Who predicted that a comment has a cyclic repeating pattern

When the mass of Earth doubles, acceleration due to gravity doubles as well.

Apply Newton's second law of motion;

F = mg --- (1)

where;

F = GmM/R²  --- (2)

where;

Solve (1) and (2)

mg =  GmM/R²

g = GM/R²

when the mass of Earth doubles, acceleration due to gravity becomes;

g' = G(2M)/R²

g' = 2(GM/R²)

g' = 2g

Thus, when the mass of Earth doubles, acceleration due to gravity doubles as well.

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The type of circuit which have been made is the series circuit in this scenario.

This is a complete path which involves the whole electric current flowing through the various parts such as resistor etc..

There is only one path of current in which it does not undergo any form of split during motion.

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Since acceleration is constant, the average and instantaneous accelerations are the same, so that

[tex]a = a_{\rm ave} = \dfrac{\Delta v}{\Delta t} = -\dfrac{v_i}{10.0\,\rm s}[/tex]

By the same token, we have the kinematic relation

[tex]v^2 - {v_i}^2 = 2a\Delta x[/tex]

where [tex]v[/tex] is final speed, [tex]v_i[/tex] is initial speed, [tex]a[/tex] is acceleration, and [tex]\Delta x[/tex] is displacement.

Substitute everything you know and solve for [tex]v_i[/tex] :

[tex]0^2 - {v_i}^2 = 2\left(-\dfrac{v_i}{10.0\,\rm s}\right)(75.0\,\mathrm m)[/tex]

[tex]\implies {v_i}^2 - \left(15.0\dfrac{\rm m}{\rm s}\right) v_i = 0[/tex]

[tex]\implies v_i \left(v_i - 15.0\dfrac{\rm m}{\rm s}\right) = 0[/tex]

[tex]\implies v_i = \boxed{15.0\dfrac{\rm m}{\rm s}}[/tex]

Answer:

Newton's corpuscular theory stated that light consisted of particles that travelled in straight lines. Huygens argued that if light were made of particles, when light beams crossed, the particles would collide and cancel each other. He proposed that light was a wave.

Answer:

Soln:Here

Given,

Distance(s)=5m

Time taken(t)=20min=2400seconds

Speed(s)=?

We know that,

Speed(s)=Distance/timetaken

=5m/2400sec

580m/s

Thus the average speed of his bicycle is 580 m/s..

Thank you...

The will dog catch up with the rabbit in 6 minutes assuming both their velocities remain constant during the chase.

The time that the dog will catch up with the rabbit is given as follows:

Let the distance covered by the rabbit be x.

Distance covered by dog = x + 30

The time taken will be the same T

Equating both times.

(x + 30)/10 = x/5

x = 30 m

Solving for T in equation (ii);

T = 30/5 = 6 minutes

In conclusion, time is obtained as a ratio of distance and speed.

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50 V to dissipate a power of 10 W

A resistor dissipates 2.00 W when the rms voltage of the emf is 10.0 V.

[tex]$\mathrm{V}_{r m s}=\mathrm{P}$[/tex]

10 v provides 2 W of power whereas V provides 10 W of power.

10 V = 2 W

x V=10 W

By using cross multiplication

2 x = 100

x = 50 V

So, 50 V to dissipate a power of 10 W

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The friction force does the greatest magnitude of work on the crate

Consider all four forces. The normal force does no work at all, since there is no motion in the direction of that force, perpendicular to the ramp. The force of gravity is smaller than the force of friction, since you still need to push the crate to get constant velocity. The force of you pushing is also smaller than the force of friction, since you are moving down a ramp, and are therefore assisted by gravity. Therefore the force doing greatest magnitude of work is the force of friction. Note that, even though the frictional work is negative, it still has the greatest magnitude

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If the primary wire's power is 10 A and one branch's power is 4 A, another branch's power will be 6A.

According to Kirchhoff's current law (KCL), the total current flowing through a parallel route circuit's junction equals the total current flowing away from it.

Provided that one of the two branches through which power exits the intersection has a flow of 4A, and also that the junction's overall flow entering it is 10A, the entire current going the junction should be 10A.

Consequently, the second wire's power may be expressed as;

I = I1+ I2 [ where I= total current (10A);

                I1= current in one branch (4A) &

                I2= current in another branch]

⇒I2 = I - I1

⇒I2 = 10A - 4A

⇒I2 = 6A

Therefore, it can be concluded that when the primary wire bears 10A power having 4A in one of its branches, another branch carries 6A power.

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Answer:

Point A

Explanation:

Because it reaches maximum height

a) The motion along the vertical direction and the motion along the horizontal direction.

b) The object remains in the air for a time period of 2usin(θ)/g.

Any object that is thrown in the air when gravity is acting on it is called a projectile. The motion of this projectile is called projectile motion.

When the projectile is thrown in the air at some angle θ, then there are two independent motions taking place at the same time. First is the component of motion along the vertical direction along which gravity acts. Second is the component of motion along the horizontal direction along which the object moves with a constant velocity. No force acts along the horizontal direction. The horizontal motion does not affect the vertical motion and the converse is also true. So these are independent of each other.

The time of flight is the time during which a projectile remains in the air. This time of flight is calculated using the formula,

T=2usin(θ)/g

where T is the time of flight, u is the initial velocity and g is the acceleration due to gravity.

Hence, the object remains in the air for a time period of 2usin(θ)/g.

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Answer:

In pair production, after the loss of Kinetic energy, the angular separation between the two photons is 180°.

Explanation:

Therefore, the angle of separation between the two photons is 180°.

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The magnitude and sign of the charge are 0.8 MC and negative respectively.

To find the answer, we need to know about the electric potential of a point charge.

From the expression of electric potential, charge is

q= (potential ×r)/K

= (-3.5×0.002)/ (9×10^(-9))

= -0.8 mega coulomb.

Thus, we can conclude that the magnitude and sign of the charge are 0.8 MC and negative respectively.

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The speed of tsunami is a.0.32 km. 

Steps involved  :

The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?

Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32

As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.

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The work that is done when twice the load is lifted twice the distance is

four times as much

The net work performed by forces acting on an object equals the change in kinetic energy, according to the work-energy theorem.

when an item slows down, the net work applied to it decreases, its change in kinetic energy is negative, and its ultimate kinetic energy is less than its starting kinetic energy. When an item accelerates, positive net work is done on it. All the forces acting on an item must be taken into consideration when determining the net work. You will obtain an incorrect result if you exclude any forces that affect an item or if you add any forces that do not affect it.

Hence The work that is done when twice the load is lifted twice the distance is four times as much

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The initial velocity of the hockey puck is obtained as 22 m/s.

The frictional force of the hockey puck is the force that causes it to stop. Now;

Ff = μmg

Ff = 0. 51 * 0.16 Kg * 9.8 m/s^2

Ff = 0.8 N

Now;

F = mv^2/2x

Where;

m = mass

v = velocity

x = distance

v =√ 2xF/m

v = √ 2 * 47.7 * 0.8 / 0.16

v = 22 m/s

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D. The efficiency of the ramp is 54.6 %.

V.R = distance moved by effort/distance moved by load = L/h

V.R = (6.12)/(1.53) = 4

M.A = Load/Effort

M.A = (451 x 9.8)/(2025)

M.A = 2.183

E = (M.A / V.R) x 100%

E = (2.183 / 4) x 100%

E = 54.6 %

Thus, the efficiency of the ramp is 54.6 %.

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The time required to reach the velocity is 15.9s.

The acceleration of the plane is 5.67m/s²

Acceleration is solved by the given formula.

Acceleration = velocity ÷ Time

Finding Time

Time = distance ÷ speed

Time = 1435m ÷ 90.1m/s = 15.9s

Finding Acceleration

Acceleration = velocity ÷ time

Acceleration = 90.1m/s ÷ 15.9s = 5.67m/s²

(A) The angle of this in polar coordinates is 30⁰.

(B) The North/South and East/West components of its displacement is 250 m and 433 m respectively.

60° N of W lies in the fourth quadrant;

θ = 90 - 60 = 30⁰

in polar coordinate = (r,θ) = (500 m, 30⁰)

Dy = D sinθ

Dy = 500 m x sin(30) = 250 m

Dx = D cosθ

Dx = 500 m x cos(30) = 433 m

Thus, the angle of this in polar coordinates is 30⁰.

The North/South and East/West components of its displacement is 250 m and 433 m respectively.

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Option (a) B3 supergiant is correct.

Just before it exploded, the star that became supernova 1987A was a B3 supergiant.

Any star with extremely high intrinsic luminosity and relatively vast size is referred to as a supergiant star.

B3 supergiant is a blue supergiant.

An OB supergiant, often known as a blue supergiant (BSG), is a hot, brilliant star like Rigel.

The hottest stars in the universe are blue supergiants, with temperatures ranging from 10,000 K to 50,000 K or higher.

A type II supernova known as SN 1987A occurred in the Milky Way's dwarf satellite galaxy, the Large Magellanic Cloud.

Red supergiants are the most frequent supernova progenitors, and it was once thought that only red supergiants could explode as supernovae. However, because SN 1987A's parent star, was a B3 blue supergiant, this notion had to be reexamined.

Hence, just before it exploded, the star that became supernova 1987A was a B3 supergiant.

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The magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.

Given values:

Length of non-conducting rod, l = 1.20 m

Charge on positive rod, +Q = +2.50 μC = +2.50 × 10⁻⁶ C

Charge on negative rod, -Q = -2.50 μC = -2.50 × 10⁻⁶ C

Distance from point P of each rod, x = 60 cm = 0.60 m

Calculation of Net electric force exerted on point P:

Consider an electron released at point P, then the net electric force exerted will be given as:

F = e. E_net       - ( 1 )

Step 1:

The net electric field value is given as:

E_net  = E₁ cos Φ + E₂ cos Φ      

           = 2E₁ cos Φ                  -( 2 )

where, E₁ & E₂ are electric fields due to positive and negative rod                

            respectively.

            Φ is phase angle

Step 2:

The electric field due to positive rod is given as:

E₁ = k (λ/r)             - ( 3 )

where, k is Coulomb's force constant

            λ is linear charge density

            r is distance between point P and half of the rod.

Now, the linear charge density is given as:

λ = Charge/length = Q/x

The value of r is given as:

r = √x²-a²

where, x is length of rod

           a is half length of rod

Applying values in above equation, we get:

r = √x²-(x/2)²

r = √(1.20 m)²-(1.20/2)²

  = √1.08

  = 1.04 m

Substituting all the determined values in equation 3 we get:

E₁ = k (λ/r)

   = k [(Q/x)/r]

   = k [ Q/xr ]

   = (9×10⁹ Nm²/C²) [ |+2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

   = 1.803×10⁴ N/C

Step 3:

Similarly, the electric field due to negative rod is given as:    

E₂ = k [ Q/xr ]

    = (9×10⁹ Nm²/C²) [ |-2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

    = 1.803×10⁴ N/C

Step 4:

Consider equation 2:

E_net  = 2E₁ cos Φ

From the figure we get the phase angle as:

Φ = tan⁻¹ (0.60 m/0.60 m)

   = tan⁻¹ ( 1 )

   = π/4  

Now, the electric field produced due to each rod is equal and mutually perpendicular. Thus, the net electric field after applying values can be calculated as:

E_net = 2(1.803×10⁴ N/C) cos π/4

          = 2(1.803×10⁴ N/C) (0.5)

          = 18030 N/C

Step 5:

Consider equation 1 :

F = e. E_net

where, e is charge on an electron

Applying values in above equation we get:

F = (1.6 × 10⁻¹⁹ C)(18030 N/C)

  = 2.885 × 10⁻¹⁵ N

Therefore, the magnitude of the net force that the rods exert after an electron is released at point P is  2.885 × 10⁻¹⁵ N.

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The number of oscillations completed by the pendulum is 2736.

The amplitude of the pendulum is 3.47 m.

The given motion is an underdamped motion. So its frequency will be similar to that of a simple harmonic motion.

The frequency of oscillation is defined as the number of oscillations completed in unit time. It is calculated using the formula.

f=(1/2π)*√(l/g)

where f is the frequency, l is the length of the pendulum, and g is the acceleration due to gravity.

Given the length of the wire l=13.9 m and acceleration due to gravity g=9.8 m/s^2. The frequency of oscillation is:

f=(1/(2*3.14)) * √(13.9/9.8)

f=0.19 Hz (approximately)

Since the pendulum started oscillating at 8:00 am, 4 hours has been passed when it shows 12:00 pm. So time t=4 hours or t=4*3600. Hence t=14400 s. The total number of oscillations is then given by the formula,

n=ft

where n is the number of oscillations.

n=0.19*14400=2736.

In damping motion, the amplitude of the pendulum decreases with time. The amplitude of the pendulum is given by the formula,

A' = A exp (-b*t)

where A' is the amplitude after time t, A is the initial amplitude, b is the damping constant, and t is the time.

Here A=1.2 m, b=0.010 kg/s and t=14400 s.

A' = 1.2 exp (-0.010*14400)

A'=3.47 m (approximately)

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Answer:

(OC) a nucleus of protons and neutrons surrounded by a cloud of electrons - modern description of an atom - where the nucleus may be hundreds of times smaller in diameter than the cloud of electrons

Answer:

measured at constant temperature and pressure

Hope this helps :)

Answer:

The speed registered by the speedometer depends on the rotations of the axle to which it is connected -

Larger diameter tires will contribute to a larger distance traveled in one rotation of the axle and hence a larger speed and the  speed shown on the speedometer will be less than the actual speed.

Answer:

final velocity=u+at 15+2×5=25

Answer: A positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod, we can conclude that the object is positively charged.

Explanation: To find the answer, we need to know more about the Electric charges and its properties.

Thus, from the above, we can conclude that, when a positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod, we can conclude that the object is positively charged.

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Answer:

positively charged

Explanation:

when a positively charged object is being repelled, that means the charge should be the same for both.

While moving over a separate electric field line, however, has a non-zero value. If the test charge is transferred in a direction perpendicular to E, no work is done and the electric potential does not change. A smooth equipotential surface characterizes every such path.

Generally, The equation of the potential difference is mathematically given as

[tex]V = Ed cos \theta.[/tex]

Where

[tex]\theta=Angle[/tex]

E=electric field

d= the varing distance

When the charge to be tested is transferred via an electric field line. Since =0, the maximum possible velocity (Vmax) is calculated.

In conclusion, Since no work is done when the test charge is shifted in a direction perpendicular to E, the electric potential does not shift in this case. Any such route is a smooth equipotential surface.

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Class 1 (Explosives) is the class of hazards that is characterized by thermal and mechanical hazards in the form of blast pressure waves, shrapnel and fragmentation, and incendiary thermal effects.

There are different classes of Hazards

Class 1 - Explosives

Class 2 - Gases

Class 3 - Flammable liquids

Class 4 - Flammable solids

Class 5 - Oxidizers

Class 6 - Toxic materials

Class 7 - Radioactive materials

Class 8 - Corrosive materials

Class 9 - Miscellaneous dangerous goods

Any substance or item, including a gadget, that is intended to function by explosion or that, through a chemical reaction inside itself, is capable of functioning similarly even if not intended to function by explosion, falls within the category of explosive materials (class 1).

Hence, Class 1 (Explosives) is the class of hazards that is characterized by thermal and mechanical hazards.

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