An electron orbits a proton in a hydrogen atom at a speed of 2.9 x 10^6 m/s. If it takes the electron 1.6 x 10^-15 s to complete one orbit, what is the circumference of the atom?

Answer:

The velocity is 2661.5 m/s.

Explanation:

Radius, horizontal distance, d = 309 km

height, h = 25 km

acceleration due to gravity on moon, g =3.71 m/s^2

Let the time taken is t and the horizontal velocity is u.

horizontal distance = horizontal velocity x time

309 x 1000 = u t .... (1)

Use second equation of motion in vertical direction.

\(h = u_yt +0.5 gt^2\\\\25000 = 0 + 0.5\times 3.71\times t^2\\\\t =116.1 s\)  

So, put in (1)

309 x 1000 = u x 116.1

u = 2661.5 m/s

Answer:

A) how your body uses oxygen

Analysing the Question:

We know that equilibrium is the state of a body when it has equal and opposite forces being applied on it

In this case, a net downward force of 496N is being applied and a net upward force of (106 + 106 + 142 + x) N

Finding the missing force:

Since we have to achieve equilibrium, the net upward forces have to be equal to the net downward forces

So,  (106 + 106 + 142 + x) = 496

354 + x = 496

x = 496 - 354

x = 142 N

Therefore, the missing force is 142 N

The speed of the truck just before braking is 23.24 m/s.

The speed of the truck before braking is calculated by applying the third kinematic equation as shown below.

v² = u² - 2as

where;

When the truck stops, the final velocity = 0

0 = u² - 2as

u²  = 2as

u = √2as

u = √ ( 2 x 6 x 45 )

u = 23.24 m/s

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The force that results in the decrease in speed from the midpoint to the end of the track is friction. The friction force slows down the vehicle because it acts in the opposite direction of the car's motion.

The force that would cause the Hot Wheels car to slow down from the midpoint of the track to the end of the track is friction between the car's wheels and the track.

Friction is a force that opposes motion between two surfaces in contact.

In this case, the wheels of the car and the surface of the track are in contact, and the friction force acts in the opposite direction of the car's motion, which slows it down.

As the Hot Wheels car travels down Track #2 during the Speed Lab activity, its initial velocity decreases due to friction.

Friction is a resistance force that opposes motion.

It is caused by the interaction between the surfaces in contact. In this case, the surface of the track and the wheels of the car are in contact.

When the car is moving, there is friction between the two surfaces.

The direction of the friction force is opposite to the direction of motion of the car.

This means that the friction force slows the car down.

In conclusion, the force that results in the decrease in speed from the midpoint to the end of the track is friction.

The friction force slows down the vehicle because it acts in the opposite direction of the car's motion.

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Answer:

Dunno

Explanation:

Dunno

Since, density of the bracelet is not equal to the density of gold, then, the bracelet is not pure gold.

To know if the bracelet is pure gold, we calculate the density of the bracelet and compare it to the density of pure gold (19.3 g/cm³).

That is, for the bracelet to pure gold,

Density of bracelet ≈ 19.3 g/cm³

Density can be defined as the ratio of the mass and the volume of a substance.

The formula of Density is give as

⇒ Where:

From the question,

⇒ Given:

⇒ Substitute these values into equation 1

Hence, since the density of the bracelet is not equal to the density of gold, then, the bracelet is not pure gold.

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Answer:

C. 4736J

Explanation:

The mechanical energy of ball is the sum of its kinetic energy and potential energy at any point;

 Mechanical energy  = kinetic energy + potential energy

Given:

kinetic energy  = 2000J

potential energy  = 2376J

So;

 Mechanical energy  = 2000J + 2376J  = 4736J

Answer:

I believe the answer is C.

Answer:

C. The ball's kinetic energy increases as it falls and its potential energy decreases.

The density of the Gasoline is 721.1302 kg/m3. This is the answer for the given question.

100 cc of gasoline should be added to the cylinder before weighing it on the scale in grams. When the cylinder is filled with gasoline, deduct the cylinder's mass from the cylinder's overall mass. This is the gasoline's mass. To determine the density, divide this number by the volume, which is 100 ml. As an alternative, use a hydrometer.

Gasoline is a mixture of many different chemicals, including hydrogen and carbon (hydrocarbons). A typical gasoline blend contains about 150 hydrocarbons. Gasoline has a density of ρ = 700–800 kg/m3, while diesel fuel varies between ρ = 830–950 kg/m3.

Given,

mass(m) = 58.7 kg

volume(V) = 0.0814 m3  

Formula for density = mass/volume

Density = 58.7 / 0.0814

Density = 721.1302 kg/m3.

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Answer:

B

Explanation:

i took the test

Answer:

The velocity of the box is related to the angular velocity of the spool, which is given by the equation:

v = r * ω

where r is the radius of the spool and ω is the angular velocity of the spool. The angular velocity of the spool, in turn, is related to the torque applied to the spool by the tension in the string, which is given by the equation:

τ = I * α

where τ is the torque, I is the moment of inertia of the spool, and α is the angular acceleration of the spool.

The tension in the string is equal to the weight of the box, which is given by:

T = m * g

Putting all of these equations together, we can solve for the time it takes for the box to reach the floor. Here's how:

First, we can find the angular acceleration of the spool using the torque equation:

τ = I * α

T = m * g = τ

m * g = I * α

α = (m * g) / I

α = (35.30 kg * 9.81 m/s^2) / 4.00 kg*m^2

α = 86.53 rad/s^2

Next, we can find the angular velocity of the spool using the kinematic equation:

ω^2 = ω_0^2 + 2 * α * θ

where ω_0 is the initial angular velocity (which is zero), θ is the angle through which the spool has turned (which is equal to the distance the box has fallen divided by the radius of the spool), and ω is the final angular velocity (which is what we want to find). Solving for ω, we get:

ω^2 = 2 * α * θ

ω = sqrt(2 * α * θ)

ω = sqrt(2 * 86.53 rad/s^2 * (3.50 m / 0.10 m))

ω = 166.6 rad/s

Finally, we can find the time it takes for the box to reach the floor using the equation:

v = r * ω

v = 0.10 m * 166.6 rad/s

v = 16.66 m/s

t = d / v

t = 3.50 m / 16.66 m/s

t = 0.21 s

Answer:

The current I1 through the resistor R1 is:

I1 = (E1 - E2) / (R1 + R2 + R3 + R4 + R5 + R6 + R7 + R8 + R9)

  = (3 - 6) / (22 + 33 + 1.1 + 3.3 + 33 + 11 + 33 + 1.1 + 3.3)

  = -3 / 116.8

  = -0.02563 A

Explanation:

Answer:

20 kg*m/s

Explanation:

we're looking for the change in the cars momentum . the formula to identify the cars momentum is Δp=m*vf-m*vi . In this case m= mass , vf= velocity final, and vi= velocity initial . The mass is how much the car weighs , which is 4 kg . Velocity final would be 8 m/s since that's how fast it last was . Velocity initial would be 3 m/s since it had first started at that speed . Now we can jus plug it into the formula .

m= 4 kg

vf= 8 m/s

vi= 3 m/s

Δp= (4)*(8) - (4)*(3)

Δp= 32-12

Δp= 20 kg*m/s

hope this helps .

Explanation:

It is given that, the radius of a ball is \((5.2\pm 0.2)\ cm\).

We need to find the percentage  error in the volume of the ball. The volume of a sphere is : \(V=\dfrac{4}{3}\pi r^2\)

The percentage  error is given by :

\(\dfrac{\Delta V}{V}=3\dfrac{\Delta r}{r}\times 100\)

We have, \(\Delta r=0.2\) and r = 5.2

So,

\(\dfrac{\Delta V}{V}=3\times \dfrac{0.2}{5.2}\times 100\\\\\%=11\%\)

So, the percentage  error in the volume of the ball is 11%

The coefficient of static friction between the block and the table is 0.07.

The coefficient of static friction between the block and the table is determined by Newton's second law of motion as shown below;

Fs = μFₙ

where;

μ = Fs/Fₙ

μ = Fs/mg

μ = 2/(3 x 9.8)

μ = 0.07

Thus, the coefficient of static friction between the block and the table is 0.07.

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Answer:

B

Explanation:

ANSEWER :B IN ROWS ONLY

Answer

.0003

Explanation:

Answer:

1) t = 3.45 s, 2)  x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,

5) θ = -40.2º

Explanation:

This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.

1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff

      y = y₀ + \(v_{oy}\) t - ½ g t²

When leaving the cliff the speed is horizontal  v_{oy}= 0 and at the bottom of the cliff y = 0

      0 = y₀ - ½ g t2

      t = √ 2y₀ / g

      t = √ (2 60 / 9.8)

      t = 3.45 s

2) The horizontal distance traveled

     x = v₀ₓ t

     x = 40 3.45

     x = 138 m

3) The vertical velocity at the point of impact

     v_{y} = I go - g t

     v_{y} = 0 - 9.8 3.45

     v_{y} = -33.81 m /s

the negative sign indicates that the speed is down

4) the resulting velocity at this point

   v = √ (vₓ² + v_{y}²)

   v = √ (40² + 33.8²)

   v = 52.37 m / s

5) angle of impact

    tan θ = v_{y} / vx

    θ = tan⁻¹ v_{y} / vx

    θ = tan⁻¹ (-33.81 / 40)

    θ = -40.2º

6) sin (-40.2) = -0.6455

7) tan (-40.2) = -0.845

8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis

The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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The gap the boat travels earlier than slowing down is 2.303m/\(αe^2\) v0 ln(10)

The equation of movement for the boat is given through:

m dv/dt = -\(αe^2v\)

in which m is the mass of the boat, v is the rate of the boat, and α is a steady that relies upon at the residences of the water.

To resolve for the rate of the boat as a feature of time, we are able to separate the variables and combine each facets of the equation:

\(m \int\limits v0^v dv / (v*√(1/v)) = -αe^2 ∫ zero^t dt\)

Simplifying the left-hand aspect the usage of the substitution u = 1/v and \(du = -dv/v^2\), we get:

\(-m ∫ 1/v0^1/v du = -αe^2 t\\-m ln(v/v0) = -αe^2 t\\ln(v/v0) = αe^2 t / m\\v(t) = v0 * e^(-αe^2 t / m)\)

To discover the time at which the boat stops, we are able to set v(t) identical to 0 and resolve for t:

\(v(t) = v0 * e^(-αe^2 t / m) = zero\\e^(-αe^2 t / m) = zero\\-αe^2 t / m = -∞\\t = ∞\)

v = zero.1v0 We can resolve for t in this example through putting v(t) = zero.1v0:

\(v(t) = v0 * e^(-αe^2 t / m) = zero.1v0\\e^(-αe^2 t / m) = zero.1\\-αe^2 t / m = ln(zero.1)\\t = -m/αe^2 ln(zero.1)\)

Finally, to discover the gap the boat travels earlier than it slows down, we are able to combine the rate equation from v0 to zero.1v0:

∫ \(x0^x dt = ∫ zero^t v(t) dt\)

in which x0 is the preliminary function of the boat. Substituting for v(t), we get: ∫\(x0^x dx / v(x) = -m/αe^2 ln(zero.1)\)

Using the substitution u = v/v0, we get: ∫ \(1^zero.1 du / u = -m/αe^2\)ln(zero.1)

\(ln(10) = -m/αe^2 ln(zero.1)\\x - x0 = -m/αe^2 ln(zero.1) * ln(10) / v0\\x - x0 = 2.303m/αe^2 v0 ln(10)\)

Therefore, the gap the boat travels earlier than slowing down is 2.303m/\(αe^2\) v0 ln(10). 

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Answer:

D

Explanation:

Answer:

A meteorologist is a scientist who studies weather Explanation:

A meterologist is a scientist who studies the weather condition of an environment.

A meteorologist is a scientist that specifies in the act of understanding and forecasting the atmospheric condition or weather of the Earth surface.

The branch of science that studies weather is called meteorology and a meterologist uses several techniques and tools to explain weather.

Therefore, a meterologist is a scientist who studies the weather condition of an environment.

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Lenses, which are pieces of curved, clear glass, were employed in early telescopes to focus light.

Curved mirrors are used by the majority of telescopes nowadays to collect light from the night sky. Light is focused by a telescope's mirror or lens' shape.

Astronomers use a telescope to observe distant things. Curved mirrors are used by the majority of telescopes, including all large telescopes, to collect and concentrate light from the night sky.

The original telescopes employed lenses, which are simply curved pieces of clear glass, to focus light. The "optics" of a telescope are the mirrors or lenses. Strong telescopes may view objects that are extremely faint and incredibly far away.

Therefore, Lenses, which are pieces of curved, clear glass, were employed in early telescopes to focus light.

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Answer:

the reaction is reversible

Answer:

the Earth's crust is broken into about 12 plates that float on hotter, softer rocks in the underlying mantle

Explanation:

Answer:Hi, Sam!

Okay, so let's call the 3 consecutive even integers, X, X + 2 , X +4

So the sum of the smaller two (X +X + 2) = X + 4 +6

2X +2 = X + 10

X = 8

Therefore,

X = 8

X + 2 = 10

X + 4 + 12

These are your integers

Explanation:

The evaluation of the capacitor (in series) network is as follows;

Part A

Q = 3.2 μC

Part B

Q₁ = 3.2 μC

Part C

Q₂ = 3.2 μC

Part D

U = 76.8 μJ

Part E

U₁ = 34 2/15 μJ

Part F

U₂ = 53 1/3 μJ

Part G

V₁ = 21 1/3 V

Part H

V₂ = 26 2/3 V

A capacitor consists of pairs of conductors separated by insulators. Capacitors are used to store electric charge.

The specified parameters are;

The voltage across ab = 48 V

The capacitance of the first capacitor, C₁ = 150 nF

Capacitance of the second capacitor, C₂ = 120 nF

Part A

The total charge in a capacitor network can be found as follows;

\(C_{eq} = \left(\dfrac{1}{150} + \dfrac{1}{120} \right)^{-1} nF = \left(\dfrac{3}{200} \right)^{-1}nF\)

\(C_{eq} =\left(\dfrac{3}{200} \right)^{-1}nF=66\frac{2}{3} \, nF\)

\(Q_{eq} = C_{eq}\times V_{ab}\)

Therefore;

\(Q_{eq}\) = 66 2/3 nF × 48 V = 3,200 × 10⁻⁹ C = 3.2 μC

Part B

The charge in the 150 nF capacitor is obtained from the formula for the charge in a capacitor; Q = C × V as follows;

Q = C₁V₁ = C₂V₂

The charge in the capacitors, C₁ and C₂ are the same as the total charge of 3.2 μC

The charge, Q₁ on the 150 nF capacitor, C₁ is therefore, 3.2 nC

Part C

The capacitors, C₁ and C₂ are in series, therefore, the charge in each capacitor is equivalent to the charge in the circuit, which is 3.2 μC.

Therefore, the charge, Q₂, in the 120 nF capacitor, C₂ is 3.2 μC

Q₂ = 3.2 μF

Part D

The total energy stored in the network can be obtained using the formula;

U = (1/2)·C·V²

Where;

U = The energy in the capacitor

C = The equivalent capacitance of the network = 66 2/3 nF

V = The voltage

Therefore;

\(U = \dfrac{1}{2} \times C_{eq}\times V^2\)

\(U = \dfrac{1}{2} \times 66\frac{2}{3} \times 10^{-9}\times 48^2 = 76.8\)

Part E

The energy stored in the 150 nF capacitor is found as follows;

\(Q_{eq}\) = Q₁ = C₁ × V₁

V₁ = \(Q_{eq}\) ÷ C₁

Therefore;

V₁ = 3.2 μC ÷ 150 nF = \(21\frac{1}{3}\) V

U₁ = 0.5×C₁×V₁²

Part F

The energy stored in the 120 nF capacitor, U₂, can be found as follows;

V₂ = 3.2 μC ÷ 120 nF = \(26\frac{2}{3}\) V

U₂ = 0.5 × 150 nF × \(\left(26\frac{2}{3} \, V\right)^2\) = \(53\frac{1}{3}\, \mathrm{ \mu J}\)

Part G;

The potential difference across the 150 nF, obtained in Part E, is 21 1/3 V

Part H

The potential difference across the 120 nF, obtained in part F, is 26 2/3 V

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Answer:

2,760watts

Explanation:

Power of the kettle = current * voltage

P = IV

Given

I = 12A

V = 230V

Substitute into the formula

Power of the kettle = 12*230

Power of the kettle = 2,760watts

The minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.

To maintain equilibrium, the torques acting on the meter stick must balance each other. The torque is given by the formula:

τ = r * F * sin(θ)

where τ is the torque, r is the distance from the pivot point to the point where the force is applied, F is the force applied, and θ is the angle between the force vector and the lever arm.

In this case, the meter stick is in equilibrium when the torques on both sides of the pivot point cancel each other out. The torque due to the weight of the meter stick itself is acting at the center of mass of the meter stick, which is at the 50.0 cm mark.

Let's denote the mass to be placed on the 0.00 cm mark as M. The torque due to the weight of M can be calculated as:

τ_M = r_M * F_M * sin(θ)

where r_M is the distance from the pivot point to the 0.00 cm mark (which is 30.0 cm), F_M is the weight of M, and θ is the angle between the weight vector and the lever arm.

Since the system is in equilibrium, the torques on both sides of the pivot point must be equal:

τ_M = τ_stick

r_M * F_M * sin(θ) = r_stick * F_stick * sin(θ)

Substituting the given values:

30.0 cm * F_M = 20.0 cm * (0.0400 kg * 9.8 m/s^2)

Solving for F_M:

F_M = (20.0 cm / 30.0 cm) * (0.0400 kg * 9.8 m/s^2)

F_M = 0.0264 kg * 9.8 m/s^2

F_M = 0.25872 N

Finally, we can convert the force into mass using the formula:

F = m * g

0.25872 N = M * 9.8 m/s^2

M = 0.0264 kg

Therefore, the minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.

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