The cart now moves toward the right with an acceleration toward the right of 2.50 m/s2. What does spring scale Fz read? Show your calculations, and explain.

Answers

Answer 1

Complete Question

The  complete question is shown on the first uploaded image

Answer:

The spring scale [tex]F_2[/tex] reads  [tex]F_2 = 2.4225 \ N[/tex]

Explanation:

From the question we are told that

      The first force is  [tex]F_1 = 10.5 \ N[/tex]

      The acceleration by which the cart moves to the right is  [tex]a = 2.50 \ m/s^2[/tex]

      The mass of the cart is  m  = 3.231  kg

       

Generally the net force on the cart is  

       [tex]F_{net} = F_1 - F_2[/tex]

This net force is mathematically represented as

      [tex]F_{net} = m * a[/tex]

So  

        [tex]m* a = 10 - F_2[/tex]

        [tex]F_2 = 10.5 - 2.5 (3.231)[/tex]

        [tex]F_2 = 2.4225 \ N[/tex]

 

The Cart Now Moves Toward The Right With An Acceleration Toward The Right Of 2.50 M/s2. What Does Spring

Related Questions

A car travels at 100 km / h, collides head-on against a pole. Assuming the vehicle stopped at 2.2 seconds after impact, calculate the magnitude of the deceleration suffered by the driver.

Answers

Answer:

12.6 m/s²

Explanation:

First, convert to m/s.

100 km/h × (1000 m/km) × (1 hr / 3600 s) = 27.8 m/s

a = Δv / Δt

a = (0 m/s − 27.8 m/s) / 2.2 s

a = -12.6 m/s²

A 7.0-kg shell at rest explodes into two fragments, one with a mass of 2.0 kg and the other with a mass of 5.0 kg. If the heavier fragment gains 100 J of kinetic energy from the explosion, how much kinetic energy does the lighter one gain?

Answers

Answer:

39.94m/s.

Explanation:

Kinetic energy is expressed as KE = 1/2 mv² where;

m is the mass of the body

v is the velocity of the body.

For the heavier shell;

m = 5kg

KE gained = 100J

Substituting this values into the formula above to get the velocity v;

100 = 1/2 * 5 * v²

5v² = 200

v² = 200/5

v² = 40

v = √40

v = 6.32 m/s

Note that after the explosion, both body fragments will possess the same velocity.

For the lighter shell;

mass = 2.0kg and v = 6.32m/s

KE of the lighter shell = 1/2 * 2 * 6.32²

KE of the lighter shell = 6.32²

KE of the lighter shell= 39.94m/s

Hence, the lighter one gains a kinetic energy of 39.94m/s.

The gain in the kinetic energy of the smaller fragment is 249.64 J.

The given parameters;

Mass of the shell, m = 7.0 kgMass of one fragment, m₁ = 2.0 kgMass of the second fragment, m₂ = 5.0 kgKinetic energy of heavier fragment, K.E₁ = 100 J

The velocity of the heavier fragment is calculated as follows;

[tex]K.E = \frac{1}{2} mv^2\\\\mv^2 = 2K.E\\\\v^2 = \frac{2K.E}{m} \\\\v= \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2 \times 100}{5} }\\\\v = 6.32 \ m/s[/tex]

Apply the principle of conservation of linear momentum to determine the velocity of the smaller fragment as;

[tex]m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\-6.32(5) \ + 2u_2 = 0(7)\\\\-31.6 + 2u_2 = 0\\\\2u_2 = 31.6\\\\u_2 = \frac{31.6}{2} \\\\u_2 = 15.8 \ m/s[/tex]

The gain in the kinetic energy of the smaller fragment is calculated as follows;

[tex]K.E_2 = \frac{1}{2} mu_2^2\\\\K.E_2 = \frac{1}{2} \times 2 \times (15.8)^2\\\\K.E_2 = 249.64 \ J[/tex]

Thus, the gain in the kinetic energy of the smaller fragment is 249.64 J.

Learn more about conservation of linear momentum here: https://brainly.com/question/7538238

A 25 kg box sliding to the left across a horizontal surface is brought to a halt in a distance of 15 cm by a horizontal rope pulling to the right with 15 N tension.

Required:
a. How much work is done by the tension?
b. How much work is done by gravity?

Answers

B)is pills everything to the surface of the earth not sure about A

The work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.

The given parameters;

mass of the box, m = 25 kgdistance traveled by the box, d = 15 cm = 0.15 mtension on the rope, T = 15 N

The work done by the tension is calculated as follows;

W = Fd

W = 15 x 0.15

W = 2.25 J

The work done by gravity is calculated as;

W = (25 x 9.8) x 0.15

W = 36.75 J

Thus, the work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.

Learn more here: https://brainly.com/question/19498865

Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

Answers

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

What is the relationship between the surface area of a parachute and the amount of air resistance it builds up when it is deployed by a sky diver?

Answers

Answer:

An open parachute increases the cross-sectional area of the falling skydiver and thus increases the amount of air resistance which he encounters. Once the parachute is opened, the air resistance overwhelms the downward force of gravity.

Explanation:

The larger a parachute, the greater the force.

Hope it helps you in a little way.

Two long parallel wires are separated by 11 cm. One of the wires carries a current of 54 A and the other carries a current of 45 A. Determine the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current.

Answers

Explanation:

It is given that,

The separation between two parallel wires, r = 11 cm = 0.11 m

Current in wire 1, [tex]q_1=54\ A[/tex]

Current in wire 2, [tex]q_2=45\ A[/tex]

Length of wires, l = 4.3 m

We need to find the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current. The magnetic force per unit length is given by :

[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 54\times 45\times 4.3}{2\pi \times 0.11}\\\\F=0.0189\ N[/tex]

So, the magnetic force on a 4.3 m length of the wire  on both of currents is F=0.0189 N.

In your own words, discuss how energy conservation applies to a pendulum. Where is the potential energy the most? Where is the potential energy the least? Where is kinetic energy the most? Where is kinetic energy the least?

Answers

Answer:

Explanation:

Energy conservation applies to the swinging of pendulum . When the bob is at one extreme , it is at some height from its lowest point . So it has some gravitational potential energy . At that time since it remains at rest its kinetic energy is zero or the least . As it goes down while swinging , its potential energy decreases and kinetic energy increases following conservation of mechanical energy . At the At the lowest point , its potential energy is least  and kinetic energy is maximum .

In this way , there is conservation of mechanical energy .

On a certain planet a body is thrown vertically upwards with an initial speed of 40 m / s. If the maximum height was 100 m, the acceleration due to gravity is

a) 15 m / s 2
b) 12.5 m / s 2
c) 8 m / s 2
d) 10 m / s 2

Answers

Answer:

C) 8 m/s²

Explanation:

Given:

v₀ = 40 m/s

v = 0 m/s

Δy = 100 m

Find: a

v² = v₀² + 2aΔy

(0 m/s)² = (40 m/s)² + 2a (100 m)

a = -8 m/s²

g A tube open at both ends, resonated at it's fundamental frequency, to a sound wave traveling at 330m/s. If the length of the tube is 4cm, find the frequency of the sound wave.

Answers

Answer:

frequency =4125Hz

Explanation:

L = 4cm = 0.04m

f =v/2L

f = 330/2 x 0.04

f = 4125Hz

As more energy from fossil fuels and other fuels is released on Earth, the overall temperature of Earth tends to rise. Discuss how temperature equilibrium explains why Earth’s temperature cannot rise indefinitely.

Answers

Answer:

processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.

Explanation:

The temperature of planet Earth is due to two main types of process, internal and external.

Internal processes are all chemical processes that occur that release heat into the environment or due to gases that trap heat on the planet, greenhouse effect

External processes is heating due to energy coming from the Sun. This includes direct heating of the surface by the absorption of energy and reflects of energy in different atmospheric layers.

These are the two terms that heat the Earth

In addition there are several processes so the planet loses energy,

* energy radiation to outer space that is a few degrees kelvin, for which there is a permanent emission

* endothermic processes that need to absorb heat to perform, this lowers the temperature of the system

* liquid (water) system that absorbs large amounts of heat to change state and temperature.

These processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.

Therefore it is impossible for the temperature to increase indefinitely since the emission would increase by decreasing the value

A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is:
A. 0
B. 50 km/hr
C. 100 km/hr
D. 200 km/hr
E. cannot be calculated without knowing the acceleration

Answers

Answer:

The average velocity for this trip is 0 km/hr

Explanation:

We know that average velocity = total displacement/total time.

Now, its displacement is d = final position - initial position.

Since the  car starts and ends at its initial position at Hither, if we assume its initial position is 0 km, then its final position is also 0 km.

So, its displacement is d = 0 km - 0 km = 0 km.

Since the total time for the round trip is 2 hours, the average velocity is

total displacement/ total time = 0 km/2 hr = 0 km/hr.

So the average velocity for this trip is 0 km/hr  

At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is increasing at a rate of 3.0 J/s, how fast is the current changing

Answers

Answer:

The current is changing at the rate of 0.20 A/s

Explanation:

Given;

inductance of the inductor, L = 5.0-H

current in the inductor, I = 3.0 A

Energy stored in the inductor at the given instant, E = 3.0 J/s

The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

E = 22.5 J/s

This energy is increased by 3.0 J/s

E = 22.5 J/s + 3.0 J/s = 25.5 J/s

Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

I² = 10.2

I = √10.2

I = 3.194 A/s

The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

≅0.20 A/s

Therefore, the current is changing at the rate of 0.20 A/s.

The rate at which the current is changing is;

di/dt = 0.2 A/s

We are given;

Inductance; L = 5 H

Current; I = 3 A

Rate of Increase of energy; dE/dt = 3 J/s

Now, the formula for energy stored in inductor is given as;

E = ¹/₂LI²

Since we are looking for rate at which current is changing, then we differentiate both sides of the energy equation to get;

dE/dt = LI (di/dt)

Plugging in the relevant values gives;

3 = (5 × 3)(di/dt)

di/dt = 3/(5 × 3)

di/dt = 0.2 A/s

Read more at; https://brainly.com/question/13112120

A cylinder is closed by a piston connected to a spring of constant 2.20 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. The piston has a cross sectional area of 0.0100 m^2 and negligible mass. What is the pressure of the gas at 250 °C?

Answers

Answer:

1.3515x10^5pa

Explanation:

Plss see attached file

The intensity of sunlight at the Earth's distance from the Sun is 1370 W/m2. (a) Assume the Earth absorbs all the sunlight incident upon it. Find the total force the Sun exerts on the Earth due to radiation pressure. N (b) Explain how this force compares with the Sun's gravitational attraction.

Answers

Answer:

F= 3.56e22N

Explanation:

Using the force of radiation acting on the earth which is

force = radiation pressure x area = (intensity/c)xpi R^2

force = 1370W/m^2 x pi x( 6.37x10^6m)^2/3x10^8m/s

force = 5.82x10^8 N

But the sun's gravitational attraction means the magnitude of the solar gravitational force on earth: If that's the case, the answer is approx 10^22 N:

F=GMm/r^2

G=6.67x10^(-11)=6.67e-11

M=mass sun = 2x10^30kg=2e30

m=mass earth = 6x10^24kg

r=earth sun distance = 1.5x10^11m

F=(6.6e-11)(2e30)(6e24)/(1.5e11)^2 =

F= 3.56e22N

A person can survive a feet-first impact at a speed of about 12 m/s (27 mi/h) on concrete, 15 m/s (34 mi/h) on soil, and 34 m/s (76 mi/h) on water. What is the reason for the different values for different surfaces

Answers

Answer:

Different surfaces have different impact force during collision which depends on the time it takes the person to come to rest after collision.

Explanation:

Given;

speed on concrete = 12 m/s (27 mi/h)

speed on soil = 15 m/s (34 mi/h)

speed on water = 34 m/s (76 mi/h)

The impact force on this person during collision is rate of change of momentum;

[tex]F = \frac{\delta P}{\delta t}[/tex]

During collision, the force exerted on this person depends on how long the collision lasts; that is, how long it takes for this person to come to rest after collision with each of the surfaces.

The longer the time of collision, the smaller the force exerted by each.

It takes shorter time for the person to come to rest on concrete surface than on soil surface, also it takes shorter time for the person to come to rest on soil surface than on water surface.

As a result of the reason above, the force exerted on the person during collision by the concrete surface is greater than that of soil surface which is  greater than that of water surface.

An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal butopposite charge on its plates. All the geometric parameters of the capacitor (plate diameter andplate separation) are now DOUBLED. If the original energy stored in the capacitor was U0, howmuch energy does it now store?

Answers

Answer:

U_f = (U_o)/2)

Explanation:

The capacitance of a given capacitor is given by the formula;

C = ε_o•A/d

While energy stored in plates capacitor is given as; U_o = Q²/2C

Now,we are told that that all the dimensions of the capacitor plate is doubled. Thus, we now have;

C' = ε_o•4A/2d

Hence, C' = 2C

so capacitance is now doubled

Thus, the final energy stored between the plates of capacitor is given as;

U_f = Q²/2C'

From earlier, we saw that C' = 2C.

Thus;

U_f = Q²/2(2C)

U_f = Q²/4C

Rearranging, we have;

U_f = (1/2)(Q²/2C)

From earlier, U_o = Q²/2C

Hence,

U_f = (1/2)(U_o)

Or

U_f = (U_o/2)

Consider the following three objects, each of the same mass and radius:
(1) a solid sphere
(2) a solid disk
(3) a hoop
All three are released from rest at the top of an inclined plane. The three objects proceed down the incline undergoing rolling motion without slipping. Use work-kinetic energy theorem to determine which object will reach the bottom of the incline first.
a) 1, 2, 3
b) 2, 3, 1
c) 3, 1, 2
d) 3, 2, 1
e) All three reach the bottom at the same time.

Answers

Answer:

Explanation:a 1

Suppose you are looking into the end of a long cylindrical tube in which there is a uniform magnetic field pointing away from you. If the magnitude of the field is decreasing with time the direction of the induced magnetic field is

Answers

Answer:

If the magnitude of the field is decreasing with time the direction of the induced magnetic field is CLOCKWISE

Explanation

This is because If the magnetic field decreases with time, the electric field will be produced in order to oppose the change in line with lenz law. Thus The right hand rule can be applied to find that the direction of electric field is in the clockwise direction.

Each of the boxes starts at rest and is then pulled for 2.0 m across a level, frictionless floor by a rope with the noted force. Which box has the highest final speed

Answers

Answer:

Explanation:

d

which of the following statements is not true Negatively charged objects attract other negatively charged objects. Positively charged objects attract negatively charged objects. Positively charged objects attract neutral objects. Negatively chargers objects attract neutral objects.

Answers

Answer:

negatively charged object attract other negatively objects

Explanation:

opposites attract

Answer:

negativelycharged objects attract other negatively charged objects

Explanation:

unlike charges attract like charges repel

You throw a stone vertically upward with a speed of 26.0 m/s. (a) How fast is it moving when it reaches a height of 15.0 m? (b) How much time is required to reach this height when it's falling down? a. 19.5 m/s , b. 4.51 s a. 17.9 m/s , b. 0.620 s a. 19.5 m/s , b. 0.800 s a. 17.9 m/s , b. 4.28 s a. 380 m/s , b. 8 s

Answers

Answer:

ok well

Explanation:

teghe

Answer:

v = 19.5 m/s

t = 4.51 s

Explanation:

a)

given:

height is 15m from the ground

initial velocity Vi = 26 m/s

acceleration a or g = 9.81 m/s²

formula: Vf² = Vi² + 2aΔy

26² = Vi² + 2 (9.81) 15

Vi = 19.5 m/s

now you can calculate the time by using the equations below:

Δy = 1/2 (Vi + Vf) t

Vf = Vi + a t

Δy = Vi t + 1/2 a t

time must be 4.51 s

If a 20kg mass hangs from a spring, whose elastic constant is 1800 N / m, the value of the spring elongation is

Answers

Explanation:

F = kx

mg = kx

(20 kg) (10 m/s²) = (1800 N/m) x

x = 0.11 m

Si se deja caer una piedra desde un helicóptero en reposo, entonces al cabo de 20 s cual será la rapidez y la distancia recorrida por la piedra

Answers

Answer:

La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.

Explanation:

Si se excluye los efectos del arrastre por la viscosidad del aire, la piedra experimenta un movimiento de caída libre, es decir, que la piedra es acelerada por la gravedad terrestre. La distancia recorrida y la rapidez final de la piedra pueden obtenerse con la ayuda de las siguientes ecuaciones cinemáticas:

[tex]v = v_{o} + g\cdot t[/tex]

[tex]y - y_{o} = v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex]

Donde:

[tex]v[/tex], [tex]v_{o}[/tex] - Rapideces final e inicial de la piedra, medidas en metros por segundo.

[tex]t[/tex] - Tiempo, medido en segundos.

[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo al cuadrado.

[tex]y[/tex]. [tex]y_{o}[/tex] - Posiciones final e inicial de la piedra, medidos en metros.

Si [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 0\,m[/tex], entonces:

[tex]v = 0\,\frac{m}{s} +\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)[/tex]

[tex]v = -196.14\,\frac{m}{s}[/tex]

[tex]y-y_{o} = \left(0\,\frac{m}{s} \right)\cdot (20\,s) + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)^{2}[/tex]

[tex]y-y_{o} = -1961.4\,m[/tex]

La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.

which of the following is a physical change?

A. a newspaper burns when placed in a fire.
B.an iron chair rusts when left outside
C.a sample of water boils and releases gas.
D.a plant changes carbon dioxide and water into sugar

Answers

It’s C. This is because all of the other options you can’t turn back but if you boil water, the gas can turn back by condensation. If a newspaper burns it’s gone. If the chair rust you can only scrape off the rust. If the plant changes the CO2 you aren’t getting it back. At least not easily

A string of mass 60.0 g and length 2.0 m is fixed at both ends and with 500 N in tension. a. If a wave is sent along this string, what will be the wave's speed? A second wave is sent in the string, what is the new speed of each of the two waves?

Answers

Answer:

a

The  speed of  wave is   [tex]v_1 = 129.1 \ m/s[/tex]

b

The new speed of the two waves is [tex]v = 129.1 \ m/s[/tex]

Explanation:

From the question we are told that

    The mass of the string is  [tex]m = 60 \ g = 60 *10^{-3} \ kg[/tex]

    The length is  [tex]l = 2.0 \ m[/tex]

    The tension is  [tex]T = 500 \ N[/tex]

Now the velocity of the first wave is mathematically represented as

     [tex]v_1 = \sqrt{ \frac{T}{\mu} }[/tex]

Where  [tex]\mu[/tex] is the linear density which is mathematically represented as

      [tex]\mu = \frac{m}{l}[/tex]

substituting values    

     [tex]\mu = \frac{ 60 *10^{-3}}{2.0 }[/tex]

     [tex]\mu = 0.03\ kg/m[/tex]

So

   [tex]v_1 = \sqrt{ \frac{500}{0.03} }[/tex]

   [tex]v_1 = 129.1 \ m/s[/tex]

Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )

     

You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after pushing it across some distance d, and the block leaves your hand with a velocity of 0.85 m/s. While you are pushing, the work done by friction between the ice and the ground is 3 Nm (3 J). Assuming that the ice block was stationary before you push it, find d.

Answers

Answer: d = 33 cm or 0.33 m

Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:

W = F.d.cosθ

F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.

For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:

W = F.d.cos(0)

W = F.d

To determine d:

d = [tex]\frac{W}{F}[/tex]

d = [tex]\frac{3}{9}[/tex]

d = 0.33 m

The distance d the block ice moved is 33 cm.

A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 26 m/s when a 60 kg skydiver drops out by releasing his grip on the glider.
What is the glider's speed just after the skydiver lets go?

Answers

Answer:

The glider’s speed after the skydiver lets go is 26 m/s

Explanation:

To calculate the glider’s speed just after the skydiver lets go, we will need to use the conservation of momentum

Mathematically;

mv = mv + mv

so 680 * 26 = (680-60)v + 60 * 26

17680 = 620v + 1560

17680-1560 = 620v

16120 = 620v

v = 16120/620

v = 26 m/s

A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside the solenoid near its center?

Answers

Answer:

Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

Explanation:

Given;

number of turns of solenoid, N = 269 turn

length of the solenoid, L = 102 cm = 1.02 m

radius of the solenoid, r = 2.3 cm = 0.023 m

current in the solenoid, I = 3.9 A

Magnitude of the magnetic field inside the solenoid near its centre is calculated as;

[tex]B = \frac{\mu_o NI}{l} \\\\[/tex]

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

[tex]B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T[/tex]

Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

Why would physics be used to study light emitted by a star?
O A. Stars form interesting shapes in the sky.
B. Light is very pretty.
O C. The positions of stars control our lives.
O D. Light is a form of energy.

Answers

Answer:

O D.

Explanation:

Physics has an aspect that deals with the study of energy

Answer:

D. Light is a form of energy

Explanation:

A scientist is testing the seismometer in his lab and has created an apparatus that mimics the motion of the earthquake felt in part (a) by attaching the test mass to a spring. If the test mass weighs 13 N, what should be the spring constant of the spring the scientist use to simulate the relative motion of the test mass and the ground from part (a)?

Answers

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

 [tex]a_{max} = 0.00246 \ m/s^2[/tex]

b

   [tex]k =722.2 \ N/m[/tex]

Explanation:

From the question we are told that

     The  amplitude is [tex]A = 1.8 \ cm = 0.018 \ m[/tex]

     The period is [tex]T = 17 \ s[/tex]

    The test weight is  [tex]W = 13 \ N[/tex]

Generally the radial acceleration is mathematically represented as

        [tex]a = w^2 r[/tex]

at maximum angular acceleration

       [tex]r = A[/tex]

So  

       [tex]a_{max} = w^2 A[/tex]

Now [tex]w[/tex] is the angular velocity which is mathematically represented as

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

Therefore

       [tex]a_{max} = [\frac{2 * \pi}{T} ]^2 * A[/tex]

substituting values

       [tex]a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018[/tex]

       [tex]a_{max} = 0.00246 \ m/s^2[/tex]

Generally this test weight is mathematically represented as

     [tex]W = k * A[/tex]

Where k is the spring constant

Therefore

        [tex]k = \frac{W}{A}[/tex]

substituting values        

      [tex]k = \frac{13}{0.018}[/tex]

     [tex]k =722.2 \ N/m[/tex]

Other Questions
Compare 6 108 to 3 106. Which best describes what happens if two waves meet and build on each other?A. constructive interferenceB. destructive interferenceC. reflectionD. absorption Exercise 10-6 Direct Materials and Direct Labor Variances [LO10-1, LO10-2] Huron Company produces a commercial cleaning compound known as Zoom. The direct materials and direct labor standards for one unit of Zoom are given below: Standard Quantity or Hours Standard Price or Rate Standard Cost Direct materials 7.40 pounds $ 2.60 per pound $ 19.24 Direct labor 0.45 hours $ 8.00 per hour $ 3.60 During the most recent month, the following activity was recorded: 12,100.00 pounds of material were purchased at a cost of $2.50 per pound. All of the material purchased was used to produce 1,500 units of Zoom. 575 hours of direct labor time were recorded at a total labor cost of $5,750. Required: 1. Compute the materials price and quantity variances for the month. 2. Compute the labor rate and efficiency variances for the month. Please answer this correctly without making mistakes Simplify the correct answer Humans developed this because it helped them stay alive by restricting the amount of information processed by the brain and scanning the scene before them quickly, looking for movement or physical threats. Group of answer choices Stereoscopic Visiion Binocular Vision Selective Vision Linear Perspective Given: F(x) = 2x - 4; GX) = 3x + 2; Hpx)=x2Find FG H(2)0 12127071 Below are 5 sets of potential solutes for you to compare. Both members of each pair are very soluble in water. If you had equal molar concentrations of each solution, which member of each pair would theoretically be the better conductor of electricity?A. CsCl, CaCl2.B. CaS, Li2S.C. KBr, AlCl3.D. AlCl3, MgC2.E. KI, K2S. During the current month, Grey Company transferred 60,000 units of finished production out of the Mixing Department at a cost of $6 each. They were transferred to finished goods. The journal entry to record the transfer would be which of the following?a. Finished Goods 360,000Work in Process 360,000b. Finished Goods 360,000Cost of Goods Sold 360,000c. Work in Process 600,000Finished Goods 600,000d. Work in Process 600,000Cost of Goods Sold 600,000 Winona's Fudge Shoppe is maximizing profits by producing 1,000 pounds of fudge per day. If Winona's fixed costs unexpectedly increase and the market price remains constant, then the short run profit-maximizing level of outputa. is less than 1,000 pounds. b. is still 1,000 pounds. c. is more than 1,000 pounds d. becomes zero. Garcia Company issues 10%, 15-year bonds with a par value of $230,000 and semiannual interest payments. On the issue date, the annual market rate for these bonds is 8%, which implies a selling price of 117 1/4. The effective interest method is used to allocate interest expense. 1. Using the implied selling price of 117 1/4, what are the issuer's cash proceeds from issuance of these bonds. 2. What total amount of bond interest expense will be recognized over the life of these bonds? 3. What amount of bond interest expense is recorded on the first interest payment date? You are mandated to pick 45 units per hour. You work 8.5 hours a day (minus a 1/2 hour lunch), Monday to Friday. How many units should you be picking each week? La _____ and el _____ are both basic parts of a dinner meal.A. carne, ensaladaB. bistec, ensaladaC. carne, bistecD. pollo, bistec What is the measure of ACE in the diagram below?A. 46B. 104C. 29D. 58 In Aryan culture, what are the war leaders called?A. The war leaders in Aryan culture are called rajas.B. The war leaders in Aryan culture are called caste.C. The war leaders In Aryan culture are calledcharge.D. The war leaders in Aryan culture are calledpeasants. Suppose Miss Roxanne Davenport is 25 years old right now and puts away $1,800 per quarter in an account that returns 6% interest. a.) How much will be in the account when she turns 65? b.)What is her total contribution to the account? A student builds a rocket-propelled cart for a science project. Its acceleration is not quite high enough to win a prize, so he uses a larger rocket engine that provides 39% more thrust, although doing so increases the mass of the cart by 13%. By what percentage does the cart's acceleration increase? In the redox reaction AgNO3 + Na NaNO3 + Ag, which element has beenreduced? A broker-dealer and its associated persons may be subjected to sanctions for violations of FINRA and SEC rules. Which of the following penalties can be levied against both the member firm AND its associated persons?A. CensureB. Limits placed on activitiesC. ImprisonmentD. Loss of SIPC coverage Question of 20Which row on the diagram best represents the Maryland Court of SpecialAppeals?MoreauthorityRow ARow BRow CUTILITI TILRow DLessauthority A Three-Point Turn Chapter 1 "You know, hardly anyone ever needs to do a three-point turn anymore," said Justin, trying to help Becky calm down. "Oh, so it's not a useful skill AND I am probably going to fail the driving test because I can't do it anyway," Becky said, raising her voice for emphasis. "That should make me feel like a million bucks when I flunk." Justin was riding with Becky so she could take her driving test. He had volunteered for the job because he thought she would be less nervous with him than with their mom, but so far, he wasn't sure he was making any difference. "Slow down, your turn is coming up here," he said, looking ahead. "I know, I know," she replied, "I've been here before rememberthe last time I flunked." Justin was pretty sure if he had let her miss the turn, things would only have deteriorated further, but he wasn't sure he was fond of being the scapegoat for Becky's anxiety. "Listen, you need to take a few deep breaths," he said, hoping he could help her at least relax a bit. "Being nervous won't help you with the three-point turn or anything else you have to do. Hey, did you just take that turn without your turn signal on?" This was going to be harder than he thought. "Stop yelling at me," Becky replied, clearly frustrated, "I can't concentrate." "Look, you need to stop and get yourself together here," Justin started. "It is not just about passing the driving test. I don't want to get in an accident, so pull into that parking lot." Becky drove into the office building's parking lot where Justin was pointing. Justin knew they were less than a mile from the licensing office, and if she continued in this condition, he'd be having this same discussion three months from now when she tried the test again for the third time. "You need to get a grip," he started after she put the car in park, "because you have studied and practiced driving all year. You know this stuff inside and out, backwards and forwards. What are you so nervous about?" "I don't know, I don't know," Becky wailed, resting her head on the steering wheel. "I just get so tired of failing." Listening quietly as Becky sobbed, Justin realized this was about much more than a driving test. He also knew if he didn't find a way to help Becky things would just get worse. Chapter 2 Justin took a deep breath and collected his thoughts. Becky was an unbelievably consistent straight-A student. It was Justin who got the bad grades in school, and Justin who had to repeat every math class he'd ever taken. It was Justin who wished he could get the grades Becky got. Some things came easier for Justin: He was athletic, handy with tools, and good at making the best of whatever life threw at him. Mom called him her "lemons into lemonade" kid. But for the most part, Becky succeeded easily, whereas Justin had to work and work to just get a passing grade. Rather than having Becky catalogue all the things she supposedly "failed" at, Justin decided to try an alternative approach, one that wouldn't remind him of all the ways he had failed. "Okay, Becky, let's assume for a moment you fail this test again. What is the worst thing that could happen?" he asked. "I would be the oldest kid at school without a license and be humiliated," she replied. Justin thought he heard a bit of panic in her voice but continued with his plan. "Yes, but won't we still have to drive to school together for at least one more year anyway?" he asked. "Yes, but..." she started. "And who will know, if you don't tell anyone except your friends, that you don't have your license? You know Mom can't afford another car just for you, right?" "Yes," she said quietly. "So what difference does it make, really," he said. "Another three months to wait in the grand scheme of your life doesn't seem like all that long, right?" "I suppose not," she said. Justin could tell she was breathing more slowly now. "Besides," he said, "I would miss all the practice driving with you," and for good measure he reached over and pinched her arm. "Ow," she said, hitting back at him, "that hurt." "So let's go do this, okay? "Okay," she said. Becky cranked up the car, backed slowly out of the parking spot and drove up to the parking lot's exit. Justin noticed, as they waited for the traffic to clear, that she had remembered the turn signal. Which is true of both Chapter 1 and Chapter 2? (5 points) 1.) Both include characters who speak very little. 2.) Both include large sections of descriptive text. 3.) Both use mainly dialogue to develop the characters. 4.) Both use the narrator to explain the conflict.