Can anyone help with this?

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 .

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:

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

Answer:

c

Explanation:

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:

when a plastic rod is rubbed with a duster, electrons are transferred from one material to the other. The material that gains electrons becomes negatively charged. The material that loses electrons becomes positively charged.

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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Explanation:

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

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:

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.

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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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:

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:

1: True

2: False

3: True

4: False

5: True

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:

it is a

Explanation:

dermis,epidermis,hypodermis

Answer:

B. hypodermis, dermis, epidermis

this is because it is inside to outside

hypodermis is deeper tissue, dermis is beneath epidermis, and epidermis is the outermost layer

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:

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  .

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

Answer:

r = 1.63×10^5 mi

Explanation:

Let r = distance of object from earth

Rs = distance between earth and sun

Ms = mass of the sun

= 3.24×10^5 Me (Me = mass of earth)

At a distance R from earth, the force Fs exerted by the sun on the object is equal to the force Fe exerted by the earth on the object. Using Newton's universal law of gravitation,

Fs = Fe

GmMs/(Rs - r)^2 = GmMe/r^2

This simplifies to

Ms/(Rs - r)^2 = Me/r^2

(3.24×10^5 Me)/(Rs - r)^2 = Me/r^2

Taking the reciprocal and then its square root, this simplifies further to

Rs - r = (569.2)r ----> Rs = 570.2r

or

r = Rs/570.2 = (9.3×10^7 mi)/570.2

= 1.63×10^5 mi

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:

the answer can be probably true since light can be refracted

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:

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 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?

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:

All 3 are critical for habitability