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VikaD [51]
2 years ago
3

Radiometric dating of a magnetic anomaly stripe of rock that is 225 km away from the mid-ocean ridge axis gives an age of 9 mill

ion years. Assuming a constant rate, seafloor spreading in this area occurs at a rate of __________.
A. 5 cm per year
B. 1,012.5 km per year
C. 20,000 cm per year
D. 50 km per year
Physics
1 answer:
VashaNatasha [74]2 years ago
7 0

Answer:

The seafloor spreading in this area occurs at a rate of 5 cm per year.

Explanation:

It is given that, the radiometric dating of a magnetic anomaly stripe of rock that is 225 km away from the mid-ocean ridge axis gives an age of 9 million years.

We need to find the rate at which the seafloor is spreading in this area. The total distance will be, d = 225 km × 2 = 450 km

Time, t = 9 million years

The rate at which the seafloor is spreading in this area is given by :

v=\dfrac{d}{t}

v=\dfrac{4.5\times 10^7\ cm}{9\times 10^6\ year}

v=5\ cm/year

So, the seafloor spreading in this area occurs at a rate of 5 cm per year. Hence, this is the required solution.

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Two 8.0 Ω lightbulbs are connected in a 12 V series circuit. What is the power of both glowing bulbs?
V125BC [204]

Answer:

18 W

Explanation:

Applying,

P = V²/R.................. Equation 1

Where P = Power of both glowing bulbs, V = Voltage, R = Combined Resistance of both bulbs

Since: It is a series circuit,

Then,

R = R1+R2............. Equation 2

Where R1= Resistance of the first bulb, R2 = Resistance of the second bulb

Given: R1 = R2 = 8 Ω

Substitute into equation 1

R = 8+8

R = 16 Ω

Also Given: V = 12 V

Substitute into equation 1

P = 12²/8

P = 144/8

P = 18 W

7 0
2 years ago
A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu
Illusion [34]
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
v=\omega r= (12 rad/s)(0.5 m)= 6 m/s

Then we can calculate the angular momentum of the stone about the center of the circle, given by
L=mvr
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}
3 0
2 years ago
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik
Lady bird [3.3K]

Answer:

The options are approximations of the exact answers:

A) 1\times10^6N/C

B) 2\times10^{-13}N

C) 1\times10^{-2}V

D) Toward the inner wall

E) 3\times10^{-17}J

Explanation:

A) The electric field in a parallel plate capacitor is given by the formula E=\frac{\sigma}{k\epsilon_0}, where \epsilon_0=8.85\times10^{-12}C/Vm and in our case \sigma=10^5C/m^2 and, for air,k=1.00059, so we have:

E=\frac{10^5C/m^2}{(1.00059)(8.85\times10^{-12}C/Vm)}=1.13\times10^6N/C

B) The K+ ion has one elemental charge excess, so its charge is q=1.6\times10^{-19}C, and the force a charge experiments under an electric field E is given by F=qE, so we have:

F=(1.6\times10^{-19}C)(1.13\times10^6N/C)=1.81\times10^{-13}N

C) The potential difference between two points separated a distance d under an uniform electric potential E is given by \Delta V=dE, so we have:

\Delta V=(10\times10^{-9}m)(1.13\times10^6N/C)=1.13\times10^{-2}V

D) The electic field goes from positive to negative charges, so it goes towards the inner wall.

E) The work done by an electric field through a potential difference \Delta V on a charge Q is W=Q\Delta V, and is equal to the kinetic energy imparted on it, so we have:

K=(3\times10^{-15}C)(1.13\times10^{-2}V)=3.39\times10^{-17}J

5 0
2 years ago
A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the
timurjin [86]

Answer:

option C

Explanation:

given,

energy dissipated by the system to the surrounding = 12 J

Work done on the system = 28 J

change in internal energy of the system

Δ U = Q - W

system losses energy = - 12 J

work done = -28 J

Δ U = Q - W

Δ U = -12 -(-28)

Δ U = 16 J

hence, the correct answer is option C

6 0
2 years ago
1. When the fission of uranium-235 is carried out, about 0.1 percent of the mass of the reactants is lost during the reaction. W
DerKrebs [107]

The correct answer to the question is that the lost mass has been converted into energy.

EXPLANATION:

From Einstein's theory, we know that energy and mass are inter convertible .

When some amount of mass is lost, same amount of energy equivalent to mass is produced.

Let us consider m is the mass lost during any reaction. Hence, the amount of energy produced will be-

                                 Energy E = mc^2                

Here, c is the velocity of light i.e c = 3\times 10^8\ m/s

As per the question, uranium-235 undergoes fission. The amount of mass defect is 0.1 %.

The mass defect is defined as the difference between mass of reactants and products. During the fission, energy is produced.

The energy produced in this reaction is nothing else than the energy equivalent to mass defect.  Approximately 199.5 Mev of energy equivalent to this mass defect is produced in this reaction.

6 0
2 years ago
Read 2 more answers
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