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The lead block is moving at a speed of 1 m/s after the bullet rebounds off of it. The correct answer is (c) -5 m/s.  

First, we need to find the momentum of the bullet before it hits the block:

p_bullet = m_bullet * v_bullet = 50 * 500 = 25,000 g

We also need to find the momentum of the block before the bullet hits it:

p_block = m_block * v_block = 2 * 2000 = 4000 g

After the bullet hits the block, it rebounds with a speed of 300 m/s. The momentum of the bullet after the collision is:

p_bullet_after = m_bullet * v_bullet_after = 25,000 * 300 = 750,000 g

The momentum of the block after the collision is:

p_block_after = m_block * v_block_after = 4000 * 300 = 120,000 g

The momentum of the system (bullet + block) before the collision is:

p_system_before = p_bullet + p_block = 750,000 + 120,000 = 870,000 g

The momentum of the system after the collision is:

p_system_after = p_bullet_after + p_block_after = 750,000 + 120,000 = 870,000 g

We want to find the speed of the block after the collision, so we can set the total momentum of the system before and after the collision equal to each other:

p_system_before = p_system_after

870,000 g = 750,000 g + 120,000 g

870,000 g = 870,000 g

Dividing both sides by 870,000 g gives us:

v_after = 1

Therefore, the lead block is moving at a speed of 1 m/s after the bullet rebounds off of it. The correct answer is (c) -5 m/s.  

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Explanation:

Given data

velocity v= 25m/s

The time it takes to put on brake t= 0.3s

the distance covered when the  brake was put on is

v=s/t

s= v*t

s= 25*0.3s

s= 7.5m

hence the distance covered is 7.5m

Also the rate of decrease in aceleration is 5m/s^2

we can also calculate the distance covered at this rate

v^2=u^2+2as

25^2= 0+2*5*s

625=10s

divide both sides by 10

s=625/10

s= 62.5m

The total  distance covered between putting on the brakes and decelareation is  7.5+62.5= 70m

Given that the tree is 75m ahead, the car would not hit the tree

Answer:

"electric or magnetic"

F = q E    force on charge q due to electric field E

F = q v x B       force on charge q due to magnetic field B

Tech a says that tinted safety glasses can be worn when working outside. tech b says that welding can cause a sunburn. Both technician a and b are correct.

According to Technician a, tinted safety glasses can be worn when working outside, it is true the tinted lenses should be used for outdoor or job specific use only. Tinted lenses will not be worn indoors unless designed for a specific radiant energy hazard, because,

The tints are to protect workers from such hazards as welding, brazing, blast furnaces and other high intensity light sources.

According to Technician b, Welding can cause a sunburn, Hence, it is true because UV radiation in a welding arc will burn unprotected skin just like UV radiation in sunlight.

This hazard is true for direct exposure to UV radiation as well as radiation that is reflected from metal surfaces, walls, and ceilings.

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Cage 2 will reach the bottom first. The time taken by cage one is 114.76 seconds and cage 2 is 0.32 seconds.

Given,

Mine depth Y = 2072 m

Cage speed 1 v = 65.0 km/h

Acceleration of Cage 2,

a = 4.00 102 m/s2

Solution:

We shall utilize the international system of measurements (SI) to interchange magnitudes using the unit system, which allows us to do so precisely and without the need for antiquities.

  v = 65 km / h (1000m/1Km) (1hr/3600s)

     = 18.06 m / s

The relationships between location, velocity, and acceleration can be discovered using kinematics.

Work each cage separately in this practice since there are two of them.

Cage 1

They suggest that it is moving at a constant speed, which we can determine using the laws of uniform motion.

v = d/t

v = d/v

t1 = 2072/18.06

t₁ = 114.76 s

Cage 2

Stop so that your starting velocity is zero.

y = v₀ t + ½ a t²

y = 0 + ½ a t²

\(t = \sqrt{2y/a} \\\\t2 = \sqrt{2*2072/4*10^{2} } \\\)

t₂ = 0.32 s

On comparing the time taken by cage 1 and cage 2, we can say that cage 2 will reach the ground first.

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The discovery that electrons move around the nucleus billions of times in one second can be attributed to multiple scientists. However, the most notable contribution came from Niels Bohr.

In 1913, Bohr proposed his atomic model, which introduced the concept of quantized energy levels for electrons. According to this model, electrons orbit the nucleus in specific, well-defined energy levels or shells.

These energy levels are characterized by their respective energy values, and electrons can transition between them by either absorbing or emitting energy. Bohr's model successfully explained phenomena like atomic spectra, and his work laid the foundation for our current understanding of atomic structure.

Overall, Bohr's model revealed the dynamic nature of electrons in their orbits and their rapid motion around the nucleus.

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As of my knowledge cutoff in September 2021, there is no current or planned space observatory specifically dedicated to observing electromagnetic radiation in the mid-infrared range, roughly from 5 to 20 micrometers (μm) in wavelength. This range is often referred to as the "thermal infrared" as it corresponds to the thermal radiation emitted by objects at room temperature or higher.

While space observatories like the Spitzer Space Telescope have made significant contributions to mid-infrared astronomy, their missions have either ended or are focused on other wavelength ranges. The James Webb Space Telescope (JWST), set to launch in late 2021, will have some capability in the mid-infrared range but will primarily operate in the near-infrared.

However, it's worth noting that the field of mid-infrared astronomy is not neglected. Ground-based observatories and instruments, such as those equipped with adaptive optics or interferometry, continue to advance our understanding of the mid-infrared universe. Additionally, future mission concepts and technology developments may lead to dedicated space observatories for the mid-infrared range.

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Answer:

An electrically polarized object can have zero net charge, while a charged object cannot have zero net charge.

I hope this helps :D

ANSWER:

Distance = 9 km

Displacement = 9 km east

STEP-BY-STEP EXPLANATION:

When we speak of distance we refer to the length between any two points in space, while displacement refers to the distance from an initial position to a final position regardless of the path.

In this case, being in the same direction, the displacement and distance are the same (in magnitude).

Therefore:

The speed of the wind is 17 mph and the speed of the plane in still air is 119 mph.

Let the speed of the plane be x and the speed of the wind be y. Then, the speed of the plane with the wind becomes x + y, while the speed of the plane against the wind is x - y.

The distance traveled with the wind in 3 hours is 408 miles.

Therefore, we can write the equation as:

3(x + y) = 408

Divide both sides by 3:

x + y = 136  .... (1)

The distance traveled against the wind in 4 hours is also 408 miles.

Therefore, the equation can be written as:

4(x - y) = 408

Divide both sides by 4:

x - y = 102  .... (2)

Now we can solve these two equations using the elimination method.

Add equations (1) and (2):

x + y + x - y = 136 + 1022x = 238x = 119 mph

Therefore, the speed of the plane in still air is 119 mph.

Now, substitute this value of x in equation (1):

119 + y = 136y = 17 mph

Therefore, the speed of the wind is 17 mph.

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The correct question is:

A plane flies 408 mi with the wind in 3 hr. The return trip takes 4 hr. What is the speed of the wind and the speed of the plane in still air?

Answer: What is the force that opposes motion between two surfaces in contact with each ... How is the relationship among mass, force and acceleration explained? ... When a force is exerted on a box, an equal and opposite force is exerted by the box.

Answer:

elp me please with the question  p osted rn im

taking a test and need it

Explanation:

Liquids are composed of molecules that are weakly attached and free to move, but they are relatively closer together than in gases and have a definite volume.

Volume is a physical quantity that measures the amount of space occupied by an object or a substance. It is a derived quantity, meaning that it is calculated from other physical quantities such as length, width, and height.

Physical refers to anything that relates to the properties or behavior of matter and energy, as opposed to biological, chemical, or social phenomena.

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A. the change in volume of the cube is approximately 0.120601 cm^3. B. the average rate of change of volume of the cube with respect to an edge length is approximately 12.0601 cm^3/cm. C. the equation of the line that passes through the point (1, 5) and is parallel to the tangent line to f(x) = 1/x at x = 3 is y = (-1/9)x + 14/9.

(a) To find the change in volume of the cube as x increases from 2.00 to 2.01 centimeters, we need to calculate the difference in volume between these two values.

The volume of a cube is given by V = x^3, where x is the edge length.

For x = 2.00 cm, the volume V1 = (2.00 cm)^3 = 8.00 cm^3.

For x = 2.01 cm, the volume V2 = (2.01 cm)^3 = 8.120601 cm^3.

The change in volume is ΔV = V2 - V1 = 8.120601 cm^3 - 8.00 cm^3 ≈ 0.120601 cm^3.

Therefore, the change in volume of the cube is approximately 0.120601 cm^3.

(b) The average rate of change of volume of the cube with respect to an edge length as x increases from 2.00 to 2.01 centimeters can be calculated by dividing the change in volume by the change in edge length.

ΔV = 0.120601 cm^3 (from part a)

Δx = 2.01 cm - 2.00 cm = 0.01 cm

The average rate of change of volume is ΔV/Δx = 0.120601 cm^3 / 0.01 cm ≈ 12.0601 cm^3/cm.

Therefore, the average rate of change of volume of the cube with respect to an edge length is approximately 12.0601 cm^3/cm.

(c) The instantaneous rate of change of volume of the cube with respect to an edge length at the instant when x = 2 centimeters can be found by taking the derivative of the volume function V = x^3 with respect to x and evaluating it at x = 2.

dV/dx = 3x^2

At x = 2 cm, the instantaneous rate of change of volume is dV/dx evaluated at x = 2:

dV/dx = 3(2 cm)^2 = 12 cm^2.

Therefore, the instantaneous rate of change of volume of the cube with respect to an edge length at x = 2 centimeters is 12 cm^2.

To find the equation of the line that passes through the point (1, 5) and is parallel to the tangent line to f(x) = 1/x at x = 3, we need to determine the slope of the tangent line.

The derivative of f(x) = 1/x is given by f'(x) = -1/x^2.

At x = 3, the slope of the tangent line is f'(3) = -1/(3^2) = -1/9.

Since the line we want to find is parallel to the tangent line, it will have the same slope. So the slope of the line is -1/9.

Using the point-slope form of a linear equation, we can write the equation of the line as:

y - y1 = m(x - x1),

where (x1, y1) is the given point (1, 5) and m is the slope.

Substituting the values, we have:

y - 5 = (-1/9)(x - 1).

Expanding and rearranging the equation, we get:

y = (-1/9)x + 14/9.

Therefore, the equation of the line that passes through the point (1, 5) and is parallel to the tangent line to f(x) = 1/x at x = 3 is y = (-1/9)x + 14/9.

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Answer:

a.ionic bond is kinds of bonds make up a formula unit of sodium chloride.

Answer:

A

Explanation:

The distance of the test charge from the source charge is 1.1 m. Option A is correct.

It states that the electric flux across any closed surface is directly proportional to the net electric charge enclosed by the surface.\(R= \sqrt \dfrac {Qk^}E\)

Where,

\(E\) = electric force = 3. 93 × 10⁵  N/C

\(k\) = Coulomb constant = \(8.99 \times 10^9 \rm\ N m ^2 /C ^2\)

\(Q\\ \)     = charges =  5.0 µC

\(r\) = distance of separation =?

Put the values in the formula,

\(R= \sqrt \dfrac {5\times 10^{-6}8.99 \times 10^9}{3. 93 \times 10^5}}\\\\ R = 1.1 \rm \ m\)

Therefore, the distance of the test charge from the source charge is 1.1 m.

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Answer:

Install sound absorbers such as foam or acoustical panels

Explanation:

They could install some sound absorbers such as foam or acoustical panels in the walls, floors, and ceiling of the conference room.

The net force acting on the object is 703.125 N

In the case of circular motion, there is a centripetal acceleration acting on the object given by:

\(a_{r}=\frac{v^{2} }{r}\)

where, v is the velocity of the object = 15 m/s,

and r is the radius of the path = 1.6 m

The net force F:

\(F=ma_{r}=m\frac{v^{2} }{r}\)  

\(F=5*\frac{15^{2} }{1.6}\\F=703.125 N\) is the net force acting on the object.

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To obtain the station pressure, one must usually make corrections for temperature and elevation.

The temperature and elevation corrections are used to transform the pressure to the station pressure by taking into account the fact that the pressure measured at a location is not just a function of the air pressure at that location, but also the air pressure at nearby locations and the air temperature and elevation at that location.

Station pressure is the atmospheric pressure measured at the location of the weather station. However, in order to obtain station pressure, temperature and elevation corrections must usually be made. The reason is that the pressure that is measured at a location is not only a function of the air pressure at that location but also of the air pressure at nearby locations, as well as the air temperature and elevation at that location.

Temperature Correction:

The temperature of the air affects the atmospheric pressure and is therefore taken into account while making corrections. The temperature of the air affects the air density. An increase in temperature results in a decrease in air density and thus atmospheric pressure. Similarly, a decrease in temperature results in an increase in air density and thus atmospheric pressure. Therefore, temperature corrections must be made to account for the impact of air temperature on atmospheric pressure.

Elevation Correction:

In addition to the temperature correction, an elevation correction must be made to account for the impact of elevation on atmospheric pressure. The air pressure decreases with an increase in altitude, which implies that atmospheric pressure varies with elevation. Therefore, it is necessary to make a correction to account for the change in elevation. A correction factor is used to account for this impact, which is dependent on the elevation of the location where the pressure is being measured.

In conclusion, station pressure is the atmospheric pressure measured at the location of the weather station. However, to obtain this pressure, temperature and elevation corrections must usually be made. The temperature correction is needed to account for the impact of air temperature on atmospheric pressure, while the elevation correction is necessary to account for the impact of elevation on atmospheric pressure.

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Answer:

Location

Explanation:

Potential energy is the energy stored in an object due to its position. The most common type of potential energy - gravitational potential energy - is the energy stored in an object due to its vertical position relative to the ground or some zero level.

"An object can store energy as the result of its position. For example, the heavy ball of a demolition machine is storing energy when it is held at an elevated position." This stored energy of position is referred to as potential energy. Similarly, a drawn bow is able to store energy as the result of its position.

"Energy is the capacity for doing work. It may exist in potential, kinetic, thermal, electrical, chemical, nuclear, or other forms."

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Answer:

Below

Explanation:

Potential Energy = m g h

                           = 15 500  kg  *   9.81 m/s^2   * 215 m = 32 691 825 J

  (or   32 692 kJ )

It 4.5 hours for the light bulb to consume 8100 watt-hours of power.

The energy consumed by the light blub = 8100 watt-hours.

Let the time taken be t hours.

We know that;

Energy = power × time

Energy (E) =  8100 watt-hours

Power (P) = 1800 watt

Time (t) =?

8100 =  1800 × t

= 8100 / 1800

= 4.5 hours

The time taken by the light bulb is 4.5 hours.

Therefore, it could take 4.5 hours for the light bulb to consume 8100 watt-hours.

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Answer:

B

Explanation:

cause it be like that

B. The type of liquid inside the boiling vessel.

Paraffin oil is used for boiling factor willpower as it has a very excessive boiling factor and so it could be used to hold a high temperature within the boiling factor and melting factor equipment without lack of the substance.

Boiling is the process by which a liquid will become a vapor whilst it's miles heated to its boiling point. The alternative from a liquid segment to a gaseous segment occurs while the vapor pressure of the liquid is identical to the atmospheric stress exerted on the liquid.

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Answer:

Weight = 3217.68

Explanation:

N times Gravity = weight

Gravity= 9.81

The weight of the emperor penguin whose mass is 328 Kg on is 3214.4 Newtons.

Weight is the force exerted by the earth on an object at the surface of the earth. Mathematically -

W = m x g

where -

[m] is the mass of object

[g] is acceleration due to gravity

Given is the mass of an emperor penguin as 328 Kg.

We can write -

Mass [m] = 328 kg

Acceleration due to gravity [g] = 9.8 m/s²

Therefore, the weight of the emperor penguin will be

W[P] = m x g = 328 x 9.8 = 3214.4 Newtons

Hence, the weight of the emperor penguin whose mass is 328 Kg on is 3214.4 Newtons

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Answer:

η = 0.882 = 88.2 %

Explanation:

The efficiency of the pulley system can be given as follows:

\(\eta = \frac{W_{out}}{W_{In}}\\\\\)

where,

η = efficiency of pulley system = ?

W_out = Output Work = (600 N)(0.6 m) = 360 J

W_in = Input Work = (35.7 N)(11.43 m) = 408.051 J

Therefore,

\(\eta = \frac{360\ J}{408.051\ J}\)

η = 0.882 = 88.2 %

According to Newton's Law of Universal Gravitation, the force between to bodies of masses m and M separated by a distance r is:

Where G is the gravitational constant:

In all cases, use M=60kg, and replace the values of the second mass m and the distance r accordingly.

A) m=80kg, r=1.4m

B) m=130,000kg, r=10m

C) m=5.22*10^9kg, r=1000m

D) m=0.045kg, r=0.95m

Answer:

its velocity, wavelength and frequency all change

Explanation:

Whenever a wave crosses the boundary between two media of different densities, its velocity, frequency and wavelength changes. This appears as a bending of the wave as it crosses the boundary from one medium to another.

Refraction is a fundamental property of waves. Hence when light is travelling from air into water, it wavelength, speed and frequency all changes at the interface between the two media.

For general projectile motion with no air resistance, the horizontal component of a projectile's acceleration is always zero.

The horizontal component of a projectile's acceleration is always zero in ordinary projectile motion with no air resistance. This is because there is no horizontal force exerted on the projectile once it is in motion.

The only horizontal force acting on the projectile is its initial velocity, which remains constant throughout its travel.

As a result, the horizontal component of acceleration remains zero throughout the projectile's trajectory.

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Answer:

Explanation:

Answer:

There was no anwser

Explanation:

because dylan did not anwser

We are asked to determine the magnitude of the force that acts parallel to the nail. To do that we will add the torque that acts on the point of contact. First, we draw a free-body diagram of the situation:

We have decomposed the force of the nail into its horizontal and vertical components. The horizontal component does no torque since there is no distance parallel to the force to the point of contact.

Now, we add the torques. We consider counterclockwise as positive:

Since we consider the moment before there is no angular acceleration the sum of torques adds up to zero:

Now, we determine the value of Ry as a function of "R" using the trigonometric function cosine:

Now, we multiply both sides by "R":

Now, we substitute in the sum of torques:

Now, we solve for "R". First, we add "30F" to both sides:

Now, we divide both sides by 5cm:

Now, we divide both sides by cosine:

Now, we substitute the values:

Solving the operations:

Therefore, the force on the nail is 1042.8 Newtons.