A mass of slug, when attached to a spring, stretches it feet and then comes to rest in the equilibrium position. Starting at , an external force equal to is applied to the system. Find the equation of motion if the surrounding medium offers a damping force that is numerically equal to times the instantaneous velocity.

Answer:

 I₂ = 25 W / m²

Explanation:

Intensity is defined as the relationship between power and area

       I = P / A

the power emitted by the sun is constant

      P = I A

for the two points of interest

      I₁ A₁ = I₂ A₂

energy is distributed on the surface of a sphere

      A = 4π R²

      I₁ R₁² = I₂ R₂²

      I₂ = [tex]\frac{R_1^2}{R_2^2} \ I_1[/tex]

let's calculate

      I₂ = [tex]\frac{R^2}{(4R)^2} \ 400[/tex]    

      I₂ = 25 W / m²

Answer: yes

Explanation:

Weight is the gravitational force experienced on a body. If you move up to higher altitudes, the distance between you and earth increases. ... Yes, weight drops as you go up in altitude (because of diminishing gravity), though your mass remains the same. However, the effect is not huge.

Answer:

uh.

Explanation:

Answer:

Convection

Explanation:

three types of heat transfer

Heat is transfered via solid material (conduction), liquids and gases (convection), and electromagnetical waves (radiation).

Answer:

M = 117.6 J

Explanation:

Given that,

The mass of a ball, m = 1kg

The height of the ball, h = 12 m

At point A, its initial velocity, v = 0

The mechanical energy is the sum of kinetic and potential energy such that,

[tex]M=\dfrac{1}{2}mv^2+mgh\\\\M=0+mgh\\\\M=1\times 9.8\times 12\\\\M=117.6\ J[/tex]

So, the mechanical energy is equal to 117.6  J.

Answer:

Newtons second law

Explanation:

Depends on mass

Answer:

2nd law

Explanation:

:))))))))))))))

Answer74.3

:

Explanation:

Answer:

a) α = 1.875 [tex]\frac{rad}{s^{2} }[/tex]

b) t = 8 s

Explanation:

Given:

ω1 = 0 [tex]\frac{rad}{s}[/tex]

ω2 = 15 [tex]\frac{rad}{s}[/tex]

theta (angular displacement) = 60 rad

*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*

a) α = ?

(ω2)^2 = (ω1)^2 + 2α(theta)

[tex]15^{2}[/tex] = [tex]0^{2}[/tex] + 2(α)(60)

225 = 120α

α = 1.875 [tex]\frac{rad}{s^{2} }[/tex]

b)

α = (ω2-ω1)/t

t = (ω2-ω1)/α = (15-0)/1.875 = 8

t = 8 s

Answer:

a)   y = 0.35 m,    b) hydrostatic balance

Explanation:

a)  For this fluid mechanics exercise, let's use that the pressure at a given level is the same, let's set a level on the bell shape.

The pressure inside is

             P_interior = P₀ + ρ g h ’

The pressure outside

             P_exterior = Pₐ + ρ g h

as the point is at the same level the pressures are equal

             P_interior = P_exterior

             P₀ + ρ g h ’= Pₐ + ρ g h

             h ’= (Pₐ- P₀)  + ρ g h  

To calculate P₀ they indicate that the pressure increases 10⁵ Pa for every 10 m, we use a direct rule of proportions or rule of three

            P₀ = 10⁵ (40 + h ’) / 10 = 4 10⁵ + h’ 10⁴

the positive sign is because the water inside the hood also increases the air pressure.

we substitute

             (4 10⁵ + h’ 10⁴) + ρ g h’ = Pₐ + ρ g h

             h’ (ρ g + 10⁴) = Pₐ - 4 10⁵ + ρ h h

            h’ (1000 9.8 + 10⁴) = (1 10⁵ -4 10⁵) + 1000 9.8 40

            h' (1.98 10⁴) = -3 105 + 3.92 10⁵

            h’ =  [tex]- \frac{0.92 \ 10^5 }{1.98 \ 10^4 }[/tex]

            h ’= -4.65 m

as the hood is only 5 m high, the free air space is

            Y = 5 - 4.65

             y = 0.35 m

it is very little free space

B) The pressure outside and inside the hood is the same, the water rises inside the hood until the pressures equalize and at this point the force is equal and in the opposite direction, which is why the system is in hydrostatic balance.

Answer:

d

Explanation:

Explanation:

Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.

Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.

Weight of the air displaced = density of air × volume

The density of air at 1 atm pressure and 20º C is 1.2 kg/m³  

the volume V = 20,000/(1.2×9.8) =  1700 m³

The pencil is clearly an opaque object because we cannot see through it and therefore forms a shadow.

Answer:

The galaxies outside of our own are moving away from us, and the ones that are farthest away are moving the fastest. This means that no matter what galaxy you happen to be in, all the other galaxies are moving away from you

Explanation:

The hypothesis about the movement of the galaxy is that galaxies are moving far from each other continuously.

The milky way galaxy is a galaxy that contains over a hundred billion stars and it also includes our solar system.  Its name describes its appearance when viewed from the earth. All the individual stars in the whole sky are a portion of the Milky Way Galaxy, the term "Milky Way" is because of the band of light.

The astronomer has discovered a new galaxy which means our universe is continuously expanding. This is because the universe encloses everything that exists.

Galaxies are moving in space and since the universe space is continuously expanding so the galaxies continuously move from each other. The farther the galaxy is from the milky way which is an observable part, the faster will be moving the galaxy and the closer the galaxy is to the milky way, the slower will be movement of the galaxy.

Learn more about the milky way galaxy, Here:

https://brainly.com/question/2905713

#SPJ5

Answer:

The object is moving at constant speed.

Explanation:

The spaces between the dots are equal.

Answer:

True

Explanation:

yes we can see that we are helping animals but we create pollution which is very bad

Answer:

no

Explanation:

so basically I am domb so I can not help you

Answer:

c

Explanation:

wavelength = speed of light/ frequency

= (3x 10^8 m/s)/(5.0 x 10^14 Hz)

= 6.0 x 10^-7 m

Answer:

a)  v = 3.116 m / s,  b)  μ = 1.65 10⁻²

Explanation:

a) to find the velocity of the set, let's define a system formed by the person and the sled, so that the forces during the collision are internal and the moment is conserved

initial instant. Before the crash

        p₀ = M v₀

final instant. After the crash

        p_f = (M + m) v

the moment is preserved

       M v₀ = (M + m) v

       v = [tex]\frac{M}{M+m} \ v_o[/tex]

let's calculate

      v = [tex]\frac{41.6}{41.6 + 14.6} \ 4.21[/tex]

      v = 3.116 m / s

b) for this part let's use the relationship between work and kinetic energy

        W = ΔK

as the body has its final kinetic energy is zero

the work of the friction forces is

          W = - fr x

the negative sign is because the friction forces always oppose the movement

let's write Newton's second law

Y axis  

        N - W_sled -W_person = 0

        N = mg + M g

        N = (m + M) g

X axis

        fr = ma

the friction force has the expression

        fr = μ N

        fr = μ g (m + M)

we substitute

        - μg (m + M) x = 0- ½ (m + M) v²

          μ = [tex]\frac{1}{2} \ \frac{v^2 }{g \ x }[/tex]

let's calculate

        μ = [tex]\frac{1}{2} \ \frac{3.116^2}{9.8 \ 30.0}[/tex]

        μ = 0.0165

        μ = 1.65 10⁻²

Answer:

a) m = 69.0 kg

b) release some gas in the opposite direction to the astronaut's movement

Explanation:

a) Let's use Newton's second law

         F = m a

         m = F / a

         m = 60.0 / 0.870

         m = 69.0 kg

b) when we exert a force on the astronaut it acquires a momentum po, as the astronaut system plus spacecraft is isolated, the momentum is conserved

         p₀ = p_f

         m v = M v '

         v ’= [tex]\frac{m}{M} \ v[/tex]

so we see that the ship is moving backwards, but since the mass of the ship is much greater than the mass of the astronaut, the speed of the ship is very small.

One method to avoid this effect is to release some gas in the opposite direction to the astronaut's movement so that the initial momentum of the astronaut plus the gas is zero and therefore no movement is created in the spacecraft.

Answer:

d

Explanation:

Answer:

32.52 N

Explanation:

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

Where F = Force needed to pull the block, m = mass of the block, a = acceleration of the block, F' = Frictional force of the surface acting on the blcok

F = ma+F'

But,

F' = μmg

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

F = ma+μmg.................... Equation 2

Given: m = 6.0 kg, a = 1.5 m/s², μ = 0.40

Constant: g = 9.8 m/s²

Substitute these values into equation 2

F = 6(1.5)+6(9.8)(0.4)

F = 9+23.52

F = 32.52 N

Slipping doesn't occur between the belt and cylinder B because the force of static friction is greater than force exerted on cylinder B.

Given the following data:

Mass A = 5 lb to kg = 2.27 kg.

Mass B = 20 lb to kg = 9.02 kg.

Force = 3.6 lb to N = 16.02 Newton.

In order to calculate the angular acceleration of each cylinder, we would take moment about the two cylinders.

For cylinder A:

[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_A\alpha _A = F_A (0.1)\\\\(\frac{m_Ar^2}{2}) \alpha _A = F_A (0.1)\\\\(\frac{2.27 \times 0.1^2}{2}) \alpha _A = F_A (0.1)\\\\0.1F_A=0.01135\alpha _A\\\\F_A=\frac{0.01135\alpha _A}{0.1} \\\\F_A= 0.1135\alpha _A[/tex]

For cylinder B:

[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_B\alpha _B = F_B (0.2)\\\\(\frac{m_Br^2}{2}) \frac{\alpha _A}{2} = F_B (0.2)\\\\(\frac{9.02 \times 0.1^2}{4}) \alpha _A = F_B (0.2)\\\\0.1F_B=0.02255\alpha _A\\\\F_B=\frac{0.02255\alpha _A}{0.2} \\\\F_B= 0.1128\alpha _A[/tex]

For the belt, we have

[tex]\sum F_A =0\\\\P-F_B-F_A=0\\\\16.02-0.1128\alpha _A-0.1135\alpha _A=0\\\\16.02=0.2263\alpha _A\\\\\alpha _A=\frac{16.02}{0.2263} \\\\\alpha _A=70.79 \;rad/s^2[/tex]

Also, we would determine the angular acceleration of cylinder B:

[tex]0.1\alpha _A=0.2\alpha _B\\\\0.1 \times 70.79 = 0.2\alpha _B\\\\7.079= 0.2\alpha _B\\\\\alpha _B=\frac{7.079}{0.2} \\\\\alpha _B=35.40\;rad/s^2[/tex]

Next, we would calculate the forces acting on the cylinders:

[tex]F_A = 0.1135\alpha _A\\\\F_A = 0.1135 \times 70.79\\\\F_A = 8.04 \;Newton[/tex]

[tex]F_B = P-F_A\\\\F_B = 16.02 - 8.04\\\\F_B = 7.98\;Newton[/tex]

Next, we would determine the force of static friction:

[tex]F_s = \mu_s N = \mu_s m_B g\\\\F_s = 0.50 \times 9.02 \times 9.8\\\\F_s=44.198\;Newton[/tex]

From the above calculation, we can deduce that the force of static friction is greater than force exerted on cylinder B. Therefore, slipping doesn't occur between the belt and cylinder B.

Read more on angular acceleration here: https://brainly.com/question/6860269

Answer:

B

Explanation:

Initially, the energies are:

[tex]U_{i}=-\frac{G M_{\varepsilon} m}{r_{e}} \\ =K_{i}=\frac{1}{2} m v_{0}^{2}[/tex]

At final point, the energies are:

[tex]U_{f}=-\frac{G M_{\varepsilon} m}{r_{e}+h} \\ K_{f}=\frac{1}{2} m(0)^{2}=0[/tex]

Using conservation law of energy,

[tex]-\frac{G M_{e} m}{r_{e}}+\frac{1}{2} m v_{0}^{2} &=-\frac{G M_{e} m}{r_{\varepsilon}+h} \\ -\frac{G M_{e}}{r_{e}}+\frac{v_{0}^{2}}{2} &=-\frac{G M_{e}}{r_{e}+h} \\ \frac{-2 G M_{e}+r_{e} v_{0}^{2}}{2 r_{e}} &=-\frac{G M_{e}}{r_{e}+h} \\ \frac{r_{e}+h}{G M_{e}} &=\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}[/tex]

The equation is further simplified as:

[tex]r_{e}+h &=\left(\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}\right) G M_{e} \\ h &=\frac{2 r_{e} G M_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}-r_{e} \\ &=\frac{2 r_{e} G M_{e}-2 r_{e} G M_{e}+r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}} \\ & h=\frac{r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}}[/tex]

Answer:

C = 2.9 10⁻⁵ F = 29 μF

Explanation:

In this exercise we must use that the voltage is

          V = i X

          i = V/X    

where X is the impedance of the system

in this case they ask us to treat the system as an RLC circuit in this case therefore the impedance is

          X = [tex]\sqrt{R^2 + ( wL - \frac{1}{wC})^2 }[/tex]

tells us to take inductance L = 0.

The angular velocity is

         w = 2π f

the current is required to be half the current at high frequency.

Let's analyze the situation at high frequency (high angular velocity) the capacitive impedance is very small

         [tex]\frac{1}{wC}[/tex] →0       when w → ∞

therefore in this frequency regime

         X₀ = [tex]\sqrt{R^2 + ( \frac{1}{2\pi 2 10^4 C} )^2 } = R \sqrt{ 1+ \frac{8 \ 10^{-10} }{RC} }[/tex]

the very small fraction for which we can despise it

        X₀ = R

to halve the current at f = 200 H, from equation 1 we obtain

         X = 2X₀

let's write the two equations of inductance

          X₀ = R                                    w → ∞

          X= 2X₀ = [tex]\sqrt{R^2 +( \frac{1}{wC} )^2 }[/tex]        w = 2π 200

we solve the system

         2R = \sqrt{R^2 +( \frac{1}{wC} )^2 }

         4 R² = R² + 1 / (wC) ²

         1 / (wC) ² = 3 R²

          w C = [tex]\frac{1}{\sqrt{3} } \ \frac{1}{R}[/tex]

          C = [tex]\frac{1}{\sqrt{3} } \ \frac{1}{wR}[/tex]

let's calculate

           C = [tex]\frac{1}{\sqrt{3} } \ \frac{1}{2\pi \ 200 \ 9}[/tex]

           C = 2.9 10⁻⁵ F

           C = 29 μF

Answer:

[tex]1.84[/tex]×[tex]10^{-26}[/tex]

Explanation:

specific charge = [tex]\frac{charge}{mass}[/tex]  so by rearranging for mass we get

mass= [tex]\frac{charge}{sepcifc charge}[/tex]

[tex]\frac{1.6×10^{-26} }{1.7×10^{-27} }[/tex] = answer in kg