A 65 kg bungee jumper leaps from a bridge. She is tied to a bungee cord that is 12 m long when unstretched, and falls a total of 33 m.

Requried:
a. Calculate the spring constant of the bungee cord assuming Hooke's law applies.
b. Calculate the maximum acceleration she experiences.

Answer:

Option A. 11

Explanation:

The atomic number of an element does not change.

Recall:

Atomic number = proton number

If the atom is neutral, then,

Proton number = electron number

Since the element is sodium, then, the atomic number of sodium–25 is 11.

Also, we were told to obtain the electrons of a neutral atom of sodium–25

Therefore,

Atomic number = proton number = 11

Since the atom is neutral,

Proton number = electron number = 11

Answer:

A. 11

Explanation:

A neutral atom of sodium-25 has the same number of protons and electrons. Since it has 11 protons, it also must have 11 electrons!

Answer:

the other force= (16.3i + 14.6j)N

EXPLANATION:

Given:

Mass=2.80-kg

t= 1.2s

Since the object started from rest, the origin is (0,0) which symbolize the the object's initial position.

We will need to calculate the magnitude of the displacement using the below formula;

d = (1/2)at2 + v0t + d0

But note that

d0 = 0,( initial position)

v0 = 0( initial position)

a is the net acceleration

d = √[4.202 + (-3.30)2] m = 5.34 m

Hence, the magnitude of the displacement is 5.34 m, then we can make 'a' the subject of formula in the above expression in order to calculate the value for acceleration, note that d0 = 0,( initial position) and v0 = 0( initial position)

d = (1/2)at2

a = 2d/t2 = 2(5.34)/(1.20)2 m/s2 = 7.42 m/s2

the net acceleration is 7.42 m/s2

Acceleration in terms of the vector can be calculated as

a=2(ri - r0)/t^2

Where t =1.2s which is the time

a= 2(4.2i - 3.30j)/ 1.2^2

a=( 5.83i - 4.58j)m/s

now the net force can now be calculated since we have known the value of acceleration, using the formula below;

F(x) = ma - mg

Where a = 5.83i - 4.58j)m/s and g= 9.8m/s

2.8(5.83i - 4.58j)m/s - (2.80 × 9.8)m/s^2

Therefore, the other force= (16.3i + 14.6j)N

Answer:

so initial speed of the rock is 30.32 m/s

correct answer is b. 30.3 m/s

Explanation:

given data

h = 15.0m

v = 25m/s

weight of the rock m = 3.00N  

solution

we use here work-energy theorem that is express as here

work = change in the kinetic energy    ..............................1

so it can be written as

work = force × distance     ...................2

and

KE is express as

K.E = 0.5 × m × v²  

and it can be written as

F × d = 0.5 × m × (vf)² - (vi)²      ......................3

here

m is mass and vi and vf is initial and final velocity

F = mg = m  (-9.8)  , d = 15 m and v{f} = 25 m/s

so put value in equation 3 we get

m  (-9.8) × 15 = 0.5 × m × (25)² - (vi)²

solve it we get

(vi)² =  919

vi = 30.32 m/s

so initial speed of the rock is 30.32 m/s

Answer:

The  frequencies are [tex](f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)[/tex]

Explanation:

From the question we are told that

  The  length of the ear canal is  [tex]l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m[/tex]

   The  speed of sound is assumed to be  [tex]v_s = 344 \ m/s[/tex]

Now  taking look at a typical  ear canal  we see that we assume it is  a  closed pipe

   Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

            [tex]f = \frac{v_s}{4 * l }[/tex]

 substituting values  

          [tex]f = \frac{344}{4 * 0.013 }[/tex]

         [tex]f = 6615.4 \ Hz[/tex]

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

        [tex]f_1 = \frac{3v_s}{4 * l}[/tex]

 substituting values  

       [tex]f_1 = \frac{3 * 344}{4 * 0.013}[/tex]

       [tex]f_1 = 19846.2 \ Hz[/tex]

Given that sound would be loudest in the pipe at the frequency, it implies that the child  will have an increased audible sensitivity at this  frequencies

Answer:

a) 1.95 V

b) 1.87 mA

c) 0.115 A

Explanation:

Given that

Number of primary turns, N(p) = 480

Number of secondary turns, N(s) = 7.8

Velocity of primary turns, V(p) = 120 V

Velocity of secondary turns, V(s) = ?

Current in the primary, I(p) = ?

Current in the secondary, I(s) ?

To solve this question, we would be using the formula

V(s)/V(p) = N(s)/N(s), now substituting the values, we have

V(s) / 120 = 7.8 / 480

V(s) = (7.8 * 120) / 480

V(s) = 936 / 480

V(s) = 1.95 V

To find the current in the primary, remember ohms law?

I = V/R

I(s) = V(s) / R(s)

I(s) = 1.95 / 17

I(s) = 0.115 A

Now, remember the relationship between current and voltage

I(p)/I(s) = V(s)/V(p)

I(p) / 0.115 = 1.95 / 120

I(p) = (1.95 * 0.115) / 120

I(p) = 0.22425 / 120

I(p) = 0.00187 A

I(p) = 1.87 mA

Answer:

126.01 rad/s^2

Explanation:

since it starts from rest, initial angular speed ω' = 0 rad/s

angular speed N = 477 rev/min

angular speed in rad/s ω = [tex]\frac{2\pi N}{60}[/tex] =  [tex]\frac{2*3.142* 477}{60}[/tex] = 49.95 rad/s

angular displacement ∅ = 1.5758 rev

angular displacement in rad/s = [tex]2\pi N[/tex] = 2 x 3.142 x 1.5758 = 9.9 rad

angular acceleration [tex]\alpha[/tex] = ?

using the equation of angular motion

ω^2 = ω'^2 + 2[tex]\alpha[/tex]∅

imputing values, we have

[tex]49.95^{2} = 0^{2} + (2 *\alpha*9.9 )[/tex]

2495 = 19.8[tex]\alpha[/tex]

[tex]\alpha[/tex] = 2495/19.8 = 126.01 rad/s^2

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± [tex]\frac{1}{2} at^2[/tex]               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± [tex]\frac{1}{2} at^2[/tex]               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

From the question;

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - [tex]\frac{1}{2} at^2[/tex]

h = 26.54(1.7) - [tex]\frac{1}{2} (10)(1.7)^2[/tex]

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

Answer:

A) 8π ft²/ft

B) 24π ft²/ft

C) 48π ft²/ft

Explanation:

Surface area of the spherical balloon is not clear here but it is supposed to be;

S = 4πr²

where:

r is the radius of the spherical balloon

So thus, the rate of change of the surface area of the spherical balloon by its radius will be:

dS/dr = 8πr

A) at r = 1ft;

dS/dr = 8 × π × 1

dS/dr = 8π ft²/ft

B) at r = 3 ft;

dS/dr = 8 × π × 3

dS/dr = 24π ft²/ft

C) at r = 6ft;

dS/dr = 8 × π × 6

dS/dr = 48π ft²/ft

Answer:

Explanation:

Answer:

a

    The apparent weight is  [tex]N = 424.5 N[/tex]

b

    The apparent weight is  [tex]N = 358.5 N[/tex]

Explanation:

From the question we are told that

    The diameter of the Ferris wheel is d  =  80ft

    The period of the Ferris wheel is  [tex]T = 24 \ s[/tex]

     The mass of the passenger is  [tex]m = 40 \ kg[/tex]

The radius of the Ferris wheel is evaluated as  [tex]r = \frac{d}{2}[/tex] substituting values

          [tex]r = \frac{80}{2}[/tex]

         [tex]r = 40 \ ft[/tex]

converting to meters

           [tex]r = 40 * 0.3048 = 12.20 \ m[/tex]

The angular velocity of the Ferris wheel is mathematically represented as  

                [tex]w = \frac{2 \pi}{T}[/tex]

substituting values  

                [tex]w = \frac{2* 3.142 }{24}[/tex]

                [tex]w = \frac{2* 3.142 }{24}[/tex]

                [tex]w = 0.2618 \ rad/s[/tex]

   At the lowest the point the  apparent weight of the passenger is  equal to the normal force on the chair which is  mathematically represented as

             [tex]N = m (g + a_c)[/tex]

Where  [tex]a_c[/tex] is the centripetal acceleration which is mathematically represented  as  

             [tex]a = w^2 R[/tex]

So  

          [tex]N = m (g + w^2 R)[/tex]

substituting values  

              [tex]N = 40 (9.8 + (0.2618)^2 * 12.2)[/tex]

              [tex]N = 424.5 N[/tex]

   At the highest the point the  apparent weight of the passenger is  equal to the normal force on the chair which is  mathematically represented as

         [tex]N = m (g - a_c)[/tex]

          [tex]N = m (g - w^2 R)[/tex]

          [tex]N = 40 (9.8 - (0.2618)^2 * 12.2)[/tex]

          [tex]N = 358.5 N[/tex]

Answer:

D

Explanation:

W = P∆V

Use the above equation and substitute, thanks

Centripetal acceleration = (speed squared) / (radius)

Centripetal acceleration = (10 m/s)² / (1.0 m)

Centripetal acceleration = (100 m²/s²) / (1.0 m)

Centripetal acceleration = 100 m/s²

Answer:

[tex]E = 55.9583\ Volts/meter[/tex]

Explanation:

First let's find the electric potential using y = 22.5:

[tex]V(y) = 1.69y^2 +15.6y+52.5[/tex]

[tex]V(22.5) = 1.69(22.5)^2 + 15.6*22.5 + 52.5[/tex]

[tex]V(22.5) = 1259.0625\ Volts[/tex]

Then, to find the magnitude of the electric field, we just need to divide the electric potential by the distance y:

[tex]E = V/d[/tex]

[tex]E = 1259.0625/22.5[/tex]

[tex]E = 55.9583\ Volts/meter[/tex]

Answer:

a)      I = 1.29 10⁻⁴ A , b) the current has the opposite direction in each cable, therefore the correct answer must be 1

Explanation:

a) Let's analyze this exercise, when a conductor carries a current the elctoenes generate a magnetic field around the cable and this field can interact with the electrons that carry a current in a nearby cable, the expression for force is

          F₁ = I₁ L B₂

where F₁ is the force on a cable through the field created by elotorcable.

The field sent by the other cable is

         B₂ = μ₀ I² / 2πa

we substitute

            F₁ = μ₀ I₁ I₂ / 2πa    L

suppose the current in the two cables is the same

            I₁ = I₂ = I

            F₁ / L = μ₀ I₂ / 2πa

            i = √ [(F₁ / L) 2πa / μ₀]

let's calculate

           I = √ [0.335 2 π / 4π 10⁻⁷]

           I = √ [0.335 / 2 10⁷]

           I = 1.29 10⁻⁴ A

b) if the two cables have the current in the same direction the force is attractive and if it has the current in the opposite direction it is negative

the two possible answers are 1 and 2

In general, to start a car, one of the cables has a current from the battery at start-up and the other cable comes from the start to the battery, therefore the current has the opposite direction in each cable, therefore the correct answer must be 1

Answer:

a) The heavier piece has a translational kinetic energy of 4647.857 joules, b) The lighter piece has a translational kinetic energy of 2582.143 joules.

Explanation:

a) The object breaking can be described by means of the Principle of Energy Conservation, knowing that heavier piece has 1.8 times the mass of the lighter ([tex]m_{h} = 1.8\cdot m_{l}[/tex]), both are modelled as particle due to the absence of rotation and that energy liberated by explosion is transform into kinetic energy, the equation that describes the phenomenon is:

[tex]E_{ex} = K_{h} + K_{l}[/tex]

Where:

[tex]E_{ex}[/tex] - Energy liberated by the explosion, measured in joules.

[tex]K_{h}[/tex], [tex]K_{l}[/tex] - Translational kinetic energies of the heavier and lighter piece, respectively.

This expression is expanded by using the definition of translational kinetic energy and supposing the both parts are liberated at the same initial speed ([tex]v_{o}[/tex]). Then:

[tex]E_{ex} = \frac{1}{2}\cdot (m_{h} + m_{l})\cdot v_{o}^{2}[/tex]

As can be seen, the energy liberated by expression is directly proportional to the mass of the system. Hence, the kinetic energy can be estimated by simple rule of three:

[tex]K_{h} = \frac{m_{h}}{m_{h}+m_{l}}\times E_{ex}[/tex]

If [tex]m_{h} = 1.8\cdot m_{l}[/tex] and [tex]E_{ex} = 7230\,J[/tex], then:

[tex]K_{h} =\frac{1.8\cdot m_{l}}{2.8\cdot m_{l}}\times E_{ex}[/tex]

[tex]K_{h} = \frac{9}{14}\cdot (7230\,J)[/tex]

[tex]K_{h} = 4647.857\,J[/tex]

The heavier piece has a translational kinetic energy of 4647.857 joules.

b) The translational kinetic energy of the lighter piece is calculated by using the equation derived from the Principle of Energy Conservation:

[tex]K_{l} = E_{ex} - K_{h}[/tex]

Given that [tex]E_{ex} = 7230\,J[/tex] and [tex]K_{h} = 4647.857\,J[/tex], the translational kinetic energy of the lighter piece is:

[tex]K_{l} = 7230\,J - 4647.857\,J[/tex]

[tex]K_{l} = 2582.143\,J[/tex]

The lighter piece has a translational kinetic energy of 2582.143 joules.

Answer:

The  change  in angular momentum is [tex]\Delta L = 0.0016 \ kgm^2/s[/tex]

Explanation:

From the question we are told that

      The initial angular velocity of the spinning top is  [tex]w_1 = 16 \ rad/s[/tex]

      The angular velocity after it slow down is  [tex]w_2 = 12 \ rad/s[/tex]

      The time  for it to slow down is  [tex]t = 18 \ s[/tex]

       The rotational inertia due to friction is  [tex]I = 0.0004 \ kg \cdot m^2[/tex]

 Generally the change in the angular momentum is  mathematically represented as  

         [tex]\Delta L = I *(w_1 -w_2)[/tex]

substituting values  

        [tex]\Delta L = 0.0004 *(16 -12)[/tex]

       [tex]\Delta L = 0.0016 \ kgm^2/s[/tex]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

[tex]K_{1} + W_{F} = K_{2}[/tex]

Where:

[tex]W_{F}[/tex] - Work done by the external force, measured in joules.

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

[tex]W_{F} = K_{2} - K_{1}[/tex]

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

[tex]W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})[/tex]

Where:

[tex]m[/tex] - Mass of the object, measured in kilograms.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

[tex]v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}[/tex]

[tex]v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}[/tex]

[tex]v_{1} \approx 25.060\,\frac{m}{s}[/tex]

Final velocity

[tex]v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}[/tex]

[tex]v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}[/tex]

[tex]v_{2} \approx 33.121\,\frac{m}{s}[/tex]

Finally, if [tex]m = 3.5\,kg[/tex], [tex]v_{1} \approx 25.060\,\frac{m}{s}[/tex] and [tex]v_{2} \approx 33.121\,\frac{m}{s}[/tex], then the work done by the force is:

[tex]W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right][/tex]

[tex]W_{F} = 820.745\,J[/tex]

The work done by the force is 820.745 joules.

Answer:

The angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s

Explanation:

Given;

moment of inertia of a skater with arms out, [tex]I_{arms \ out}[/tex] = 3.1 kg.m²

moment of inertia of a skater with arms in, [tex]I_{arms \ in}[/tex] = 0.9 kg.m²

inward angular speed, [tex]\omega _{in}[/tex] = 4 rev/s

The angular momentum of the skater when her arms are out and one leg extended is equal to her angular momentum when her arms and legs are in.

[tex]L_{out} = L_{in}[/tex]

[tex]I_{out} \omega_{out} = I_{in} \omega_{in}\\\\\omega_{out} = \frac{ I_{in} \omega_{in} }{I_{out} } \\\\\omega_{out} = \frac{0.9*4}{3.1} \\\\\omega_{out} = 1.161 \ rev/s[/tex]

Therefore, the angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s

Answer:

d. may be either greater, smaller, or equal to that observed inside the bus.

Explanation:

Answer:

El trabajo realizado por la fuerza de rozamiento sobre el bloque tras recorrer este último una distancia de 5 metros sobre el plano es de 500.566 joules.

Explanation:

El fenómeno alrededor del bloque puede ser modelado por el Principio de Conservación de la Energía y el Teorema del Trabajo y la Energía. Al descender lentamente, significa que la aceleración neta experimentada por el bloque es aproximadamente cero. El diagrama de cuerpo libre del bloque se presenta a continuación como archivo adjunto. Las ecuaciones de equilbrio del sistema son:

[tex]\Sigma F_{x'} = P\cdot \cos \theta + m\cdot g \cdot \sin \theta - f = 0[/tex]

[tex]\Sigma F_{y'} = N + P\cdot \sin \theta -m\cdot g\cdot \cos \theta = 0[/tex]

Donde:

[tex]P[/tex] - Fuerza externa aplicada a la caja, medida en newtons.

[tex]m[/tex] - Masa del bloque, medida en kilogramos.

[tex]g[/tex] - Aceleración gravitacional, medidas en metros sobre segundo al cuadrado.

[tex]f[/tex] - Fuerza de rozamiento, medida en newtons.

[tex]N[/tex] - Fuerza normal del plano sobre la caja, medida en newtons.

[tex]\theta[/tex] - Ángulo de inclinación del plano, medido en grados sexagesimales.

Dado que todas las fuerzas son constantes, se puede emplear la definición de trabajo como el producto de la fuerza paralela a la dirección del movimiento y la magnitud de distancia recorrida en el movimiento, entonces la primera ecuación de equilibrio queda así al multiplicar cada lado por la distancia recorrida:

[tex]P\cdot \Delta s \cdot \cos \theta + m\cdot g \cdot \Delta s \cdot \sin \theta - W_{f} = 0[/tex]

Ahora, la cantidad de trabajo realizado por la fuerza de rozamiento es:

[tex]W_{f} = (P\cdot \cos \theta+m\cdot g\cdot \sin \theta)\cdot \Delta s[/tex]

Si [tex]P = 50\,N[/tex], [tex]m = 10\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}}[/tex], [tex]\theta = 37^{\circ}[/tex] and [tex]\Delta s = 5\,m[/tex], entonces el trabajo realizado por la fuerza de rozamiento es:

[tex]W_{f} = \left[(50\,N)\cdot \cos 37^{\circ}+(10\,kg)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot \sin 37^{\circ}\right]\cdot (5\,m)[/tex]

[tex]W_{f} = 500.566\,J[/tex]

El trabajo realizado por la fuerza de rozamiento sobre el bloque tras recorrer este último una distancia de 5 metros sobre el plano es de 500.566 joules.

Answer:

The tangential speed of Fayetteville, NC is 1372.2 km/h.

Explanation:

To find the tangential speed at the latitude 35 °N, first, we need to calculate the radius at that latitude [tex]r_{lat}[/tex]:

[tex] r_{lat} = r_{eq}*cos(\theta) [/tex]

Where:

[tex]r_{eq}[/tex]: is the equatorial radius = 6380 km

θ = 35 °

[tex] r_{lat} = r_{eq}*cos(\theta) = 6380 km*cos(35) = 5226.2 km [/tex]

Now, the tangential speed is:

[tex] v = \omega*r = \frac{2\pi}{23.93 h}*5226.2 km = 1372.2 km/h [/tex]

Therefore, the tangential speed of Fayetteville, NC is 1372.2 km/h.

I hope it helps you!

Answer:

The angular momentum of the particle about the origin is [tex]\vec l = -19.305\,k\,\left[kg\cdot \frac{m}{s} \right][/tex].

Explanation:

Vectorially speaking, the angular momentum is given by the following cross product:

[tex]\vec l = \vec r \times m\vec v[/tex]

This cross product can be solved with the help of determinants and its properties, that is:

[tex]\vec l = \left|\begin{array}{ccc}i&j&k\\r_{x}&r_{y}&0\\m\cdot v_{x}&m\cdot v_{y}&0\end{array}\right|[/tex]

[tex]\vec l = m\left|\begin{array}{ccc}i&j&k\\r_{x}&r_{y}&0\\v_{x}& v_{y}&0\end{array}\right|[/tex]

The 3 x 3 determinant is solved by the Sarrus Law:

[tex]\vec l = m \cdot (r_{x}\cdot v_{y} - r_{y}\cdot v_{x})k[/tex]

If [tex]m = 1.30\,kg[/tex], [tex]\vec r = 1.50\,i + 2.20\,j\,[m][/tex] and [tex]\vec v = 4.50\,i-3.30\,j\,\left[\frac{m}{s} \right][/tex], the angular momentum of the particle about the origin is:

[tex]\vec l = (1.30\,kg)\cdot \left[\left(1.50\,m\right)\cdot\left(-3.30\,\frac{m}{s} \right)-\left(2.20\,m\right)\cdot\left(4.50\,\frac{m}{s} \right)\right]k[/tex]

[tex]\vec l = -19.305\,k\,\left[kg\cdot \frac{m}{s} \right][/tex]

The angular momentum of the particle about the origin is [tex]\vec l = -19.305\,k\,\left[kg\cdot \frac{m}{s} \right][/tex].

Answer:

6.22²

Explanation:

Given that

Mass of the skydiver, m = 80 kg

Terminal speed of the skydiver, v(f) = 50 m/s

Speed of the skydiver, v(i) = 30 m/s

Acceleration of the skydiver, a = ?

To solve this, we use the formula

W - k v² = ma, where

W = weight of the skydiver

k = constant

v = speed of the skydiver

m = mass of the skydiver

So, if we substitute the values into it we have

W = mg = 80 * 9.8 = 784 N

784 - k 50² = 80 *0

784 - 2500k = 0

784 = 2500k

k = 0.3136

Now, we use this value of k to find the needed acceleration using the same formula at a speed of 30 m/s

784 - 0.3136 * 30² = 80 * a

784 - 0.3136 * 900 = 80a

784 - 282.24 = 80a

497.76 = 80a

a = 497.76 / 80

a = 6.22 m/s²

Thus, we can conclude that the acceleration when the speed of the skydiver is 30 m/s, is 6.22 m/s²

Answer:

A wave is transporting energy from left to right. The particles of the medium are moving back and forth in a leftward and rightward direction.

Explanation:

No, that's silly.

Waves carry energy. So they travel in the same direction, and wherever the wave goes, the energy goes.

Explanation:

a)

θ = 4.91 + 9.7t + 2.06t² when t = 0

θ = 4.91 rad

θ = 4.91 + 9.7t + 2.06t²

ω = dθ/dt = 9.7 + 2.06t, when t =0

ω = dθ/dt = 9.7 + 0

ω = 9.7 rad/s

α = d²θ/dt² = 2.06

α= 2.06 rad/s²

b) please use same method above for t = 2.94 s

At zero seconds, the angular position is 4.91 rad, the angular velocity is  9.7 radian/second and the angular acceleration is 2.06 rad/s².

When the angular velocity change in relation to time, then it results in angular acceleration. It comes with direction, so it is a vector quantity.

Given: Angular position equation (θ) = 4.91 + 9.7t + 2.06[tex]t^{2}[/tex]

If the angular velocity is ω, angular acceleration is α, and t is the time then

(A) At time t = 0 second

θ = 4.91 + 9.7 × 0 + 2.06 × [tex]0^{2}[/tex]

θ = 4.91 rad

Angular velocity (ω) = dθ/dt

ω = dθ/dt,

ω = 9.7 + 2.06t   (at t = 0)

ω = 9.7 radian/second

Angular acceleration (α) = d²θ/dt²

α= 2.06 rad/s²

Similarly, at t= 2.94 sec

θ = 51.23 rad

ω = 15.76 rad/s

α = 2.06 rad/s²


To know more about Angular acceleration:

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Answer:

Therefore, the answer is E. They strike the plane at the same time.

Explanation:

Here, it is seen that the time depends only on acceleration due to gravity (which is a constant) and vertical displacement, and not on velocity of the bullets or mass of the bullets.

Hence, the bullets that are fired simultaneously parallel to the horizontal plane will strike the plane at the same time.

using equation of motion for displacement

s= ut + ¹/₂gt²

here, g is the acceleration due to gravity along y- direction

U along y is 0

s = (0)t + ¹/₂gt²

s=¹/₂gt²

make t the subject of formula =  [tex]\sqrt{\frac{2s}{g} }[/tex]

Answer:

432 J

Explanation:

When moving linearly:

Kinetic Energy = (1/2)mV^2

So here you have:

KE=(1/2)(6)(12^2)=(1/2)(6)(144)=432

The unit for energy is Joules (J), so your answer would be 432 J.

Answer:

The potential enwrgy associated with charge decreases.

The ele ric field does negative work on the charge.

Explanation:

Answer:

The potential energy associated with the charge decreases

The electric field does positive work on the charge.

Answer:

540C.

Explanation:

A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;

Q = CV          ----------(i)

From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.

Where;

C = 6F

V = 90V

Substitute these values into equation (i) as follows;

Q = 6 x 90

Q = 540 C

Therefore, the initial charge on the capacitor is 540C.

Answer:

The air-water interface is an example of boundary. The transmitted portion of the initial wave energy is way smaller than the reflected portion. This makes the boundary  wave hard to hear.

When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can travel directly to your ear.

Explanation:

The air-to-water sound wave transmission is inhibited because more of reflection than transmission of the wave occurs at the boundary. In the end, only about 30% of the sound wave eventually reaches underwater. For sound generated underwater, all the wave energy is transmitted directly to the observer. Sound wave travel faster in water than in air because, the molecules of water are more densely packed together, and hence can easily transmit their vibration to their neighboring molecules, when compared to air.

Answer: The air-water interface is an example of boundary. The (transmitted) portion of the initial wave is way smaller than the (reflected) portion. This makes the (transmitted) wave hard to hear.

When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can (travel directly to your ears.)

Explanation:

The part of the sound wave that is transmitted across the boundary between air and water is much smaller than the part of the wave that is reflected. This is what makes it hard to hear your friend knocking two rocks together above the surface.

When you and the rocks are underwater, the sound that comes from knocking the rocks together can travel directly to your ears rather than having to be transmitted across mediums.