Two objects are moving at equal speed along a level, frictionless surface. The second object has twice the mass of the first object. They both slide up the same frictionless incline plane. Which object rises to a greater height?

a. Object 1 rises to the greater height because it weighs less.
b. Object 2 rises to the greater height because it possesses a larger amount of kinetic energy.
c. Object 2 rises to the greater height because it contains more mass.
d. Object 1 rises to the greater height because it possesses a smaller amount of kinetic energy.
e. The two objects rise to the same height.

The magnitude of the frictional force between the person and the wall is

[tex]\[ f = \frac{mv^2}{r} \][/tex]

To find the magnitude of the frictional force between the person and the wall, we can use the concept of centripetal force.

First, let's define the variables:

m = mass of the person

v = tangential velocity of the person

R = radius of the circular path (distance from the person to the axis of rotation, in this case, the distance from the person to the wall)

The centripetal force [tex](\( F_c \))[/tex] required to keep the person moving in a circular path is given by:

[tex]\[ F_c = \frac{{mv^2}}{R} \][/tex]

Since the person is held against the wall, the frictional force [tex](\( F_f \))[/tex] provides the centripetal force. Therefore,

[tex]\[ F_f = F_c \][/tex]

[tex]\[ F_f = \frac{{mv^2}}{R} \][/tex]

Therefore, the magnitude of the frictional force between the person and the wall is [tex]\( \frac{{mv^2}}{R} \)[/tex].

Answer:

φ = -7.16 × 10⁴ Nm²/C

Explanation:

q = -3.80 × 10⁻⁶ C

ε₀ = 8.85 10⁻¹² C²/Nm²

By Gauss's law

φ = q/ε₀

as the cube has 6 faces so for one side the flux will

φ =1/6 ( q/ε₀)

φ = 1/6 ( -3.80 × 10⁻⁶ C / 8.85 10⁻¹² C²/Nm²)

φ = -71,563.088 Nm²/C

φ = -7.16 × 10⁴ Nm²/C

-7.00 x 10⁴Nm²/C

Let the single point charge of -3.80µC be Q

i.e

Q = -3.80µC = -3.80 x 10⁻⁶C

From Gauss's law, the total electric flux,φ, through an enclosed surface is equal to the quotient of the enclosed charge Q, and the permittivity of free space, ε₀. i.e;

φ = Q / ε₀     ----------------------(i)

Where;

ε₀ = known constant = 8.85 x 10⁻¹²F/m

Now, If the total flux through the enclosed cube is given by equation (i), then the flux, φ₁, through one of the six sides of the cube is found by dividing the equation by 6 and is given as follows;

φ₁ = Q / 6ε₀     --------------------(ii)

Substitute the values of  Q and ε₀ into equation (ii) as follows;

φ₁ = -3.80 x 10⁻⁶ / (6 x 8.85 x 10⁻¹²)

φ₁ = -0.07 x 10⁶

φ₁ = -7.00 x 10⁴

Therefore, the electric flux through one side of the cube is -7.00 x 10⁴ Nm²/C

Note:

The result (flux) above is negative to show that the electric field lines from the point charge points radially inwards in all directions.

Answer:

Surface charge density = 1.43 × 10⁶μC/m²

Explanation:

surface charge density = Q /A_____(1)

where charge Q is the uniformly distributed on surface area A and d surface charge density σ

The electric field due to uniformly charge sphere of charge Q a distance r from the center of the sphere

[tex]E = k\frac{Q}{r^2}[/tex]______(2)

where k is 8.988 × 10⁹N.m²C²

The surface area of the sphere is 4πR² ______(3)

The Capacitance is 4πε₀R

where ε₀ = 8.8542 × 10⁻¹²C/Nm² is a constant

Given that,

R = 11cm = 0.11m

E = 4.90 ✕ 10⁴ N/C

r = 20.0cm = 0.20m

substitute for Q in eqn(2) and for A in eqn(3)

surface charge density = [tex]\frac{Er^2 }{k(4\pi R^2)} \\\frac{(4.9 * 10^4)(0.20)^2}{4\pi (8.988 * 10^9)(0.11)^2 }[/tex]

= 1.43 * 10⁶C/m²

Surface charge density = 1.43 μC/m²

An isolated, charged conducting sphere of radius 11.0 cm creates an electric field of 4.90×10⁴ N/C at a distance 20.0 cm from its center. (a) What is its surface charge density? (b) What is its capacitance?

(a) 1.47 x 10⁻⁶ C/m² or 1.47 μC/m²

(b) 12.07 x 10⁻¹² F  or 12.07 pF

The surface charge density, σ, of a surface (sphere, in this case) of area A which has a charge Q uniformly distributed on it is given by;

σ = [tex]\frac{Q}{A}[/tex]          -----------------(i)

Also, the electric field, E, due to the charge Q, at a distance r, from the center of the sphere to another point on the sphere is given as;

E = k x [tex]\frac{Q}{r^2}[/tex]        --------------(ii)

Where;

k = Coulomb's constant = 8.99 x 10⁹Nm²/C²

(a) i. First calculate the charge on the sphere as follows;

From the question;

r = 20.0cm = 0.20m

E = 4.90 x 10⁴ N/C

Substitute these values into equation (ii) as follows;

4.90 x 10⁴ = 8.99 x 10⁹ x [tex]\frac{Q}{0.20^2}[/tex]

4.90 x 10⁴ = 8.99 x 10⁹ x [tex]\frac{Q}{0.04}[/tex]

4.90 x 10⁴ = 224.75 x 10⁹ x Q

Q = [tex]\frac{4.90*10^4}{224.75*10^9}[/tex]

Q = 0.022 x 10⁻⁵

Q = 0.22 x 10⁻⁶ C

(a) ii. Also calculate the area A, of the sphere as follows;

A = 4π R²

Where;

R = radius of the sphere = 11.0cm = 0.11m

Substitute this value into equation above;

A = 4π (0.11)²             [Take π = 3.142]

A = 4(3.142)(0.0121)

A = 0.15m²

(a) iii. Now calculate the surface charge density, of the sphere as follows;

Substitute the values of A and Q into equation (i) as follows;

σ = [tex]\frac{0.22 * 10^{-6}}{0.15}[/tex]

σ = 1.47 x 10⁻⁶C/m²

Therefore the surface charge density is 1.47 x 10⁻⁶C/m²

==============================================================

(b) The capacitance C, of an isolated charged sphere with radius R, is given by;

C = Aε₀ / R           ----------------(iii)

Where;

R = 11.0cm = 0.11m

A =  area of the sphere = 0.15m²                                 [as calculated above]

ε₀ = permittivity of free space = 8.85 x 10⁻¹² C/Nm²   [a known constant]

Substitute these values into equation (iii) as follows;

C = 0.15 x 8.85 x 10⁻¹² / 0.11

C = 12.07 x 10⁻¹²F

Therefore, the capacitance of the charged sphere is 12.07 x 10⁻¹²F

Answer:

(a) 380.96 pC

(b) 3.25V

Explanation:

(a) Before immersion,

[tex]C_{air}[/tex] = [tex]\frac{E_{0}A }{d}[/tex]

⇒(8.85E-12× 25E-4× 260) ÷(0.0151)

= 380.96 pC

(b)   Charge on the plates after immersion can be calculated by,

  Q = ΔV×C

      = Δ[tex]V_{air}[/tex] ÷ K

where K is the constant for distilled water

      = 260 ÷ 80

      = 3.25V

Answer:

Explanation:

(a) Charge  on the plates before immersion

as we know that,

C = εA/d

C = 8.85×[tex]10^{-12}[/tex]× 25×[tex]10^{-4}[/tex]/ 1.51×[tex]10^{-2}[/tex]

   = 1.46×[tex]10^{-12}[/tex]

Q = CV

   = (1.46×[tex]10^{-12}[/tex]) (260)

   = 3.796×[tex]10^{-10}[/tex] C

(b) Charge on the plates after immersion

Q = 3.796×[tex]10^{-10}[/tex] C

The charge will remain the same, as the capacitor was disconnected before it was immersed.

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

[tex]v_{1}=\dfrac{3.0\times10^{3}}{3600}[/tex]

[tex]v_{1}=0.833\ m/s[/tex]

The force F₁is constant acceleration is also a constant.

[tex]F_{1}=ma_{1}[/tex]

We need to calculate the acceleration

Using formula of acceleration

[tex]a_{1}=\dfrac{v}{t}[/tex]

[tex]a_{1}=\dfrac{0.833}{10}[/tex]

[tex]a_{1}=0.083\ m/s^2[/tex]

Similarly,

[tex]F_{2}=ma_{2}[/tex]

For total force,

[tex]F_{3}=F_{2}+F_{1}[/tex]

[tex]ma_{3}=ma_{2}+ma_{1}[/tex]

The speed of second tugboat is

[tex]v=\dfrac{11\times10^{3}}{3600}[/tex]

[tex]v=3.05\ m/s[/tex]

We need to calculate total acceleration

[tex]a_{3}=\dfrac{v}{t}[/tex]

[tex]a_{3}=\dfrac{3.05}{10}[/tex]

[tex]a_{3}=0.305\ m/s^2[/tex]

We need to calculate the acceleration a₂

[tex]0.305=a_{2}+0.083[/tex]

[tex]a_{2}=0.305-0.083[/tex]

[tex]a_{2}=0.222\ m/s^2[/tex]

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

[tex]\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}[/tex]

[tex]\dfrac{F_{1}}{F_{2}}=3.7[/tex]

[tex]F_{1}=3.7F_{2}[/tex]

Hence, The magnitude of F₁ is 3.7 times of F₂

Answer:

a) Height of the antenna (in m) for a radio station broadcasting at 604 kHz = 124.17 m

b)Height of the antenna (in m) for radio stations broadcasting at 1,710 kHz =43.86 m

Explanation:

(a) Radiowave wavelength= λ = c/f

As we know, Radiowave speed in the air = c = 3 x 10^8 m/s

f = frequency = 604 kHz = 604 x 10^3 Hz

Hence, wavelength = (3x10^8/604x10^3) m

λ = 496.69 m

So the height of the antenna BROADCASTING AT 604 kHz =  λ /4 = (496.69/4) m

= 124.17 m

(b) As we know , f = 1710 kHz = 1710 x 10^3 Hz  (1kHZ = 1000 Hz)

Hence, wavelength =  λ = (3 x 10^8/1710 x 10^3) m

 λ= 175.44 m

So, height of the antenna =  λ /4 = (175.44/4) m

= 43.86 m  

Answer:

TRUE.

Explanation:

The first part of the statement is understood as certain since the definition of the electric field warns that its existence is dependent on the charge of the source. At the same time, we know that for there to be an electric force it is necessary to consider two charges. Either for the generation of the force of repulsion or for the force of attraction.

The statement is true. An electric field can exist due to a single source charge, while an electric force requires a test charge to respond to the field generated by the source charge.

The statement is

TRUE

. The phenomenon takes place within the study of

electromagnetism

. An

electric field

indeed can exist solely due to a source charge. The source charge creates an electric field in the space around it. However, to cause an electric force, there must be at least one other charge (test charge) that is capable of sensing the presence of this field. The greater the source charge, the stronger the electric field. The test charge responds to the field by experiencing a force. The direction of the electric force is the same as the direction of the electric field if the test charge is positive, but opposite if the test charge is negative.

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Answer:

ρ

=

55.0

π

⋅

15.0

2

⋅

62.0

Explanation:

Answer:

Explanation:

In standing wave pattern we find region of nodes where vibration is minimum or cold spots . The distance between any two consecutive node is half the wave length . There are 5 cold spot or node in between which is equal to 4 half wave length .

width of oven = 4 x half wave length

= 4 x (12 / 2 )

= 24 cm

This answer explains the calculation of turns in a transformer's secondary winding based on given voltage outputs and primary turns, along with determining maximum output currents for different voltage levels.

Input Voltage: 240 V

Primary Coil Turns: 280 turns

Secondary Parts Turns: 5.60 V - 77 turns, 12.0 V - 56 turns, 480 V - 7 turns

Maximum Output Currents: 5.60 V - 5 A, 12.0 V - 2.1 A, 480 V - 0.042 A

Final answer:

The question delves into advanced transformer analysis, focusing on impedance during a short circuit, input parameters at full load, and how voltage regulation varies with load. However, specific calculations for these scenarios are complex and require more detailed analysis beyond the provided data.

Explanation:

The student's question involves a complex analysis of a 60-Hz single-phase transformer under different loading conditions, focusing on electrical parameters such as impedance, current, power, and voltage regulation. It covers several fundamental concepts in electrical engineering, particularly those relating to transformers' performance and efficiency under load.

Total Input Impedance

To calculate the total input impedance when the secondary is shorted, we must consider the impedances in parallel and series as per the equivalent circuit of the transformer. However, given the specific data provided, a direct calculation isn't presented here.

Input Current, Voltage, Power, and Power Factor at Full Load

At full load (150 kW) with a power factor of 0.83 and a secondary voltage of 230 V, the input current, voltage, power, and power factor can be derived from the transformer's ratings and load characteristics. However, to accurately calculate these values, one would typically apply the transformer's equivalent circuit model, considering real and reactive power components.

Voltage Regulation vs. Load

The voltage regulation versus load plot and the determination of the load that produces 5% regulation require an analysis of how the secondary voltage varies with load, against rated conditions. This involves calculations and assumptions not fully detailed in the question.

Answer:0.161 m

Explanation:

Given

mass of cube  [tex]m=0.49\ kg[/tex]

Spring constant [tex]k_1=606\ N/m[/tex]

compression in the spring [tex]x_1=0.1\ m[/tex]

When this cube is released then it will compress another spring of spring constant [tex]k_2=233\ N/m[/tex]

Conserving energy

[tex]\frac{1}{2}k_1x^2=\frac{1}{2}mv^2=\frac{1}{2}k_2x'^2[/tex]

[tex]\frac{x'}{x}=\sqrt{\frac{k_1}{k_2}}[/tex]

[tex]x'=0.1\times \sqrt{\frac{606}{233}}[/tex]

[tex]x'=0.1\times 1.61[/tex]

[tex]x'=0.161\ m[/tex]

1 hour = 3600 seconds

2 hrs = 7200 sec

(24 frame/sec) x (72 sec) =

172,800 frames.

If each frame of a motion picture film is 35 mm high, and 24 frames go by in a second, then the number of frames needed to show a two-hour-long movie would be 172800 frames

Finding the product of two or more numbers in mathematics is done by multiplying the numbers. It is one of the fundamental operations in mathematics that we perform on a daily basis.

As given in the problem If each frame of a motion picture film is 35 mm high, and 24 frames go by in a second,

The number of seconds in 2 hours = 2 ×3600

                                                        =7200 seconds

The number of frames goes by in one second =  24 frames

the number of frames in a 2-hour movie = 24×7200

                                                                =172800 frames

Thus, the number of frames needed to show a two-hour-long movie would be 172800 frames

Learn more about multiplication here,

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Answer:

16.5 m

Explanation:

Given,

Magnetic field = 0.02 T

radius of electron = 9 mm

speed of electron  = speed of proton

radius of proton = ?

We know,

[tex]F_e = q_e vB[/tex].........(1)

[tex]F_p = q_p vB[/tex]

using newton second law

[tex]F = m a = m\dfrac{v^2}{r}[/tex]

equating Force due to electron and proton

[tex]F_e = F_p[/tex]

[tex]\dfrac{m_ev^2}{r_e}=\dfrac{m_pv^2}{r_p}[/tex]

[tex]\dfrac{m_e}{r_e} = \dfrac{m_p}{r_p}[/tex]

m_ e = 9.1 x 10⁻³¹ Kg    and m_p = 1.67 x 10⁻²⁷ Kg

[tex]r_p = \dfrac{m_p}{m_e}\times r_e[/tex]

[tex]r_p = \dfrac{1.67\times 10^{-27}}{9.1 \times 10^{-31}}\times 9 \times 10^{-3}[/tex]

[tex]r_p = 16.5 m[/tex]

Hence, the radius of proton is equal to 16.5 m.

Answer:

0.2v

Explanation:

Data given,

Diameter=18.5cm

Hence we can calculate the radius as D/2=18.5/2=9.25cm

radius=9.25cm/100=0.0925m

The area is calculated as

[tex]area=\pi r^{2}\\Area=0.0925^{2}*\pi \\Area=0.02688m^{2}\\[/tex]

magnetic field, B=1.3T

time,t=0.18s

The flux is expressed as

[tex]flux=BAcos\alpha \\[/tex]

since the loop is parallel, the angle is 0

Hence we can calculate the flux as

[tex]flux=1.3*0.02688cos(0)\\flux=0.0349Wb\\[/tex]

to determine the emf induced in the loop, we use Faraday law

[tex]E=-N\frac{d(flux)}{dt}\\ E=-0.0349/0.18\\E=0.19V\\E=-0.2v[/tex]

Note the voltage is not negative but the negative sign shows the current flows in other to oppose the flux

Answer:

U = 12 J.

Explanation:

The potential energy in a spring is given by the following formula

[tex]U = \frac{1}{2}kx^2[/tex]

where k is the spring force constant and x is the displacement from the equilibrium.

If U = 3 J when x = 0.05 m, then k is

[tex]3 = \frac{1}{2}k(0.05)^2\\k = 2400~N/m[/tex]

Using this constant, we can calculate the potential energy at x = 0.10 m:

[tex]U = \frac{1}{2}(2400)(0.1)^2 = 12 ~J[/tex]

The potential energy when the mass is at [tex]\( x = 0.10 \)[/tex] m is 12 J.

The potential energy (U) of a mass-spring system is given by the equation:

[tex]\[ U = \frac{1}{2} k x^2 \][/tex]

Given that the potential energy is 3.0 J when \( x = 0.050 \) m, we can use this information to find the spring constant \( k \). Plugging the values into the equation, we get:

[tex]\[ 3.0 \, \text{J} = \frac{1}{2} k (0.050 \, \text{m})^2 \] \[ k = \frac{2 \times 3.0 \, \text{J}}{(0.050 \, \text{m})^2} \] \[ k = \frac{6.0 \, \text{J}}{0.0025 \, \text{m}^2} \] \[ k = 2400 \, \text{N/m} \][/tex]

Now that we have the spring constant, we can find the potential energy when [tex]\( x = 0.10 \)[/tex]m using the same formula:

[tex]\[ U = \frac{1}{2} k x^2 \] \[ U = \frac{1}{2} (2400 \, \text{N/m}) (0.10 \, \text{m})^2 \] \[ U = \frac{1}{2} (2400 \, \text{N/m}) (0.010 \, \text{m}^2) \] \[ U = 1200 \, \text{N/m} \times 0.010 \, \text{m}^2 \] \[ U = 12 \, \text{J} \][/tex]

Therefore, the potential energy when the mass is at[tex]\( x = 0.10 \)[/tex]m is 12 J."

Answer: 5) "The direction of the magnetic force acting on a moving charge in the magnetic field is perpendicular to the direction of the magnetic field"

3) "An electric charge moving perpendicular to the magnetic field experiences a magnetic force"

Explanation:

When a charge of magnitude q, moving with a velocity v is placed in a magnetic field of strength B, it experiences a force, the magnitude of this force F is given mathematically as

F =qvB sinx

Where x is the angle between the magnetic field and the velocity of the charge.

From this equation, we can see that the force is zero when magnetic field strength B is parallel to velocity (x=0) or when velocity v is zero.

Also the force F is maximum when the angle between magnetic field strength B and velocity is 90 ( that's they are perpendicular).

By the right hand rule, the force, velocity and strength of magnetic field are perpendicular to each other.

These points have made statement 3 and 5 of the questions to be correct.

To find the initial speed of the football, you can use the kinematic equation. Given that the football is moving upward at 0.500 m/s and the displacement is 4.00 m, the equation can be used to find the initial velocity.

To find the initial speed of the football, we can use the kinematic equation that relates the final velocity, initial velocity, acceleration, and displacement: vf = vi + at. Given that the football is moving upward at 0.500 m/s and the displacement is 4.00 m, we can plug in the values to solve for the initial velocity:

0.500 m/s = vi + (-9.8 m/s2) x t

Since the football is thrown straight up, the acceleration due to gravity is negative. Let's assume the time taken for the football to reach a height of 4.00 m is t. Since the football goes up and then comes back down, the time taken to reach the height of 4.00 m in the upward direction will be half of the total time. Let's call this time tu. Since the motion is symmetric, the time taken to reach the height of 4.00 m on the way down will also be tu. Therefore, the total time taken for the football's motion is 2tu.

Given that the total time is 2tu, we can substitute it into the equation:

0.500 m/s = vi + (-9.8 m/s2) x 2tu

Now, we can solve for vi by isolating it:

vi = 0.500 m/s - (-9.8 m/s2) x 2tu

Since the time taken for the motion is unknown, we cannot determine the precise initial velocity without more information.

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Answer: A

Explanation: We can use the concept of conservation energy which implies that the kinetic energy of the froghoppers equals it potential energy from the ground level.

Where potential energy = mgh

Where m = mass of the object , g = acceleration due gravity and h = height from ground level.

The value of potential energy will reduce when the height is inclined at an angle.

Let us assume equal mass for both froghoppers, say m , g = 10 m/s^2 and a value of h.

For the first froghopper, potential energy = m×9.8×h = 9.8 mh

For the second froghopper, potential energy = m×9.8×hsin58 ( hsin58 is the vertical componet of height h inclined at angle 58),

potential energy = 8.3109 mh

As we can see , froghopper A has more potential energy than froghoppers B which implies that A has more kinetic energy than B

Final answer:

Both froghoppers experience the same change in kinetic energy, as the change in potential energy is the same for both due to reaching the same height.

Explanation:

To determine which froghopper experiences the greatest change in kinetic energy we can consider their jumps from an energy perspective. As both froghoppers are aiming for the same leaf at a height of 35 cm, the change in gravitational potential energy (UG) from the ground to the leaf will be the same for both, since potential energy depends only on the height and mass, which are constant for both froghoppers. Assuming that both froghoppers start at rest (kinetic energy Ki = 0), the change in kinetic energy during the jump (ΔK) will equal the change in potential energy (ΔUG) at the top of their trajectories where their velocity is zero.

If we follow the work-energy principle, the work done by the froghoppers' muscles will convert to potential energy at the peak of their jumps. Thus, the greatest change in kinetic energy will be equivalent to the potential energy gained at the leaf's height, which is same for both, meaning that neither froghopper experiences a greater change in kinetic energy than the other when reaching the leaf. The takeoff angle does not affect the change in kinetic energy in this case since both reached the same height.

Answer:

0.2289

Explanation:

Power required to climb= Fv where F is force and v is soeed. We know that F= mg hence Power, P= mgv and substituting 700 kg for m, 9.81 for g and 2.5 m/s for v then

P= 700*9.81*2.5=17167.5 W= 17.1675 kW

To express it as a fraction of 75 kw then 17.1675/75=0.2289 or 22.89%

Answer:

88.1 N

Explanation:

As shown in the free body diagram attached to the question, the only forces acting on the inner block include in the required vertical direction.

- Its Weight.

- Frictional forces as a result of the Two compression forces.

If the block is not to slip off, the weights must match the two frictional forces

Let the compressive forces be F and frictional force be Fr

Fr = μ F

Sum of force acting on the inner block in the y-direction

Fr + Fr - W = 0

μ F + μ F = mg

2 μF = 118

2(0.67) F = 118

F = 118/1.34

F = 88.1 N

Each of the compression forces is 88.1 N

Hope this Helps!!!

The magnitude of one of the compression forces acting on either side of the inner board in a vice setup is 88.1 N

Given that the coefficient of friction () between the inner and the outer boards is 0.67 and the normal force (weight of the inner board) is 118 N, we can use the relationship for static friction to find the frictional force. Since the inner board is at equilibrium and not moving, the magnitude of one of the compression forces that acts on either side of the inner board must be equal to the static frictional force.

Thus,

Let the compressive forces be F and the frictional force be Fr

Fr = μ F

The sum of force acting on the inner block in the y-direction

Fr + Fr - W = 0

μ F + μ F = mg

2 μF = 118

2(0.67) F = 118

F = 118/1.34

F = 88.1 N

Answer:

In the Equator:

As far as the temperature is concerned equator is more or less the same throughout the year, however, there are some fluctuations also, I will set this device in March and September because, in this month, the Sun is exactly over the equator and I would be able to get the more results in this month.

In Antarctic:

As far as the climatic conditions of Antarctic are concerned, it is all the same while it fluctuates in December because the Sun is very much close to Antarctic that's why I will choose this month.

Explanation:

Final answer:

To find the angle that the string makes with the pole, we can use the concept of centripetal force. The tension in the string provides the centripetal force that keeps the ball in circular motion.Therefore angle is approximately 45.6°.

Explanation:

To find the angle that the string makes with the pole, we can use the concept of centripetal force. The tension in the string provides the centripetal force that keeps the ball in circular motion.

At the top of the circle, the tension in the string is equal to the weight of the ball, which is given by T = mg. We can use this equation to find the angle the string makes with the pole.

tan(theta) = T/ (m*v²/ R), where m is the mass of the ball, v is the speed of the ball, and R is the length of the string. Plugging in the values given, we get tan(theta) = (5.93 kg * 4.25 m/s²) / (5.93 kg * 9.81 m/s²). Solving for theta, we find that the angle is approximately 45.6°.

Final answer:

To find the work done when stretching or compressing a non-Hookean spring by 0.050 m, we calculate the definite integral of the force function over the respective intervals. The work done for stretching is the integral from 0 to -0.050 m, and for compression, it is the integral from 0 to 0.050 m. The ease of stretching versus compressing depends on the non-linear force constants.

Explanation:

Work Done to Stretch or Compress a Non-Hookean Spring

To compute the work (W) required to stretch or compress the spring an amount x, we need to integrate the force function Fx = kx - bx2 + cx3 from the unstretched length (0) to the final position (x).

For stretching (x < 0), substitute the given constants (k = 100 N/m, b = 700 N/m2, c = 12,000 N/m3) and integrate from 0 to -0.050 m.

For compression (x > 0), use the same constants and integrate from 0 to 0.050 m.

Analyze the dependence of Fx on x to determine which process (stretching or compressing) requires less work.

The work done for both stretching and compressing can be calculated by evaluating the definite integral of Fx over the interval [0, x].

(a) Work done to stretch:

W = ∫0-0.050 (100x - 700x² + 12,000x3) dx

(b) Work done to compress:

W = ∫00.050 (100x - 700x² + 12,000x3) dx

(c) Comparison of ease:

The comparison is based on the shape of the force-displacement curve. With the given non-linear characteristics, it might be easier or harder to stretch or compress the spring depending on the values of b and c which affect the quadratic and cubic terms, respectively.

The ice ball reaches the bottom first, when the sheet is then tipped up. In the given condition, the ball and the ice cube have the same inertia.

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

The ice ball reaches the bottom first, when the sheet is then tipped up. In the given condition, the ball and the ice cube have the same inertia.

Because the ice cube slides down with virtually no friction. Without friction, an object can easily slide with the more speed.

The ball is spherical in nature and a spherical surface is in more contact with the surface during the inclined motion.

Hence, the ice ball reaches the bottom first, when the sheet is then tipped up.

To learn more about the friction force, refer to the link;

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a) [tex]1.5\cdot 10^5 m/s^2[/tex]

b) 6.3 cm

c) 12.6 cm

Explanation:

a)

The acceleration of an object is the rate of change of its velocity; it is given by:

[tex]a=\frac{v-u}{t}[/tex]

where

u is the initial velocity

v is the final velocity

t is the time interval taken for the velocity to change from u to t

In this problem for the spore, we have:

u = 0 (the spore starts from rest)

v = 1.11 m/s (final velocity of the spore)

[tex]t=7.40\mu s = 7.40\cdot 10^{-6}s[/tex] (time interval in which the spore accelerates from zero to 1.11 m/s)

Substituting, we find the acceleration:

[tex]a=\frac{1.11-0}{7.40\cdot 10^{-6}}=1.5\cdot 10^5 m/s^2[/tex]

b)

Since the upward motion of the spore is a free fall motion (it is subjected to the force of gravity only), it is a uniformly accelerated motion (=constant acceleration, equal to the acceleration due to gravity: [tex]g=9.8 m/s^2[/tex]). Therefore, we can apply the following suvat equation:

[tex]v^2-u^2=2as[/tex]

where:

v = 0 is the final velocity of the spore (when it reaches the maximum height, its velocity is zero)

u = 1.11 m/s is the initial velocity (the velocity at which it is ejected)

[tex]a=-g=-9.8 m/s^2[/tex] is the acceleration (negative because it is downward)

s is the vertical displacement of the spore, which corresponds to the maximum height reached by the spore

Solving for s, we find:

[tex]s=\frac{v^2-u^2}{2a}=\frac{0^2-(1.11)^2}{2(-9.8)}=0.063 m = 6.3 cm[/tex]

c)

If the spore is ejected at a certain angle [tex]\theta[/tex] from the ground, then its motion is a projectile motion, which consists of two independent motions:

- A uniform horizontal motion, with constant horizontal velocity

- A uniformly accelerated motion along the vertical direction (free fall motion)

The horizontal range of a projectile, which can be derived from the equations of motion, is given by:

[tex]d=\frac{v^2 sin(2\theta)}{g}[/tex]

where

v is the initial velocity

[tex]\theta[/tex] is the angle or projection

g is the acceleration of gravity

From the equation, we observe that the maximum range is achevied when

[tex]\theta=45^{\circ}[/tex]

For this angle, the range is

[tex]d=\frac{v^2}{g}[/tex]

For the spore in this problem, the initial velocity is

v = 1.11 m/s

Therefore, the maximum range is

[tex]d=\frac{(1.11)^2}{9.8}=0.126 m = 12.6 cm[/tex]

Answer:

The speed is 24 [tex]\frac{meter}{s}[/tex]

Explanation:

A wave is a disturbance that propagates through a certain medium or in a vacuum, with transport of energy but without transport of matter.

The wavelength is the minimum distance between two successive points of the wave that are in the same state of vibration. It is expressed in units of length (m).

Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

The speed of propagation is the speed with which the wave propagates in the middle, that is, the magnitude that measures the speed at which the wave disturbance propagates along its displacement. Relate wavelength (λ) and frequency (f) inversely proportionally using the following equation:

v = f * λ.

In this case, λ= 8 meter and f= 3 Hz

Then:

v= 3 Hz* 8 meter

So:

v= 24 [tex]\frac{meter}{s}[/tex]

The speed is 24 [tex]\frac{meter}{s}[/tex]

The speed of the wave is 24 m/s.

To find the speed of a wave, you can use the relationship between speed, frequency, and wavelength.

The formula is:

Given in this problem:

We substitute these values into the formula:

Thus, the speed of the wave is:

Complete Question

The complete question is shown on the first uploaded image

Answer:

The slit width [tex]a = \frac{2L \lambda}{W}[/tex]

Explanation:

Assuming the unit on the graph is cm

Given that the slit to screen distance is D = 200 cm = 20 000 m

The wavelength [tex]\lambda[/tex]  = 633 nm = [tex]633*10^{-9}m[/tex]

                         slit width a = ?

 The width of the spot that is the width of the peak from the graph is

            W = 1.6 × 2 = 3.2 cm

Where the 1.6 is the distance from 0 to the right  end point of the peak

        The change in y i.e [tex]\Delta y[/tex] has a formula

                         [tex]\Delta y[/tex]  = Ltanθ

An the width of the spot is 2 × [tex]\Delta y[/tex]

                                       W = 2Ltanθ

Applying this formula qsinθ = m[tex]\lambda[/tex]

   where m = 1 because we a focused on the first zeros ,using small angle approximation we have y

              [tex]a\theta = (1) \lambda[/tex]

               [tex]\theta = \frac{\lambda}{a}[/tex]

Substituting this into W = 2ltanθ

         Using small angle approximation

                  W = 2ltanθ = 2Lθ

                  [tex]W = 2L\frac{\lambda}{a}[/tex]

                   [tex]a = \frac{2L \lambda}{W}[/tex] and this is the slit width

Answer:

Potential difference = 245 V

Explanation:

First, we have to find velocity at which ionized molecules are accelerated into a magnetic field. Here, magnetic force acts as a centripetal force due to circular path of molecules towards detector slot.

Now, the magnetic force to the centripetal force equates as:

qvB = [tex]\frac{mv2}{r}[/tex] = [tex]\frac{mv2}{d/2}[/tex]

For Velocity (v) :

v = [tex]\frac{qBd}{2m}[/tex]

According to law of conservation of energy, we know that the initial potential energy of the molecules is converted into kinetic energy as they enter into the magnetic field. So,

[tex]qV=\frac{1}{2}mv2[/tex]

Substitute the value of v in above equation:

qV = [tex]\frac{1}{2}[/tex] m [tex]\frac{q2B2d2}{4m2}[/tex]

V = [tex]\frac{qB2d2}{8m}[/tex]

V= 245 V

Answer:

Explanation:

30 N force is pulling total mass of 15 kg , so acceleration in the system of masses

= 30 / 15

= 2 m / s²

Let us now consider forces acting on 9 kg . 30 N is pulling it in forward direction . Tension T in the string attached to it is pulling it in reverse direction

so net force on it

30 - T

Applying Newton's law of motion on it

30 - T = mass x acceleration

30 - T = 9  x 2

30 - 18 = T

T = 12 N

Using Newton’s second law, the applied force is used to find the acceleration of the whole system. We then calculate the force (tension) required to move the 6 kg block using this acceleration which is 12N.

In Physics, specifically Newtons’ second law of motion, the tension in the string between the two blocks can be calculated by using the equation F=ma, where F is the force, m is the mass, and a is the acceleration. The force applied on the 9 kg block can be considered to cause an acceleration in the entire system (both blocks) given the fact that they are connected by a string. First, find the acceleration of the whole system by using the formula a = F/total mass = 30N/(9kg+6kg) = 2 m/s². Then, to find the tension in the string between the blocks, we calculate the force required to move the 6 kg block with that acceleration: T = ma = 6kg * 2 m/s² = 12 N. Therefore, the tension in the string between the blocks is 12N.

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