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Lisa [10]
3 years ago
12

1) The current in a light bulb is 0.335 Amps. How long does it take for a total charge of 2.76 C to

Physics
1 answer:
uysha [10]3 years ago
7 0

Answer:

Time=8.23880597 seconds

Explanation:

Quantity of charge(q)=2.76c

Current(I)=0.335A

Time(t)=?

t=q/I

t=2.76/0.335

t=8.23880597seconds

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garri49 [273]
1)
p = 2.4 * 10^5 Pa
T = 18° C + 273.15 = 291.15 k
r = 0.25 m => V = [4/3]π(r^3) = [4/3]π(0.25m)^3 = 0.06545 m^3 = 65.45 L

Use ideal gas equation: pV = nRT => n = pV / RT = [2.4*10^5 Pa * 0.06545 m^3] / [8.31 J/k*mol * 291.15k] = 6.492 mol

Avogadro number = 1 mol = 6.022 * 10^23 atoms

Number of atoms = 6.492 mol * 6.022 *10^23 atom/mol = 39.097 * 10^23 atoms = 3.91 * 10^24 atoms

2) Double atoms => double volume

V2 / V1 = r2 ^3 / r1/3

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r2 = [∛2]r1

The factor is ∛2
5 0
3 years ago
You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner
omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

Q_1 = Q_2

V_1A_1=V_2A_2

Our values are given as,

A_1=\frac{1}{2}^2*\pi=0.785 in^2

A_2=\frac{5}{8}^2*\pi=1.227 in^2

Re-arrange the equation to find the first ratio of rates we have:

\frac{V_1}{V_2}=\frac{A_2}{A_1}

\frac{V_1}{V_2}=\frac{1.227}{0.785}

\frac{V_1}{V_2}=1.56

The second ratio of rates is

\frac{V2}{V1}=\frac{A_1}{A2}

\frac{V2}{V1}=\frac{0.785}{1.227}

\frac{V2}{V1}=0.640

3 0
3 years ago
If two objects at different temperatures are in contact with each other, what happens to their temperatures?
miv72 [106K]

In that case, heat energy flows from the warmer object to the cooler one. 
As heat flows from one to the other, the temperature of the warmer object
falls, and the temperature of the cooler object rises. When the temperatures
are equal, the flow of heat energy from one to the other stops.

6 0
2 years ago
19. Calculate the work done when a 30 N force pushes a rock 10 m. a) 3J b) 0.33 J c) 40 J d) 300 J
aksik [14]

19. Calculate the work done when a 30 N force pushes a rock 10 m. a) 3J b) 0.33 J c) 40 J d) 300 J

20. How far did a rock move if a force of 200 N moved it doing 100 J of work? a) 2 m b) 0.5 m c) 20,000 m d) 300 m

21. An explosion of gas from a Hawaiian volcano blows a rock into the sky. the rock has 2000J of kinetic energy. What was its mass if the rock had a velocity of 20m/s. a) 200Kg b) 20,000Kg c) 10Kg d) 200Kg

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25. A force of 100 newtons was necessary to lift a rock. A total of 150 joules of work was done. How far was the rock lifted? a) 1.5 m b) 15 m c) 0.15 m d) 150 m19. Calculate the work done when a 30 N force pushes a rock 10 m. a) 3J b) 0.33 J c) 40 J d) 300 J

20. How far did a rock move if a force of 200 N moved it doing 100 J of work? a) 2 m b) 0.5 m c) 20,000 m d) 300 m

21. An explosion of gas from a Hawaiian volcano blows a rock into the sky. the rock has 2000J of kinetic energy. What was its mass if the rock had a velocity of 20m/s. a) 200Kg b) 20,000Kg c) 10Kg d) 200Kg

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21. An explosion of gas from a Hawaiian volcano blows a rock into the sky. the rock has 2000J of kinetic energy. What was its mass if the rock had a velocity of 20m/s. a) 200Kg b) 20,000Kg c) 10Kg d) 200Kg

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25. A force of 100 newtons was necessary to lift a rock. A total of 150 joules of work was done. How far was the rock lifted? a) 1.5 m b) 15 m c) 0.15 m d) 150 m

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5 0
2 years ago
A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha
Gelneren [198K]

Answer:

a_{acceleriation}=-4cm/s^{2}\\ or\\  a_{acceleriation}=-0.04m/s^{2}

Explanation:

Given data

velocity v₀=20 cm/s at time t=3s

velocity vf=0 at time t=8 s

To find

Average Acceleration at time=3s to 8s

Solution

As we know that acceleration is first derivative of velocity with respect to time

a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\  a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\  a_{acceleriation}=-4cm/s^{2}\\ or\\  a_{acceleriation}=-0.04m/s^{2}

8 0
2 years ago
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