A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. Assume the spring is compressed 6 cm and released from rest. (c) Determine the speed of the object as it passes the point 1.9 cm from the equilibrium position

Answer:

a) T = 0.5 s

b) v = 1.2π m/s ≈ 3.77 m/s

Explanation:

It makes two revolutions in one second so makes one revolution in ½ second

circumference of the circle is

C = 2πr = 0.6π m

which it traverses in one time period

0.6π m / 0.5 s = 1.2π m/s

To solve this, we must be knowing each and every concept related to speed and its calculations. Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.

Speed may be defined as the distance traveled by an item in the amount of time it requires to travel that distance. In other words, it measures how rapidly an item travels but does not provide direction.

Speed may be calculated in Science. The speed equation is a scientific formula that is used to calculate various types of speed.

Mathematically, the formula for speed can be given as

speed= distance/time

Values that are given

Time period= 0.5 s

Circumference = 2πr = 0.6π m

substituting all the given values in the above equation, we get

speed     =0.6π m / 0.5 s

On calculations, we get

              = 1.2π m/s

              =3.77 m/s

Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.

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Answer:

100Hz

Explanation:

In a full wave rectifier, the fundamental frequency of the ripple is twice that of input frequency. Given the input frequency of 50 Hz, the fundamental frequency will be 2 × 50 = 100Hz

Answer:

HZ 100 is the right answer hope you like it

Answer:

Minimum area of rectangle = 24 centimeter²

Explanation:

Given:

Length of rectangle = 6 centimeter

Width of rectangle = 4 centimeter

Find:

Minimum area of rectangle

Computation :

Minimum area of rectangle = Length of rectangle x Width of rectangle

Minimum area of rectangle = 6 x 4

Minimum area of rectangle = 24 centimeter²

Answer:

the strength of the magnetic field is 3 x 10⁻⁵ T

Explanation:

Given;

velocity of the cosmic ray, v = 5 x 10⁷ m/s

force experienced by the ray, f = 1.7 x 10⁻¹⁶ N

angle between the ray's velocity and the magnetic field, θ = 45⁰

The strength of the magnetic field is calculated as;

[tex]F = qvB \ sin(\theta)\\\\B = \frac{F}{qv\times sin(\theta)} \\\\where;\\\\B \ is \ the \ strength \ of \ the \ magnetic \ field\\\\q \ is \ the \ charge \ of \ the \ cosmic \ ray \ proton = 1.602 \times 10^{-19} \ C\\\\B = \frac{1.7\times 10^{-16}}{(1.602 \times 10^{-19})\times (5\times 10^7) \times sin \ (45)} \\\\B = 3 \times 10^{-5} \ T[/tex]

Therefore, the strength of the magnetic field is 3 x 10⁻⁵ T

Answer:

circles

A=7.07multiple 10-6 m2

I hope you understand and help

The area of the circular cross-section will be 7068 × [tex]10^{-6}m^{2}[/tex].

The measurement that represents the size of a region on a plane or curved surface is called an area.

A cross-section would be the non-empty point where a solid body intersects a plane in three dimensions or its equivalent in higher dimensions.

Given data:

Diameter = 3 × [tex]10^{-3} m[/tex].

It is known that. Diameter = 2 radius.

The area can be calculated by using the formula:

A = 1/4 [tex]\pi[/tex][tex]d^{2}[/tex] = 1/4 (3.14) [tex](3 * 10^{-3})^{2}[/tex]= 7068 × [tex]10^{-6}m^{2}[/tex].

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Answer:

a. solve for a

[tex]vf ^{2} = vo ^{2} + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2} - vo ^{2} \\ a = \frac{vf^{2} - vo^{2} }{2(xf - xo)} \\ a = \frac{vf ^{2} - vo ^{2} }{2xf - 2xo} [/tex]

I hope I helped you ^_^

The least distance up to which we can see the objects clearly without any strain is called least distance of distinct vision. Least distance of distinct vision for a normal human being is 25cm. For young people, the least distance of distant vision will be within 25cm which however it varies with age.

Answer:

25 you said ? thats incorecct

Explanation:

Answer:

the lipid bilayer would not be able to hold its shape in water and the cell membrane would disassemble. Only a lipid monolayer would be possible. The fatty acid "tails' would fall off The phospholipids would flip upside down.

Explanation:

hope this helps

2. two dimensional motion

Because it has just 2 dimensions x and y

Answer:

the force is known as gravity or gravitational force

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

Answer:

7.28×10⁻⁵ T

Explanation:

Applying,

F = BILsin∅............. Equation 1

Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.

make B the subject of the equation

B = F/ILsin∅.................. Equation 2

From the question,

Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°

Substitute these values into equation 2

B = 0.16/(68×34×sin72°)

B = 0.16/(68×34×0.95)

B = 0.16/2196.4

B = 7.28×10⁻⁵ T

Answer:

(a) 2.3 x 10^-16 J

(b) 1.1 x 10^7 K

Explanation:

charge, q = 1.6 x 10^-19 C

distance, r = 10^-12 m

(a) Let the potential energy is U.

[tex]U = \frac{k q q}{r^2}\\\\U = \frac{9\times 10^{9}\times 1.6\times 1.6\times 10^{-38}}{10^{-12}}\\\\U = 2.3\times 10^{-16} J[/tex]

(b) Let the temperature is T.

[tex]U = K = \frac{3}{2} kT\\\\2.3 \times 10^{-16} = 1.5\times 1.38\times 10^{-23} T\\\\T = 1.1\times 10^7 K[/tex]

Answer:

a) required area is 1.1318 m²

b) the maximum potential difference that can be applied across the compactor is 1931.1 V

Explanation:

Given the data in the question;

dielectric constant εr = 2.35

distance between plates ( thickness ) d = 0.0785 mm = 7.85 × 10⁻⁵ m

dielectric strength = 49.5 MV/m

a)

given that capacity capacitor C = 0.3 uF = 0.3 × 10⁻⁶ F

To find the Area, we use the following the expression.

C = ε₀εrA / d

we know that The permittivity of free space, ε₀ = 8.854 x 10⁻¹²  (F/m)

we substitute

0.3 × 10⁻⁶ = [ (8.854 x 10⁻¹²) × 2.35 × A  ] /  7.85 × 10⁻⁵

A = [ (0.3 × 10⁻⁶) × (7.85 × 10⁻⁵) ] / [ 2.35 × (8.854 x 10⁻¹²) ]

A = 2.355 × 10⁻¹¹ / 2.08069 × 10⁻¹¹

A = 1.1318 m²

Therefore, required area is 1.1318 m²

b)

the maximum potential difference that can be applied across the compactor.

We use the following expression;

⇒ 1/2 × dielectric strength × thickness d

we substitute

⇒ 1/2 × ( 49.5 × 10⁶ V/m ) × ( 7.85 × 10⁻⁵ m )

⇒ 1931.1 V

Therefore, the maximum potential difference that can be applied across the compactor is 1931.1 V

b. solve for Vo

[tex] vf ^{2} = vo^{2} + 2a(xf - xo) \\ vf ^{2} = vo ^{2} + 2axf - 2axo\\ vo ^{2} = vf ^{2} - 2axf + 2axo \\ vo = \sqrt{vf ^{2} - 2axf + 2axo } \\ vo = - \sqrt{vf^{2} - 2axf + 2axo } [/tex]

I hope I helped you ^_^

Solution :

We know that :

Formula for Gravitational force is given by :

[tex]$F_g=\frac{Gmn}{r^2}$[/tex]

where, G is the gravitational constant

            M is the mass of the bigger body

            m is the mass of the smaller body

            r  is the distance between the two bodies.

And the formula for the centripetal force is given by :

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

where, m is the mass of the rotating body

            v is the velocity

             r is the radius of rotation of the body.

We know that mathematically, the gravitational force is equal to the centripetal force of the body.

Therefore,

[tex]$F_g=F_c$[/tex]

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

[tex]$\sqrt{\frac{GM}{r}}=v$[/tex]

Hence derived.

Answer:

A

Explanation:

respect the opinion of your friends

There are broadly 3 states of matter (there are 5, but they don't teach 2 of them at school).

1. Solid

Examples: Iron, wood, steel, ice, paper

2. Liquid

Examples: Water, mercury, milk, soup, juice

3. Gas

Examples: Oxygen, Chlorine, Carbon dioxide, Sulphur dioxide, Nitrogen

Answer:

The final pressure in bar will be "[tex]\frac{10}{3} \ Bar[/tex]".

Explanation:

As we know,

PV = nRT

[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2} =CONST[/tex]

then,

⇒ [tex]\frac{2 \ bar}{300 \ K} = \frac{P_2}{500 \ K}[/tex]

⇒ [tex]P_2=(\frac{2}{300}\times 500 )Bar[/tex]

        [tex]=\frac{10}{3} \ Bar[/tex]

Thus the above is the correct answer.

Answer:

The wavelength stays the same.

Explanation:

When the amplitude is increased, the wavelength stays the same.

Here the wavelength doesn't depend upon the amplitude.

Answer:

It was an apparently temporary feature about the size of Earth, similar to the Great Red Spot of Jupiter, but disappeared within a few years.

Explanation:

The Great Dark Spot of Neptune was an immense spinning storm in the southern atmosphere of Neptune. The size of the entire Earth, it had the strongest winds ever recorded on any planet in the solar system. It was discovered by the Voyager 2 spacecraft in 1989, but by 1994 the Hubble Space Telescope saw it was gone.

The Great Red Spot is a storm found in Jupiter's southern hemisphere, with similar characteristics to the Great Dark Spot.

Complete question is;

One ball is projected in the upward direction with a certain velocity ‘v’ and other is thrown downwards with the same velocity at an angle θ.

The ratio of their potential energies at highest points of their journey, will be:

Answer:

u² : (u cos θ)²

Explanation:

Maximum potential energy for the first ball will be at a maximum height of;

H = u²/2g

Thus;

PE = mg(u²/2g)

For second ball at an angle θ, maximum PE will occur at a max height of (u cos θ)²/2g

PE = mg((u cos θ)²/2g)

The ratios of the potential energies are;

mg(u²/2g) : mg((u cos θ)²/2g)

mg will will cancel out since they are of same mass.

Thus;

(u²/2g) : (u cos θ)²/2g

Again 2g will cancel out to give;

u² : (u cos θ)²

Answer:

45

Explanation:

ft/sec2

Answer:

(a) The spring constant is 59.23 N/m

(b) The total energy involved in the motion is 0.06 J

Explanation:

Given;

mass, m = 240 g = 0.24 kg

frequency, f = 2.5 Hz

amplitude of the oscillation, A = 4.5 cm = 0.045 m

The  angular speed is calculated as;

ω = 2πf

ω = 2 x π x 2.5

ω = 15.71 rad/s

(a) The spring constant is calculated as;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m[/tex]

(b) The total energy involved in the motion;

E = ¹/₂kA²

E = (0.5) x (59.23) x (0.045)²

E = 0.06 J

Answer:

(a) 1.2 rad/s

(b) 1.8 rad

Explanation:

Applying,

(a) α = (ω-ω')/t................ Equation 1

Where α = angular acceleration, ω = final angular velocity, ω' = initial angular velocity, t = time.

From the question,

Given: α = 0.40 rad/s², t = 3 seconds, ω' = 0 rad/s (from rest)

Substitute these values into equation 1

0.40 = (ω-0)/3

ω = 0.4×3

ω = 1.2 rad/s

(b) Using,

∅ = ω't+αt²/2.................. Equation 2

Where ∅ = angle turned.

Substitutting the values above into equation 2

∅ = (0×3)+(0.4×3²)/2

∅ = 1.8 rad.

Answer:

Explanation:

Kinematic equation

v = u + at

If UP is assumed to be the positive direction and we let gravity be 10 m/s² which will be in the downward direction so will be negative.

a) v = 6  + (-10)(0.5) = 1 m/s    the result is positive, so upward

b) v = 6  + (-10)(2) = -14 m/s    the result is negative, so downward

Answer:

F' = 16 F

Hence, the electric force between charges becomes sixteen times its initial value.

Explanation:

The electric force between the two charges is given by the Colomb's Law:

[tex]F = \frac{KQ_1Q_2}{R^2}[/tex] ------------------- eq(1)

where

F = electric force

K = Colomb's Constant

Q₁ = magnitude of the first charge

Q₂ = magnitude of the second charge

R = Distance between charges

Now the magnitudes of the charges are doubled and the distance between them is halved. Therefore:

[tex]F' = \frac{K(2Q_!)(2Q_2)}{(\frac{R}{2})^2}\\\\F' = 16 \frac{KQ_1Q_2}{R^2}[/tex]

using equation (1):

F' = 16 F

Hence, the electric force between charges becomes sixteen times of its initial value.