Which force does the shuttle overcome to create an unbalanced net force as it initially lifts from the ground?

Air resistance
Friction
Gravity
Thrust of rocket engine

Answer: The volume of the anchor is approx. 29.56 m^3.

               Its mass is approx. 228,511.72

Explanation: Archimedes' Law States that "When a body is submerged in a fluid, it experiences an upward buoyant force that is equal to the weight of the fluid displaced by the body".

Here Buoyant force = 290N i.e. the difference in weight due to fluid displaced.

Buoyant force = D*V*g

D= 1025 kg/m^3 for seawater

g = 9.81 m/s^2

V = 290/1025*9.81

(a) V = 0.02884 m^3

Mass of Anchor = Density of Anchor* V

Mass of Anchor = 7730*0.02884

Mass of Anchor = 222.94 kg

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Answer:

MRBEAST-

Explanation:

Answer:

MRBEAST

Explanation:

MRBEAST

At 30°C, the rms speed of a sample of oxygen with a molar mass of 16 g/mol is approximately 482.34 m/s.

The root mean square (rms) speed of a gas molecule is a measure of the average speed of the gas particles in a sample. It can be calculated using the formula:

vrms = √(3kT/m)

Where:

vrms is the rms speed

k is the Boltzmann constant (1.38 x 10^-23 J/K)

T is the temperature in Kelvin

m is the molar mass of the gas in kilograms

To calculate the rms speed of oxygen at 30°C (303 Kelvin) with a molar mass of 16 g/mol, we need to convert the molar mass to kilograms by dividing it by 1000:

m = 16 g/mol = 0.016 kg/mol

Substituting the values into the formula, we have:

vrms = √((3 * 1.38 x 10^-23 J/K * 303 K) / (0.016 kg/mol))

Calculating this expression yields the rms speed of the oxygen sample:

vrms ≈ 482.34 m/s

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The charge is obtained as 5.35 * 10^-9 C.

The electric filed is defined as the region in space where the influence of a charge is felt. In this case, we know that the magnitude of the electric charge has been given to be 4.70 N/C and the distance is 3.20 m.

Let us recall that;

E = Kq/r^2

K = electric field constant

q = magnitude of charge

r = distance of separation

E = Kq/r^2

Er^2 = Kq

q = Er^2/K

q =  4.70 N/C * (3.20 m)^2/9.0 * 10^9

q = 5.35 * 10^-9 C

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Answer:

a sample of water boils and releases gas.

If the applied force is 20 N and the displacement is from X=0m to X=5.0m, the work done is 100 Joules.

To determine the work done by a force, you need to know the displacement and the component of the force parallel to the displacement.

In this case, if the force applied is 20 N and the displacement is from X=0m to X=5.0m, you need to determine if the force is parallel to the displacement.

If the force is parallel to the displacement, then the work done is simply the product of the force and the displacement, which is:

Work = Force x Displacement

Work = 20 N x (5.0 m - 0 m)

Work = 100 Joules

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The complete question is below:

What is the work done from X =0m to 5.0m? if the applied force is 20 N

Answer:

todas las cosas que la cuerda de salto hará

Explanation:

1. Mueva suavemente las muñecas y prepárese para levantar la parte superior del cuerpo.

2.Activa tus hombros sin presionarlos.

3. Calienta todo tu cuerpo elevando tu frecuencia cardíaca y cebando para la coordinación de todo el cuerpo que necesitas para el entrenamiento de la parte superior del cuerpo.

Given:

• Mass of person, m = 27 kg

• Force = 432 N

Let's find the acceleration.

To find the acceleration, apply Newton's Second Law:

Where:

F is the Force

m is the mass

a is the acceleration.

Rewrite the formula for a, plug in the values of m and F, the solve:

Therefore, the acceleration is 16 m/s².

ANSWER:

16 m/s²

The correct statement is " A dry cell battery has magnetic properties.", The correct option is C.

A dry cell battery does generate its own magnetic field due to the flow of electric current through the battery.

The magnetic field is created by the movement of charged particles (electrons) within the battery. This magnetic field is relatively weak and is not typically strong enough to be used for practical applications outside of the battery itself.

So, the magnetic properties of the dry cell battery are important for understanding its behavior within an electrical circuit.

Therefore, The correct answer is option C.

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Answer:

First Blank. 30,000

Second Blank. 12,000

Third Blank. 20.0

Explanation:

Answer:

Conduction: Touching a stove and being burned. Ice cooling down your hand. .

Explanation:

Answer:

MgS

Explanation:

Answer: B. Energy and a larger nucleus.

Explanation:Nuclear fission products are the atomic fragments left after a large atomic nucleus undergoes nuclear fission. Typically, a large nucleus like that of uranium fissions by splitting into two smaller nuclei, along with a few neutrons, the release of heat energy (kinetic energy of the nuclei), and gamma rays.

Answer:

Factors Affecting Friction

Friction is dependent on the smoothness or roughness of the two surfaces that are in contact with each other. When the surface is smooth, the friction between the two reduces as there is not much interlocking of irregularities. While the surface is rough, friction increases.

Explanation:

Answer:

10m/s2

Explanation:

250/25=10

Sound waves are a type of mechanical wave that propagate through a medium, typically air but also other materials such as water or solids.

Frequency: the frequency of a sound wave refers to the number of cycles or vibrations it completes per second and is measured in Hertz (Hz).

Amplitude: the amplitude of a sound wave refers to the maximum displacement or intensity of the wave from its equilibrium position. It represents the loudness or volume of the sound, with larger amplitudes corresponding to louder sounds and smaller amplitudes corresponding to softer sounds.

Wavelength: the wavelength of a sound wave is the distance between two consecutive points in the wave that are in phase, such as from one peak to the next or one trough to the next. It is inversely related to the frequency of the wave.

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Answer:

acceleration an velocity are two different types of hate for each individual to choose from my favorite ones and not just one for me like you said you would have to pay for the person you want and you will need to pay for your business.

Explanation:

no exceptions

The distance between the glass to have the given heat loss is 2.54 m.

The given parameters:

The area of the glass window is calculated as follows;

\(A = 0.6 \times 0.9\\\\A = 0.54 \ m^2\)

The distance between the glass is calculated as follows;

\(Q = \frac{KA \Delta T}{\Delta x} \\\\\Delta x = \frac{kA \Delta T}{Q} \\\\\Delta x = \frac{0.8 \times 0.54 \times 24 }{4.09} \\\\\Delta x = 2.54 \ m\)

Thus, the distance between the glass to have the given heat loss is 2.54 m.

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Answer:

The correct answer is:

(a) 27.5 Joules

(b) 141.5 Joules

Explanation:

Given:

Energy,

\(Q_c = 110 \ J\)

Coefficient of performance refrigerator,

\(Cop(refrig)=4\)

(a)

As we know,

⇒ \(Cop(refrig) = \frac{Q_c}{Work}\)

or,

⇒ \(Work=\frac{Q_c}{Cop(refrig)}\)

              \(=\frac{110}{4}\)

              \(=27.5 \ Joules\)

(b)

⇒ \(Heat \ expelled = Heat \ removed +Work \ done\)

or,

⇒ \(Q_h = Q_c+Work\)

         \(=114+27.5\)

         \(=141.5 \ Joules\)

Answer:

1.  Energy

2. All of the above

3. Store and use

4.  Kinetic

5. mass and velocity

6. Potential

7. mass, height, and gravity

8. Chemical Potential

9. Elastic Potential

10.  (`⌒*)O-(`⌒´Q)

Explanation:

The mass will be moving at 14.3 m/s when it reaches the equilibrium position.

To determine the speed of the mass when it reaches the equilibrium position, we need to consider the conservation of mechanical energy, which includes both the spring potential energy and the gravitational potential energy.

The total mechanical energy (E) of the system is the sum of the potential energy and the kinetic energy:

E = PE (potential energy) + KE (kinetic energy)

At the equilibrium position, all the potential energy is converted into kinetic energy, so the total mechanical energy is entirely in the form of kinetic energy.

The potential energy stored in the spring (PE_spring) is given by Hooke's Law:

PE_spring = (1/2) * k * \(x^{2}\)

Where k is the spring constant (250 N/m) and x is the displacement from the equilibrium position (0.32 m).

PE_spring = (1/2) * 250 N/m *\((0.32 m)^2\)

PE_spring = 12.8 J

The change in gravitational potential energy (ΔPE_gravity) is given by:

ΔPE_gravity = m * g * h

Where m is the mass (0.125 kg), g is the acceleration due to gravity (9.8 m/\(s^2\)), and h is the change in height (which is zero in this case since the height doesn't change).

ΔPE_gravity = 0 J

Therefore, the total mechanical energy (E) is equal to the potential energy stored in the spring:

E = PE_spring

E = 12.8 J

Since the total mechanical energy is entirely in the form of kinetic energy at the equilibrium position, we can calculate the speed (v) using the equation:

E = (1/2) * m * \(v^2\)

Rearranging the equation, we get:

\(v^2\) = (2 * E) / m

\(v^2\) = (2 * 12.8 J) / 0.125 kg

\(v^2\) = 204.8 \(m^2\)/\(s^2\)

Taking the square root of both sides:

v = √ 204.8 \(m^2\)/\(s^2\)

v ≈ 14.3 m/s

Therefore, the mass will be moving at approximately 14.3 m/s when it reaches the equilibrium position.

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(a) The work done by the 150 N force is 877.5 Joules.

(b) The coefficient of kinetic friction between the crate and the surface is approximately 0.437.

To answer this problem, we must take into account the work done by the applied force as well as the work done by friction.

(a) The applied force's work may be estimated using the following formula:

Work = Force * Distance * cos(theta)

where the force is 150 N and the distance is 5.85 m. Since the force is applied horizontally and the displacement is also horizontal, the angle theta between them is 0 degrees, and the cosine of 0 degrees is 1.

As a result, the applied force's work is:

Work = 150 N * 5.85 m * cos(0) = 877.5 J

So, the work done by the 150 N force is 877.5 Joules.

(b) Frictional work is equal to the force of friction multiplied by the distance. The work done by friction is identical in amount but opposite in direction to the work done by the applied force since the crate travels at a constant speed.

The frictional work may be estimated using the following formula:

Work = Force of Friction * Distance * cos(theta)

The net force applied on the crate is zero since it is travelling at a constant pace. As a result, the friction force must be equal to the applied force, which is 150 N.

Thus, the work done by friction is:

Work = 150 N * 5.85 m * cos(180) = -877.5 J

Since the work done by friction is negative, it indicates that the direction of the frictional force is opposite to the direction of motion.

The coefficient of kinetic friction may be calculated using the following equation:

Friction Force = Kinetic Friction Coefficient * Normal Force

The normal force equals the crate's weight, which may be computed as:

Normal Force = mass * gravity

where the mass is 35.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2.

Normal Force = 35.0 kg * 9.8 m/s^2 = 343 N

Now, we can rearrange the equation for the force of friction to solve for the coefficient of kinetic friction:

Force of Friction = coefficient of kinetic friction * Normal Force

150 N = coefficient of kinetic friction * 343 N

coefficient of kinetic friction = 150 N / 343 N ≈ 0.437

As a result, the kinetic friction coefficient between the container and the surface is roughly 0.437.

In summary, the work done by the 150 N force is 877.5 Joules, and the coefficient of kinetic friction between the crate and the surface is approximately 0.437.

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The waveform of a signal is the shape of its graph as a function of time in the domains of electronics, acoustics, and allied sciences, regardless of its time and magnitude scales or any shift in time.  

Thus, Waveforms with periodic variations are those that recur consistently at set intervals.

The phrase is typically used in electronics to describe periodically changing voltages, currents, or electromagnetic fields. It is typically used in acoustics to describe constant periodic sounds caused by changes in air pressure or other media.

In these situations, the signal's frequency, amplitude, or phase shift have no bearing on the waveform, which is a characteristic. Additionally, non-periodic signals like chirps and pulses can be referred to by this name.

Thus, The waveform of a signal is the shape of its graph as a function of time in the domains of electronics, acoustics, and allied sciences, regardless of its time and magnitude scales or any shift in time.  

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Answer:

\(E=3.69*10^{-11}\frac{V}{m}\)

Explanation:

To solve this problem you use the following formula, for the calculation of the electric field along the axis of the dipole.

\(E=\frac{p}{2\pi \epsilon_ox^3}\)   (1)

p: dipole moment = 6.16*10^-30 Cm

x: distance to the center of mass of the dipole = 3.00*10^-9m

eo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

You replace the values of the variables in the equation (1):

\(E=\frac{6.16*10^{-30}Cm}{2\pi(8.85*10^{-12}C^2/Nm^2)(3.00*10^{-9}m)^3}\\\\E=3.69*10^{-11}\frac{V}{m}\)

a) The angular velocity of the turntable after 0.200 s is 0.430 rev/s. b) the turntable has spun through 0.0088 revolutions in this time interval. c) the tangential speed of a point on the rim of the turntable at t = 0.002 s is 0.094 m/s.

The coefficient of kinetic friction between the object and the rough surface is 0.083. We can use the work-energy principle to solve this problem. According to the principle, the work done by all forces acting on an object is equal to the change in its kinetic energy.

The work done by the force of friction is given by W_friction = -f_k * d, where f_k is the force of kinetic friction and d is the distance traveled. The change in kinetic energy of the object is given by ΔK = K_f - K_i = (1/2) * m * v_\(f^2\) - (1/2) * m * v_\(i^2.\)

Since the object is moving on a horizontal surface, the work done by gravity is zero. Therefore, we have W_friction = ΔK.

Substituting the given values, we get:

-f_k * d = (1/2) * m * v_f^2 - (1/2) * m * v_\(i^2\)

-f_k * 8.0 m = (1/2) * m * (2.0 \(m/s)^2\) - (1/2) * m * (9.0\(m/s)^2\)

Simplifying and solving for f_k, we get:

f_k = (m/8.0 m) * [(1/2) * (2.0 \(m/s)^2\)- (1/2) * (9.0 \(m/s)^2\)]

f_k = 0.813 \(m/s^2\)

The coefficient of kinetic friction is given by μ_k = f_k / N, where N is the normal force. Since the object is moving horizontally, the normal force is equal to the weight of the object, which is N = m * g, where g is the acceleration due to gravity.

Substituting the value of f_k and N, we get:

μ_k = f_k / N = 0.813 \(m/s^2\) / (m * g)

The value of g is approximately 9.8 \(m/s^2\). Therefore, the coefficient of kinetic friction is:

μ_k = 0.083

Therefore, the coefficient of kinetic friction between the object and the rough surface is 0.083.

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The gravitational potential energy paper airplane have PE = 0.735J.

To calculate the potential energy we are going to use the formula,

PE = mgh

where,

PE is gravitational potential energy

m is mass of object

g is gravitational acceleration

h is height

So, given data are :

m = 0.005Kg

g = 9.8 m/s²

h = 1.5 m

PE = ?

Now, putting the values

PE = 0.005Kg x 9.8 m/s² x 1.5 m

PE = 0.0735 J

Hence, the gravitational potential energy paper airplane have is PE = 0.0735J

The potential energy that a huge item has in relation to another massive object due to gravity is known as gravitational energy or gravitational potential energy. When two objects descend toward one another, potential energy associated with the gravitational field is released.

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The principal stresses and corresponding principal directions for the stresses in a problem can be found using various methods, one of which is Mohr's circle.

What is Mohr's circle?

Mohr's circle is a graphical representation of a two-dimensional stress state. Given the stress tensor components, the Mohr's circle can be used to visualize and determine the maximum and minimum normal stresses (i.e., the principal stresses) and the orientation of the planes on which they act (i.e., the principal directions).

It is important to note that finding the principal stresses and directions requires a thorough understanding of stress analysis, including stress transformation and the use of Mohr's circle. If you have access to the full problem statement, I would be happy to help you work through the solution.

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Answer:

a

   \(F = 67867.2 \ N\)

b

  \(\Delta L = 2.6 \ mm\)

Explanation:

From the question we are told that

      The Young modulus is  \(Y = 1.50 *10^{10} \ N/m^2\)

      The stress is  \(\sigma = 1.50 *10^{8} \ N/m^2\)

      The  diameter is  \(d = 2.40 \ cm = 0.024 \ m\)

The radius is mathematically represented as

       \(r =\frac{d}{2} = \frac{0.024}{2} = 0.012 \ m\)

The cross-sectional area is  mathematically evaluated as

        \(A = \pi r^2\)

         \(A = 3.142 * (0.012)^2\)

        \(A = 0.000452\ m^2\)

Generally the stress is mathematically represented as

        \(\sigma = \frac{F}{A}\)

=>     \(F = \sigma * A\)

=>    \(F = 1.50 *10^{8} * 0.000452\)

=>    \(F = 67867.2 \ N\)

Considering part b

      The length is given as \(L = 26.0 \ cm = 0.26 \ m\)

Generally Young modulus is mathematically represented as

           \(E = \frac{ \sigma}{ strain }\)

Here strain is mathematically represented as

         \(strain = \frac{ \Delta L }{L}\)

So    

       \(E = \frac{ \sigma}{\frac{\Delta L }{L} }\)

        \(E = \frac{\sigma }{1} * \frac{ L}{\Delta L }\)

=>     \(\Delta L = \frac{\sigma * L }{E}\)

substituting values

       \(\Delta L = \frac{ 1.50*10^{8} * 0.26 }{ 1.50 *10^{10 }}\)

       \(\Delta L = 0.0026\)

Converting to mm

      \(\Delta L = 0.0026 *1000\)

      \(\Delta L = 2.6 \ mm\)