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2 years ago

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Jake uses a fire extinguisher to put out a small fire. When he squeezes the handle, the flame rettardant is released from the ex

tinguisher with considerable force. As a result, Jake has to reinforce his grip on the extinguisher to hold it in place as it pushes back into his hands. The projectile in this scenario is the . The recoil in this scenario is the . The force on the flame rettardant is the force on the extinguisher. The mass of the flame rettardant is the mass of the extinguisher. The acceleration of the flame rettardant is the acceleration of the extinguisher.

2 answers:

Tpy6a [65]2 years ago

7 0

Can you attach a picture of the actual problem?

pogonyaev2 years ago

5 0

The projectile in this scenario is the flame r.

The recoil in this scenario is the extinguisher.

The force on the flame r is equal to the force on the extinguisher.

The mass of the flame r is less than the mass of the extinguisher.

The acceleration of the flame r is greater than the acceleration of the extinguisher.

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Bill drives and sees a red light. He slows down to a stop. A graph of his velocity over time is shown below.

antoniya [11.8K]

Answer:

-2 m/s^2

Explanation:

Acceleration is equal to the slope of the graph. You just find the slope of that section. The rise is -20 and the run is 10, so you get -2.

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2 years ago

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The position function x(t) of a particle moving along an x axis is x = 4.00 - 6.00t2, with x in meters and t in seconds. (a) at

elena-14-01-66 [18.8K]

The position function x(t) of a particle moving along an x axis is $x=4.00 - 6.00t^2$

a) The point at which particle stop, it's velocity = 0 m/s

So dx/dt = 0

0 = 0- 12t = -12t

So when time t= 0, velocity = 0 m/s

So the particle is starting from rest.

At t = 0 the particle is (momentarily) stop

b) When t = 0

$x=4.00 - 6.00*0^2 = 4m$

SO at x = 4m the particle is (momentarily) stop

c) We have $x=4.00 - 6.00t^2$

At origin x = 0

Substituting

$0 = 4.00 - 6.00t^2\\ \\ t^2 = \frac{2}{3}$

t = 0.816 seconds or t = - 0.816 seconds

So when t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.

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2 years ago

The deuterium nucleus starts out with a kinetic energy of 1.24 × 10-13 joules, and the proton starts out with a kinetic energy o

MrMuchimi

Answer:

The total kinetic energy of both particles is $2.43\times10^{-13}$

Explanation:

Given that,

Kinetic energy of nucleus $K.E= 1.24\times10^{-13}\ J$

Kinetic energy of proton $K.E= 2.47\times10^{-13}\ J$

Radius of proton $r= 0.9\times10^{-15}\ m$

We need to calculate the final potential energy

Using formula of final potential energy

$U=\dfrac{kq^2}{r}$

Put the value into the formula

$U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times0.9\times10^{-15}}$

$U_{f}=1.28\times10^{-13}\ J$

We need to calculate the initial energy of both the particles

Using formula of energy

$E_{i}=(K.E_{n}+K.E_{p})+U_{i}$

$E_{i}=1.24\times10^{-13}+2.47\times10^{-13}+0$

$E_{i}=3.71\times10^{-13}\ J$

We need to calculate the total kinetic energy of both particles

Using conservation of energy

$E_{i}=E_{f}$

$E_{i}=K.E_{f}+U_{f}$

$3.71\times10^{-13}=K.E_{f}+1.28\times10^{-13}$

$K.E_{f}=3.71\times10^{-13}-1.28\times10^{-13}$

$K.E_{f}=2.43\times10^{-13}$

Hence, The total kinetic energy of both particles is $2.43\times10^{-13}$

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2 years ago

A ball with a mass of 0.5 kilograms is lifted to a height of 2.0 meters and dropped. It bounces back to a height of 1.8 meters.

Degger [83]

Hi, thank you for posting your question here at Brainly.

To compute for the change in potential energy, the equation would be:

delta PE = mg*delta h
delta PE = 0.5*9.81*(2-1.8)
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The potential energy is converted to kinetic energy.

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2 years ago

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Anni [7]

Answer:

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Therefore the paper charge is

$q_{paper}=-4.1\times10^{-15}C$

Now the distance from the charge is

$r=1cm=0.01m$

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$mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}$

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2 years ago