Which of the following properties of a sound wave determines its loudness (or intensity): wavelength, speed, amplitude, or frequency?

Explanation:

(a) The acceleration is 9.8 m/s² downwards.

(b) At the maximum height, the velocity is 0 m/s.

(c) v = at + v₀

0 m/s = (-9.8 m/s²) (1.91 s) + v₀

v₀ = 18.7 m/s

(d) Δy = vt − ½ at²

Δy = (0 m/s) (1.91 s) − ½ (-9.8 m/s²) (1.91 s)²

Δy = 17.9 m

Answer:

A. The new capacitance is half in term of C

B. The charge remains the same in terms of  [tex]Q_{0}[/tex]

C. The potential difference is double in terms of  [tex]V_{0}[/tex]

Explanation:

The battery with a voltage of  [tex]V_{0}[/tex]  is used to charge the plates, giving it a capacitance of C.

The charging process leaves a charge of magnitude [tex]Q_{0}[/tex] on the plate

The battery is disconnected (this will leave it with a constant charge  [tex]Q_{0}[/tex])

the relationship between the charge, voltage and capacitance of the plate is

[tex]Q_{0}[/tex] = C[tex]V_{0}[/tex]   ......... equ 1

A. The relationship between capacitance and the distance of the plate is given as

C = Aε/d  ......... equ 2

where A is the area of the plate,

ε is the permeability of free space,

d is the distance between the plates

The area of the plate does not change, and permeability of free space is a constant. The combination of all these means that if the distance is doubled, then the capacitance will be halved. This is from equ 2 when the distance becomes 2d, then we have

C' = Aε/2d

==> C' = C/2

B. Since the battery is disconnected, and the capacitor is not discharged, the charge on the plate will remains the same as [tex]Q_{0}[/tex]. This is due to the conservation of charges.

C. Since the charge remains constant, and the capacitance is halved, then from equ 1, the new potential difference V will become double of the initial potential difference [tex]V_{0}[/tex]

==> V = 2[tex]V_{0}[/tex]

Answer:

speed of puck acc. to the radar gun = 138 km/h

speed of player = 15 km/h

since the player is in motion when he shoots, the speed of the puck will be the sum of the speed of the player and the speed at which he shot. so,

speed of puck = speed of player + speed of puck acc. to player

138 = 15 + speed of puck acc. to player

speed of puck acc. to player = 138 -15

speed of puck acc. to player = 123 km/h

Brainly this answer if you think it deserves it

Answer:

A. 80.0microColoumbs

B.120.0 microcoloumbs

C.37.3v

Explanation:

See attached file

a. The charge Q1 on capacitor C1 should be 80.0 μC.

b. The charge on capacitor C3 should be 120 μC.

c. The applied voltage should be 37.5 v.

a. We know that

v = q/c

Q1/c1 = Q2/C2

Q1 = (C1/C2) Q2

= (6.00/3.00) * 40

= 80.0

b.

Q3 = Q1 + Q2

= (80 + 40)

= 120

c.

The voltage should be

= 80/ 6+ 120/5

= 37.5 V

hence,

a. The charge Q1 on capacitor C1 should be 80.0 μC.

b. The charge on capacitor C3 should be 120 μC.

c. The applied voltage should be 37.5 v.

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Answer:

Only the horizontal component of a projectile’s momentum is conserved. Where as The vertical component of the momentum is not conserved, because the net vertical force Fy–net is not zero

Explanation:

Answer:

Fundamental frequency = 1376Hz

Fundamental frequency of it is 86 Hz.

In physics, frequency is the number of waves that pass a fixed point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time.

After moving through a sequence of situations or locations and then returning to its initial position, a body in periodic motion is said to have experienced one cycle or one vibration.

Given parameters:

Length of the one end closed tube; L = 1 m.

Velocity of sound; v = 344 m/s.

We have to find, fundamental frequency of it: f₀ = ?

We know that  frequency of one end closed tube is given by: f = (2n−1)v/4L where; n = 1, 2, 3, 4, .....

So, for n = 1; fundamental frequency is given by: f₀ = v/4L = 344/(4×1) Hz. = 86 Hz.

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Answer:

a) 23.51 m/s

b) 1.07 kg

Explanation:

Parameters given:

Kinetic energy, K = 295 J

Momentum, p = 25.1 kgm/s

a) The kinetic energy of a body is given as:

[tex]K = \frac{1}{2} mv^2[/tex]

where m = mass of the body and v = speed of the body

We know that momentum is given as:

p = mv

Therefore:

K = 1/2 * pv

=> v = 2K / p

v = (2 * 295) / 25.1 = 23.51 m/s

The velocity of the body at that instant is 23.51 m/s.

b) Momentum is given as:

p = mv

=> m = p / v

m = 25.1 / 23.51  = 1.07 kg

The mass of the body at that instant is 1.07 kg

Answer:

[tex]\large \boxed{\text{3.3 cm}^{3}}[/tex]

Explanation:

Assume the stone consists of basalt, which has a density of 3.0 g/cm³.

[tex]\rho = \text{10 g}\times\dfrac{\text{1 cm}^{3}}{\text{3.0 g}} = \text{3.3 cm}^{3}\\\\\text{The volume of the stone is $\large \boxed{\textbf{3.3 cm}^{3}}$}[/tex]

Answer:

Explanation:

Given that

speed u=4*10^6 m/s

electric field E=4*10^3 N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates

acceleration a=qE/m=1.6*10^-19*4*10^3/9.1*10^-31=0.7*10^15 =7*10^14 m/s

now we find the horizantal distance travelled by electrons hit the plates

horizantal distance X=u[2y/a]^1/2

=4*10^6[2*2*10^-2/7*10^14]^1/2

=3*10^-2=3 cm

now we find the velocity f the electron strike the plate

v^2-(4*10^6)^2=2*7*10^14*2*10^-2

v^2=16*10^12+28*10^12

v^2=44*10^12

speed after hits =>V=6.6*10^6 m/s

Complete question:

A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct?

A. the interior field points in a direction parallel to the exterior field

B. There is no electric field on the interior of the conducting sphere.

C. The interior field points in a direction perpendicular to the exterior field.

D. the interior field points in a direction opposite to the exterior field.

Answer:

B. There is no electric field on the interior of the conducting sphere.

Explanation:

Conductors are said to have free charges that move around easily. When the conductor is now placed in a static electric field, the free charges react to attain electrostatic equilibrium (steady state).

Here, a solid conducting sphere is placed in an external uniform electric field. Until the lines of the electric field are perpendicular to the surface, the free charges will move around the spherical conductor, causing polarization. There would be no electric field in the interior of the spherical conductor because there would be movement of  free charges in the spherical conductor in response to any field until its neutralization.

Option B is correct.

There is no electric field on the interior of the conducting sphere.

Answer:

A pie chart is a type of graph in which a circle is divided into sectors that each represents a proportion of the whole

Explanation:

pie charts are a useful way to organize data in order to see the size of components relative to the whole.

Answer:

a. 70 rad/s²

b. 5000 rev

Explanation:

As we know,

[tex]\omega = 20000\frac{rev}{min}\frac{2 \pi rad}{1 \ rev}\frac{1 \ min}{60 \ sec}[/tex]

then,

[tex]\omega=2100 \ rad/s[/tex]

a...

⇒  [tex]\bar{\alpha}=\frac{\omega-\omega_{0}}{\Delta t}[/tex]

On substituting the values, we get

⇒      [tex]=\frac{2100}{30}[/tex]

⇒      [tex]=70 \ rad/s^2[/tex]

b...

⇒  [tex]\theta=\theta_{0}=\omega_{0}t+\frac{1}{2}\alpha t^2[/tex]

       [tex]=\frac{1}{2}\alpha t^2[/tex]

       [tex]=\frac{1}{2}\times 70\times (30)^2[/tex]

       [tex]=31500 \ rad[/tex]

       [tex]=31500 \ rad\frac{1 \ rev}{2\pi rad}[/tex]

       [tex]=5000 \ rev[/tex]

(a) The average angular acceleration will be 70 rad/s².

(b) 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.

Angular acceleration is defined as the pace of change of angular velocity with reference to time. It is denoted by α. Its unit is rad/s².

The given data in the problem is;

n is the revolution of centrifugal rotor =  20000 rpm

t is the time interval= the 30s

(α) is the Angular acceleration=?

n₁ is the revolution when the acceleration is constant =?

(a) The average angular acceleration will be  70 rad/s².

The value of the angular velocity is given by

[tex]\rm \omega_f = \frac{2\pi N}{60} } \\\\ \rm \omega_f = \frac{2 \times 3.14 \times 20000}{60} \\\\ \rm \omega_f= 2100\ rad/sec.[/tex]

The formula for angular acceleration is guven by;

[tex]\rm \alpha =\frac{ \omega_f-\omega_i}{dt} \\\\ \rm \alpha =\frac{ 2100-0}{3}\\\\ \rm \alpha =70\ rad/sec^2[/tex]

Hence the average angular acceleration will be 70 rad/s².

(b 5000 revolutions have the centrifuge rotor turned during its acceleration period.

[tex]\rm \theta= \theta_0+\frac{1}{2} \alpha t^2 \\\\ \rm \theta= \frac{1}{2} \times 70 \times (30)^2 \\\\ \rm \theta=31500\ rad[/tex]

As we know that the angular velocity is given by

[tex]\rm \omega = \frac{\theta}{t} \\\\ \rm \omega = \frac{31500}{30} \\\\ \rm \omega = 1050 \ rad/sec[/tex]

The relation of angular velocity and revolution will be

[tex]\rm n= \frac{ \omega \times 60}{2\pi} \\\\ \rm n= \frac{ 2100 \times 60}{2\times 3.14 } \\\\ \rm n = 20063.69 \ rev[/tex]

Hence 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.

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Answer:

[tex]\frac{50}{\pi }[/tex]Hz

Explanation:

In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;

V(t) = V sin (ωt + Ф)            -----------------(i)

Where;

V = amplitude value of the voltage

ω = angular frequency = 2 π f        [f = cyclic frequency or simply, frequency]

Ф = phase difference between voltage and current.

Now,

From the question,

V(t) = 230 sin (100t)              ---------------(ii)

By comparing equations (i) and (ii) the following holds;

V = 230

ω = 100

Ф = 0

But;

ω = 2 π f = 100

2 π f = 100             [divide both sides by 2]

π f = 50

f = [tex]\frac{50}{\pi }[/tex]Hz

Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz

Answer:

Explanation:

A converging lens id also known as a convex lens. A convex lens has a positive focal length.

Using the lens formula to determine the image distance from the lens for each object distance.

1/f = 1/u + 1/v

f = focal length

u = object distance

v = image distance

For an object placed at distance of 18.0cm with focal length 14.0cm,

1/v = 1/f-1/u

1/v = 1/14 - 1/18

1/v = 9-7/126

1/v = 2/126

v = 126/2

v = 63cm

Since the image distance is positive for an object 18cm from the lens, the image formed by the object at this distance is a real and inverted image.

The magnification = v/u = 63/18 = 3.5

Similarly for an object placed at distance of 7.0cm with focal length 14.0cm,

v = uf/u-f

v = 7(14)/7-14

v= -14.0 cm

Since the image distance is negative for an object placed 7.0 cm from the lens, the image formed by the object at this distance is a virtual and upright image.

The magnification = v/u = 14/7 = 2

Note that the negative value is not taken into account when calculating magnification. The negative value only tells us the nature of the image formed.

Answer:

Radius r = 20.34 cm

The radius that can produces such a disk is 20.34 cm

Explanation:

Area of a circle;

A = πr^2

A = area

r = radius

Making r the subject of formula;

r = √(A/π) ........1

Given;

A = 1300 cm^2

Substituting into the equation 1;

r = √(1300/π)

r = 20.34214472564 cm

r = 20.34 cm

The radius that can produces such a disk is 20.34 cm

Answer:

-1748*10^N/C

Explanation:

See attached file

Answer:

An increase in frequency will cause an increase in the number of lines per centimeter and a smaller distance between each consecutive line. That is the distance between each trough

Explanation:

This is because higher frequency light source should produce an interference pattern with more lines per centimeter in the pattern and a smaller spacing between lines.

Answer:

208 V

Explanation:

resistor voltage = Vr = 124 V

capacitor voltage Vc = 167 V

the total voltage in the RC circuit is the resultant voltage of the resistor and the capacitor

total voltage i= [tex]\sqrt{Vr^{2} + Vc^{2} }[/tex]

==> [tex]\sqrt{124^{2} + 167^{2} } =[/tex] 208 V

Answer:

21 m/s

Explanation:

There are three forces on the pendulum:

Weight force mg pulling down,

Vertical tension component Tᵧ pulling up,

and horizontal tension component Tₓ pulling towards the center of the curve.

Sum of forces in the y direction:

∑F = ma

Tᵧ − mg = 0

T cos 37° = mg

T = mg / cos 37°

Sum of forces in the centripetal direction:

∑F = ma

Tₓ = mv²/r

T sin 37° = mv²/r

(mg / cos 37°) sin 37° = mv²/r

g tan 37° = v²/r

v = √(gr tan 37°)

v = √(9.8 m/s² × 60 m × tan 37°)

v = 21 m/s

Answer:

-i - 7j

Explanation:

The computation of components of the force is shown below:-

torque T = r cross F

T = (4.00 i + 5.00 j + 0 k) X (1.00 i + 7.00 j)

Now we will cross multiplying vectorilly

T = 4 × (iXi) + 28 × (jXi) + 0 + 5 × (iXj)35 × (jXj) + 0

T = 4 × 0 + 28 × (-k) + 5 × (k) + 35 × 0

T = 28 × (-k) + 5 × (k) = -23k

net torque |T| = 23 N - m

direction >> negative k

Or Simply we can do by the below method

r × f

28 - 5 = 23 K

-i - 7j

Answer:

Time taken to stop = 6 sec

Explanation:

Given:

Rotation of wheel = 1.10 × 10² rev/min

Final velocity (v) = 0

Angular acceleration (a) = - 1.92 rad/s²

Find:

Time taken to stop

Computation:

Initial velocity (u) = (1.10 × 10² × 2π rad) / 60 sec

Initial velocity (u) = 11.52 rad / sec

We know that,

V = U +at

t = (v-u)a

t = (0 - 11.52) / (-1.92)

Time taken to stop = 6 sec

Answer:

2.45 m/s

Explanation:

kinetic energy = 1/2 * m * v^2

then, 0.5 * 2500 * x^2 = 0.5 * 67 * 15^2

by solving for x, X = 2.45 m/s

Answer:

x(t) = -3sin2t

Explanation:

Given that

Spring force of, W = 720 N

Extension of the spring, s = 4 m

Attached mass to the spring, m = 45 kg

Velocity of, v = 6 m/s

The proper calculation is attached via the image below.

Final solution is x(t) = -3.sin2t

Answer:

Explanation:

1 )

angular velocity of merry go round = 2π n

= 2π  x 3 / 60

ω = .1 x π radian / s

This will be the rotational speed of the whole system including that of Cora and Cameron . It will not depend upon their relative position with respect to

the centre of the merry go round .

So rotational speed of Cora = Rotational speed of Cameron

option ( d ) is correct .

2 )

Tangential speed  v = ω R where R is the distance from the centre of merry go round .

Tangential speed of Cora = .1 x π x 1

= .314 m /s

Tangential speed of Cameron = .1 x π x 2

= .628 m /s .

So tangential speed of Cameron is twice that of Cora .

Option ( e ) is correct .

Answer:

aaaaaaaaaaaaaaaaaaa

Explanation:

Answer:

STATIC,  STATIC

KINETIC friction is less than static friction

Explanation:

In this exercise you are asked to complete the sentences with the correct words.

STATIC friction prevents the relative movement of two surfaces in contact.

For moving surfaces the friction is STATIC is greater than the kinetic friction.

For the last two sentences I think they are misspelled, the correct thing is

KINETIC friction is less than static friction

Answer:

for the first interference m = 1   y = 2,839 10-3 m

for the second interference m = 2   y = 5,678 10-3 m

Explanation:

The double slit interference phenomenon, for constructive interference is described by the expression

                d sin θ = m λ

where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.

In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles

           tan θ = y / L

with the angle it is small,

          tan θ = sin θ / cos θ = sin θ

we substitute

         sin θ = y / L

         d y / L = m λ

the distance between the central maximum and an interference line is

        y = m λ L / d

let's reduce the magnitudes to the SI system

     λ = 546 nm = 546 10⁻⁹ m

     d = 0.25 mm = 0.25 10⁻³ m

let's substitute the values

      y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³

      y =  m 2,839 10⁻³

the explicit value for a line depends on the value of the integer m, for example

for the first interference m = 1

the distance from the central maximum to the first line is y = 2,839 10-3 m

for the second interference m = 2

the distance from the central maximum to the second line is y = 5,678 10-3 m

Answer:

[tex]v_i = 1.44\frac{m}{s}[/tex]

Explanation:

The computation of the initial velocity of the fly is shown below:-

But before that we need to do the following calculations

For 4.22 seconds

[tex]\bar v = \frac{-5.80 m}{4.22 s}[/tex]

[tex]= -1.37\frac{m}{s}[/tex]

For uniform acceleration

[tex]\bar v = \frac{v_i +v_f}{2}[/tex]

[tex]= v_i + v_f[/tex]

[tex]= -2.74\frac{m}{s}[/tex]

With initial and final velocities

[tex]= -1.33\frac{m}{s^2}[/tex]

[tex]= \frac{v_i +v_f}{4.22s}[/tex]

[tex]= -v_i + v_f[/tex]

[tex]= -5.61\frac{m}{s}[/tex]

So, the initial velocity is

[tex]v_i = 1.44\frac{m}{s}[/tex]

We simply applied the above steps to reach at the final solution i.e initial velocity

Answer:

(a)    f = 0.58Hz

(b)    vmax = 0.364m/s

(c)    amax = 1.32m/s^2

(d)    E = 0.1J

(e)    [tex]x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calulated by using the following formula:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]          (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz[/tex]

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by:

[tex]v_{max}=\omega A=2\pi f A[/tex]     (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

[tex]v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}[/tex]

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by:

[tex]a_{max}=\omega^2A=(2\pi f)^2 A[/tex]

[tex]a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}[/tex]

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is:

[tex]E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J[/tex]

The total energy is 0.1J

(e) The displacement as a function of time is:

[tex]x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]

Answer:

because it has gravity in it . as it has more mass which can attract the things like we human are also attracted towards its centre and also mass is directly proportional to gravity which leads it to be round in shape.

hope u will get it..

Answer:

its round because of gravity

Explanation:

Answer:

A. 1.24micro farad

B. 1.20x10^7V/m

Explanation:

See attached file

1.24micro farad is the capacitance per square centimeter of such a cell wall

In its normal resting state, a cell has a potential difference of 85mV across its membrane. the electric field inside this membrane is  1.20x10^7V/m.

The cell membrane  is a selectively semi-permeable membrane  surrounds the cytoplasm, the primary function of the  membrane is to provide protection the integrity of the interior of the cell.

The cell membrane is called as multifaceted membrane which envelopes a cell's cytoplasm and help the cell to maintain the its shape.

Cell membrane is composed of Proteins and lipids where the mix or ratio of proteins and lipids can vary depending on the function of a specific cell.

Phospholipids are the crucial components of cell membranes which are arranged to form a lipid bilayer and certain substances can diffuse through the membrane to the cell's interior.

some cell organelles are surrounded by cell membranes, The nucleus and mitochondria have its own membrane.

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