Answer:
5.5g of ice melts when a 50g chunk of iron at 80°C is dropped into a cavity
Explanation:
The concept to solve this problem is given by Energy Transferred, the equation is given by,

Where,
Q= Energy transferred
m = mass of water
c = specific heat capacity
Temperature change (K or °C)
Replacing the values where mass is 50g and temperature is 80°C to 0°C we have,



Then we can calculate the heat absorbed by m grams of ice at 0°C, then

How Q_1=Q_2, so



Then 5.5g of ice melts when a 50g chunk of iron at 80°C is dropped into a cavity
Answer:
The boundaries cause the waves to change direction an effect called <u>refraction.</u>
Explanation:
When a wave crosses a boundary between different materials, the speed of the wave and its wavelength changes.When passing from air to water the two properties (speed and wavelength) decreases, and the wave is observed to change direction as it crosses the boundary between the two material.The bending of the wave is called refraction.
Answer:
The electrical potential energy is 0.027 Joules.
Explanation:
The values from the question are
charge (q) = 
Electric Field strength (E) = 
Distance from source (d) = 0.030 m
Now the formula for the electrical potential energy (U) is given by

So now insert the values to find the answer

On further solving

Answer:
perpendicular to
Explanation:
it means perpendicular to .....should u come across something like this / / , this one means parallel to .....
1). The equation is: (speed) = (frequency) x (wavelength)
Speed = (256 Hz) x (1.3 m) = 332.8 meters per second
2). If the instrument is played louder, the amplitude of the waves increases.
On the oscilloscope, they would appear larger from top to bottom, but the
horizontal size of each wave doesn't change.
If the instrument is played at a higher pitch, then the waves become shorter,
because 'pitch' is directly related to the frequency of the waves, and higher
pitch means higher frequency and more waves in any period of time.
If the instrument plays louder and at higher pitch, the waves on the scope
become taller and there are more of them across the screen.
3). The equation is: Frequency = (speed) / (wavelength)
(Notice that this is exactly the same as the equation up above in question #1,
only with each side of that one divided by 'wavelength'.)
Frequency = 300,000,000 meters per second / 1,500 meters = 200,000 per second.
That's ' 200 k Hz ' .
Note:
I didn't think anybody broadcasts at 200 kHz, so I looked up BBC Radio 4
on-line, and I was surprised. They broadcast on several different frequencies,
and one of them is 198 kHz !