Are the objects described here in static equilibrium, dynamic equilibrium, or not equilibrium at all? Explain.

a. a 200 pound barbell is held above your head
b. a girder is being lifted at a constant speed by a crane
c. a girder is being lowered into place. its is slowing down
d. a jet plane has reached its cruising speed an altitude
e. a box in the back of a truck doesn't slide as the truck stops

If you're approaching an intersection where the traffic signal is out, treat it as a four-way stop. Come to a full stop and proceed when it's your turn and it's safe. Some laws require drivers to treat broken traffic signals as stop signs.

When approaching an intersection where the traffic signal is out, you should treat the situation as you would a four-way stop. This means that you should make a full stop at the intersection and proceed only when it's your turn and it's safe to do so. This rule is in place to prevent accidents from occurring.

Many vehicles speed through intersections without regard to the posted speed limit, especially when traffic signals are out. However, state laws typically require drivers to think of broken traffic signals as stop signs. If two vehicles arrive at the intersection at the same time, the vehicle on the left should yield to the vehicle on the right.

Furthermore, if the intersection is busy and multiple vehicles are waiting to proceed, drivers should take turns going in the order in which they arrived at the intersection. This is essential for maintaining order and preventing accidents at intersections where the traffic signal is out.

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The de Broglie wavelength of an electron can be calculated using the de Broglie equation that relates the wavelength of a particle to its momentum. The given velocity and known constants are used in this calculation.

To calculate the de Broglie wavelength of an electron, we apply the de Broglie equation which states λ = h / p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle. For an electron moving at non-relativistic speeds, the momentum p is calculated using p = mv, where m is the mass of the electron and v is the velocity.

In this case, the electron's velocity v is provided as 1.34×10⁵ m/s. We know the mass m of the electron to be approximately 9.11 × 10^-31 kg, and the Planck's constant h to be 6.626 × 10^-34 kg m²/s from quantum theories. Plugging these values into the equations, we can find the de Broglie wavelength.

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The de Broglie wavelength of an electron travelling at 1.34 × 10⁵ m/s is approximately 5.41 nm, calculated using the formula λ = h / (mv). Here, h is 6.626 × 10⁻³⁴ Js and m is 9.109 × 10⁻³¹ kg.

The de Broglie wavelength of a particle is given by the formula λ = h / (mv), where λ is the wavelength, h is Planck's constant (6.626 × 10⁻³⁴ Js), m is the mass of the particle, and v is its velocity. For an electron travelling at a speed of 1.34 × 10⁵ m/s, we can calculate the wavelength as follows:

We plug these values into the formula:

λ = h / (mv) = 6.626 × 10⁻³⁴ Js / (9.109 × 10⁻³¹kg × 1.34 × 10⁵ m/s) ≈ 5.41 × 10⁻⁹ meters

Therefore, the de Broglie wavelength of an electron travelling at 1.34 × 10⁵ m/s is approximately 5.41 nm.

Answer:

1837.5

Explanation:

work will be equal to the potential energy gained by the person in climbing the stairs. work= potential energy gained = mgh W= 75kg*9.8m/s2*2.50m= 1837.5 J

A 75 kg person climbs stairs gaining 2.50 meters in height, then the work done to accomplish this task is 1837.5 J.

When we apply force "F" to a block, the body moves with some acceleration or, additionally, its speed increases or decreases depending on the direction of the force. The system's kinetic energy changes as speed increases or decreases. Since we are aware that energy cannot be created or destroyed, it must be changed into another form. This perspective refers to it as completed work. When negative energy is finished, the energy declines, and when positive energy is finished, the energy rises. Now, we'll see how to gauge the amount of work done.

According to the question, the given values are :

Mass, m = 75 kg

Height, h = 2.50 m

The work done will be equal to the potential energy gained by the person.

W = mgh

W = 75 × 9.8 × 2.50

W = 1837.5 J.

Hence, the work done will be 1837.5 J

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The total force to slide the bank safe is 795 N. The normal force exerted by the safe on the floor is 2943 N. Hence, the coefficient of kinetic friction on the bank floor is 0.27.

To find the coefficient of kinetic friction, we first need to determine the total force used to move the safe. Bonnie and Clyde are both applying their forces to move the safe in the same direction. So, we add Clyde’s force to Bonnie's force. This gives us a total force of 795 N (445 N + 350 N).

Since the safe slides at a constant speed, it means the applied force is equal to the friction force. Therefore, the friction force is also 795 N. We calculate the normal force which is the weight of the safe, that is mass times the gravity. The mass of the safe is 300 kg and the gravity is approximately 9.81 m/s². Therefore, the normal force is 2943 N (300 kg×9.81 m/s²).

Now, we use the formula of kinetic friction, which is the friction force divided by the normal force, to find the coefficient of kinetic friction. Therefore, the coefficient of kinetic friction on the bank floor is 0.27 (795 N / 2943 N).

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It is given that,

Speed of Barbara w.r.t ground, [tex]v_b=-5.9\ m/s[/tex] ( in south)

Speed of neil w.r.t ground, [tex]v_n=-1.4\ m/s[/tex] ( in west)

So, Neil's velocity as seen by Barbara is :

[tex]v^2=\sqrt{(1.4\ m/s)^2+(5.9\ m/s)^2}[/tex]

[tex]v=6.06\ m/s[/tex]

Direction relative to due wet is :

[tex]tan\theta=\dfrac{5.9\ m}{1.4\ m}[/tex]

[tex]\theta=tan^{-1} (4.21)[/tex]

[tex]\theta=76.6\ ^0[/tex]

So, with respect to barbara, Neil is travelling with velocity of 6.06 m/s making an angle of [tex]76.6^0[/tex] due west.

Hence, this is the required solution.

The displacement of the volleyball is approximately [tex]\(1.428 \, \text{m}\)[/tex] downward.

To calculate the displacement of the volleyball, we'll need to use the kinematic equation that relates initial velocity [tex](\(v_0\))[/tex], final velocity [tex](\(v\))[/tex], acceleration [tex](\(a\))[/tex], and displacement [tex](\(d\))[/tex]:

[tex]\[ v^2 = v_0^2 + 2ad \][/tex]

Given:

Initial velocity [tex](\(v_0\))[/tex] = 5.4 m/s downward (from the problem statement)

Final velocity [tex](\(v\))[/tex] = 1.1 m/s upward

Acceleration [tex](\(a\)) = \( g \)[/tex], acceleration due to gravity [tex]\(= 9.8 \, \text{m/s}^2\)[/tex]

We need to find the displacement [tex](\(d\))[/tex].

First, let's convert the initial velocity to be consistent with the direction of the final velocity. Since upward is taken as positive, we'll change the initial velocity's sign:

[tex]\[ v_0 = -5.4 \, \text{m/s} \][/tex]

Now, let's plug the given values into the kinematic equation:

[tex]\[ (1.1)^2 = (-5.4)^2 + 2 \times 9.8 \times d \][/tex]

[tex]\[ 1.21 = 29.16 + 19.6d \][/tex]

Now, let's solve for [tex]\( d \)[/tex]:

[tex]\[ 19.6d = 1.21 - 29.16 \][/tex]

[tex]\[ 19.6d = -27.95 \][/tex]

[tex]\[ d = \frac{-27.95}{19.6} \][/tex]

[tex]\[ d \approx -1.428 \, \text{m} \][/tex]

The negative sign indicates that the displacement is downward.

Therefore, the displacement of the volleyball is approximately [tex]\(1.428 \, \text{m}\)[/tex] downward.

The displacement of the volleyball is [tex]\( 0.06173 \, \text{m} \)[/tex] downward.

To calculate the displacement of the volleyball, we can use the equation of motion:

[tex]\[ v^2 = u^2 + 2as \][/tex]

where:

-v is the final velocity (1.1 m/s upward),

- u is the initial velocity,

- a is the acceleration, and

-s is the displacement.

Given that the initial velocity is 0 m/s (as the ball is thrown upward), we can rearrange the equation to solve for s:

[tex]\[ s = \frac{{v^2 - u^2}}{{2a}} \][/tex]

Since the ball is moving upward, acceleration due to gravity will act in the opposite direction, and its magnitude is [tex]\( 9.8 \, \text{m/s}^2 \) (assuming no air resistance). So, \( a = -9.8 \, \text{m/s}^2 \).[/tex]

Now, let's plug in the values:

[tex]\[ s = \frac{{(1.1 \, \text{m/s})^2 - (0 \, \text{m/s})^2}}{{2 \times (-9.8 \, \text{m/s}^2)}} \][/tex]

[tex]\[ s = \frac{{1.21 \, \text{m}^2/s^2}}{{-19.6 \, \text{m/s}^2}} \]\\ s = -0.06173 \, \text{m} \][/tex]

Since displacement is a vector quantity, the negative sign indicates that the displacement is in the opposite direction to the direction of acceleration (downward in this case). So, the displacement of the volleyball is [tex]\( 0.06173 \, \text{m} \)[/tex] downward.

Complete question:

Calculate the displacement of the volleyball  when the volleyball’s final velocity is 1.1 m/s upward.

Answer:

C. course conditions

Explanation:

In the game of the golf the ground/court in which it is played is called as golf course. It has various obstacles such as grassy land, water holes, trees, shrubs etc. The landscape is also uneven and has lot of inclines and declines.

The parameters which will control the distance a golf ball will travel will include course conditions in addition to the ones mentioned in the question. For example, if it drops on a declined slope it will travel farther than its drop point. Other options will have not impact on the distance the ball will travel.

This question is about the physics of a car's stopping distance and the effect of reaction time on it.

The subject of this question is Physics and it is appropriate for high school level students.

In this scenario, the driver is traveling at a speed of 25m/s when they see a hole 55m ahead of them. Since their reaction time is zero, meaning they immediately apply the brakes, the total displacement during the reaction time is 15.0m greater than if they reacted instantly. The stopping distance, in this case, would be the same for both dry and wet concrete, so we need to calculate the distance the car travels during the reaction time and add that to the stopping distance.

To summarize, this question is about the physics of a car's stopping distance and the effect of reaction time on it.

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The speed required to escape Earth's gravitational pull, 11,180 meters per second, is equivalent to 11.18 kilometers per second.

To convert the speed from meters per second to kilometres per second, you need to know that 1 kilometre is equivalent to 1000 meters. So, if an object needs to reach a speed of 11,180 meters per second to escape Earth's gravitational pull, we divide 11,180 by 1000 to get the speed in kilometres per second.

11,180 m/s ÷ 1000 = 11.18 km/s

This results in a speed of 11.18 kilometers per second.

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Answer:

Explanation:

A protostar forms once the nebular cloud condenses and the core begins HEAT.

A protostar is a very young star that is still gathering mass from its parent molecular cloud. The protostellar phase is the earliest one in the process of stellar evolution. A protostar is the earliest stage in star formation in the universe.  When the gas cloud begins to collapse on its own weight, the core begins to heat up. The heat is due to the force of compression of the dust and gas under its gravity. When the pressure and temperatures become high enough, the hydrogen in the core begins to fuse.

Answer:

D.55.86 N

Explanation:

Answer:

The correct answer is 3.5 m/s²

Explanation:

To determine the average acceleration of the rock,

the change in velocity is divided by the time interval (in seconds)

change in velocity = V₂ - V₁

where V₁ is the initial velocity (2.0 m/s) and V₂ is the final velocity (44.0 m/s)

change in velocity = 44 - 2 = 42 m/s

The time interval is 12 seconds (no need for conversion as it is in seconds already)

average acceleration of the rock = 42 ÷ 12 = 3.5 m/s²

oragnisms ?????????????????????

Answer:

C.

one inclined plane and two wedges

Explanation:

You need to find the mass of water in the pool.
Find the volume (10 x 4 x 3) = 120 m3

Water has a density of 1000g/m3,so 120 m3 = 120 x 1000 = 120 000 kg

You then use the heat of combustion knowing that each mole of methane releases 891 kJ of heat so if you divide 891 into the previous answer, you will get the number of moles of CH4

Answer:

119.69 kilo liter volume of methane measured at STP, must be burned.

Explanation:

Length of the pool= 10.0 m

Breadth of the pool = 4.0 m

Height upto which the pool will be filled= 3.0 m

Volume of the water in the pool = volume of the pool :

= 10.0 m × 4.0 m × 3.0 m = [tex]120.0 m^3[/tex]

Density water = [tex]997 kg/m^3[/tex]

Mass of the water(m) = [tex]Density \times volume =997 kg/m^3\times 120 m^3=119,640 kg[/tex]

Energy required to raise the temperature = Q

Mass of the water = m = 119,640 kg =[tex]119,640\times 1,000 g=119,640,000 g[/tex]

Specific heat of water ,c= 4.186 J/g °C

Change in temperature, [tex]\Delta T=29.7^oC-20.2^oC=9.5 ^oC[/tex]

[tex]Q=mc\Delta T[/tex]

[tex]Q=119,640,000 g\times 4.186J/g ^oC\times 9.5 ^oC[/tex]

[tex]Q = 476,113.36\times 10^4J=476,1133.6 kJ[/tex]

Heat required to raised the temperature is 476,1133.6 kJ.

Heat evolved when 1 mol of methane = -891 Kj

Number of moles of methane giving (- 476,1133.6) kJ of heat:

[tex]\frac{-476,1133.6 kJ.}{-891}=5,343.58 mol[/tex]

At STP, 1 mol occupies = 22.4 L

Then 5,343.58 mol will occupy :[tex]22.4\times 5,343.58 L=119,696.19 l=119.69 kilo liter[/tex]

119.69 kilo liter volume of methane measured at STP, must be burned.

Answer:

As we know that the Earth core is at high temperature and due to this high temperature the core of Earth contains large amount of molten metal.

Due to this large amount of molten metal there exist huge amount of charges or ions along with the molten metal. Now these molten metal or ions revolve around the axis of rotation of earth with the rotation of Earth around its own axis.

Due to this revolution of molten metal there exist a huge current in the earth due to motion of ions.

Now we know that magnetic field due to rotating current carrying coil is given by

[tex]B = \frac{\mu_0 Ni}{2R}[/tex]

so here this will produce magnetic field due to rotation of earth about its own axis.

Final answer:

Uranus and Neptune are the two adjacent planets with the largest distance between them, approximately 11 AU apart, unlike Jupiter and Saturn or Mars and Jupiter which are closer together.

Explanation:

The two adjacent planets in our solar system with the largest distance between them are Uranus and Neptune. These outer planets, also known as ice giants, have vastly different compositions compared to their neighbouring gas giants. While Jupiter and Saturn share many similarities, Uranus and Neptune differ significantly in composition and structure. The distance between Uranus and Neptune is greater due to their position in the solar system, with Uranus orbiting at approximately 19 AU (Astronomical Units) and Neptune at about 30 AU from the Sun. This indicates that the distance separating them is roughly 11 AU, which is more than that between Mars and Jupiter or Jupiter and Saturn.

Furthermore, the study of the giant planets — which include Jupiter, Saturn, Uranus, and Neptune — has been enhanced by spacecraft explorations, most notably by Voyager 2 and the missions Galileo and Cassini. These have provided us with invaluable information on the composition, atmospheres, and internal structures of these distant worlds.

Answer:

Distance moved = 9.0 m

Displacement magnitude = 4.6 m

Direction : Northwards

Explanation:

the quarter back pedals 2.2 m South

Then he pedal 6.8 m North

So here we can say that total distance moved by the quarterback is

[tex]Distance = 6.8 + 2.2 = 9.0 m[/tex]

Now to find the displacement we need to add them with direction

so we have

[tex]\vec d = 6.8 - 2.2 = 4.6 m[/tex]

Direction : Towards North

The total distance covered by the cyclist is 9 m, while the displacement is 4.6 m northwards.

The given parameters;

Distance is scalar quantity. That is, distance can be represented in a complete sense with magnitude only.

The total distance covered during the motion = 2.2 m + 6.8 m = 9m

The direction describes the position of the total displacement.

The displacement = (6.8 m - 2.2 m) northward

The displacement = 4.6 m northward.

Thus, we can conclude that the total distance covered by the cyclist is 9 m, while the displacement is 4.6 m northwards.

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The centripetal acceleration (a ) of an object moving in a circular path of radius  r at a constant speed and v is  = a = v²/r

The riders are subjected to a centripetal acceleration that is 1.50 times the acceleration due to gravity, a = 1.50 × g.

a = r  ×ω ²  

1.50 g = r  ×ω ²  

ω ²   = √1.50 g / r

rpm =  ω/2π  ×60

rpm=  √1.50⋅g/ r/2 π  ×60

rpm = 9.50

Therefore,  the riders will be subjected to approximately 9.50 revolutions per minute.

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