g an intergalactic spaceship arrives at a distant planet that rotates on its axis with a period of t

Answer:

hmmm thats too hard for me.

Explanation:

Answer:

the frequency of revolution of the second particle is f

Explanation:

centripetal force is balanced by the magnetic force for object under magnetic field is given as

Mv²/r= qvB

But v= omega x r

Omega= 2pi x f

f= qB/2pi x M

So since frequency does not depend on the velocity.therefore the frequency of revolution of the second particle remains the same and its equal to f

Answer:

  (b)  Friction increases as the normal force (weight) increases, and also depends on the surface

Explanation:

The coefficient of friction relates the normal force between surfaces to the friction force opposing motion. The coefficient depends on the surface.

The appropriate choice is ...

  Friction increases as the normal force (weight) increases, and also depends on the surface

Answer:

Products are formed from reactants by the breaking and forming of new bonds

Explanation:

In a chemical reaction, products are formed from reactants by the breaking and forming of new bonds.

Here, a chemical change takes place.

Answer:Products are formed from reactants by the breaking and forming of new bonds

Explanation:

Answer:

I think cultural value is not the benefits of biodiversity.

Answer:

see below

Explanation:

closing   1 and 2   will get the A part

close 6 and 3 to get the parallel B part

leave all of the other switches open

The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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The magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹ (Option E).

The electric flux through a surface is given by the product of the electric field strength and the area of the surface projected perpendicular to the electric field.

In this case, the electric field strength is 18 N C⁻¹, and the area of the sheet projected perpendicular to the electric field is 0.450 m²

(since the normal to the sheet makes an angle of 60° with the electric field). Multiplying these values gives the electric flux:

Electric flux = Electric field strength × Area

Electric flux = 18 N C⁻¹ × 0.450 m²

Electric flux = 8.1 N m² C⁻¹

In summary, the magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹. This value is obtained by multiplying the given electric field strength by the projected area of the sheet perpendicular to the electric field.

The angle of 60° is taken into account to determine the effective area for calculating the flux.(Option E).

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Based on the densities of the two liquids, the height of the light liquid in the right arm of the U-tube is 0.203 cm.

Let's first consider the situation before the light liquid is added. At this point, the heavy liquid fills both arms of the U-tube to the same height, h.

The pressure at point A is equal to the pressure at point B

Therefore:

P₀ + ρgh = P₀ + ρgh

where P₀ is the atmospheric pressure, ρ is the density of the heavy liquid, and g is the acceleration due to gravity.

Simplifying this equation, we get:

ρgh = ρgh

Canceling out the ρ and solving for h, we get:

h = h

In other words, the height of the heavy liquid is the same in both arms of the U-tube.

Now let's consider the situation after the light liquid is added to the right arm of the U-tube. We want to find the height, L, of the light liquid in the right arm.

Since the pressure at any two points in a connected vessel is the same, the pressure at point B (the top of the heavy liquid in the right arm) must be equal to the pressure at point C (the top of the light liquid in the right arm).

Therefore, we can write:

P₀ + ρgh = P₀ + ρg(L+h)

where L is the height of the light liquid in the right arm.

Simplifying this equation, we get:

ρgh = ρgL + ρgh

Canceling out the ρgh and solving for L, we get:

L = (ρ/ρ₀)h

where ρ₀ is the density of the light liquid.

Substituting the given values, we get:

L = (0.92 g/cm³ / 13 g/cm³)h

L = 0.070769h

Now we need to find h. We can use the fact that the volume of the heavy liquid in the left arm is equal to the volume of the heavy liquid plus the light liquid in the right arm.

The volume of the heavy liquid in the left arm is:

V₁ = Ah = (13.2 cm²)(h cm)

V₁ = 13.2h cm³

The volume of the heavy liquid plus the light liquid in the right arm is:

V₂ = A(L+h) = (2.11 cm²)(L+h cm)

V₂ = 2.11(L+h) cm³

Since these volumes are equal, we can set them equal to each other and solve for h:

13.2h = 2.11(L+h)

13.2h = 2.11L + 2.11h

11.09h = 2.11L

h = (2.11/11.09)L

Substituting this into our expression for L, we get:

L = 0.070769(2.11/11.09)L

L = 0.01345L

L = 0.01444h

Substituting the given value for the density of the heavy liquid, we get:

L = 0.01444h = 0.01444(13 g/cm³)/(0.92 g/cm³)

L = 0.203 cm

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Complete question:

A heavy liquid with a density 13 g/cm³ is poured into a U-tube as shown in the left- hand figure below. The left-hand arm of the tube has a cross-sectional area of 13.2 cm², and the right-hand arm has a cross-sectional area of 2.11 cm². A quantity of 90.2 g of a light liquid with a density 0.92 g/cm³ is then poured into the right-hand arm as shown in the right-hand figure below.

Determine the height L of the light liquid in the column in the right arm of the U-tube, as shown in the second figure above. Answer in units of cm.

A. The average speed you can use to pull the safe is 1.02 m/s

B. The time needed to pull the safe up is 17.65 s

First, we shall obtain the force. This is shown below:

F = mg

F = 60 × 9.8

F = 588 N

Finally, we shall obtain the average speed. Details below:

Power = force × average speed

600 = 588 × average speed

Divide both sides by 588

Average speed = 600 / 588

Average speed = 1.02 m/s

The time needed to pull the safe up can be obtained as follow:

Time = Total distance / average speed

Time = 18 / 1.02

Time = 17.65 s

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Complete question:

You know you can provide 600 W

of power to move large objects. You need to move a 60-kg

safe up to a storage loft, 18 m

above the floor.

Part A

With what average speed can you pull the safe straight up?

Part B

What is the time needed to pull the safe up?

Recall that

\({v_f}^2-{v_i}^2=2a\Delta x\)

where \(v_i\) and \(v_f\) are initial and final velocities, respectively; \(a\) is acceleration; and \(\Delta x\) is the net displacement, or distance if the object is moving in a single direction.

The car has initial speed 20 m/s and acceleration -3.25 m/s². It comes to a stop, so it has 0 final speed. Then

0² - (20 m/s)² = 2 (-3.25 m/s²) ∆x

∆x = (20 m/s)² / (7.5 m/s²) ≈ 53.3 m

Answer:

m5=555 5/9 kg

Explanation:

Consider the system formed by the blocks m1, m2 and m3.

Let the forces acting in the direction of the force F be positive.

There is a total of 4 forces acting on the system: the force F and the weighs of blocks m1, m2 and m3.

The force F and the weight of m1 are positive, while the weighs of blocks m2 and m3 are negative.

The weighs of blocks m1, m2 and m3 can be found by multiplying the masses m1, m2 and m3 by the acceleration of gravity, g=9.81m/s^2:

The net force acting on the system is:

The total mass of the system is:

According to Newton's Second Law of Motion, the acceleration of the system is given by:

Notice that none of the options coincides with this answer. Nevertheless, the option 1.33m/s^2 is closest to 1.46m/s^2. We can justify that answer if we consider g=10m/s^2, which leads us to the following calculations:

Therefore, the correct choice is: 1.33 m/s^2.

The gravitational potential energy (PE) of an object can be calculated using the formula:

PE = mgh

Where:

PE is the gravitational potential energy

m is the mass of the object

g is the acceleration due to gravity

h is the height or distance above the reference point (usually the ground)

In this case, you provided the weight of the object, which is 24 N. The weight of an object is the force due to gravity acting on it, and it can be calculated using the formula:

Weight = mg

Where:

Weight is the force due to gravity

m is the mass of the object

g is the acceleration due to gravity

Since you haven't provided the mass of the object, we can find it by rearranging the weight formula:

m = Weight / g

Now, we can calculate the mass of the object:

m = 24 N / 9.8 m/s² ≈ 2.45 kg

Now we have the mass (m = 2.45 kg), the height (h = 3.2 m), and the acceleration due to gravity (g ≈ 9.8 m/s²). Plugging these values into the formula for gravitational potential energy:

PE = mgh

= 2.45 kg × 9.8 m/s² × 3.2 m

≈ 76.736 J

Therefore, the gravitational potential energy of the object at a height of 3.2 m from the ground is approximately 76.736 joules (J).

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The work done by gas is 0.753 J and change in internal energy is 4.247J

So we are given that mass is 2kg , radius 1 cm and the amount of heat is 5 cm

The piston raised by 2.4cm

As we know that Work done is PΔV

Where ΔV is change in volume

Therefore ΔV =  πr^2 h = π x (.01)^2 x .024 =7.53×10^(-6)m^3

Here pressure is 10^5 pa

So W = \(10^5\times7.53\times10^-(6)\)

Therefore W = 0.753 J

Now coming to change in internal energy

Change in Internal Energy = Heat Added - Energy lost in work

∴ 5J - 0.753 J = 4.247J

Hence the change in internal energy is 4.247 J

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It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.

2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.

3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.

4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.

5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.

6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".

7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).

8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).

9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.

10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.

11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).

12. Evaluating the expression, we find: t ≈ 546.

13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

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The right sphere will acquire an equal and opposite net positive charge to balance the negative charge on the left sphere.

Since the left sphere is attracted to the positively charged rod, it means that the left sphere acquires a temporary negative charge due to induction.

The positive charge on the rod repels electrons in the left sphere, causing them to move away from the rod side and accumulate on the opposite side, resulting in a net negative charge on the left sphere.

According to the principle of charge conservation, the net charge on the system must remain zero. Therefore, the right sphere acquires an equal and opposite net positive charge to balance the negative charge on the left sphere.

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Answer:

If you meant 8 m per second then

Explanation:

If you divide 8 by 200 that equals 25 so that would make the runner finish in 25 seconds!

Your welcome!

Answer:

maximum velocity,then u draw your diagram then u calculate for max velocity then the answer u get use it for total shape

Answer:

Useable energy, electricity

The orbital period of the rocks in the asteroid belt is approximately 5.73 years.

What is the average distance, a, between the sun and the asteroid, expressed in astronomical units?

The distance from the Sun to the asteroid belt is between 2-4 AU, or between 300 and 600 million kilometers (186-272 million mi).

Using Kepler's Third Law, we have:

P² = a³

where P is the orbital period of the rocks in years, and a is the semi-major axis of their orbit in astronomical units (AU).

Substituting a = 3.2 AU, we get:

P² = (3.2 AU)³

P² = 32.768 AU³

P = \(\sqrt{32.768}\) years

P ≈ 5.73 years

Therefore, the orbital period of the rocks in the asteroid belt is approximately 5.73 years.

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The paragraph primarily discusses the concept of overwinding old clocks and its consequences, indicating that overwinding clocks used to be a common mistake. Here option D is the correct answer.

The paragraph explains that overwinding occurs when the mainspring is almost fully wound, but the operator continues to turn the winding key, resulting in the spring coiling too tightly or even breaking.

This suggests that overwinding was a mistake commonly made in the past when operating old-fashioned clocks. The other options, such as clocks changing over the years or clocks becoming fragile with age, are not directly addressed in the paragraph and are therefore less supported.

The option e. "time flies when you're having fun" is unrelated to the paragraph and can be disregarded as an irrelevant answer choice. Hence option D is the correct answer.

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The wavelength of light that should be used to obtain fringes that are 0.0045 m wide after reducing the distance between the screen and the slit by half is 2.25 * 10^7 Å.

Huygens's Principle states that every point on a wavefront can be considered as a source of secondary spherical wavelets that spread out in all directions with the same speed as the original wave. The new wavefront is formed by the envelope of these secondary wavelets at a later time.

Now, let's consider a Young's double-slit experiment. In this experiment, when light passes through two narrow slits, it creates an interference pattern on a screen behind the slits. The fringe width is the distance between two consecutive bright or dark fringes in the pattern.

Given that the fringe width obtained is 0.6 cm and the wavelength of light used is 4500 Å (Angstroms), we can calculate the wavelength of light required to obtain fringes that are 0.0045 m wide.

We can use the formula for fringe width in Young's double-slit experiment:

w = (λ * D) / d

Where:

w is the fringe width,

λ is the wavelength of light,

D is the distance between the screen and the double slits, and

d is the distance between the two slits.

Let's calculate the value of D/d using the given information:

D/d = w / λ

= 0.006 m / 4500 Å (1 m = 10^10 Å)

= 0.006 * 10^10 / 4500 m^-1

Now, if the distance between the screen and the slit is reduced by half, the new value of D/d would be:

(D'/d) = (0.006/2) * 10^10 / 4500 m^-1

Now, we can rearrange the equation to solve for the new wavelength (λ'):

(λ' * D') / d = (D/d)

λ' = (D/d) * d / D

= [(0.006/2) * 10^10 / 4500] * (4500 / 0.006) Å

= 0.0045 m * 10^10 / 2 Å

= \(0.00225 * 10^{10\) Å

=\(2.25 * 10^7\)Å

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Answer:

the energy of the spring at the start is 400 J.

Explanation:

Given;

mass of the box, m = 8.0 kg

final speed of the box, v = 10 m/s

Apply the principle of conservation of energy to determine the energy of the spring at the start;

Final Kinetic energy of the box = initial elastic potential energy of the spring

K.E = Ux

¹/₂mv² = Ux

¹/₂ x 8 x 10² = Ux

400 J = Ux

Therefore, the energy of the spring at the start is 400 J.

Answer:

they are quantities with magnitude without direction e.g weight,

When the teacher asked the student to make words out of the jumbled word gzeysktqix, the student was being tested on his ability to unscramble words. Unscrambling words is the process of taking a word or series of letters that are out of order and rearranging them to form a word that makes sense.

When trying to unscramble a word, it is important to look for any patterns that can help identify smaller words within the jumbled letters. This can help make the process easier and quicker. For example, in the jumbled word gzeysktqix, one might notice that the letters "sktqix" appear together.

This could indicate that these letters could potentially form a word. By looking at the remaining letters, one could notice that the letters "g", "z", "e", and "y" could also form smaller words. After some rearranging, the letters can be unscrambled to form the words "sky", "zig", "sex", and "yet". These are just a few examples, as there are likely many other words that can be formed from this jumbled word.

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Given,

The resistance of the resistor, R=25 Ω

The current travelling through the resistor, I=8 A

From Ohm's law, the voltage across a resistor is proportional to the current through the resistor.

Thus the voltage across the given resistor is,

On substituting the known values,

Thus the voltage of the battery of the circuit is 200 V

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: True

Explanation: