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4 years ago

Please help me answer this

Physics

1 answer:

yanalaym [24]4 years ago

6 0

Answer:

care of it and I will be there at work and I will be there at work

Explanation:

g the day off and I will be there at work and I will be there at work and I will be there

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The battery is made up of three 1.5 V cells.

IrinaVladis [17]

Answer: 4.5 V

Explanation:

Given

The voltage of each cell is 1.5 V

If they are connected end to end their potential added to give a higher potential i.e. 1.5+1.5+1.5=4.5 V

6 0

3 years ago

How much net force is required to accelerate a 5 kg toy car, initially at rest to a velocity of 3 m/s in 6 s?

inna [77]

Answer:

force = 0.20N .F = m ×a .& a = v/t then the f = m×v/t.

Explanation:

4 0

3 years ago

A charge q of 1.3 × 10-16 coulombs moves from point A to a lower potential at point B in an electric field of 3.2 × 102 newtons/

LUCKY_DIMON [66]

Explanation :

It is given that,

Charge, $q=1.3\times 10^{-16}\ C$

Electric field, $E=3.2\times 10^2\ N/C$

Distance, $d=1.1\times 10^{-2}\ m$

The work done is stored in the form of potential energy.

$W=F.d$

$\because F=qE$

So, $W=qE\ d$

$W=1.3\times 10^{-16}\ C\times 3.2\times 10^2\ N/C\times 1.1\times 10^{-2}\ m$

$W=4.576\times 10^{-16}\ J$

Hence, this is the required solution.

5 0

3 years ago

Read 2 more answers

A 40.0-μFcapacitor is connected across a 60.0 Hz generator. An inductor is then connected in parallel with the capacitor. What i

amm1812

Answer:

The value of the inductance is 175.9 mH.

Explanation:

Given that,

Capacitor $C= 40.0\ \muF$

Frequency = 60.0 Hz

The inductor and capacitor is connected in parallel, the voltage across each of these elements is the same.

We have,

$V_{L}=V_{C}$

Using ohm's law

$I_{rms}\times X_{L}=I_{rms}\times X_{C}$

$X_{L}=X_{C}$

$2\pi f L=\dfrac{1}{2\pi f C}$

$L=\dfrac{1}{4\pi^2\times f^2\times C}$

Put the value into the formula

$L=\dfrac{1}{4\times\pi^2\times60.0^2\times40\times10^{-6}}$

$L=0.1759\ H$

$L=175.9\times10^{-3}\ H$

$L=175.9\ mH$

Hence, The value of the inductance is 175.9 mH.

5 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) How much heat energy must be removed from your two liters of beverage?

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) 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?

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) 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 $)?

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