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4 years ago

A car with a mass of 600 kg is traveling at a velocity of 10 m/s. How much kinetic energy does it have?

Physics

1 answer:

Viefleur [7K]4 years ago

5 0

Answer:

KE = 30,000 J

Explanation:

KE = $\frac{1}{2}$ mv²

KE = $\frac{1}{2}$ (600)(10)²

KE = $\frac{1}{2}$ (600)(100)

KE = $\frac{1}{2}$ (60000)

KE = 30,000 J

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At 13:20 on the last Friday in September, 1989 a frantic call was received at the local police station. There had been a serious

astraxan [27]

The kinematics and conservation of momentum relationships allow to find the results for the questions about the measurements and conditions to know the accident conditions are:

1) The principles of conservation of momentum and kinematics.

2) $\theta = tan^{-1} \frac{P_{01}}{p_{02}}$

3) The factors that influence the direction and distance traveled are:

The initial speed of the vehicles

The mass of the vehicles

Pavement conditions: dry, wet, humid, with dirt or loose sand.

1) Kinematics studies the movement of bodies, they look for relationships between position, speed and acceleration.

To analyze the accident, you must answer a question such as what speed the vehicles had before the accident, this is the initial speed of each one and their directions.

When reviewing the kinematics relationships we have.

$v^2 = v_o^2 - 2 a x$

Where the acceleration is given by Newton's second law.

fr = m a

The friction force is given by the expression.

fr = μ N

fr = μ mg

Let's Substitute.

μ m g = m a

a = μ g

The friction coefficient is tabulated, for different types of rubber, pavement and wet, dry conditions, etc.

Therefore, the researcher must measure the braking distance of the vehicles, which can be taken from the mark of the tires on the road and note the condition of the pavement if it is dry or wet and the visibility of the day.

This is the speed of the vehicles just after the impact, using the law of conservation of momentum you can find the speed of the vehicles before the impact

$p_o = p_f$ \f

Where this is a vector expression, which in general is solved with the components of each directional.

Consequently to answer the questions you must use the principles of conservation of momentum and kinematics.

2) If the vehicles travel at right angles and their masses are similar, the direction after the collision in each direction.

$tan \theta = \frac{p_{01}}{p_{02}}$

$\theta = tan ^{-1} \frac{p_{01}}{p_{02}}$

3) The factors that influence the direction and distance traveled are:

The initial speed of the vehicles

The mass of the vehicles

Pavement conditions: dry, wet, humid, with dirt or loose sand.

In conclusion, using the kinematics and conservation of momentum relationships, we can find the answers to the question about measurements and conditions to know the accident conditions are:

1) The principles of conservation of momentum and kinematics.

2) $\theta = tan^{-1} \frac{P_{01}}{p_{02}}$

3) The factors that influence the direction and distance traveled are:

The initial speed of the vehicles

The mass of the vehicles

Pavement conditions: dry, wet, humid, with dirt or loose sand.

Learn more about kinematics and momentum here: brainly.com/question/17304001

5 0

3 years ago

Two wires, each of length 1.3 m, are stretched between two fixed supports. On wire A there is a second-harmonic standing wave wh

UNO [17]

Answer:

Explanation:

Given

Length of each wire $L=1.3\ m$

On wire A second harmonic frequency is given by

$f_2_{a}=2\times (\frac{v}{2L})$

where f=frequency

v=velocity of wave

L=length of wire

$v_a=f_2\times L$

$v_a=640\times 1.3=832\ m/s$

For wire B third harmonic is given by

$f_3_{b}=3\times (\frac{v}{2L})$

$v_b=\frac{2L}{3}\cdot f_3_{b}$

$v_b=\frac{2\times 1.3}{3}\times 640=554.66\ m/s$

3 0

4 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:

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

8 0

3 years ago

A ball is thrown up into the air. When it falls back down and reaches the

bezimeni [28]

Answer:

A. 3.3 m

Explanation:

Here, we have to use conservation of energy principle.

When the ball is at maximum height, the instantaneous velocity at that point is 0 m/s. So, the kinetic energy of the ball is also 0 at the maximum height. Thus, at maximum height, the energy possessed by the ball is gravitational potential energy only.

Now, when the ball reaches the ground, all the gravitational potential energy changes into kinetic energy because of the conservation of energy.

Therefore, the energy transformation can be given as:

Decrease in potential energy = Increase in Kinetic energy

Decrease in potential energy is given as:

$\Delta U=mg(h-0)=mgh$

Increase in kinetic energy is given as:

$\Delta K=\frac{1}{2}mv^2-0=\frac{1}{2}mv^2$

Therefore,

$\Delta U=\Delta K\\mgh=\frac{1}{2}mv^2\\h=\frac{v^2}{2g}$

Now, plug in 8 for $v$ , 9.8 for $g$ and solve for height $h$ . This gives,

$h=\frac{8^2}{2\times 9.8}\\h=\frac{64}{19.6}=3.265\approx 3.3\ m$

Therefore, the maximum height reached by the ball is 3.3 m.

4 0

4 years ago

A horizontal line on the velocity vs. time graph indicates a constant, positive

Gala2k [10]

Answer:

False

Explanation:

When the line is horizontal in a graph of velocity vs time, it implies that the velocity is constant as time keeps moving. Whenever the velocity is constant then it means there is no change in velocity. Since acceleration is rate of change of velocity per unit time, then here acceleration is zero.

3 0

3 years ago