The figure(Figure 1) is a graph of Ex. The potential at the origin is -180 V . What is the potential at x=3.0m?

Answer:

The answer is "Option A".

Explanation:

Please find the complete question in the attached file.

The strength is equal due to the third law of Newton, and [tex]Fsp = Fps = \frac{Gmsmp}{r^2}\[/tex], therefore the force increases with decreased separation r.

Answer:

Fault

Explanation:

it only carries a current if there is a fault.

Answer:

Explanation:

In a quiet forest, you can sometimes hear a single leaf fall to the ground. ... greater its pressure amplitude, the more the air is compressed in the sound it creates. ... Graphs of the gauge pressures in two sound waves of different intensities. ... The sound intensity level β in decibels of a sound having an intensity I in watts per .

Answer:

1: True

2: False

3: True

4: False

5: True

Answer:

the answer can be probably true since light can be refracted

The net work that is needed to accelerate a 2.0 kg mass from rest to velocity 2v is 4.0 m/s.

When a force acts on a body at rest or at uniform motion in a straight line, then it starts to accelerate.

Given that, the velocity is 5.0

Time is 2 sec

The given mass is 2.0 kg

First, we calculate the acceleration

Acceleration = v / t = 10 / 5.0 = 2

From Newton's second law, we know that

force = mass x acceleration

F = ma

F = 2.0 x 2 = 4.0 m/s

Therefore, the net work that is needed to accelerate a 2.0 kg mass from rest to velocity 2v is 4.0 m/s.

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The question is incomplete. Your most probably complete question is given below:

What net force is needed to accelerate a 2.0 kg mass from rest to a velocity of 10 m/s in 5.0 s?

Ans-The density of states function describes the number of states that are available in a system and is essential for determining the carrier concentrations and energy distributions of carriers within a semiconductor.

Answer:

density equal to mass upon volume

simple definition

Explanation:

This is because of a phenomenon called refraction which effects the direction of sound.

Answer:

c

Explanation:

Answer:

f = 4 Hz

Explanation:

Given that,

The speed of a wave, v = 12 m/s

The wavelength of a wave, [tex]\lambda=3\ m[/tex]

We need to find the frequency of the wave. We know that the speed of wave is equal to the product of wavelength and frequency. So,

[tex]v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{12}{3}\\\\f=4\ Hz[/tex]

So, the frequency of the wave is equal to 4 Hz.

Answer:

Imag distance = 28.9 cm

Explanation:

Given that,

Object distance, u = -65 cm

The focal length of the convex les, f = +20 cm

We need to find the image distance. The lens formula gives the relation between focal length, object distance and image distance is given by :

[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{-65}\\\\v=28.9\ cm[/tex]

So, the image distance is 28.9 cm

If force is doubled, acceleration will also be doubled.

In this given question, force is doubled. We know that force is directly proportional to acceleration. As we increase the force on an object the acceleration increases proportionally. Therefore, if you double the force you double the acceleration.

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Answer:

What is the mass tho?

You need the mass to answer this question

Answer:

[tex]distance = ut + \frac{1}{2} g {t}^{2} \\since \: the \: fruit \: is \: from \: rest : \: initial \: velocity \: is \: 0 \: (u = 0) \\ 5 = (0 \times t) + ( \frac{1}{2} \times 9.81 \times {t}^{2} ) \\ 5 = 4.905 {t}^{2} \\ t = \sqrt{1.0194} \\ t = 1 \: second[/tex]

Answer:

area is the space bounded by certain lines, the SI unit of area is m²

area of rectangle =l×w×h

square=l²

Answer:

Explanation:

Using law of conservation of momentum during collision , velocity after collision can be found .

common velocity V = m v / ( m + M )

= .030 x 300 / ( .030 + 1 )

V= 8.7378 m /s

The kinetic energy of the bullet and block will be stored as elastic energy

1/2 ( M + m ) V² = 1/2 k A² where k is force constant and A is maximum compression in the spring .

( M + m ) V² =  k A²

1.030 x 8.7378² = 2000 x A²

A² = .0393

A= .1982 m .

= 19.82 cm

Answer:

Explanation:

Emf induced e = nBAω sinωt where n is no of turns , B is magnetic field, A is area of the coil ω is angular frequency of rotation .

maximum emf =  nBAω

Given n = 27 , B = 74 x 10⁻⁶ T , A = .12 m x .09 m = 108 x 10⁻⁴ m²

ω = 2π x 57/60 = 5.966 rad/s .

Putting the values

maximum emf  = 27 x 74 x 10⁻⁶ T x 108 x 10⁻⁴ m² x 5.966 rad/s .

= 1.29 x 10⁻⁴ V  .

The relation between a frog's maximum height and maximum range is that when the frog jumps at an angle of 45°, its maximum height is half of the horizontal range.

The maximum height refers to the maximum vertical distance that an object can travel.

The maximum range is the maximum horizontal distance that an object can travel.

The maximum height, H that the projectile reaches when the range, R, is at its greatest is H = Rmax /4.

R = 4Hcot is the relationship between the horizontal range and maximum height.

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The horizontal component of the cat's displacement, Dx = 0 m

The vertical component of the cat's displacement, Dy = 12 m

The horizontal component of a vector is the value of that vector acting in the horizontal direction.

The horizontal component of a vector is given as Dx.

Dx = Dcosθ

where;

D is the resultant of the vector

θ is the angle of the vector from the horizontal

The horizontal component of the cat's displacement is:

Dx = 12 * 0

Dx = 0 m

The vertical component of a vector is the value of that vector acting in the vertical direction.

The vertical component of a vector is given as Dy.

Dy = Dsinθ

where;

D is the resultant of the vector

θ is the angle of the vector from the horizontal

The vertical component of the cat's displacement is:

Dy = 12 * 1

Dy = 12 m

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Answer:

C

Explanation:

it depends only on the forces acting along the y- axis

Answer:

The statement that is true for the motion of sliding the object on the table is : its independent of all forces acting on the object

Explanation:

Answer:

0.00296

Explanation: This answer works for Acellus! :)

The voltage applied to the capacitor is approximately 2.964 x 10⁻³ C/F.

Capacitance is a fundamental electrical property that measures the ability of a system to store an electric charge when a potential difference (voltage) is applied across it. It is denoted by the symbol "C" and is measured in farads (F). A capacitor is a device commonly used to create capacitance in an electrical circuit.

In simple terms, capacitance is a measure of how much electric charge can be stored on an object (usually a capacitor) for a given electric potential difference. It is directly proportional to the amount of charge stored and inversely proportional to the applied voltage. Capacitance plays a crucial role in various electronic circuits and systems, including filtering, energy storage, and timing applications.

To calculate the voltage applied to the capacitor, we can use the formula:

Q = C × V

Where:

Q is the charge on the plates of the capacitor

C is the capacitance

V is the voltage applied to the capacitor

Given:

Capacitance (C) = 3.89 x [tex]10^{(-8)[/tex] F

Charge on the plates (Q) = 1.15 x [tex]10^{(-10)[/tex]C

Rearranging the formula, we can solve for V:

V = Q / C

Substituting the given values:

V = (1.15 x [tex]10^{(-10)[/tex] C) / (3.89 x [tex]10^{(-8)[/tex] F)

Dividing the numerator and denominator by [tex]10^{(-8)[/tex]:

V = (1.15 / 3.89) x [tex]10^{(-10 - (-8))[/tex] C/F

V = (1.15 / 3.89) x [tex]10^{(-2)[/tex] C/F

V = 0.2964 x [tex]10^{(-2)[/tex] C/F

Using scientific notation, we can express the voltage as:

V = 2.964 x [tex]10^{(-3)[/tex]C/F

Therefore, the voltage applied to the capacitor is approximately 2.964 x [tex]10^{(-3)[/tex] C/F.

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Answer:

Explanation:

4

The magnitude and direction of the resultant force on q3 is;

F_net = 32.2 N in the direction of q3

We are given;

q1 = +2 μC = 2 × 10^(-6) C

q2 = -3 μC = -3 × 10^(-6) C

q3 = +4 μC = 4 × 10^(-6) C

Now,from the given image, charges q1 and q2 attract charge q3.

Formula for force on a charge is;

F = (kq1 * q2)/r²

Where;

k = 9 × 10^(9) N.m²/C²

Thus;

F_y = kq1*q3/(r1)²

We have;

q1 = 2 × 10^(-6) C

r1 = 5 cm = 0.05 m

q3 = 4 × 10^(-6) C

Thus;

F_y = (9 × 10^(9) × 2 × 10^(-6) × 4 × 10^(-6))/(0.05^(2))

F_y = 28.8 N

Similarly;

F_x = kq2*q3/(r2)²

r2 = 5/(tan 30)

r2 = 8.66 cm = 0.0866 m

F_x = (9 × 10^(9) × -3 × 10^(-6) × 4 × 10^(-6))/(0.0866^(2))

F_x = -14.4 N

Thus, magnitude of resultant force is;

F_net = √((28.8)² + (-14.4)²)

F_net = √1,036.8

F_net = 32.2 N

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Solution :

Mass of the sphere, m = 10 kg

Diameter, D = 300 mm = 0.3 m

Volume of the sphere is [tex]$V=\frac{4}{3} \pi \left(\frac{0.3}{2}\right)^3$[/tex]

                                       [tex]$V=0.01414 \ m^3$[/tex]

So the volume of displaced water, [tex]$V_w=V = 0.01414 \ m^3$[/tex]

Additional mass, [tex]$m_w = \frac{1}{2} \ \rho_w\times v_w$[/tex]

                                  [tex]$=\frac{1}{2} \times 1000 \times 0.01414$[/tex]

                                  [tex]$=7.0686 \ kg$[/tex]

So the total mass, [tex]$M = m_w+m$[/tex]

                                   = 7.0686 + 10

                                   = 17.0686

Force required, F = Ma

[tex]$F=17.0686 \times 10$[/tex]

   = 170.686 N

The force that is necessary to accelerate when the acceleration rate of 10 m/s2 in the horizontal direction should be 170.686 N.

Since Mass of the sphere, m = 10 kg

Diameter, D = 300 mm = 0.3 m

Now volume of the sphere is

V = 4/π (diameter/3)^3

V = 4/3π(0.3/2)^3

= 0.01414m^3

Now the additional mass is

= 1/2* 1000 * volume

= 1/2 *1000*0.01414

= 7.0686

So, the total mass is

= Additional mass + force

= 7.0686 + 10

= 17.0686

Now the force is

= 17.0686 * 10

= 170.686

Hence, The force that is necessary to accelerate when the acceleration rate of 10 m/s2 in the horizontal direction should be 170.686 N.

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Answer: [tex]243.10\ mm\ \text{of}\ Hg[/tex]

Explanation:

Given

The initial volume of gas [tex]V_1=5\ ml[/tex]

Initial pressure is [tex]P_1=141\ mm\ \text{of}\ Hg[/tex]

Final volume is [tex]V_2=2.9\ ml[/tex]

As it is given [tex]P_1V_1=P_2V_2[/tex]

insert the values

[tex]141\times 5=P_2\cdot 2.9\\P_2=243.10\ mm\ \text{of}\ Hg[/tex]

Answer:

Opening stock details of raw material, semi-finished goods and finished goods.

Closing stock details of raw material, semi-finished goods, and finished goods.

Total purchases of goods fewer Purchase Returns.

Total sales of goods fewer Sales Returns.

Answer:

0.15 m/s²

Explanation:

Acceleration is rate of change of velocity-time function. We can find acceleration by finding the slope. The slope of graph can be found by:

[tex] \displaystyle{a = \dfrac{ \Delta v}{ \Delta t} \: \to \: \dfrac{v_f - v_i}{t_f- t_i}}[/tex]

Substitute in:

[tex] \displaystyle{a = \dfrac{5 - 2}{40 - 20}} \\ \\ \displaystyle{a = \dfrac{3}{20}} \\ \\ \displaystyle{a = 0.15 \: \text{m/s}^{2} }[/tex]

Therefore, the acceleration is 0.15 m/s²

The mass of the moving item is referred to as momentum. There is magnitude and direction to momentum. The object has a momentum of 7.2 g-m/s.

Momentum is a vector quantity that is created during motion by an object's mass and velocity. Thus, momentum will be expressed in units of kg-m/s.

Considering that item has motion. Likewise, 1 kilogram equals 1000 g. Therefore, the momentum will transform into units of grams when we enter the value of 1 kg.

Momentum = 7.2 x 10⁻³ 1000g-m/s

Momentum = 7.2 g-m/s.

Thus, we may deduce that the object's momentum, measured in grams per second, is 7.2 g-m/s.

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Since the efficiency of the pulley is 75% and the single movable pulley is used to lift 4000N load to the height of 50 cm, the effort distance, effort applied, output work and mechanical advantage are as follows:

Effort Distance = 50 cm

Effort Applied = 5333.33 N

Output work = 2000 J

Mechanical Advantage = 0.75.

When a single movable pulley is used to lift a load, the effort applied is equal to the weight of the load being lifted, and the effort distance is the distance the effort is applied.

Load (output force) = 4000 N

Height lifted = 50 cm = 0.5 m

Efficiency of the pulley = 75%

Given the above information, we can calculate the effort applied, output work, and mechanical advantage as follows:

Effort applied (input force) = Load ÷ Efficiency

= 4000 N ÷ 0.75

= 5333.33 N

Output work = Load x Height lifted

= 4000 N x 0.5 m

= 2000 J

Effort distance = Height lifted

= 50 cm

Mechanical advantage = Load ÷ Effort applied

= 4000 N ÷ 5333.33 N

= 0.75

It is worth noting that pulley efficiency is the ratio of output work to the input work. And since the efficiency of the pulley is 75% so the input work is greater than the output work, which means some of the energy is wasted in overcoming friction and other losses.

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Answer:

Explanation:

All 3 are critical for habitability