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3 years ago

The frequency of a sound wave is determined by the ????

Physics

1 answer:

Alja [10]3 years ago

4 0

The frequency of a sound is determined by its source. It can't change once it leaves there.

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What happens to the brightness of bulb a when the switch is closed and bulb b lights up?

eimsori [14]

The problem describes the relationship of "bulb a" and "bulb b" to be in connected in series. When the switch is open then no current can flow, on the other hand, when it is closed, current will pass through.

When only "bulb a" is connected to the battery then more current is flowing to "bulb a" causing it to be bright.

Closing the switch would mean that "bulb b" is already included in the circuit and the battery will push small current to flow around the whole circuit. The more bulbs are connected, the harder for the current to flow because the resistance will be very high.

So the light of "bulb a" will be dimmer.

8 0

4 years ago

What is the relative energy expense of galloping rather than trotting at 3.5 m/s?

labwork [276]

Answer: Trotting uses only 75 percent of the energy as galloping

Explanation: Trotting is only 300 J/m, whereas galloping is roughly 400 J/m

7 0

3 years ago

In Europe the standard voltage in homes is 220 V instead of the 120 V used in the United States. Therefore "100-W" European bulb

Maslowich

Explanation:

In Europe the standard voltage in homes is 220 V instead of the 120 V used in the United States.

100-W European bulb would be intended for use with a 220-V potential difference.

(a) If V = 220 V and P = 100 W

Power :

$P=\dfrac{V^2}{R}\\\\R=\dfrac{V^2}{P}\\\\R=\dfrac{(220)^2}{100}\\\\R=484\ \Omega$

If you bring a "100-W" European bulb home to the untied States, what should be its US power rating. So,

$P=\dfrac{V^2}{R}\\\\P=\dfrac{(120)^2}{484}\\\\P=29.75\ W$

P = 29.8 watts

(b) Let I is the current will the 100-W European bulb draw in normal use in the United States. So,

$I=\dfrac{V}{R}\\\\I=\dfrac{120}{484}\\\\I=0.248\ A$

Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers

What causes the phases of our moon

Helga [31]

sunlight, and the position of the earth

4 0

4 years ago

Is it possible for the momentum of a system consisting of two carts on a low-friction track to be zero even if both carts are mo

tamaranim1 [39]

Answer:

Yes, if the carts are travelling into opposite directions

Explanation:

The total momentum of a two carts system is the sum of the momenta of the individual carts:

$p=p_1 +p_2 = m_1 v_1 + m_2 v_2$

where

m1, m2 are the masses of the two carts

v1, v2 are the velocities of the two carts

In order to have a total momentum of zero, we must have

$m_1 v_1 + m_2 v_2 = 0\\m_1 v_1 = - m_2 v_2$ (1)

Let's remind that velocity (and so, momentum as well) is a vector quantity: this means that it has a direction, so when summing together the momenta, we must also consider the sign of the velocity, depending on its direction.

Therefore, if the two carts are moving in opposite directions, the signs of the two velocities will be opposite. For example, we can have

$v_1 > 0 \\v_2 < 0$

This means that the condition in eq.(1) can be satisfied, provided that the two carts are travelling into opposite directions.

8 0

3 years ago