Mass on a Spring: A 0.150-kg air track cart is attached to an ideal spring with a force constant (spring constant) of 3.58 N/m and undergoes simple harmonic oscillations. What is the period of the oscillations

Answer:

The average energy delivered to a surface is 19.116 W/m².

Explanation:

Given;

maximum electric field, E₀ = 120 v/m

The average energy delivered by the wave to a surface is given by

[tex]I_{avg} = \frac{c\epsilon_ o E_o^2}{2}[/tex]

where;

c is the speed of light, = 3 x 10⁸ m/s

ε₀ is the permittivity of free space = 8.85 x 10⁻¹² c²/Nm²

[tex]I_{avg} = \frac{c\epsilon_ o E_o^2}{2} \\\\I_{avg} = \frac{(3*10^8)(8.85*10^{-12})( 120)^2}{2}\\\\ I_{avg} =19.116 \ W/m^2[/tex]

Therefore, the average energy delivered to a surface is 19.116 W/m².

Answer:

Explanation:

Given,

F = 100 , M = 25

Now, let's find the value of A

[tex] \sf{F = MA}[/tex]

plug the values

⇒[tex] \sf{100 = 25 A}[/tex]

Swap the sides of the equation

⇒[tex] \sf{25A = 100}[/tex]

Divide both sides of the equation by 25

⇒[tex] \sf{ \frac{25A}{25} = \frac{100}{25} }[/tex]

Calculate

⇒[tex] \sf{A = 4}[/tex]

Hope I helped!

Best regards!!

Given:-

To find out:-

Calculate the acceleration, a ?

Formula applied:-

F = m × a

Solution:-

We know,

[tex] \sf{F = m × a}[/tex]

Substituting the values of mass and acceleration,we get

[tex] \sf\implies \: 100 = 25 \times a[/tex]

[tex] \sf \implies a = \cancel \dfrac{100}{25} [/tex]

[tex] \sf \implies a = 4 \: ms {}^{ - 1} [/tex]

Answer:

a = (v² – v₀²)/ 2(s – s₀)

Explanation:

v² = v₀² + 2a (s – s₀)

We can make 'a' the subject of the above expression as follow:

v² = v₀² + 2a (s – s₀)

Subtract v₀² from both side

v² – v₀² = v₀² + 2a (s – s₀) – v₀²

v² – v₀² = v₀² – v₀² + 2a (s – s₀)

v² – v₀² = 2a (s – s₀)

Divide both side by (s – s₀)

(v² – v₀²)/ (s – s₀) = 2a

Divide both side by 2

(v² – v₀²)/ (s – s₀) ÷ 2 = a

(v² – v₀²)/ (s – s₀) × 1/2 = a

(v² – v₀²)/ 2(s – s₀) = a

a = (v² – v₀²)/ 2(s – s₀)

Answer:

Explanation:

SI unit of

length=meter

volume =dm^3

mass =kilogram

time=second

energy= joule

temperature =kelvin

1/200 that should get your

answer

Answer:

a = 264.14 m/s²

Explanation:

From the question;

Initial velocity; u will be 0 m/s since the ball will start from rest.

Final velocity; v = 43 m/s

distance covered by the motion; s = 3.5m

To get the acceleration, we will make use of Newton's third equation of motion which is;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging in the relevant values to give;

a = (43² - 0)/(2 × 3.5)

a = 264.14 m/s²

The average acceleration of the ball during the throwing motion is 265.14m/s².

In order to get the acceleration, the Newton's third law of motion will be used. This will be:

v² = u² + 2as

We'll make a to be the subject of the formula and this will be:

a = (v² - u²) / 2s

We'll plug in the value into the equation and this will be:

a = (43² - 0) / (2 × 3.5)

a = 1849 / 7

= 264.14 m/s²

Therefore, the acceleration is 265.14m/s.

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Answer and Explanation: Centripetal Acceleration is the change in velocity caused by a circular motion. It is calculated as:

[tex]a_{c}=\frac{v^{2}}{r}[/tex]

v is linear speed

r is radius of the curve the object in traveling along

For its first lap:

[tex]a_{c}_{1}=\frac{v_{i}^{2}}{R}[/tex]

After a while:

[tex]a_{c}_{2}=\frac{(4v_{i})^{2}}{R}[/tex]

[tex]a_{c}_{2}=\frac{16v_{i}^{2}}{R}[/tex]

Comparing accelerations:

[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=\frac{16.v_{i}^{2}}{R}.\frac{R}{v_{i}^{2}}[/tex]

[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=\frac{16.v_{i}^{2}}{R}.\frac{R}{v_{i}^{2}}[/tex]

[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=16[/tex]

[tex]a_{c}_{2}=16a_{c}_{1}[/tex]

With linear speed 4 times faster, centripetal acceleration is 16 times greater.

Answer:

1999.46 mm

45.59 s

Explanation:

given that

cylindrical water tank with diameter, D = 3 m

Height of the tank above the ground, h = 2 m

Depth of the water in the tank, d = 2 m

Diameter of hole, d = 0.420 cm

We start by calculating the volume of water in the tank, which is given as

Volume = πr²h

V = (πD²)/4 * h

V = (3.142 * 3²)/4 * 2

V = 28.278/4 * 2

V = 7.07 * 2

V = 14.14 m³

If 1.0 gal of water is equal to 0.0038m³, then

1 gal is 0.0038 = A * h

the area of the tank is 7.07 m²

therefore, 0.0038 = 7.07 * h

h₁ =0.00054 m = 0.54 mm is the height of water that flow out

the change in height of water in the tank = h - h₁ = 2 - 0.00054 = 1.99946 m

b)

Like we stated earlier, 1.0 gal of water is 0.0038m³

to solve this we use the formula

Q = Cd * A * √2gH

where Cd is a discharge coefficient, and is given by 0.9 for water

A is the area of the small hole

A = (πD²)/4

A = (π * 0.0042²)/4

A = 5.54*10^-5 / 4

A = 1.39*10^-5 m²

H= height of the hole from the tank water level = 2m - 0.0042 = 1.9958 m

g = 9.8 m/s²

Q = 0.9 * 1.39*10^-5 m² * √2 * 9.8 * 1.9958

Q = 1.251*10^-5 * 6.25

Q = 7.82*10^-5 m³/s

Q = V/t

t = V/Q = 0.0038m³ / 7.82*10^-5 m³/s

t = 45.59 s

Explanation:

Pressure = force / area

P = (7400 kg × 10 m/s²) / (8.00 m × 6.80 m)

P = 1360 Pa

Answer: 90 m/s

Explanation:

The formula for speed is distance/time. The distance is 180 m and the time it took for the ball to travel is 2 s.

180 m/2 s = 90 m/s

Answer:

The wavelength is  [tex]\lambda = 10.01 *10^{-15} \ m[/tex]

Explanation:

From the question we are told that

  The  speed is  [tex]v = 0.132 c[/tex]

Where c is the speed of light with value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

Generally the wavelength is mathematically represented as

    [tex]\lambda = \frac{h}{m* v }[/tex]

where m is the mass of the proton with the value  [tex]m = 1.6726 ^{-27} \ kg[/tex]

           h is the Planck's constant with value  [tex]h = 6.626 *10^{-34} \ J\cdot s[/tex]

=>    [tex]\lambda = \frac{6.626 *10^{-34}}{1.6726 *10^{-34}* 0.132*3.0*10^8 }[/tex]

=>   [tex]\lambda = 10.01 *10^{-15} \ m[/tex]

Answer:

89.6 cm

Explanation:

From the question,

Volume of the rectangular object = Mass/Density.

V = m/D.................. Equation 1

Given: m = 1.278 kg, D = 4.98 g/cm³ = 4980 kg/m³

Substitute into equation 1

V = 1.278/4980

V = 2.57×10⁻⁴ m³.

But,

V = lwh............... Equation 2

Where l = length of the rectangular object, w = width of the rectangular object, h = height of the rectangular object.

make h the subject of the equation

h = V/lw........... Equation 3

Given: V = 2.57×10⁻⁴ m³, l = 0.047 m, w = 0.061 m.

Substitute into equation 3

h = 2.57×10⁻⁴/(0.047×0.061)

h = 0.896 m

h = 89.6 cm

Answer:

16 is D, 17 is A, 18 is C, 19 is B    ( would approve of brainliest)

Explanation:

16 is D, 17 is A, 18 is C, 19 is B

In the First option, distance is constant so D shows the correct graph,

In the second option, distance is increasing with time so, the velocity graph A is correct,

In the third option distance is constantly increasing with time, so C is the correct option.

In the last option distance is decreasing with time, so option B is the correct.

A distance-time graph is defined as how far an object has traveled in a given amount of time which is a simple line graph that shows the plot of distance versus time on a graph. Distance is plotted on the Y-axis while time is plotted on the X-axis.

The graphs which is shown in the question is distance-time graphs for various types of body motion.

Thus, the correct options for 16, 17, 18 and 19 are D, A, C and B respectively.

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Answer:

      θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

Explanation:

The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.

The first minimum occurs for m = 1, so the diffraction equation of a slit remains

        a sin θ = λ

in general, the diffraction patterns occur at very small angles, so

        sin θ = θ

          θ = λ / a

in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.

        θ = 1.22 λ /a

In this exercise we are told that the opening changes

         a’ = 2 a

we substitute

          θ ‘= 1.22  λ / 2a

          θ' = (1.22 λ / a) 1/2

          θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

Answer:

(a) -136.8 Ns.

(b) -1.135 N

Explanation:

(a)

Impulse: This can be defined as the change in momentum.

From the question,

I = mv-mu.................. Equation 1

Where I = impulse, m = mass of the hammer, v = final velocity, u = initial velocity.

Given: m = 18 kg, u = 7.6 m/s, v = 0 m/s (to rest)

Substitute these values into equation 1

I = 18(0)-18(7.6)

I = -136.8 Ns.

(b)

Average force = It.............. Equation 2

Where t = time.

Given: t = 8.3 ms = 0.0083 s.

Average force = -136.8(0.0083)

Average force = -1.135 N

Answer:

[tex] \boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise} [/tex]

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine ([tex] \sf \eta [/tex])

Explanation:

[tex] \boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}[/tex]

[tex] \sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v} [/tex]

Substituting values of r, ρ, σ, v & g in the equation:

[tex] \sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25} [/tex]

[tex]\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25} [/tex]

[tex]\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25} [/tex]

[tex]\sf \implies \eta = \frac{2}{9} \times 64.7196[/tex]

[tex]\sf \implies \eta = 2 \times 7.191[/tex]

[tex]\sf \implies \eta = 14.382 \: poise[/tex]

Answer:

Explanation:

A ) radius of tank r = 1.2 m

depth of water in the tank = 2 m

1 gal of water =  1 / 264.17 m³

= 3.785 x 10⁻³ m³

Let h be the change in height of water in the tank .

volume of water flowing out

= π r² x h = 3.785 x 10⁻³

3.14 x 1.2² x h = 3.785 x 10⁻³

h = 83.71 x 10⁻⁵ m

= .84 mm .

B )

change in height is negligible .

velocity of efflux of water from the hole at the bottom

v = √ 2 gh

h is height of water level which is 2 m

v = √ (2 x 9.8 x 2 )

= 6.26 m / s

radius of hole = .3 x 10⁻² m

area of cross section

= π r²

= 3.14 x ( .3 x 10⁻² )²

= 28.26 x 10⁻⁶ m²

volume of water flowing through the hole per unit area

= area of cross section x velocity of efflux

= 28.26  x 10⁻⁶ x 6.26

If t be the time required ,

28.26  x 10⁻⁶ x 6.26 x t   = 3.785 x 10⁻³

t = 21.4 s

Answer:

  b = 0.6487 kg / s

Explanation:

In an oscillatory motion, friction is proportional to speed,

               fr = - b v

where b is the coefficient of friction

when solving the equation the angular velocity has the form

               w² = k / m - (b / 2m)²

In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction

let's call

               w₀² = k / m

               w² = w₀² - b² / 4m²

               b² = (w₀² -w²) 4 m²

Let's find the angular velocities

             w₀² = 5 / 0.1

             w₀² = 50

             w = 2π f

             w = 2π 1

             w = 6.2832 rad / s

we subtitute

               b² = (50 - 6.2832²) 4 0.1²

               b = √ 0.42086

                b = 0.6487 kg / s

The coefficient friction of the mass during the measurement is 0.648 kg/s.

The given parameters;

During oscillatory motion, friction is directly proportional to speed.

[tex]F_k = -vb[/tex]

where;

The angular velocity is given as;

[tex]\omega ^2 = \frac{k}{m} - \frac{b^2}{4m^2} \\\\\omega ^2 = \omega _0^2 - \frac{b^2}{4m^2}\ \ ---\ (1)[/tex]

From the equation above, we will have the following;

[tex]\omega_0^2 = \frac{k}{m} \\\\\omega_0^2 = \frac{5}{0.1} \\\\\omega_0^2 = 50[/tex]

Also, the instantaneous angular speed is calculated as;

[tex]\omega = 2\pi f\\\\\omega = 2\pi \times 1\\\\\omega = 2\pi\\\\\omega = 6.284 \ rad/s[/tex]

From equation (1), the coefficient of friction is calculated as follows;

[tex]\omega ^2 = \omega ^2_0 - \frac{b^2}{4m^2} \\\\ \frac{b^2}{4m^2} = \omega ^2_0 - \omega ^2 \\\\b^2 = 4m^2( \omega ^2_0 - \omega ^2)\\\\b= \sqrt{ 4m^2( \omega ^2_0 - \omega ^2)}\\\\b = \sqrt{ 4\times 0.1^2\times ( 50 - 6.284^2)}\\\\b = 0.648 \ \ kg/s[/tex]

Thus, the coefficient friction of the mass during the measurement is 0.648 kg/s.

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Answer:

The factors of production are resources that are the building blocks of the economy; they are what people use to produce goods and services. Economists divide the factors of production into four categories: land, labor, capital, and entrepreneurship

Answer:

Explanation:

Given the following data

initial temperature T1= 150 °C

final temperature T2= 250 °C

specific heat of gold c= 0.13 J/g°C

mass of gold m= 75.0 grams

we can use the expression stated below to solve for the quantity of heat

[tex]Q= mc(T2-T1)---------1[/tex]

Substituting our known data into the expression we can solve for the value of Q

[tex]Q= 75*0.13(250-150)---------1\\\\Q= 75*0.13(100)\\\\Q= 975 Joules[/tex]

The quantity of heat need to raise the temperature from 150°C to 250°C  is 975 J

Answer:

B. 980 joules

Explanation:

Answer:

Δx = 2.5 x 10⁻³ m = 2.5 mm

Explanation:

The distance between two consecutive fringes, also known as fringe spacing, in Young's Double Slit Experiment, is given as follows:

Δx = λL/d

where,

Δx = distance between consecutive fringes = ?

λ = wavelength of light = 500 nm = 5 x 10⁻⁷ m

L = Distance between slits and screen = 5 m

d = slit separation = 1 mm = 1 x 10⁻³ m

Therefore,

Δx = (5 x 10⁻⁷ m)(5 m)/(1 x 10⁻³ m)

Δx = 2.5 x 10⁻³ m = 2.5 mm

The separation between adjacent bright fringes on a screen 5 m from the slits is: 2.5 mm

We are given;

Wavelength of light; λ = 500 nm = 500 × 10⁻⁹ m

Distance of slit separation; d = 1mm = 0.001 m

Distance between slit and the screen; D = 5 m

Now, formula for fringe width is;

β = λD/d

Plugging in the relevant values gives;

β = (500 × 10⁻⁹ × 5)/0.001

β = 2.5 × 10⁻³ m

Converting to mm gives;

β = 2.5 mm

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Answer:

C. All waves have the same speed.

Explanation:

Wave equation is given as;

V = fλ

where;

V is the speed of the wave

f is the frequency of the wave

λ is the wavelength

The speed of the wave depends on both wavelength and frequency

The speed of the electromagnetic waves in a vacuum is 3 x 10⁸ m/s, this also the speed of light which is constant for all electromagnetic waves.

Therefore, the correct option is "C"

C. All waves have the same speed.

Answer:

 P = 2923.89 W

Explanation:

Power is

     P = F v

so This exercise will solve them in parts using the conservation of momentum and then using the conservation of energy

To use the conservation of the momentum we must define a system, formed by the bodies, so that the forces during the collision have internal forces and the moment is conserved

initial instant, before the crash

        p₀ = m v

final instant. Right after the crash

       [tex]p_{f}[/tex] = (M + m) v₂

       p₀ =p_{f}

       m v = (M + m) v₂

       v₂ = m / (m + M) v

this is the speed with which two come out, now we can apply the conservation of energy to the system formed by the two bodies together

Starting point. Lower

        Em = K = ½ (M + m) v²

Final point. Highest point

        Em = U = (M + m) g h

        Eo₀ = [tex]Em_{f}[/tex]

         ½ (M + m) v2 = (M + m) g h

          h = 1/2 v2 / g

          h = ½ [m / (m + M) v] 2 / g

          h = 1/2 (m / m + M) 2 / g we must calculate the force, let's use Newton's second law, let's set a coordinate system with a parallel axis flat and the other axis (y) perpendicular to the plane

X Axis

         Fe - Wₓ = 0

         F = Wₓ

Y Axis

         N - [tex]W_{y}[/tex] = 0

let's use trigonometry for the components of the weight

         sin 6 = Wₓ / W

         cos 6 = [tex]W_{y}[/tex] / W

         Wx = W sin 6

         W_{y}= W cos 6

          F = mg cos 6

          F = 75 9.8 cos 6

          F = 730.97 N

let's calculate the power

        P = F v

        P = 730.97 4.0

        P = 2923.89 W

Although they're all 'close', none of the planets orbits in the same plane as any other planet.  They're all in slightly different planes.

The farthest out compared to all the others is Pluto, with an orbit inclined about 17 degrees compared to the ecliptic plane (Earth's orbit).  But Pluto is officially not a planet, so I don't think it's a good answer.

The next greatest inclination compared to Earth's orbit is Mercury.  That one is about 7 degrees.

The other six planets are all in different orbital planes inclined less than 7 degrees compared to Earth's orbit.

Answer:

2 oxygen 1 carbon

Explanation:

Answer:

2 oxygen, 1 carbon

Explanation:

Answer:

a) 800N × 20 cm = 1100N × x cm

16000= 1100x

x= 14.5

therefore it must be placed on the (50 + 14.5)cm mark

= 64.5 cm mark

b) 800N × 25 cm = (1100N × 10 cm)+(450N × x cm)

20000 = 11000 + 450x

450x = 9000

x = 20 cm

therefore it must be placed on the (50 + 20)cm mark

= 70 cm mark

a) The distance at which the mass m₂(110 g) should be suspended is 64.54 cm.

b) The distance at which mass m₃(45 g) should be suspended is 70 cm.

According to the Principle of Moments, when a body is balanced or is at equilibrium, the total clockwise and anticlockwise moments about a given point are equal.

a) m₁ = 80 g

m₂ = 110 g

r₁ = 30 cm

According to the Principle of Moments,

m₁r₁ = m₂r₂

Therefore, the distance,

r₂ = m₁r₁/m₂

r₂ = 80 x 20/110

r₂ = 14.54 cm

So, the distance at which mass m₂ should be suspended is,

r' = 50 + 14.54

r' = 64.54 cm

b) m₁ = 80 g

m₂ = 110 g

m₃ = 45 g

r₁ = 25 cm

r₂ = 60 cm

According to the Principle of Moments,

m₁r₁ = m₂r₂ + m₃r₃

80 x 25 = (110 x 10) + (45 x r₃)

45 x r₃ = 2000 - 1100

r₃ = 900/45

r₃ = 20 cm

So, the distance at which mass m₃ should be suspended is,

r' = 50 + 20

r' = 70 cm.

Hence,

a) The distance at which the mass m₂(110 g) should be suspended is 64.54 cm.

b) The distance at which mass m₃(45 g) should be suspended is 70 cm.

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Answer:

Option (A) : Zero

Explanation:

Electric field Intensity inside a metallic body is always ZERO.

Answer:

The  ratio is  [tex]\frac{w_1}{w_2} = \frac{R_2}{R_1}[/tex]

Explanation:

From the question we are told that

   The first radius is [tex]R_1[/tex]

    The  second radius is  [tex]R_2[/tex]

Generally the angular speed of the turntable is mathematically represented as

       [tex]w_1 = \frac{ v_k }{R_1 }[/tex]

Generally the angular speed of the rubber roller is mathematically represented as

        [tex]w_2 = \frac{ v_k }{R_2 }[/tex]

Where [tex]v_k[/tex] is the velocity of both turntable and  rubber roller

So

    [tex]\frac{w_1}{w_2} = \frac{\frac{v_k}{R_1} }{\frac{v_k}{R_2} }[/tex]

    [tex]\frac{w_1}{w_2} = \frac{R_2}{R_1}[/tex]

Answer:

0.276

0.9

0.756

Explanation:

Given that

Wavelength of the light, λ = 588 nm

Distance from the slit to the screen, L = 2.7 m

Width of the slit, a = 0.0351 mm

a point on the screen, y = 1.3 cm = 0.013 m

Sinθ = y/L

Sinθ = 0.013/2.7

sinθ = 0.0081

θ = sin^-1 0.00481

θ = 0.276°

α = (π.a.sinθ)/λ

α = (3.142 * 3.51*10^-5 * sin 0.276) / 588*10^-9

α = 5.3*10^-7 / 588*10^-9

α = 0.9 rad

I/i(m) = ((sinα)/α)²

I/I(m) = ((sin 0.9) / 0.9)²

I/I(m) = (0.783/0.9)²

I/I(m) = 0.87²

I/I(m) = 0.756

Note, our calculator has to be set in Rad instead of degree for part C, to get the answer

Answer:

3.7 eV

5.92*10^-19 J

Explanation:

Given that.

Potential difference of the metal, V = 3.7 V

Wavelength of the light, n = 235 nm

maximum kinetic energy given to the electrons is giving them the formula

K(max) = e.V(s), where

KE(max) is the maximum kinetic energy needed

V = potential difference of the metal

KE(max) = e * 3.7

KE(max) = 3.7eV

converting our answer to Joules, we have

3.7eV = 3.7eV * 1.6*10^-19 J/eV

3.7eV = 5.92*10^-19 J

Therefore, the maximum kinetic energy in both eV and Joules is 3.7eV and 5.92*10^-19 Joules respectively

Answer:

Explanation:

d dnnd