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2 years ago

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An unmarked police car traveling a constant 85 km/h is passed by a speeder. Precisely 2.00 s after the speeder passes, the polic

e officer steps on the acceleratorIf the police car accelerates uniformly at 2.00 m/s2 and overtakes the speeder after accelerating for 9.00 s , what was the speeder's speed?.

1 answer:

steposvetlana [31]2 years ago

6 0

Answer:

observe o gráfico abaixo que mostra o número de internação em um hospital no município no período de 1960

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The tops of the towers of the Golden Gate Bridge in San franciscoo , are 227 m above the water. Suppose a worker drops a 655 g w

yaroslaw [1]

Answer:

Kinetic energy is 1425.11 J.

Explanation:

Given:

Mass of the wrench is, $m=655\ g=0.655\ kg$

Height of fall is, $h=227\ m$

Force of resistance is, $F=0.141\ N$

Now, the total energy at the top is equal to the potential energy of the wrench at the top since the kinetic energy at the top is 0.

Now, potential energy at the top is given as:

$U=mgh\\U=0.655\times 9.8\times 227\\U=1457.113\ J$

Now, the potential energy at the top is converted to kinetic energy at the bottom and some energy is wasted in overcoming the resistance force by air.

Potential Energy = Kinetic energy + Energy to overcome resistance.

⇒ Kinetic energy = Potential Energy - Energy to overcome resistance.

Energy to overcome resistance force is the work done by the wrench against the resistance force and is given as:

$W_{res}=Fh=0.141\times 227=32.007\ J$

Therefore, Kinetic energy at the bottom is given as:

$K=U-W_{res}\\\\K=1457.113\ J - 32.007\ J\\\\K=1425.106\approx1425.11\ J$

Hence, the kinetic energy of the wrench be when it hits the water is 1425.11 J.

7 0

3 years ago

At what speed, as a fraction of c , is a particle's total energy twice its rest energy

WINSTONCH [101]

The rest energy of a particle is
$E_0=m_0 c^2$
where $m_0$ is the rest mass of the particle and c is the speed of light.

The total energy of a relativistic particle is
$E=mc^2 = \frac{m_0 c^2}{ \sqrt{1- \frac{v^2}{c^2} } }$
where v is the speed of the particle.

We want the total energy of the particle to be twice its rest energy, so that
$E=2E_0$
which means:
$\frac{m_0c^2}{ \sqrt{1- \frac{v^2}{c^2} } }=2m_0 c^2$
$\frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }=2$
From which we find the ratio between the speed of the particle v and the speed of light c:
$\frac{v}{c}= \sqrt{1- (\frac{1}{2})^2 } =0.87$
So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.

3 0

2 years ago

Chlorine contains 17 protons and 19 neutrons, and 18 electrons. What is the net charge of the ion?

Ymorist [56]

Answer:

Net Charge is -1.6 x 10 (to the negative 19th power) C. l

Explanation:

4 0

2 years ago

What do electricians call disconnects or disconnecting?

muminat

Service disconnect
this is the first disconnect device after the utility meter.

4 0

1 year ago

A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The

lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s

Explanation:

1) We can find the speed of the object by conservation of energy:

$E_{i} = E_{f}$

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s$

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s$

3) When the object is at the equilibrium position, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s$

I hope it helps you!

8 0

2 years ago