Is a man kicking ball potential or kinetic?

lyudmila [28]

Answer: A negatively charged particle that is found in atoms.

4 0

3 years ago

The burner on an electric stove has a power output of 2.0 kW. A 710 g stainless steel tea kettle is filled with 20∘C water and p

asambeis [7]

Answer:

The volume of water that was in the kettle is 1170 $cm^{3}$

Explanation:

Given:

Power, P = 2.0 kW = 2000 W, Mass of stainless steel, $m_{s}$ = 710 g = 0.71 kg at temperature of $20^{0} C$

Part A:

If it takes time, t = 3.5 minutes to reach boiling point of water $100^{0} C$ , then from conservation of energy,

Total energy supplied by the burner = Total heat gained by the water and the stainless steel to rise from $20^{0} C$ to $100^{0} C$

i.e. Pt = $m_{s}$ $c_{s}$ (100 - 20 ) + $m_{w}$ $c_{w}$ (100 - 20 )

$m_{w}$ = $\frac{pt - 80m_{s} C_{s} }{80c_{w} } = \frac{2000*3.5*60 - 80*0.71* 450}{80*4200}$

$m_{w}$ = 1.17 kg

where $c_{w}$ = 4200 J/Kgk (specific heat capacity of water), $c_{s}$ = 450 J/Kgk (specific heat capacity of steel)

But volume of water in the the kettle, v = $\frac{mass}{density} = \frac{1.17}{1000}= 1.17 *10^{-3}$ $m^{3}$

∴ v = 1170 $cm^{3}$

4 0

4 years ago

Can you complete and EXPLAIN please?????:(

jonny [76]

Answer:

Explanation:

Well you have the voltages right, and that is no trivial matter. Each one of the resistors in a parallel circuit sees the same input voltages (in this case 6).

Now I think it would be a good idea to fill in the the resistance column.

R1 = 3 ohms

R2 = 6 ohms

R3 = 2 ohms

The total resistance can be calculated in two ways. I'll get around to doing both of them but I'll do the conventional way first. One hint: the total resistance must be smaller than the smallest resistor. Read that sentence over a couple of times. What it means is that it must be less than 2 ohms in a parallel circuit.

1/r1 + 1/r2 + 1/r3 = 1/rt

1/3 + 1/6 + 1/2 = 1/rt

Change all the denominators to 6ths.

2/6 + 1/6 + 3/6 = 1/rt

(2 + 1 + 3)/6 = 6/6 = 1

rt = 1

====================

So the current I is V/R

V = 6

R = 1

Current = V/R = 6/1 = 6 amps.

====================

The current in each resistor is

I1 = V / R1

I1 = 6/3 = 2 amps

I2 = V/R2

I2 = 6/6 = 1 amp

I3 = 6/2 = 3 amps

The total is I1 + I2 + I3 = 2 + 1 + 3 = 6 amps.

======================

Remember I said there was 2 ways of figuring out the total resistance. I did one of them about. Here's the other.

R = V / It

R = 6 / 6

R = 1 ohm just what you got before.

====================

Power

P = V * I

P1 = 6 * 2 = 12 watts

P2 = 6*1 = 6 watts

P3 = 6*3 = 18 watts

Pt = 36 watts.

Pt can be done by using the voltage * the total current

Pt = 6 volts * 6 amps = 36 watts, just what you would expect.

3 0

3 years ago

If one sound wave has twice the amplitude of another the first sound is

nikdorinn [45]

Answer:

occurs when path-length difference is:

Δr=mlambda and m=0,1,2,3...

Explanation:

I think thats the answer.

5 0

4 years ago

Three children are riding on the edge of a merry-go-round that is 182 kg, has a 1.60 m radius, and is spinning at 15.3 rpm. The

Korolek [52]

Answer:

The new angular velocity of the merry-go-round is 18.388 revolutions per minute.

Explanation:

The merry-go-round can be represented by a solid disk, whereas the three children can be considered as particles. Since there is no external force acting on the system, we can apply the principle of angular momentum conservation:

$\left(\frac{1}{2}\cdot M + m_{1}+m_{2} + m_{3} \right)\cdot R^{2}\cdot \dot n_{o} = \left(\frac{1}{2}\cdot M + m_{1} + m_{3})\cdot R^{2}\cdot \dot n_{f}$ (1)

Where:

$M$ - Mass of the merry-go-round, in kilograms.

$m_{1}$ , $m_{2}$ , $m_{3}$ - Masses of the three children, in kilograms.

$R$ - Radius of the merry-go-round/Distance of the children with respect to the center of the merry-go-round, in meters.

$\dot n_{o}$ , $\dot n_{f}$ - Initial and final angular speed, in revolutions per minute.

If we know that $M = 182\,kg$ , $m_{1} = 17.4\,kg$ , $m_{2} = 28.5\,kg$ , $m_{3} = 32.8\,kg$ , $R = 1.60\,m$ and $\dot n_{o} = 15.3\,\frac{rev}{min}$ , then the final angular speed of the system is:

$\dot n_{f} = \dot n_{o}\cdot \left(\frac{\frac{1}{2}\cdot M + m_{1} + m_{2} + m_{3} }{\frac{1}{2}\cdot M + m_{1} + m_{3} } \right)$

$\dot n_{f} = \left(15.3\,\frac{rev}{min} \right)\cdot \left[\frac{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg +28.5\,kg + 32.8\,kg }{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg + 32.8\,kg } \right]$

$\dot n_{f} = 18.388\,\frac{rev}{min}$

The new angular velocity of the merry-go-round is 18.388 revolutions per minute.

7 0

3 years ago