Which statement describes a situation with a displacement of zero check all tha apply

Answer:

I believe the answer is C. 1 and 3

Explanation:

I'm 100% sure that 1 is correct, from what I remember learning.

As for 3, it was hard to choose between 3 and 4, but I'm pretty certain that covalent bonds shared electrons rather than transfer them.

The work can be seen as a transfer of energy, and can be calculated with the formula below:

Where W is the work in Joules, F is the force in Newtons and d is the distance in meters.

When we have a positive work, that means the body or system is using its own energy and force to move a certain distance, for example.

A negative work means the system is receiving energy from an external source.

Therefore the statement is TRUE.

Answer:to measure distances between objects and to study the early universe

Explanation:

I had this question and I got it right

Answer:

A charged atom or particle is called an ion :)

A)The total mechanical energy of the two blocks prior to being released from rest can be found by adding the gravitational potential energy of mA and the pulley to zero.

B).The gravitational potential energy of mB and the pulley is(3.30 kg + mp) × 9.81 m/s² × 0 m = 0 J,where mp is the mass of the pulley.The total mechanical energy of the two blocks prior to being released from rest is54.33 J + 0 J = 54.33 J

C) The rotational kinetic energy of the pulley just before mA hits the ground is(0.178 mp) J.

D) The mass of the pulley ismp = (1/2)mpr²/R² =(1/2)(0.020 kg)(0.100 m)²/(0.200 m)² = 0.001 kg = 1 g.r = (1/2)R.

The Atwood's machine shown in Figure 1 consists of two masses mA = 6.50 kg and mB = 3.30 kg. The height of mA above the floor is 0.865 m. When mA hits the floor, its velocity is 1.89 m/s. The pulley has a moment of inertia (1/2)mpr². We have to find the total mechanical energy of the two blocks before they are released, the total mechanical energy when mA hits the ground, the rotational kinetic energy of the pulley just before mA hits the ground, and the mass of the pulley. Let's solve these one by one. Part A The total mechanical energy of the two blocks prior to being released from rest can be found by adding the gravitational potential energy of mA and the pulley to zero.

The equation for gravitational potential energy is mgh. The gravitational potential energy of mA and mB is mAg(h-hB)where h is the height of mA above the floor and hB is the height of mB above the floor. Since the pulley is at the same height as mB, its gravitational potential energy ismBg(h-hB).The gravitational potential energy of mA is6.50 kg × 9.81 m/s² × 0.865 m = 54.33 J.The gravitational potential energy of mB and the pulley is(3.30 kg + mp) × 9.81 m/s² × 0 m = 0 J,where mp is the mass of the pulley.The total mechanical energy of the two blocks prior to being released from rest is54.33 J + 0 J = 54.33 J.Part BThe total mechanical energy of the two blocks when mA hits the ground can be found by adding the kinetic energy of mA, the kinetic energy of mB, and the rotational kinetic energy of the pulley to the gravitational potential energy of mB and the pulley. The equation for kinetic energy is (1/2)mv². The kinetic energy of mA is(1/2) × 6.50 kg × (1.89 m/s)² = 11.54 J.The kinetic energy of mB is(1/2) × 3.30 kg × 0 m/s² = 0 J, since it is at rest.The gravitational potential energy of mB and the pulley is(3.30 kg + mp) × 9.81 m/s² × 0 m = 0 J.The rotational kinetic energy of the pulley is(1/2) × (1/2)mp × R² × ω²,where R is the radius of the pulley and ω is its angular velocity just before mA hits the ground. We can use the fact that the linear speed of the rope is the same on both sides of the pulley to find ω. The equation for linear speed is v = Rω. When mA hits the ground, its speed is 1.89 m/s. The speed of mB is zero. Since the rope is inextensible, the speed of the rope is also 1.89 m/s.

Therefore, the speed of the pulley is also 1.89 m/s. We can find the angular velocity of the pulley by dividing the linear velocity by the radius.ω = v/R = 1.89 m/s ÷ (0.200 m/2) = 18.9 rad/s.The rotational kinetic energy of the pulley is(1/2) × (1/2)mp × R² × ω² =(1/4)mpR²ω² =(1/4)mp(0.200 m)²(18.9 rad/s)² =(0.178 mp) J.The total mechanical energy of the two blocks when mA hits the ground is11.54 J + 0 J + 0 J + (0.178 mp) J = 11.72 J + (0.178 mp) J.Part CThe rotational kinetic energy of the pulley just before mA hits the ground is(0.178 mp) J.Part DWe can find the mass of the pulley by using the moment of inertia of a disk and the mass of the pulley. The moment of inertia of a disk is (1/2)mr². Therefore,(1/2)mpR² = (1/2)mpr²,where R is the radius of the pulley and r is the radius of gyration of the pulley. The radius of gyration of a disk is (1/2)R.

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Answer:

volume is 0.1 L

Explanation:

you can use the equation density=mass/volume

100 = 1000 / v

divide by 1000 on both sides

0.1 = v

Answer:

Explanation:

according to the graph at B the potential energy of the particle is 2J

therefore we can use the kinetic energy equation to calculate the particle's velocity or speed.

\(E_{k} =1/2mv^{2}\)

2J= 1/2*1/2kg*v^2

8=v^2

v= 2√2 ms-1

Answer:

Follows are the solution to the given question:

Explanation:

In this question the missing file of the circuit is not be which is defined in attached file please find it.

In Option 1, this statement is true because the current is on \(R_3 \ and \ R_4\), that is the same.

In option 2, this statement is false because \(Rcd=R_3+R_4\) therefore it implies that Rcd is always larger then \(R_3\).

In option 3,  this statement is true because the voltage of \(R_1 \ and \ R_2\) is always equal.

In option 4, this statement is true because \(Rab= \frac{1}{(\frac{1}{R1}+\frac{1}{R2})}=\frac{R1R2}{(R1+R2)}. \frac{R2}{(R1+R2)}\)is always smaller then 1 therefore,\(R_1 \times (\frac{R2}{(R1+R2)})\) is always equal to R1.

Answer:

The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.

Explanation:

The object is not under the influence of any external force, meaning that Principle of Momentum Conservation to calculation of the velocity of the third fragment:

\((m_{1}+m_{2}+m_{3})\cdot \vec {v}_{o} = m_{1}\cdot \vec v_{1} + m_{2}\cdot \vec v_{2} + m_{3}\cdot \vec v_{3}\) (1)

Where:

\(m_{1}\), \(m_{2}\), \(m_{3}\) - Masses of the first, second and third fragments, in kilograms.

\(\vec v_{o}\) - Initial velocity of the object, in meters per second.

\(\vec v_{1}\), \(\vec v_{2}\), \(\vec v_{3}\) - Velocities of the first, second and third fragments, in meters per second.

If we know that \(m_{1} = 0.5\,kg\), \(m_{2} = 1.3\,kg\), \(m_{3} = 1.2\,kg\), \(\vec v_{o} = (0,0)\,\left[\frac{m}{s} \right]\), \(\vec v_{1} = \left(-2.8, 0\right)\,\left[\frac{m}{s} \right]\) and \(\vec v_{2} = \left(0,-1.5\right)\,\left[\frac{m}{s} \right]\), the velocity of the third fragment is:

\((-1.4,0) + (0,-1.95) + 1.2\cdot \vec v_{3} = (0,0)\)

\(1.2\cdot \vec v_{3} = (1.4,1.95)\)

\(\vec v_{3} = (1.167, 1.625)\,\left[\frac{m}{s} \right]\)

The speed of the third fragment is the magnitude of the result found above:

\(v_{3} = 2\,\frac{m}{s}\)

And the direction of the third fragment is:

\(\theta_{3} = \tan^{-1} \left(\frac{1.625}{1.167}\right)\)

\(\theta_{3} \approx 54.316^{\circ}\)

The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.

Answer:

122.5m.

Explanation:

I think it's right..

Answer:

16m/s

Explanation:

\(v_{f}=v_{i}+at\)

\(v_{f}=0+2\cdot8\)

\(v_{f}=16\ \frac{m}{s}\)

Therefore,  the speed after 8 seconds is 16m/s

The graph of the velocity and the time  can be shown by option D.

We know that the movement of an object as it is falling under gravity would have a constant acceleration. The constant acceleration means that the velocity of the object is also held a constant.

We now have to look at the graphs as we have them here. The graph as it has been shown has the the velocity on the vertical axis and it has the time on the horizontal axis. The gradient of the slope is what we would refers to as the acceleration of the body.

We also need to recall that the acceleration has to be a constant since tje tow object would have to reach the ground at the same time if they are released from the same height at the same time.

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The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.

The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.

The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.

The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.

The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).

The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).

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Answer: In ridge push, the mantle wells upward because of the convection and elevates the edges of spreading oceanic plates. Because these plates are higher at the spreading center, they are forced downhill due to gravity and eventually flatten out to the ocean floor.

the answer I need points for my math test

Answer:

Light energy

Explanation:

The light bulbs converted the electric energy to light energy.

a) The multimeter in figure 2(a) is set up to measure DC voltage. The setting is indicated by the V with a straight line above it, which stands for "DC voltage".

b) The multimeter in figure 2(a) reads 25.6 V (volts).

c) The multimeter in figure 2(b) is set up to measure DC current. The setting is indicated by the A with a straight line above it, which stands for "DC current".

d) The multimeter in figure 2(b) reads 0.40 A (amperes).

What is multimeter?

A multimeter is an electronic instrument that is used to measure electrical parameters such as voltage, current, and resistance. Multimeters can measure AC and DC voltage, AC and DC current, resistance, continuity, capacitance, and frequency.

What is DC volage?

DC voltage refers to the level or strength of a direct current (DC) electrical signal. Direct current is a type of electrical current that flows in one direction only and does not alternate in polarity. DC voltage can be measured using a multimeter or other voltage measurement instrument, and is typically expressed in volts (V). DC voltage is commonly used in many electronic devices and systems, such as batteries, power supplies, and DC motors.

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Answer:

The laws of reflection govern how incident light rays are reflected on reflecting surfaces such as mirrors, smooth metal surfaces, and clear water. The laws of reflection are as follows: The incident ray reflected ray, and normal to the mirror’s surface are all in the same plane. The incidence angle equals the reflection angle.

Explanation:

Answer:

0.79 cm

Explanation:

The computation is shown below:-

Particle acceleration is

\(a = \frac{qE}{m}\)

We will take d which indicates distance as from the negative plate, so the travel by proton is 0.800 cm - d at the same time

\(d = \frac{1}{2} a_et^2\\\\0.800 cm - d = \frac{1}{2} a_pt^2\\\\\frac{d}{0.800 cm - d} = \frac{a_e}{a_p} \\\\\frac{d}{0.800 cm - d} = \frac{m_p}{m_e} \\\\\frac{d}{0.800 cm - d} = \frac{1836m_e}{m_e}\)

After solving the equation we will get 0.79 cm from the negative plate.

Therefore it is 0.79 cm far from the negative pate i.e the point at which the electron and proton pass each other

The point at which the electron and proton pass each other will be 0.79 cm.

When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.

The given data in the problem is;

d' is the distance between the two parallel plates= 0.800 cm

The acceleration is given as;

\(\rm a= \frac{qE}{m} \\\\\)

The distance from Newton's law is found as;

\(d = ut+\frac{1}{2} at^2 \\\\ u=0 \\\\ d= \frac{1}{2} at^2 \\\\ d-d' = \frac{1}{2} a_pt^2 \\\\ 0.800-d= \frac{1}{2} a_pt^2 \\\\\ \frac{d}{0.800-d} =\frac{a}{a_p} \\\\ \frac{d}{0.800-d} =\frac{m_p}{m} \\\\ \frac{d}{0.800-d} =\frac{1836m_e}{m_e} \\\\ d=0.79 \ cm\)

Hence the point at which the electron and proton pass each other will be 0.79 cm.

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Answer:   Option C

Magnetic energy  |  Electromagnetic energy

Explanation:

The energy of the system before the collision must equal the energy after the collision.

After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is

\(\frac{18.41 kg (8.8 m/s)^2}{2} = 713 J\)

Before the collision only the bullet has kinetic energy.

So we can now determine the speed of the bullet using

\(\frac{0.11kg (v^2)}{2} = 713 J\\v = 114 m/s\)

Given:

The force applied to the box, F=6 N

The distance of the walk, d=8 m

To find:

The work done.

Explanation:

The work can be described as the transfer of energy from a source to an object through the application of force to move the object.

Thus the work done in carrying the box for the given distance is calculated using the formula,

On substituting the known values,

Final answer:

The work done is 48 J.

The speed of the puck is 3.67 m/s.

To find the speed of the puck, we can use the concept of centripetal force. The tension in the string provides the necessary centripetal force to keep the puck moving in a circle. At the same time, the tension in the string also supports the weight of the suspended mass.

Using Newton's second law, we can write two equations of motion: one for the puck and one for the suspended mass. For the puck, the net force acting on it is the tension in the string, which is equal to the centripetal force required to keep it moving in a circle. Thus, we can write:

= m1 * v^2 / R

where T is the tension in the string, v is the speed of the puck, and R is the radius of the circle.

For the suspended mass, the net force acting on it is its weight minus the tension in the string, which must be zero since the mass is in equilibrium. Thus, we can write:

T = m2 * g

where g is the acceleration due to gravity.

Combining these two equations, we can solve for the speed of the puck:

v = sqrt(T * R / m1) = sqrt(m2 * g * R / m1)

Substituting the given values, we get:

v = sqrt(1.0 kg * 9.81 m/s^2 * 0.9 m / 0.21 kg) = 3.67 m/s

Therefore, the speed of the puck is 3.67 m/s.

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Because of friction and air resistance, every moving object will eventually come to rest unless a continuous net force is applied to it.

Friction is the opposition offered to motion of a body over another body.

Friction can either be:

The law of inertia states that an object will remain in its state of rest or state of motion in a straight line unless acted upon by an external force.

However, due to friction and air resistance, every moving object will eventually come to rest unless a continuous net force is applied to it.

In conclusion, moving objects come to rest because of friction and air resistance.

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PV=nRT

Here

P=Pressure

V=Volume

n=Molarity

R=universal gas constant

T=Temperature.

LHS

\(\\ \tt\bull\leadsto PV\)

\(\\ \tt\bull\leadsto [ML^2T^{-2}][M^0L^3T^0]\)

\(\\ \tt\bull\leadsto [ML^5T^{-2}]\)

RHS

\(\\ \tt\bull\leadsto nRT\)

\(\\ \tt\bull\leadsto [M^0L^{3}T^0][M^1 L^2 T^{-2}K^{-1}][M^0L^0T^0K^1]\)

\(\\ \tt\bull\leadsto [ML^5T^{-2}]\)

LHS=RHS

hence verified

The mass of the rule is 3.36 kg and the weight of the rule is 33.6 N.

The butt of the blade is designed to balance the majority of fine chef's knives. This is due to the fact that a pinch grip is used in both Western and Eastern cutting styles for improved control. Therefore it makes sense that you'd want your knife to be balanced where you'll be holding it.

To balance the meter rule, Assume the mass of the rule "M".

It is possible to determine the rule's magnitude and weight, which act downward:

weight of rule = M * g

where g = acceleration due to gravity.

weight of rule * (50 cm) = (20 g) * (100 cm - 58 cm)

M * g * (50 cm) = (20 g) * (42 cm)

M = (20 g * 42 cm) / (50 cm * g)

M = 33.6 g / g

M = 33.6 g / 10 m/s^{2}

M = 3.36 kg

(ii) The weight of the rule:

weight of rule = M * g

weight of rule = 3.36 kg * 10 m/s^{2}

weight of rule = 33.6 N

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Answer:

  c)  480 mL

Explanation:

16 fluid ounces is 473.176473 mL exactly. The closest answer choice is ...

  480 mL

_____

1 gallon = 128 fl oz = 3.785411784 liters exactly

Answer:

(1.8 × 10^9) meters in one minute

Explanation:

To determine how far light can travel in one minute, we need to multiply its speed by the number of seconds in a minute.

The speed of light is 3 × 10^8 meters per second.

There are 60 seconds in a minute.

Therefore, the distance light can travel in one minute is:

Distance = Speed × Time

Distance = (3 × 10^8 meters per second) × (60 seconds)

Calculating this, we get:

Distance = 3 × 10^8 meters/second × 60 seconds

Distance = 18 × 10^8 meters

Distance = 1.8 × 10^9 meters

So, light can travel approximately 1.8 billion (1.8 × 10^9) meters in one minute.

1) The force acting on the object during the time interval 0-3 seconds is 1/3 N.

2) The force acting on the object during the time interval 6-10 seconds is -0.5 N.

3) There is no time interval in which no force acts on the object.

(i) Force acting on the object in time interval 0-3 seconds. Force acting on the object is equal to the product of its mass and acceleration, i.e.,F = ma.

In the given velocity-time graph, the acceleration of the object can be determined by determining the slope of the velocity-time graph from 0 to 3 seconds.

Slope = (change in velocity) / (change in time)= (20-0) / (3-0) = 20/3 m/s^2

Acceleration, a = slope= 20/3 m/s^2

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × 20/3= 1/3 N.

Therefore, the force acting on the object during the time interval 0-3 seconds is 1/3 N.

(ii) Force acting on the object in time interval 6-10 seconds. Similar to the first question, the force acting on the object in time interval 6-10 seconds can be determined by determining the acceleration of the object during this time interval.

The slope of the velocity-time graph from 6 seconds to 10 seconds can be determined as follows:

Slope = (change in velocity) / (change in time)= (-20-20) / (10-6) = -40/4= -10 m/s^2 (negative sign indicates that the object is decelerating)

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × (-10)= -0.5 N.

Therefore, the force acting on the object during the time interval 6-10 seconds is -0.5 N.

(iii) Time interval in which no force acts on the object. There is no time interval in which no force acts on the object. This is because, as per Newton's first law of motion, an object will continue to remain in a state of rest or uniform motion along a straight line unless acted upon by an external unbalanced force.In other words, if the object is moving with a constant velocity, there must be a force acting on the object to maintain its motion.

Therefore, there is no time interval in which no force acts on the object.

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The current remains the same in all parts of a series circuit.

The correct option is B.

A circuit is a complete path for the flow of electric current.

A series circuit is a circuit that the component of the circuit are connected end-to-end and the flow of current is in one direction.

Another form of connection is the parallel connection in which current flows through alternate paths.

In a series circuit, the flow of current is constant and remains the same in all parts of a series circuit. However, the voltage across the circuit varies.

In a parallel circuit,  the voltage is constant and remains the same in all parts of a parallel circuit.  However, the flow of current across the circuit varies.

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