if the electric field is suddenly turned off, what is the speed of the sphere at the bottom of its swing (in m/s)?

The friction which acts on the object that is not moving is static friction, so, option A is correct.

Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.

Static friction is 0 for an object at rest on a level surface. Static friction generates an equal and opposite force that holds the book at rest if you push horizontally with a small force.

The static friction force grows to meet the force as you push harder. The book eventually moves when the maximum static friction force is reached.

Thus, the friction is static.

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Answer:

Its angular acceleration is zero.

Explanation:

If the angular velocity of an object (in this case the toy) is constant, then its angular acceleration will be zero. Why? Because angular acceleration is the time rate change of angular velocity. Since there is no change, this brings the answer to zero.

(a). the upward force experienced by the rocket is 16.065 N. (b). the upward force experienced by the rocket is 7 N when the acceleration is reduced to 20 m/s^2.

(a) Using Newton's second law, we know that the upward force experienced by the rocket is equal to the product of its mass and acceleration: F = ma

F = 0.35 kg x 45.9 m/s^2

F = 16.065 N

Therefore, the upward force  16.065 N.

(b) We can calculate the new upward force when the acceleration is reduced to 20 m/s^2:

F = 0.35 kg x 20 m/s^2

F = 7 N

Therefore, the upward force is 7 N when acceleration is reduced to 20 m/s^2.

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If there is a decrease in the angle of inclination, the acceleration would decrease accordingly.

We know that the acceleration is the change in the velocity of the object with time. Given that the mass of the pulley remains the same and the puly is being moved up the incline as we can see.

We have to also recall that we have been told that the incline is frictionless and as a result we are not going to consider the frictional force. In any case, the force that moves the pully along the incline stull depends on the angle of inclination.

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The ratio of the tension Tt in the string when the ball is at the top of its path to Tb when the ball is at the bottom of its path is given by the expression TbTt = 1 - v²/gL. Therefore, the correct option is D.

When a ball of mass m is swung in a vertical circle with uniform speed, it follows a circular path with radius R, which is equal to the length of the string L. The gravitational force provides the centripetal force required to keep the ball moving in a circle.The tension T in the string can be calculated at the top and bottom of the circular path by resolving the forces acting on the ball. At the top, the weight of the ball acts downwards, and the tension acts upwards.

Hence, the net force is the centripetal force, which is given by

mv²/R - mg = mv²/L

The tension Tt can be calculated by resolving the forces vertically, which givesTt - mg = mv²/LTt = mv²/L + mg ------------ (1)

At the bottom of the path, the weight of the ball acts downwards, and the tension acts upwards. Hence, the net force is the centripetal force, which is given bymv²/R + mg = mv²/LThe tension Tb can be calculated by resolving the forces vertically, which gives

Tb + mg = mv²/LTb = mv²/L - mg ------------- (2)

Taking the ratio of equations (1) and (2), we get

TbTt = (mv²/L - mg)/(mv²/L + mg)

TbTt = (v²/gL - 1)/(v²/gL + 1) TbTt = 1 - v²/gL

The expression TbTt = 1 - v²/gL gives the ratio of the tension Tt in the string when the ball is at the top of its path to Tb when the ball is at the bottom of its path. Hence, the correct option is D.

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The air pressure calculated using barometric formula at an elevation of 2000 meters is 0.068 MPa if the air pressure at sea level is 0.1 MPa.

Air pressure =  0.1 MPa

Elevation height = 2000 meters

The barometric formula is used to calculate the air pressure at an elevation of 2000 meters. Here the atmospheric pressure decreases when the elevation increases. The barometric formula is:

P = P₀ * exp(-M * g * h / (R * T))

P = pressure of altitude

P₀ = pressure at sea level

M = molar mass of Earth's atmoshpere = 0.02896 kg/mol

g =  acceleration due to gravity = 9.8 m/s²

h =  altitude height = 2000 m

R = ideal gas constant = 8.314 J/(mol·K)

T =  temperature of the air at a constant.

Substituting the values:

P = 0.1 MPa * exp(-0.02896 kg/mol * 9.8 m/s² * 2000 m / (8.314 J/(mol·K) * T))

T = 293K

P = 0.1 MPa * exp(-0.02896 kg/mol * 9.8 m/s² * 2000 m / (8.314 J/(mol·K) * 293 K))

P = 0.1 MPa * exp(0.2929808 / (8.314 J/(mol·K) * 293 K))

P =  0.1 MPa * exp(0.2929808 / 2436.002))

P  = 0.068 MPa

Therefore, we can conclude that the air pressure at an elevation of 2000 meters is 0.068 MPa.

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Answer:

I know you may not be reading this but for number 13 the answer is 2 hours because t=d/s which is 4000/2000 which equals to 2.

Explanation:I've currently doing that problem as of the time i'm writing this.

Since an ammeter is meant to measure current and is a low resistance device, it is always connected in series with the circuit element through which current is to be measured.

Current of a system is defined as the rate at which the electrons flow from a given point in the absolute electrical circuit.

a. Since, Current in two or more components connected in series is always the same. So circuit (a) will be used.

b. As per manufacturer's claim, resistance of bulb r = V /I.

= 12 /6 = 2 Ω.

Total resistance in circuit a,  R = 2 +0.1 = 2.1 A

Ammeter reading I' = V /R

= 12 /2.1 = 5.71 A

Which is close enough to claimed value of 6 A.

c. Now the bulb and ammeter are connected in parallel, so the voltage across both will be the same.

Current through ammeter I' = V /Ra = 12 /0.1 = 120 A

Which is 20 times greater than claimed value.

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The question is incomplete, but most probably the complete question is,

The manufacturer of a 12V car headlight specifies it will draw a current of 6A. You would like to check this claim with an ammeter designed to measure currents up to 10Aand having a resistance of 0.1 Ohms.

a)Which of the two figures represents a circuit where the ammeter correctly measures the current in the headlight?

b) How much current (in A) would flow in the ammeter for Circuit a?

c) How much current (in A) flows through the ammeter for Circuit b?

State function is the property do not depend on the path taken to reach that specific value.

As, we know the property or a feature which does not rely upon the course is known as a state characteristic. Work-achieved in adiabatic processes is equal to the bad of trade in inner power from the primary regulation of thermodynamics. So, it is a state function.

Path functions are houses or portions whose values rely on the transition of a machine from the preliminary kingdom to the final state. the two most common path functions are warmth and paintings.

From all of the above given thermodynamic quantities, work is the best property which depends at the course accompanied by using the system i.e. on the preliminary and the final states of the system. on the opposite, the rest all others i.e. internal energy, enthalpy and entropy are path independent.

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Answer:

Explanation:

Answer:

electrons

Explanation:

they go through the wires and power our computer tech

Answer:

Electric energy is a form of kinetic energy.

The net radiant energy lost by the 2-cm-diameter cylinder per meter of length is X Joules. The temperature of the 6-cm-diameter cylinder is Y °C.

To calculate the net radiant energy lost by the 2-cm-diameter cylinder per meter of length, we need to consider the Stefan-Boltzmann law and the emissivities of both cylinders. The formula for net radiant heat transfer is given:

Q_net = ε1 * σ * A1 * (T1^4 - T2^4)

Where:

- Q_net is the net radiant energy lost per meter of length.

- ε1 is the emissivity of the 2-cm-diameter cylinder.

- σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/(m^2·K^4)).

- A1 is the surface area of the 2-cm-diameter cylinder.

- T1 is the temperature of the 2-cm-diameter cylinder.

- T2 is the temperature of the surroundings (27 °C).

To calculate the temperature of the 6-cm-diameter cylinder, we can use the formula for the net radiant energy exchanged between the two cylinders:

Q_net = ε1 * σ * A1 * (T1^4 - T2^4) = ε2 * σ * A2 * (T2^4 - T3^4)

Where:

- ε2 is the emissivity of the 6-cm-diameter cylinder.

- A2 is the surface area of the 6-cm-diameter cylinder.

- T3 is the temperature of the 6-cm-diameter cylinder.

By solving these equations simultaneously, we can find the values of Q_net and T3.

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A long cylinder having a diameter of 2 cm is maintained at 600 °C and has an emissivity of 0.4. Surrounding the cylinder is another long, thin-walled concentric cylinder having a diameter of 6 cm and an emissivity of 0.2 on both the inside and outside surfaces. The assembly is located in a large room having a temperature of 27 °C. Calculate the net radiant energy lost by the 2-cm-diameter cylinder per meter of length. Also, calculate the temperature of the 6-cm-diameter cylinder

The amount of heat transferred from the ocean to the atmosphere as latent heat than sensible heat will be 10 times as much.

Latent heat and sensible heat are defined as follows:

Thus, the amount of heat transferred from the ocean to the atmosphere as latent heat than sensible heat will be 10 times as much.

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The question addresses the stability of the atmosphere and the factors that determine convective stability or instability. It also explains the difference between the adiabatic lapse rate of dry air and wet saturated air.

a) The stability of the atmosphere is determined by the temperature profile and relative densities of the air cell and atmosphere. If the temperature of the surrounding atmosphere decreases with altitude at a rate greater than the adiabatic lapse rate of the air cell, the atmosphere is considered convectively stable.

In this case, the air cell will return to its original position. Conversely, if the temperature of the surrounding atmosphere decreases slower than the adiabatic lapse rate of the air cell, the atmosphere is convectively unstable. The air cell will continue moving in the same direction.

b) The adiabatic lapse rate refers to the rate at which temperature decreases with altitude for a parcel of air lifted or descending adiabatically (without exchanging heat with its surroundings). The adiabatic lapse rate of dry air is higher (around \(9.8^0C\) per kilometer) compared to the adiabatic lapse rate of wet saturated air (around 5°C per kilometer).

This difference arises because when water vapor condenses during the ascent of saturated air, latent heat is released, reducing the rate of temperature decrease. A diagram can illustrate the difference between the two lapse rates, showcasing their respective slopes.

c) When wet unsaturated air rises from the ocean surface, its temperature decreases at a rate equal to the dry adiabatic lapse rate. However, if the ambient lapse rate (temperature decrease with altitude) is higher than the adiabatic lapse rate for dry air, a temperature inversion layer forms at higher altitudes.

In this inversion layer, the temperature increases with altitude instead of decreasing. A schematic diagram can depict the temperature changes of the wet air in comparison to the ambient temperature, showing the inversion layer.

Cumulus clouds form at the altitude where the rising moist air reaches the level of the temperature inversion layer. These clouds are formed due to the condensation of water vapor as the air parcel cools to its dew point temperature.

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If Robin moves the wheelbarrow twice as fast, the power  developed is 696 J/s.

If Robin moves the wheelbarrow twice as fast, the power  developed will be determined as follows;

Power = Energy / time

when moved twice as fast, new power developed  is given as;

new power = Energy / time/2

new power developed = ( Fd ) / ( t/2)

where;

new power developed = ( 145 N x 60 m ) / ( 25 s / 2 )

new power developed = 696 J/s

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Answer:

ch 46umm nym m m m   m hj hj hj

Explanation:

Answer:

A. NGC 3603 would be classified as Nebula

Explanation:

The approximate period of this pendulum on the moon where the acceleration due to gravity is roughly 1/6 that of earth is 15s. The correct option is -d. 15 s.

On Earth, we know that T=6.0 s. Let's assume the length of the pendulum remains constant.
Now, on the moon, the acceleration due to gravity is approximately 1/6 that of Earth's, so g'=g/6.

Using the same equation as before, we can find the new period T' on the moon:
T' = 2π√(L/g') = 2π√(L/(g/6)) = 2π√(6L/g)

Substituting in T=6.0 s, we have:
T' = 2π√(6L/g) = 2π√(6T^2g/L) = 2π√(6(6.0 s)^2(9.81 m/s^2)/L)

Since we are looking for an approximate answer, we can estimate L to be roughly the same on the moon as it is on Earth. Therefore, we can simplify the equation to:

T' ≈ 2π√(6(6.0 s)^2(9.81 m/s^2)/L) ≈ 2π√(216) ≈ 29.1 s
Therefore, the correct option is -d. 15 s.

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The period of a pendulum is the time it takes for the pendulum to complete one full swing. In this case, we know that a simple pendulum on earth has a period of 6.0 s. However, on the moon, the acceleration due to gravity is roughly 1/6 that of earth.Therefore, the correct answer is (b) 2.4 s.

This means that the force acting on the pendulum is much weaker on the moon than on earth. As a result, the pendulum will swing slower on the moon than on earth. To calculate the approximate period of the pendulum on the moon, we can use the formula T=2π√(l/g), where T is the period, l is the length of the pendulum, and g is the acceleration due to gravity. Plugging in the appropriate values, we get T=2π√(l/(1/6g)). Simplifying this equation, we can see that the period on the moon will be approximately 2.4 s. Therefore, the correct answer is (b) 2.4 s.

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The correct answer is "velocity." Frontal area and velocity are two factors that greatly affect air resistance on falling objects. Air resistance, is the force exerted by air molecules on an object as it moves through the air.

Frontal area refers to the cross-sectional area of the object that is directly facing the direction of motion. Objects with larger frontal areas experience more air resistance because they encounter a greater number of air molecules. However, velocity is another crucial factor.

As the velocity of the falling object increases, the air resistance it encounters also increases. This is because at higher velocities, the object pushes more air molecules aside per unit of time, leading to a greater resistance force. Therefore, both frontal area and velocity play significant roles in determining the magnitude of air resistance experienced by falling objects.

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(a)

The work done by the spring force can be calculated with the formula below:

Using k = 320 N/m and x = 0.075 m, we have:

(b)

The increase in thermal energy is given by the work done by the friction force.

To calculate this work, first let's find the friction force:

Now, calculating the work, we have:

(c)

The block speed can be found by converting the potential energy from the spring (same value of the calculated work in item a) into kinetic energy for the block:

Answer:

Approximately \(2.0\; \rm m\), assuming that \(g = 9.81 \; {\rm m \cdot s^{-2}}\) and that air resistance is negligible.

Explanation:

If air resistance is negligible, the acceleration of this grapefruit during the fall would be constantly equal to \(g\) (gravitational field strength.)

Let \(x\) denote the distance that the grapefruit travelled during the fall. The following SUVAT equation gives an expression for \(x\!\) in terms of \(a\), \(t\), and \(v_{0}\):

\(\begin{aligned}x &= \frac{1}{2}\, a\cdot t^{2} + v_{0}\, t\end{aligned}\).

In this question, the grapefruit was initially on a tree. Hence, assume that the initially velocity of this grapefruit was \(v_{0} = 0\; \rm m\cdot s^{-1}\) right before the fall started.

If there was no drag on the grapefruit during the fall, the acceleration of this grapefruit would be equal to the gravitational field strength: \(a \approx 9.81\; \rm m\cdot s^{-2}\).

The duration of the fall, \(t\), has also been given. Hence, the distance that the grapefruit travelled during the fall would be:

\(\begin{aligned}x &= \frac{1}{2}\, a\cdot t^{2} + v_{0}\, t \\ &= \frac{1}{2} \times 9.81\; \rm m\cdot s^{-2} \times (0.64\; \rm s)^{2} + 0 \\ &\approx 2.0\; \rm m\end{aligned}\).

Answer:

Absolutely no real-life phenomena/ uses are possible that are related to any branch of physics, except:

1 The device you are reading this answer on. (quantum mechanics, electronics )

2 The air conditioner in your room. ( electromagnetism, electronics )

3 The satellites and space stations in space which will help in the deciphering of the universe will help us look for a suitable planet when we render the earth useless. ( general and special relativity, classical mechanics )

4 Ohh. I forgot, do you wear clothes, well they are made in factories where factories run on the principles of thermodynamics, which is physics.

5 Ever gone somewhere? Well, the transportation uses classical mechanics, thermodynamics, electromagnetism, relativity ( if you know Global Positioning System )

6 GPS itself! It’s used in online maps, navigating, and this sort of stuff. It works on general relativity.

Honestly, this list is endless.

The kinetic energy of the spring is obtained as 12 J. Option C

We know that the kinetic energy is the energy that is required to stretch the spring. We can use the graph that has been shown to show the force constant of the spring.

We can find the force constant as the slope of the graph as follows;

K = \(y_{2} - y_{1} /x_{2} - x_{1}\)

K = 60 - 0/0.3 - 0.2

K = 60/0.1

K = 600 N/m

We then have;

W = KE = 1/2 Kx^2

Thus;

KE = 0.5 * 600 * (20 * 10^-2)^2

KE = 12 J

It would have a kinetic energy of 12 J.

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Answer:

The net force is zero because both volleyball players are hitting the ball with same force.

Explanation:

Answer:

Attenuation

Step By step Explanation:

The decrease in the intensity of light over distance, such as with increasing depth in water, is known as ______.

It is called attenuation.

The decraese in the amplitude of the signals is called attenuation.

The three product which can be potentially fail considering Load Strength graph and precautionary measure to prevent failure are as below;

Ball Pen:

1. Ink Leakage: One potential failure of a ball pen is ink leakage. This can occur due to poor sealing between the ink reservoir and the ballpoint mechanism. Ink leakage can result in messy hands, stained documents, and reduced functionality of the pen. To prevent this failure, manufacturers can improve the quality control process to ensure proper sealing and use high-quality materials for the pen's components.

2. Ballpoint Jamming: Another failure is ballpoint jamming, where the ball gets stuck and prevents smooth writing. This can be caused by a buildup of dried ink or debris inside the pen's mechanism. To prevent ballpoint jamming, regular cleaning and maintenance of the pen can be recommended. Additionally, manufacturers can design the pen with features that facilitate easy cleaning or provide instructions on how to clear any blockages.

3. Weak Barrel Construction: The barrel of the pen may also be prone to failure if it is weak or brittle. Excessive pressure or rough handling can lead to cracks or breakage, rendering the pen unusable. To prevent this, manufacturers can use durable materials for the pen barrel, such as sturdy plastics or reinforced metal, and perform quality checks to ensure structural integrity.

Room Key:

1. Keycard Malfunction: A potential failure of a room key is a malfunction in its electronic components. This can result in the keycard being unreadable by the door lock system, preventing access to the room. To prevent this failure, regular maintenance and replacement of keycard readers can be implemented. Additionally, guests should be advised to keep their keycards away from magnets and electronic devices that can interfere with the card's functionality.

2. Magnetic Strip Damage: Another failure can occur if the magnetic strip on the keycard gets damaged or demagnetized. This can happen due to exposure to magnetic fields or physical damage. To prevent this failure, keycards can be made more durable with protective coatings or alternative technologies such as RFID. Guests should also be educated on proper handling and storage of keycards to avoid damage.

3. Battery Drain: Some room keys use batteries to power their electronic components. A failure can occur if the battery drains, leading to an inactive keycard. To prevent this, low-power consumption designs can be implemented, and regular battery checks or replacements can be carried out by hotel staff. Guests should be informed about the importance of returning the keycard to the front desk for recycling or proper disposal to ensure the battery is replaced as needed.

Blender:

1. Motor Burnout: One potential failure of a blender is motor burnout due to prolonged use or overloading. Continuous operation at high speeds or attempting to blend hard or frozen ingredients beyond the blender's capacity can cause the motor to overheat and fail. To prevent motor burnout, manufacturers can provide clear guidelines on the maximum load capacity and recommended usage durations. Automatic thermal protection mechanisms can also be incorporated to shut off the blender if it detects excessive heat.

2. Blade Jamming: Another failure can occur if food particles or ingredients get jammed between the blender's blades, preventing them from spinning freely. This can happen if the blender is not properly cleaned or if ingredients are not adequately prepared before blending. To prevent blade jamming, users should be advised to clean the blender thoroughly after each use and ensure that ingredients are cut into manageable sizes. Manufacturers can also design blades with accessible mechanisms for easy cleaning or provide cleaning tools.

3. Leakage: A failure in a blender can also manifest as leakage. This can happen if the blender jar or its sealing components are damaged or improperly assembled. Liquid or food can leak out during blending, resulting in a messy and potentially unsafe situation. To prevent leakage, manufacturers should ensure proper sealing mechanisms and use high-quality materials for the blender jar and lid. Regular inspection of the sealing components can be advised,

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The magnitude of the angular acceleration of the centrifuge is approximately 68.7 rad/s^2.

To solve this problem, we can use the formula for the angular displacement of an object

θ = ω_i t + 1/2 α t^2

where θ is the angular displacement, ω_i is the initial angular velocity, α is the angular acceleration, and t is the time.

In this case, we know that the initial angular velocity ω_i is 3,800 rev/min. To use this in the formula, we need to convert it to radians per second

ω_i = (3800 rev/min) x (2π rad/rev) x (1/60 min/s) = 397.89 rad/s

We also know that the final angular velocity ω_f is 0, since the centrifuge comes to rest. Therefore, we can use the formula for the final angular velocity in terms of the initial angular velocity, angular acceleration, and time

ω_f = ω_i + α t

Since ω_f is 0, we can solve for the time it takes for the centrifuge to come to rest

0 = ω_i + α t

t = -ω_i / α

We also know that the angular displacement of the centrifuge is 48 revolutions, or 96π radians (since 1 revolution = 2π radians). Plugging in the values we know and solving for α, we get

96π = (397.89 rad/s) (-397.89/α) + 1/2 α (-397.89/α)^2

96π = -397.89^2/α + 1/2 α (-397.89)^2/α^2

96π = -397.89^2/α + 1/2 (-397.89)^2/α

α = 397.89^2 / (2 x 96π)

α ≈ 68.7 rad/s^2

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If alcohol or drugs are used in small quantities, and with little or no damaging consequences, it's not usually considered drug abuse.

The correct option is A.

A person who has completed his 18 years of age, wants to drive any vehicle whether car or bike, he must have a license called the driver's license.

Drug abuse is the aggressiveness developed in person after drinking alcohol.

In countries, when a person takes in alcohol with little or no damaging consequences, he is not to be considered as drug abuse.

The given statement is True.

Thus, the correct option is A.

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Answer:

A

Explanation:

Newton third law of motion state that action and reaction are equal and opposite

Answer:

Always true

Explanation:

I got it right on the test.