The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth? AU

A wave is propagating from left to right in a medium. The particles in the medium are also vibrating from left to right. What ki

sattari [20]

Answer:

In this case, the particles of the medium move parallel to the direction that the pulse moves. This type of wave is a longitudinal wave.

6 0

3 years ago

Suppose an element had an atomic number of 52. How many protons would that element have?

frutty [35]

The element would have 52 protons because atomic number gives protons.

4 0

3 years ago

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A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t

Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

P.E = mgh joule

Considering h = 0 for the vertical position

The h at ∅ = 42° is h = L (1 - cos∅)

P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

P.E = 25 x 9.8 x 2.2 (1 - cos42°)

= 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

∴ K.E = P.E

1/2 mv² = 138.44

1/2 x 25 x v² 138.44

v² = 11.0752

v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

Tension on string, T = Force acting on the swing, F

$W=L\int\limits^0_\phi{F} \, d \phi$

$=L\int\limits^0_\phi{mg.sin \phi} \, d \phi$

= $-Lmg[cos\phi]_{42}^{0}$

= - 2.2 x 25 x 9.8 [cos0 - cos 42°]

= - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0

3 years ago

At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does

insens350 [35]

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

$V = V_R + V_L$

The potential difference across the inductance is given by

$V_L(t) = V e^{-\frac{t}{\tau}}$ (1)

where

$\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s$ is the time constant of the circuit

At time t=0,

$V_L(0) = V e^0 = V = 20 V$

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

$V_R = V-V_L = 20 V-20 V=0$

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

$t >> \tau$

Since $\tau$ is at the order of less than milliseconds.

Using eq.(1), we see that for $t >> \tau$ , the exponential becomes zero, and therefore the potential difference across the coil is zero:

$V_L = 0$

Therefore, the potential difference across the resistor will be

$V_R = V-V_L = 20 V- 0 = 20 V$

(c) Yes

The two voltages will be equal when:

$V_L = V_R$ (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

$V=V_L +V_R$

And rewriting this equation,

$V_R = V-V_L$

Substituting into (2) we find

$V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V$

So, the two voltages will be equal when they are both equal to 10 V.

(d) at $t=5.77\cdot 10^{-4}s$

We said that the two voltages will be equal when

$V_L=\frac{V}{2}$

Using eq.(1), and this last equation, this means

$V e^{-\frac{t}{\tau}} = \frac{V}{2}$

And solving the equation for t, we find the time t at which the two voltages are equal:

$e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s$

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

$I_0 = 3.20 A$

The problem says this current is stable: this means that we are in a situation in which $t>>\tau$ , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

$V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V$

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

$V_L(0)=.V$

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

$t>>\tau$

If we use this approximation into the formula

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

We find that

$V_L = 0$

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

7 0

3 years ago

The movement of water into the atmosphere as it changes from a liquid to a gas is called _____.

nekit [7.7K]

Evaporation . Because it drys up and turns into gas

5 0

3 years ago

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