What is the primary evidence used to determine how the Moon formed? O A. Moon craters and Earth craters were caused by the same asteroid
strike.
B. Moon rocks and Earth rocks are made up of many of the same
materials.
O C. The Moon and Earth are exactly the same age.
D. The Moon and Earth have similar atmospheres.

Answer:

[tex]I=11.1A[/tex]

Explanation:

From the question we are told that:

Radius [tex]R=2.80[/tex]

Turns [tex]N=790[/tex]

Magnetic field B=0.0760

Generally the equation for Magnetic field at the center of the coil is mathematically given by

 [tex]B=\frac{\mu NI}{2r}[/tex]

 [tex]0.076=\frac{4\p*10^-^7*790*I}{2*0.028}[/tex]

 [tex]I=\frac{0.076*2*0.028}{4\p*10^{-7}*790}[/tex]

 [tex]I=11.1A[/tex]

Answer:

C

Explanation:

correct me if I'm wrong.

Answer:

A

Explanation:

i got it right on acellus

Answer:

(a) Negative Q

(b) Positive Q

Explanation:

Charge is the inherent property of matter due to the transference of electrons.

There are three methods of charging a body.

(i) Charging by friction: When two uncharged bodies rubbed together, then one body gets positive charged and the other is negatively charges it is due to the transference of electrons form one body to another.  

(ii) Conduction: when a charged body comes in contact with the another uncharged body, the uncharged body gets the same charge and the charge is distributed equally.

(iii) Induction: When a uncharged body keep near the charged body, the uncharged body gets the same amount of charge but opposite in sign.  

(a) When a small tack of charge Q is lowered into the hole, then due to the process of induction, the charge on the inner surface of the shell is - Q.

(b) Due to the process of conduction, the charge on the outer surface of the shell is Q.

The charge on the inner surface of the shell is negative whereas the charge on the outer surface of the shell is positive.

Due to the process of induction the inner surface of the shell creates negative charge because when a uncharged body bring near to the charged body, the uncharged body gets the same amount of charge but opposite in sign.

While on the other hand, there is no charge interaction with the outer surface so it remains positively charge so we can conclude that the charge on the inner surface of the shell is negative whereas the charge on the outer surface of the shell is positive.

Learn more about charge here: https://brainly.com/question/18102056

Answer:

the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors.

hope this helps

have a good day :)

Explanation:

Answer:

Potential Energy is the stored energy in an object or system because of its position or configuration.

Explanation:

Example: Water at the top of a waterfall, before the precipice.

Answer:

the answer should be D

Explanation:

Because if you want to earn your goals you must complete small goals to earn big goals

Answer:

a) size of the bright spot is proportional to the hollow size

b) as the size of the hole increases, the circular point decreases.

Explanation:

a) In this case the diameter of the hole is much greater than the wavelength, as the size of the hole is many orders greater than the wavelength we are in the part of geometric optics,

Consequently the size of the bright spot is proportional to the hollow size.

Consequently the size increases

b) in this case the hole diameter d = 100 10⁻⁶m and the wavelength that for the green color is lam = 500 nm  = 5 10⁻⁷ m

We see that angles are very small so the wavelength of the office is greater than the wavelength, but you can observe the effects of diffraction

          d sin θ = 1.22 m λ

the numerical constant appears by solving the equation in polar coorθdinates, because the hole is circular

the first zero occurs for m = 1

        sin θ = 1.22 λ / d

In these experiments the angles are small

        sin θ = θ

we substitute

       θ = 1.22 λ/ d

       θ = 1.22 500 10⁻⁹ / 100 10⁻⁶

       θ = 6.1 10⁻³

without the hole diameter increases by 20%            

       d’ = 1.2 d

we substitute

    θ'= 1.22 λ / d'

    θ’ = 1.22 λ /1.2 d

    θ‘= 1.22 λ /d  [tex]\frac{1}{1.22}[/tex]

    θ ’=  θ  0.83

    θ ’= 6.1 10⁻³ 0.83

    θ' = 5 10⁻³ rad

Therefore, the answer is that as the size of the hole increases, the circular point decreases.

a = (dx / dt)²

Explanation:  Unit of distance is m (metres) and unit of time is s (seconds) speed v is  first derivative of distance x versus time:

  v = dx / dt, unit is m/s.    Acceleration is second derivative of

 speed versus time  a = (dx / dt)² = (dv/dt) , unit is m/s²

Answer:

Explanation:

Acceleration is derived unit because it has two fundamental units involved i.e. meter and second square.

Answer:

The speed of the bullet just before the impact is 701 m/s

Explanation:

Given;

mass of the bullet, m₁ = 20 g = 0.02 kg

mass of the block, m₂ = 2.8 kg

displacement of the block, d = 9 mm = 9 x 10⁻³ m

duration of motion of the bullet, t = 5 ms = 5 x 10⁻³ s

Apply the principle of conservation of energy;

The final kinetic energy of the bullet = maximum potential energy of the block

[tex]\frac{1}{2} m_1v^2 = m_2gh\\\\v^2 = \frac{2m_2gh}{m_1} \\\\v= \sqrt{\frac{2m_2gh}{m_1} } \\\\v = \sqrt{\frac{2 \times 2.8 \times 9.8 \times (9\times 10^{-3})}{0.02} } \\\\v = 4.97 \ m/s[/tex]

Apply the principle of conservation of linear momentum, to determine the initial velocity of the bullet before the impact.

m₁u₁  +  m₂u₂  =  v(m₁  +  m₂)

where;

u₁ is the initial velocity of the bullet

u₂ is the initial velocity of the block = 0

m₁u₁  +  0  = v(m₁  +  m₂)

m₁u₁  =  v(m₁  +  m₂)

0.02u₁ = 4.97(2.8 + 0.02)

0.02u₁ = 14.02

u₁ = 14.02 / 0.02

u₁ = 701 m/s

Therefore, the speed of the bullet just before the impact is 701 m/s

Answer:

0.9378

Explanation:

Weight (W) of the rider = 100 kg;

since 1 kg = 9.8067 N

100 kg will be = 980.67 N

W = 980.67 N

At the slope of 12%, the angle θ is calculated as:

[tex]tan \ \theta = \dfrac{12}{100} \\ \\ tan \ \theta = 0.12 \\ \\ \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0[/tex]

The drag force D = Wsinθ

[tex]\dfrac{1}{2}C_v \rho AV^2 = W sin \theta[/tex]

where;

[tex]\rho = 1.23 \ kg/m^3[/tex]

A = 0.9 m²

V = 15 m/s

∴

Drag coefficient [tex]C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}[/tex]

[tex]C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}[/tex]

[tex]C_D =0.9378[/tex]

Explanation:

A socio-economically disadvantaged child is the one who is disadvantaged in terms of social position and economic position. Such children have limited resources in terms of education, money and future options. Four ways of helping such children are as follows:

1) Help them in education - You can help such children by giving them free tuition. If you belong to a well off family, you can get them admitted in schools as well. Provide them with books and uniform.

2) Encourage them to do well in school and pursue their passions.

3) Provide them with meals, if they do not have access to regular meals.

4) encourage them to go school regularly.

Answer:

Answer:

83.33% of the roof area will be occupied by the PV cells

Explanation:

Given the data in the question;

time-averaged lighting requirement [tex]P_{lighting[/tex] = 10 W/m²

the annual average solar irradiance [tex]q_{solar[/tex] = 150 W/m²

the PV efficiency η[tex]_{pv[/tex] = 10% = 0.1

battery charging/discharging efficiency η[tex]_{battery[/tex] = 80% = 0.8

we know that; Annual average power to the light = [tex]P_{lighting[/tex] × A[tex]_{roof[/tex]

Now, the electrical power delivered by the solar cell battery system will be;

⇒  [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]

[tex]P_{lighting[/tex]A[tex]_{roof[/tex] = [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]

Such that;

A[tex]_{pv[/tex] = [tex]P_{lighting[/tex]A[tex]_{roof[/tex] / [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]

A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = [tex]P_{lighting[/tex] /  [tex]q_{solar[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]

so we substitute

A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 10 W/m² / [ 150 W/m² × 0.1 × 0.8 ]

A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 10 W/m² / 12 W/m²

A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 0.8333

A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = (0.8333 × 100)%

A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 83.33%

Therefore, 83.33% of the roof area will be occupied by the PV cells.

Answer:

person that you know and like (not a member of your family), and who likes you

Answer:

Acceleration is 2.5 m/s^2

Explanation:

Let train 1 travel x meters and train 2 travels 1000-x meters

As per the las of acceleration

[tex]v^2 = u^2 + 2ax[/tex]

Substituting the given values, we get -

[tex]0 = 50^2 + 2ax\\ax = -1250\\[/tex]---Eq (1)

Similarly

[tex]0 = 50^2 + 2a(1000-x)\\a(1000-x) = 1250[/tex]-- Eq (2)

Substituting the value of ax from eq (1) into Eq(2), we get -

1000*a  - 1250 = 1250

a = 2500/1000 = 2.5 m/s^2

Answer:

s = 11250 m = 11.25 km

Explanation:

The distance covered by the sound wave while traveling from submarine to ocean floor and then back to submarine can be given as follows:

[tex]s = vt[/tex]

but, the distance between the floor and the submarine will be half of this value:

[tex]s = \frac{1}{2}vt[/tex]

where,

s = distance between submarine and ocean floor = ?

v = velocity of sound = 1500 m/s

t = time taken for the round trip = 15 s

Therefore,

[tex]s = \frac{1}{2}(1500\ m/s)(15\ s)[/tex]

s = 11250 m = 11.25 km

Answer:

120 m

General Formulas and Concepts:

Kinematics

Explanation:

Step 1: Define

Identify

[Given] v = 30 m/s

[Given] t = 4 s

Step 2: Solve

Answer:

[tex]\boxed {\boxed {\sf 120 \ meters}}[/tex]

Explanation:

Distance, or how far an object travels, is the product of velocity and time.

[tex]d= v*t[/tex]

The velocity is 30 meters per second and the time is 4 seconds.

Substitute the values into the formula.

[tex]d= 30 \ m/s * 4 \ s[/tex]

Multiply. The units of seconds (s) cancel.

[tex]d= 30 \ m * 4[/tex]

[tex]d=120 \ m[/tex]

The car travels a distance of 120 meters in 4 seconds at a velocity of 30 meters per second.

Answer:

Explanation:

The angular momentum of that same disk-sphere remains unchanged the very same way before and after the impact of the collision when the clay sphere adheres to the disk.

 [tex]\mathbf{I_w}[/tex] = constant.

The overall value of such moment of inertia is now altered when the clay spherical sticks. Due to the inclusion of the clay sphere, the moment of inertia will essentially rise. As a result of this increase, the angular speed w decreases in value.

Recall that:

The Kinetic energy is given by:

[tex]\mathbf{K = \dfrac{1}{2} Iw^2} \\ \\\mathbf{K = \dfrac{1}{2} lw*w}[/tex]

where;

[tex]\mathbf{I_w}[/tex] is constant and w reduces;

As a result, just after the collision, the system's total kinetic energy decreases.

The total kinetic energy of the system decreases after the collision.

The angular momentum of any rotating body is defined as the product of the moment of inertia of the body and the angular velocity of the body.

Now from the question, we can see that the angular momentum of the body remains constant before and after the impact of the collision when the clay sphere adheres to the disk.

So angular momentum will be

[tex]Iw[/tex] = constant.

The overall value of such a moment of inertia is now changed when the clay spherical sticks. Due to the inclusion of the clay sphere, the moment of inertia will essentially rise. As a result of this increase, the angular speed w decreases in value.

The Kinetic energy is given by:

[tex]KE=\dfrac{1}{2} Iw^2[/tex]

[tex]KE= \dfrac{1}{2} Iw\times w[/tex]

Since the angular momentum [tex]Iw[/tex] is constant and w is reducing then ultimately the energy of the system is decreasing.

Thus the total kinetic energy of the system decreases after the collision.

To know more about Angular momentum follow

https://brainly.com/question/25677703

If two objects are the SAME temperature and are physically touching,

then

. . .

. . .

. . .

The objects would be in thermodynamic equilibrium and as a result there would be no heat transfer.

Answer:yeah it A

Explanation:

Static friction cancels out the force of the push, so it also has a magnitude of 25 N.

Answer:

b. False

Explanation:

The visible matter that belongs to the Milky way Galaxy are traced out to be about 50 kpc distance from the center.

Kpc stands for kiloparsec. It is the unit of measurement of distance.

A parsec is[tex]$\text{ used to measure large distances}$[/tex] of the astronomical objects that lies [tex]$\text{outside the solar system}$[/tex], mainly where galaxies are involved.

1 kiloparsec is 1000 parsec and is equal to 3260 light years.

So the visible matter is about 163,078 light years away.

Hence the answer is FALSE.

(a) 4.20 J

(b) 16.74 J

For a parallel plate vacuum capacitor with area A and whose plates are separated by by a distance of d, its capacitance C is given by;

C = A∈₀ / d              --------------------(i)

Where;

∈₀ = constant called permittivity of vacuum.

The energy U stored in such capacitor is given by;

U = [tex]\frac{1}{2}[/tex]CV²             ----------------------(ii)

or

U =  [tex]\frac{1}{2}[/tex](Q²/C)        -------------------(**)

Where;

V = potential difference or voltage across the plates.

Q = charge on the plates.

(a) If the charge is held constant

Combine equations (i) and (**) to give;

U =  [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)     -----------------------(iii)

From the question;

The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e

U = 8.40J

d = 2.30mm = 0.023m

Substitute these values into equation (iii)

8.40 =  [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / 0.023)

8.40 =  [tex]\frac{1}{2}[/tex]Q² x (0.023 / A∈₀)

Multiply through by 2

2 x 8.40 = Q² x (0.023 / A∈₀)

16.80 = Q² x (0.023 / A∈₀)

Divide through by 0.023

16.80 / 0.023 = Q² x (0.023 / A∈₀) / 0.023

730.4 = Q² / (A∈₀)

Make Q² subject of the formula

Q² = 730.4(A∈₀)

Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e

d = 1.15mm = 0.0115m

Q = constant [this means that Q² still remains 730.4(A∈₀) ]

The energy stored is found by substituting these values of d and Q² into equation (iii) as follows;

U =   [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)

U = [tex]\frac{1}{2}[/tex](730.4(A∈₀)) / (A∈₀ / 0.0115)

U = [tex]\frac{1}{2}[/tex](730.4(A∈₀))(0.0115 / A∈₀)

U = [tex]\frac{1}{2}[/tex](730.4)(0.0115)

U = 4.20J

Therefore, the energy stored if the charge Q on the plates is held constant is 4.20 J

(b) If the voltage is held constant

Combine equations (i) and (ii) to give;

U =  [tex]\frac{1}{2}[/tex](A∈₀ / d)V²     -----------------------(iv)

From the question;

The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e

U = 8.40J

d = 2.30mm = 0.023m

Substitute these values into equation (iv)

8.40 =  [tex]\frac{1}{2}[/tex](A∈₀ / 0.023)V²

Multiply through by 2 x 0.023

2 x 0.023 x 8.40 = (A∈₀)V²

2 x 0.023 x 8.40 = (A∈₀)V²

0.385 = (A∈₀)V²

Make V² subject of the formula

V² = 0.385/(A∈₀)

Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e

d = 1.15mm = 0.0115m

V = constant [this means that V² still remains 0.385/(A∈₀) ]

The energy stored is found by substituting these values of d and V² into equation (iv) as follows;

U =  [tex]\frac{1}{2}[/tex](A∈₀ / 0.0115)[0.385/(A∈₀)]  

U = [tex]\frac{1}{2}[/tex](0.385/0.0115)

U = 16.74

Therefore, the energy stored if the voltage V across the plates is held constant is 16.74 J

Answer:

[tex]R=2.78\ \Omega[/tex]

Explanation:

Given that,

The current flowing in the circuit, I = 3 A

The power of the battery, P = 25 W

We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :

[tex]P=I^2R[/tex]

Put all the values to find R.

[tex]R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega[/tex]

So, the resistance is equal to [tex]2.78\ \Omega[/tex].

Explanation:

Answer:

Explanation:

A parachute works by forcing air into the front of it and creating a structured 'wing' under which the canopy pilot can fly. Parachutes are controlled by pulling down on steering lines that change the shape of the wing, cause it to turn. The main forces acting on a parachute are gravity and drag. When you first release the parachute, the force of gravity pulls it downward, and the parachute speeds toward the ground. The faster the parachute falls, though, the more drag it creates.

Answer:

52.5 J

Explanation:

Applying,

Hook's law,

F = ke............... Equation 1

Where F = Force, k = spring constant, e = extension.

make k the subject of the equation

k = F/e............ Equation 2

From the question,

Given: F = 350 Newtons, e = 30 cm = 0.3 m

Substitute these values into equation 2

k = 350/0.3 N/m

Also,

W = 1/2(ke²).................. Equation 3

Where W = work done in stretching the spring.

Also given: e = (50-20) cm = 30 cm = 0.3 m, k = 350/0.3 N/m

Substitute these values into equation 3

W = 1/2(350/0.3)(0.3²)

W = 350×0.3/2

W = 52.5 J

Answer:

Option 2

Explanation:

The complete question is

A 50 kg child sits on the left side of the bathtub. A small toy boat of 0.5 kg is on the right side of the bathtub. Which part of the bathtub has the greatest pressure

Solution

It is the bottom of the bucket that will high pressure because of the additional weight of 50 Kg boy along with the weight of the water and the tub itself.

Pressure acts in the down ward direction and is equal to the force/weight divided by the area.

Hence, option 2 is correct

(a) 42m/s

(b) 15m/s

Given:

The position of a particle at time t

s(t) = t³ + 3t

(i) To compute the average velocity

Average velocity ([tex]V_{avg}[/tex]) is given by the quotient of the change in position and change in time at a given interval of time. i.e

[tex]V_{avg}[/tex] = Δs / Δt

[tex]V_{avg}[/tex] = (s₂ - s₁) / (t₂ - t₁)             --------------------(ii)

Given interval of time is [2,5]

Therefore,

t₁ = 2

t₂ = 5

s₁ = position of the particle at t₁.

This is found by substituting t = 2 into equation (i)

s₁ = (2)³ + 3(2)

s₁ = 8 + 6 = 14

s₂ = position of the particle at t₂

This is found by substituting t = 5 into equation (i)

s₂ = (5)³ + 3(5)

s₂ = 125 + 15 = 140

Now, substitute t₁, t₂, s₁ and s₂ into equation (ii) as follows;

[tex]V_{avg}[/tex] = (s₂ - s₁) / (t₂ - t₁)

[tex]V_{avg}[/tex] = (140 - 14) / (5 - 2)

[tex]V_{avg}[/tex] = 126 / 3

[tex]V_{avg}[/tex] = 42

Therefore, the average velocity is 42m/s

(ii) To compute the instantaneous velocity.

The instantaneous velocity is the velocity of the particle at a given instant in time.

The given instant in time is t = 2.

To get the instantaneous velocity (V), differentiate equation (i) with respect to t as follows;

V = [tex]\frac{ds}{dt}[/tex]

V = [tex]\frac{d(t^3 + 3t)}{dt}[/tex]

V = 3t² + 3

Now substitute the value of t = 2 into the above equation

V = 3(2)² + 3

V = 12 + 3

V = 15

Therefore, the instantaneous velocity at t = 2 is 15m/s

Answer:

the distance between the submarine and the ocean floor is 11,250 m

Explanation:

Given;

speed of the wave, v = 1500 m/s

time of motion of the wave, t = 15 s

The time taken to receive the echo is calculated as;

[tex]time \ of \ motion \ (t) = \frac{total \ distance }{speed \ of \ wave} = \frac{2d}{v} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{1500 \times 15}{2} \\\\d = 11,250 \ m[/tex]

Therefore, the distance between the submarine and the ocean floor is 11,250 m