Answer:
final temperature of slug of gold is 37°C
final temperature of slug of manganese is 28.56°C
Explanation:
Hello! To solve this problem we must take into account the concept of heat capacity, this is defined as the ratio between energy and temperature rise.
In other words it is the amount of energy that is required to increase a temperature degree.
Taking into account the above we infer the following equation
where
C=heat capacity
i think so it is frequency
Answer:
The frequency increases.
Explanation:
When the Musician draws the slide in the length of the horn gets shorter, which causes a decrease in the wavelength. A decrease in the wave length results in an increase in frequency.
Note:
The diameter of the horn has an effect on frequency, so a wider horn is effectively a long horn - open end correction ( distance between the the antinode and the open end of a pipe).
Frequency also depends on how hard the musician blows the trombone. The musician can change the frequency with the lip pressure being applied.
Answer:
speed of current is 5 mile/hr
Explanation:
GIVEN DATA:
speed of motorboat = 15 miles/hr relative with water
let c is speed of current
15-c is speed of boat at upstream
15+c is speed of boat at downstream
we know that
travel time=distance/speed
150+10c+150-10c=1.5(15-c)(15+c)
300=1.5(225-c^2)
300=337.5-1.5c^2
200=225-c^2
c^2=25
c = 5
so speed of current is 5 mile/hr
Answer:
Explanation:
i )
When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2
So charged stored in it will remain unchanged .
ii )
Potential difference = charge / capacitance
in the first case potential difference = Q / C
in the second case potential difference = Q / 2C
So potential difference becomes half .
iii ) electric field = potential diff / plate separation
in the first case electric field = Q / (d x C )
in the second case electric field = 2 Q / (d x 2C)
= Q / (d x C )
So electric field remains unchanged .
iv)
energy stored in first case = Q² / 2C
In the second case energy stored = Q² / 2x2C
so energy stored becomes half .
