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3 years ago

You drop a ball from a height of 1.7 m, and it bounces back to a height of 1.2 m.

Physics

1 answer:

leva [86]3 years ago

8 0

Answer:

A) ΔEm = 0.29, B) v₁ = 5.8 m/s, c) v₂= 4.9 m / s D) the correct answer from 4

Explanation:

For this exercise we will use conservation of energy, taking care of how to choose our system

A) For this case we take two instants

starting point. When the ball goes out

Em₀ = U = m g y₁

Final point. When the ball reaches its maximum height

$Em_{f}$ = U = m g y₂

In this case we see that there is a loss of mechanical energy at the moment of rebound, therefore the fraction of energy lost is

ΔEm = Em_{f} / Em₀

ΔEm = mg y₂ / mg y₁

ΔEm = y₂ / y₁

ΔEm = 1.2 / 1.7

the lost part of energy

ΔEm = 1 -0.706

ΔEm = 0.29

B) the velocity just before the bounce

starting point. When the ball is released

Em₀ = U = m g y₁

final punot. Just wide of the bounce

Em_{f} = K = ½ m v₁²

As it has not yet rebounded, it has no energy loss, therefore the mechanical energy is conserved

Em₀ = Em_{f}

m g y₁ = ½ m v₁²

v₁ = √ 2 g y₁

let's calculate

v₁ =√ (2 9.8 1.7)

v₁ = 5.77 m / s

v₁ = 5.8 m/s

C) the velocity just after the bounce

starting point, after bounce

Em₀ = K = ½ m v₂²

final point. Maximum height after bounce

Em_{f} = U = m g y₂

as it already bounced, the energy is conserved in this interval

Em₀ = Em_{f}

½ m v₂² = m g y₂

v₂ = √ (2 g y₂)

v₂ = √ (2 9.8 1.2)

v₂ = 4.85 m / s

v₂= 4.9 m / s

D) during the time that the bounce lasts, there is a strong change in energy, part of it is transformed into thermal energy, due to several processes: friction, change in the potential energy of the molecules of the ball, change in the internal energy of the balls. molecules.

Therefore we cannot specify a single process, consequently the correct answer from 4

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3 years ago

A 2 kg blue car is moving 6 m/s to the right and collides with a 3 kg red car that is moving 2 m/s to the left. The cars collide

snow_lady [41]

Answer:

Their velocity after the collision is 1.2 m/s, to the right.

Explanation:

Given;

mass of the blue car, m₁ = 2 kg

initial velocity of the blue car, u₁ = 6 m/s

mass of the red car, m₂ = 3 kg

initial velocity of the red car, u₂ = 2 m/s

let the blue car moving to the right be in positive direction

also, let the red car moving to the left be in negative direction

Apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ - m₂u₂ = v(m₁ + m₂)

where;

v is their velocity after the collision

(2 x 6) - (3 x 2) = v(2 + 3)

12 - 6 = 5v

6 = 5v

v = 6/5

v = 1.2 m/s, to the right

Therefore, their velocity after the collision is 1.2 m/s, to the right.

7 0

3 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

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3 years ago

The practice of playing top hits several times a day on radio stations is called _____. blocking rotation circulation format dri

spin [16.1K]

The practice of playing top hits several times a day on radio stations is called rotation.

7 0

3 years ago

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