What is the acceleration of the car at segment C?

How much heat has to be added to 508 g of copper at 22.3°C to raise the temperature of the copper to 49.8°C? (The specific heat

viva [34]

Answer:

Q = 5267J

Explanation:

Specific heat capacity of copper (S) = 0.377 J/g·°C.

Q = MSΔT

ΔT = T2 - T1

ΔT=49.8 - 22.3 = 27.5C

Q = change in energy = ?

M = mass of substance =508g

Q = (508g) * (0.377 J/g·°C) * (27.5C)

Q= 5266.69J

Approximately, Q = 5267J

8 0

3 years ago

How much energy (in kilojoules) is required to convert 200 mL of diethyl ether at its boiling point from liquid to vapor if its

LiRa [457]

Answer:

55.96kJ

Explanation:

Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether

Volume (v) = 200mL, density (d) = 0.7138g/mL

Mass = d × v = 0.7138 × 200 = 142.76g

Enthalpy of vaporization of diethyl ether = 29kJ/mol

MW of diethyl ether (C2H5)2O = 74g/mol

Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g

Energy = 142.76g × 0.392kJ/g = 55.96kJ

4 0

3 years ago

A 62 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 12 m. She falls a total of 3

Andrew [12]

Answer:

k = 104.46 N/m

Explanation:

Here we can use energy conservation

so we will have

initial gravitational potential energy = final total spring potential energy

as we know that she falls a total distance of 31 m

while the unstretched length of the string is 12 m

so the extension in the string is given as

$x = L - L_o$

$x = 31 - 12 = 19 m$

so we have

$mgH = \frac{1}{2}kx^2$

$62(9.81)(31) = \frac{1}{2}k (19^2)$

$k = 104.46 N/m$

5 0

3 years ago

If we use 1 millimeter to represent 1 light-year, how large in diameter is the Milky Way Galaxy?

storchak [24]

Answer:

d.100 meters

Explanation:

The diameter of the Milky Way Galaxy is approximately 100,000 light years.

Here we are using 1 millimiter (1 mm) to represent 1 light-year (1 ly). So, we can set the following proportion:

$1 mm : 1 ly = x : 100,000 ly$

and by finding x, we find the diameter of the Milky Way Galaxy in the scale used:

$x=\frac{(1mm )(100,000 ly)}{1 ly}=100,000 mm = 100 m$

so the correct answer is

d. 100 meters

4 0

3 years ago

Submarine a travels horizontally at 11.0 m/s through ocean water. it emits a sonar signal of frequency f 5 5.27 3 103 hz in the

xeze [42]

Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ

6 0

3 years ago