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3 years ago

The kinetic energies of particles in a sample of matter are increasing. This sample is most likely

Physics

1 answer:

KIM [24]3 years ago

8 0

gaining thermal energy in form of heat

Explanation:

When the thermal energy of a matter is increasing, it will directly increase the kinetic energy of the matter.

Temperature changes causes the average kinetic energy of molecules to increase or decrease.
If the kinetic energies of particles are increasing, the temperature must have been on rise.
Increase in temperature causes molecules to start vibrating and moving about.
The gain more energy that helps them to move away from their original position

Learn more:

Kinetic energy brainly.com/question/7966903

#learnwithBrainly

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3 years ago

A 0.111 kg hockey puck moving at 55 m/s is caught by a 80 kg goalie at rest. With what speed does the goalie slide on the (frict

Andrews [41]

Answer: 0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

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3 years ago

A 2.02 nF capacitor with an initial charge of 4.55 µC is discharged through a 1.22 kΩ resistor. (a) Calculate the current in the

Mandarinka [93]

Answer:

a) 0.048A

b) 0.18µC

c) 1.85A

Explanation:

The discharged current of the capacitor as a function of time is given by:

$i=\frac{q_o}{RC}*e^{-\frac{t}{\tau}}\\where:\\\tau=RC\\$

$\tau=1.22*10^3*2.02*10^{-9}\\\tau=2.46*10^{-6}s$

$i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{9\µs}{2.46\µs}}\\\\i=0.048A$

$q=q_o*e^{-\frac{t}{\tau}}$

$q=4.55\µC*e^{-\frac{8\µs}{2.46\µs}}\\q=0.18\µC$

c) the maximum current occurs when t=0

$i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{0\µs}{2.46\µs}}\\\\i=1.85A$

6 0

3 years ago

Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th

satela [25.4K]

The tank has a volume of $\dfrac\pi3R^2H$ , where $H=6\,\rm m$ is its height and $R=\dfrac d2=2\,\rm m$ is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

$\dfrac26=\dfrac rh\implies r=\dfrac h3$

The volume of water in the tank at any given time is

$V=\dfrac\pi3r^2h$

and can be expressed as a function of the water level alone:

$V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3$

Implicity differentiating both sides with respect to time $t$ gives

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}$

We're told the water level rises at a rate of $\dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min}$ at the time when the water level is $h=2\,\mathrm m=200\,\mathrm{cm}$ , so the net change in the volume of water $\dfrac{\mathrm dV}{\mathrm dt}$ can be computed:

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}$

The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:

$\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})$

We're told the water is leaking out at a rate of $10,500\,\frac{\mathrm{cm}^3}{\rm min}$ , so we find the rate at which it's being pumped in to be

$\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}$

$\implies\text{rate in}\approx289,753\,\dfrac{\mathrm{cm}^3}{\rm min}$

4 0

3 years ago