A hiker starts a point p and walks 2.0 kilometers due east and then 1.4 kilometers due north.

What potential difference would an electron have to fall through to acquire a speed of 3.00*10^6 m/sec?

kirill [66]

Answer:

25.6 V

Explanation:

The kinetic energy of electron associated with its potential difference is given by eV which is equal to the 1/2 mv^2.

m = 9.1 x 10^-31 kg, v = 3 x 10^6 m/s, e = 1.6 x 10^-19 C

eV = 1/2 m v^2

V = mv^2 / 2 e

V = (9.1 x 10^-31) x (3 x 10^6)^2 / (2 x 1.6 x 10^-19)

V = 25.6 V

4 0

4 years ago

A two-liter bottle of your favorite beverage has just been removed from the trunk of your car. The temperature of the beverage i

Ksivusya [100]

Answer:

a) 209.3 kilojoules must be removed from two liter of beverage, b) A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles, c) Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

Explanation:

a) <em>How much heat energy must be removed from your two liters of beverage?</em>

At first we suppose that the beverage has the mass and specific heat of water and that there are no energy interactions between the bottle and its surroundings.

From the First Law of Thermodynamics and definition of sensible heat, we get that amount of removed heat ( $Q$ ), measured in kilojoules, is represented by the following formula:

$Q = \rho \cdot V\cdot c\cdot (T_{o}-T_{f})$ (Eq. 1)

Where:

$\rho$ - Density of the beverage, measured in kilograms per cubic meter.

$V$ - Volume of the bottle, measured in cubic meters.

$c$ - Specific heat of water, measured in kilojoules per kilogram-Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures, measured in Celsius.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $V = 2\times 10^{-3}\,m^{3}$ , $c = 4.186\,\frac{kJ}{kg\cdot ^{\circ}C}$ , $T_{o} = 35\,^{\circ}C$ and $T_{f} = 10\,^{\circ}C$ , then:

$Q = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (2\times 10^{-3}\,m^{3})\cdot \left(4.186\,\frac{kJ}{kg\cdot ^{\circ}C} \right) \cdot (35\,^{\circ}C-10\,^{\circ}C)$

$Q = 209.3\,kJ$

209.3 kilojoules must be removed from two liter of beverage.

b) <em>You are having a party and need to cool 10 of these two-liter bottles in one-half hour. What rate of heat removal, in kW, is required?</em>

The total amount of heat that must be removed from 10 2-L bottles is:

$Q_{T} = 10\cdot (209.3\,kJ)$

$Q_{T} = 2093\,kJ$

If we suppose that bottles are cooled at constant rate, then, rate of heat removal is determined by this formula:

$\dot Q = \frac{Q_{T}}{\Delta t}$ (Eq. 2)

Where:

$Q_{T}$ - Total heat, measured in kilojoules.

$\Delta t$ - Time, measured in seconds.

$\dot Q$ - Rate of heat removal, measured in kilowatts.

If we know that $Q_{T} = 2093\,kJ$ and $\Delta t = 1800\,s$ , we find that rate of heat removal is:

$\dot Q = \frac{2093\,kJ}{1800\,s}$

$\dot Q = 1.163\,kW$

A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles.

c) <em>Assuming that your refrigerator can accomplish this and that electricity costs 8.5 cents per kW-hr, how much will it cost to cool these 10 bottles (in $)?</em>

A kilowatt-hour equals 3600 kilojoules. The electricity cost is equal to the removal heat of 10 bottles ( $Q_{T}$ ), measured in kilojoules, and unit electricity cost ( $c$ ), measured in US dollars per kilowatt-hour. That is:

$C = c\cdot Q_{T}$

If we know that $c = 0.085\,\frac{USD}{kWh}$ and $Q_{T} = 2093\,kJ$ , the total cost of cooling 10 bottles is:

$C = \left(0.085\,\frac{USD}{kWh}\right)\cdot \left(2093\,kJ\right)\cdot \left(\frac{1}{3600}\,\frac{kWh}{kJ} \right)$

$C = 0.049\,USD$

Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

3 0

4 years ago

In physics class, Carrie learns that a force, F, is equal to the mass of an object, m, times its acceleration, a. She writes the

Phantasy [73]

Answer:

2.2 m/s^2

Explanation:

Acceleration = Force / Mass

= 7.92 / 3.6 = 2.2m/s^2

Hope this help you :3

5 0

3 years ago

Read 2 more answers

What do these circuits all have in common? A. They all contain switches. B. They are all AC. C. They are all DC. D. They all con

I am Lyosha [343]

Answer:

A

Explanation:

8 0

3 years ago

Which set of measurements best represents accuracy if the true value is 6.80 kg? a 6.80 kg, 6.70 kg and 6.90 kg b 6.75 kg, 6.80

scoundrel [369]

Answer:

Explanation:

To get the set of measurements that best represents accuracy if the true value is 6.80 kg, we will find the mean of the set of numbers and chech which is equal to 6.80kg

According to the set of data 6.8kg, 6.7kg and 6.9kg

Mean = sum of the number/sa mple size

Sum of the numbers = 6.8+6.7+6.9 = 20.4

sample size = 3

Mean of the set of measurement = 20.4/3

Mean of the set of measurement is 6.80kg

Since the answer is 6.8kg, hence set of measurements that best represents accuracy are 6.80kg, 6.70kg and 6.90kg.

8 0

3 years ago