = -4\hat i + 7\hat j + 6\hat k$

so the vector is given by above equation

Answer:

R= -4 i + 7 j + 6 k

Explanation:

Being P - Q + R = 0, solving for R you get:

R= Q - P

You know:

P=6 i - 5 j + 2 k
Q=2 i + 2 j + 8 k

Replacing the expressions of P and Q you get:

R= (2 i + 2 j + 8 k) - (6 i - 5 j + 2 k)

i, j and k indicate that the values correspond to the x, y and z components respectively. To subtract the vectors, their respective components are subtracted from each vector. So in this case:

R= (2-6) i + [2-(-5)] j + (8-2) k

So the answer is:

R= -4 i + 7 j + 6 k

Movement that occurs along a straight line is called:

Answers

linear motion! hope i helped

Make a island with all renewable energy soursws and how it works and about it but at year 9 level

Answers

Answer:

How are we expected to make an island

Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first block is pulled with a force of 100 N. What is the tension in the string connecting block 100 to block 99? What is the tension in the string connecting block 50 to block 51?

Answers

We have that for the Question "What is the tension in the string connecting block 100 to block 99? What is the tension in the string connecting block 50 to block 51?"

it can be said that

The tension in the string connecting block 100 to block 99 = $1N$

The tension in the string connecting block 50 to block 51 = $50N$

From the question we are told

Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first block is pulled with a force of 100 N.

Assuming mass of each block is 1 kg

The equation for the force is given as

$F = ma\n\na = (F)/(m)\n\n = (100)/(100*1)\n\n = 1 m/s^2$

Now, between block 100 and 99,

$F = ma\n\nF = 1*1 \n\n= 1 N$

Now between block 50 and 51. There are 50 blocks behind 51 st block, so,

$m = 50 Kg\n\na = 1 m/s2 (assuming all blocks accelerate at same rate)\n\nF = 50 * 1\n\nF= 50 N$

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Answer:

The tension in the string connecting block 50 to block 51 is 50 N.

Explanation:

Given that,

Number of block = 100

Force = 100 N

let m be the mass of each block.

We need to calculate the net force acting on the 100th block

Using second law of newton

$F=ma$

$100=100m* a$

$ma=1\ N$

We need to calculate the tension in the string between blocks 99 and 100

Using formula of force

$F_(100-99)=ma$

$F_(100-99)=1$

We need to calculate the total number of masses attached to the string

Using formula for mass

$m'=(100-50)m$

$m'=50m$

We need to calculate the tension in the string connecting block 50 to block 51

Using formula of tension

$F_(50)=m'a$

Put the value into the formula

$F_(50)=50m* a$

$F_(50)=50*1$

$F_(50)=50\ N$

Hence, The tension in the string connecting block 50 to block 51 is 50 N.

All nuclear energy results in the rapid release of energy, such as in atomic bombs. a. True
b. False

Answers

It is true that all nuclear energy results inthe rapid release of energy, such as in atomic bombs. A process wherein the atom’s nucleusis further divided into two nuclei, that is, fission products, is callednuclear fission. This happens when the nuclei of radioactive elements (e.g.Uranium or Plutonium isotopes) are able to attract neutrons. Because theseisotopes’ nuclei are unstable, an addition to their energy causes them todivide into new pieces of neutron. In nuclear reactors, fission occurs when thepower plant gathers heat from a steam in a physical process. Nuclear reactorsare designed to sustain nuclear chain reactions.

Answer:

FALSE

Explanation:

verified answer***

Nuclear energy is released from an atom through one of two processes: nuclear fusion or nuclear fission. In nuclear fusion, energy is released when the nuclei of atoms are combined or fused together. But isn't rapidly release

If the stone loses 10% of its speed in 10 s of grinding, what is the force with which the man presses the knife against the stone?

Answers

w1=200rpm*2pi/60=20.9rad/s

w2=200rpm-.10(200rpm)=180rpm

180rpm*2pi/60=18.8rad/s

a=(18.8-20.9)/10= -.21rad/s^2

I=1/2(28)(.15)^2
I=.315kg*m^2
Torque= -.21(.315)=.06615 N*m
Friction force= -.06615/.15=-.441 N
Normal force of man=.441/.2= 2.205 N
Final answer: -2.205 N

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