PLEASE HELP ASAP WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST!!!!
Because of the forces acting on the cart, it will

A. not accelerate

B. accelerate upwards

C. accelerate to the right

D. accelerate to the left

Electrical conductivity often decreases, when elements receive additional electrons in their outer shells. As a result of the valence shell being filled, the number of movable electrons decreases.

Electrons are mobile charges in the conduction band while holes are mobile charges in the valence band.

Current flow in a semiconductor is caused by charge carrier migration in both the conduction and valence bands.

Which creates a net motion of positively charged carriers in the same direction as the field and negatively charged carriers in the opposite direction as a result of the force.

The outermost electrons in metals are not bonded to each atom which is shared by many atoms. This decentralization is what allows metals to conduct electric currents.

Hence, the elements gain more electrons in their outer shells, and their electrical conductivity generally decreases.

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The following mathematical operations are and are not allowed on two quantities with different unit dimensions are C) Allowed : Multiplication, Division , Not Allowed : Addition, Subtraction, Equality

To answer the question, we have to know what mathematical operations are.

Mathematical operations are operations which are performed to change the value of a variable. The arithmetic mathematical operations we have are

Now, for two quantities with different unit dimensions, addition, subtraction and equality are not allowed because these mathematical operations require the quantities to be in the same unit dimensions

Also, for two different quantities with differnt unit dimensions, multiplication and division are allowed because these mathematical operations do not require the quantities to be in the same unit dimensions.

So, the mathematical operations that are allowed are multiplication and division while the mathematical operations that are not allowed are addition, subtraction and equality.

So, the following mathematical operations are and are not allowed on two quantities with different unit dimensions are C) Allowed : Multiplication, Division , Not Allowed : Addition, Subtraction, Equality

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Given,

The initial horizontal velocity of the projectile, u=22.2 m/s

The total displacement covered by the projectile in the x-direction, i.e., the range of the projectile, R=36.0 m

The projectile was launched horizontally, thus the horizontal component of initial velocity, uₓ=u=22.2 m/s

And the vertical component of the velocity, uy=0 m/s

And the angle of projection is θ=0°

The time interval, t, in which the projectile completes its flight is calculated from the equation,

On substituting the known values in the above equation,

The height of the building is calculated from the equation of the motion,

Where y is the final height of the projectile, i.e., y=0 m.

y₀ is the initial height of the projectile, i.e., the height of the building.

g is the acceleration due to gravity.

Let's consider the upward direction as the positive direction. That makes the height of the building a positive value and the acceleration due to gravity a negative value.

On substituting the known values,

Thus the height of the building is 12.86 m

The angle in radians through which the wheel rotates is \(14207.511^{\circ}\).

What is angular velocity?

In physics, angular velocity or rotational velocity, also known as angular frequency vector, is a pseudovector representation of how fast the angular position or orientation of an object changes with time.

Given

initial angular velocity \($\omega_1=25 \mathrm{rad} / \mathrm{s}$\)

Final Angular velocity \($\omega_2=36 \mathrm{rad} / \mathrm{s}$\)

time interval \($t=3 \mathrm{~s}$\)

using

\($$\begin{aligned}& \omega_2=\omega_1+\alpha t \\& 36=25+2 \\& \alpha=1.5 \mathrm{rad} / \mathrm{s}^2\end{aligned}$$\)

(b)Average angular speed \($\frac{\Delta \theta}{\Delta t}$\)

\($$\begin{aligned}& \theta=\omega_1 t+\frac{1}{2} \alpha t^2 \\& \theta=25 \times 8+\frac{1}{2} \times 1.5 \times 8^2 \\& \theta=200+48=248 \mathrm{rad}\end{aligned}$$\)

average angular speed \($=\frac{\Delta \theta}{\Delta t}$\)

\(=\frac{248}{8}=31 \mathrm{rad} / \mathrm{s}$$\)

(c)Angle rotated =248 radians

(d)angles in degree \($=14207.511^{\circ}$\)

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Answer:

If acceleration points in the same direction as the velocity, the object will be speeding up. And if the acceleration points in the opposite direction of the velocity, the object will be slowing down.

Answer:

The ejection of electrons from metals illuminated with different frequencies of light (photoelectric effect)

Explanation:

Ejection of electrons from metals illuminated with different frequencies of light provides evidence for the particle theory of light

The photoelectric effect proves that light has particle-like activity. The photoelectric effect happens when photons are shone on metal and electrons are ejected from the surface of that metal. The electrons that are ejected are determined by the wavelength of light which determines the energy of photons.

correct answer is

Ejection of electrons from metals illuminated with different frequencies of light

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Answer:

5 m/s

Explanation:

Total distance = 300 + 100 = 400m

Total time = 60 + 20 = 80s

Velocity = 400/80 = 5m/s

Average velocity of man is 5 meter per second.

To find the average speed of an object we divide the total distance travelled by the total time time taken by object.

Total distance travelled by man = 300 + 100 = 400 m

Total time taken by man = 60 + 20 = 80 seconds

Average velocity,\(=\frac{400}{80} =5m/s\)

Thus, Average velocity of man is 5 meter per second.

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Senen Bacani is an entrepreneur, the CEO of La Fuerta Inc which is one of the largest exporters of bananas in Philippines.

Senen Bacani was the formal sectary of Agriculture to president Corazon Aquino and during that time he visited many Philippines province and realize how poor people struggle to fend for their families and remember how his father who  was a doctor received farm produce from his patient, appreciating him for medical care they received seeing the need of the of the rural people, has compelled him to work for shifts in policy and programs as the secretary and later in 1993 after working for president Aquino administration, he establish a business in agriculture to provide employment and to alleviate poverty in the rural areas Mostly in Mindanao, where he establish business venture that brought about not only employment but peace where there had been years of unemployment, crime and inter-ethnic conflict.

Senen Bacani presently head 9 different companies and presently occupies the position of Chairman for  different Inc., in the Philippines.

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Answer:

Your answer would be (B)

The horizontal component of the vector is determined as 6i.

The horizontal component of the vector is calculated as follows;

V = (0 i, 0 j) + (6i, 8j)

V = (6i, 8j)

where;

Thus, the horizontal component of the vector is determined as 6i.

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a) The value of t u = 140/t`b.

b) The y position of the cannonball at the time t is  55.5 mc.

c) The initial speed of the projectile is 52.4 m/s.

Given that a cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 55.5 m above ground level, and the ball is fired with initial horizontal speed u. The projectile lands at a distance D = 140 m from the cliff. Assume that the cannon is fired at time t = 0 and that the cannonball hits the ground at time t.Now,We have to find the value of t, y position of the cannonball at the time t and the initial speed of the projectile.

a. To find the value of t:Here, we have to use the formula of distance

i.e.,S = ut + (1/2)gt², Where S = 140 m, u = u and g = 9.8 m/s².Hence,140 = u×t ………..(1)We know that, time taken by the cannonball to hit the ground can be calculated as,`(2H)/g`

Since the height of the cannon from the ground is 55.5m, the total height of the cannonball from the ground is

(2H) = 2 × 55.5

= 111 m`2H/g

= 111/9.8`

= 11.32653 s

From equation (1),u×t = 140u = 140/t

Therefore, `u = 140/t`b.

b)To find the y position of the cannonball at the time t:

Here, we have to use the formula of height i.e.,y = u×t – (1/2)gt²,

Where, y = height of the cannonball at time t, u = 140/t, t = time taken by the cannonball to hit the ground and g = 9.8 m/s².

We have already calculated the time taken by the cannonball to hit the ground in the previous step.`

y = 140 - (1/2) × 9.8 × t²`

On substituting the value of t as `t = 11.32653`,

we get,y = 140 - (1/2) × 9.8 × (11.32653)²= 55.5 mc.

c)  To find the initial speed of the projectile:

To calculate the initial speed of the projectile, we need to use the formula of range of projectile

.i.e.,R = u²sin2θ/g

Where R = 140 m, g = 9.8 m/s², θ = 0° (horizontal)

u² = R × g/sin2θ

   = 140 × 9.8/sin0°  

   = 2744m²/s²u  

   = \(\sqrt(2744m^2/s^2)\)

   = 52.4 m/s

Hence, the initial speed of the projectile is 52.4 m/s.

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Answer: The given statement is False.

Explanation:

Boyle's law states that pressure is inversely proportional to the volume of the gas at constant temperature and the number of moles.

Mathematically,

\(P\propto \frac{1}{V}\) (At constant temperature and number of moles)

OR

\(P_1V_1=P_2V_2\)

where,

\(P_1\text{ and }V_1\) are the initial pressure and volume of the gas

\(P_2\text{ and }V_2\) are the final pressure and volume of the gas

Hence, the given statement is False.

Answer: 4R

Explanation:R=pL/A

R second wire=p2L/1/2A

=4R

The three rules of refraction for converging lenses, also known as convex lenses, are as follows:

1. The incident ray parallel to the principal axis refracts through the lens and passes through the focal point on the opposite side of the lens.

2. The incident ray passing through the focal point refracts and emerges parallel to the principal axis.

3. The incident ray passing through the center of the lens continues undeviated.

The three rules of refraction for converging lenses, also known as convex lenses, are as follows:

1. The incident ray parallel to the principal axis refracts through the lens and passes through the focal point on the opposite side of the lens.

When a ray of light traveling parallel to the principal axis (the horizontal line passing through the center of the lens) enters a converging lens, it refracts and bends toward the principal axis. After refraction, the ray passes through the focal point on the opposite side of the lens. This rule is essential in understanding the behavior of light rays in converging lenses.

2. The incident ray passing through the focal point refracts and emerges parallel to the principal axis.

When a ray of light passes through the focal point before entering the lens, it refracts and emerges parallel to the principal axis. This rule implies that light rays that enter the lens at different angles converge and meet at the focal point on the opposite side.

3. The incident ray passing through the center of the lens continues undeviated.

When a ray of light passes through the center of a lens, it continues in a straight line without deviating from its path. This rule is derived from the fact that the refractive index of the lens material is the same throughout, resulting in no change in the direction of light passing through the center of the lens.

These rules of refraction for converging lenses help in understanding how light rays interact with the lens and how the lens focuses incoming light. By applying these rules, one can predict the behavior of light rays and determine the location and characteristics of images formed by converging lenses.

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Answer: low = 1.73 m and high = 0.129 m

Explanation: All you have to do is use the formula (Wavelength = Velocity/Frequency. Which would be 340m/s divided by 196 Hz and then 340m/s divided by 2637Hz.

The wavelength of a wave is its speed divided by frequency. The wavelength corresponds to the lower frequency of 196 Hz is 1.73 m and that for 2637 Hz is 0.12 m. Hence the wavelength range is 1.73 m to 0.12 m.

Frequency of a wave is the number of wave cycles per unit time. It is the inverse of the time period. Thus, its unit is s⁻¹ which is equal to Hz. Frequency of a wave is inversely proportional to the wavelength.

The relation between speed and frequency with wavelength of the wave is given by,

c = νλ

Given frequency  ν1 = 196 Hz.

speed of sound wave c = 340 m/s

then, wavelength at this frequency λ1 = 340 m/s / 196 Hz = 1.73 m.

For a frequency ν2 = 2637 Hz.

λ2 = 340 m/s/ 2637 Hz = 0.12 m.

Therefore, the range of wavelength of the notes from the violin will be in between 1.73 m to 0.12 m.

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a.  The force on the +3.0 mC charge is 1.35 x 10^-6 N.

b.   The electric field at (x, y) = (1.00 m, 0.50 m) is  5.49 x 10^5 N/C

c. The point where the bc is zero (other than at infinity) is located on the line that connects the two charges and at the midpoint of the line segment connecting them.

d.  The potential at the point found in part c is at point zero

An electric field is described as the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them.

a.  we need to use Coulomb's law, which states that the force between two charges is given by:

F = k * (q1 * q2) / r^2

F = k * (q1 * q2) / r^2 = 8.99 x 10^9 Nm^2/C^2 * (1.0 x 10^-3 C * 3.0 x 10^-3 C) / (2.0 m)^2 = 1.35 x 10^-6 N

b. To calculate the electric field at (x, y) = (1.00 m, 0.50 m), we use the equation for the electric field due to a point charge:

E = k * q / r^2

The distance from the point charge is:

r = sqrt((1.00 m - 2.0 m)^2 + (0.50 m - 0 m)^2) = sqrt(3.25 m^2) = 1.8 m

Therefore, the electric field at (x, y) = (1.00 m, 0.50 m) is:

E = k * q / r^2 = 8.99 x 10^9 Nm^2/C^2 * (3.0 x 10^-3 C) / (1.8 m)^2 = 5.49 x 10^5 N/C

c. This point is located on the line that connects the two charges and at the midpoint of the line segment connecting them.

d.  At the point found in c, the electric field is zero, so the work done to move a test charge from infinity to that point is also zero.

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An item at rest will remain at rest, and an object in motion will continue to move in a straight path at a constant speed, according to the first law of motion, commonly known as the law of inertia.

Newton's laws of motion govern how a baseball moves as a result of being thrown or struck. According to Newton's first law, a moving ball will continue to move in a straight line until other forces are acting on it.

The ball is severely distorted by the enormous force the bat applies to it ball being struck. The average force acting during the bat-ball collision is therefore about two tons, with a peak force of nearly four tons, during the 0.7 millisecond contact time. There's a lot of force there!

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Here

According to first equation of kinematics.

\(\\ \sf\longmapsto v=u+at\)

\(\\ \sf\longmapsto v=0+at\)

\(\\ \sf\longmapsto v=at\)

\(\\ \sf\longmapsto v\propto t\)

Answer:

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Answer:

The density of the wooden cube is   \(\delta= 0.4 \, \frac{g}{cm^3}\)

Explanation:

Recall that the weight of the entire piece of wood that is floating must equal the weight of the displaced water in the case we have an object under flotation.

At the same time. we have a relationship between Volume (V), Weight (w) and density \((\delta)\)  given by the formula:

\(w=\delta\,*\,V\)

Since both weights must be the same, we can write the following equation:

\(w_{wood} = w_{water}\\\delta *\,(5\,\,cm)^3= (\frac{1\,g}{cm^3} )\,(2\,*5^2) cm^3\\\delta= \frac{2}{5} \, \frac{g}{cm^3} \\\delta= 0.4 \, \frac{g}{cm^3}\)

Answer:

Positive zero error: When the jaws of the caliper are closed, the zero mark on the vernier scale is to the right of the zero mark on the main scale. This means that the caliper reads a value greater than the actual value, leading to a positive zero error.

Negative zero error: When the jaws of the caliper are closed, the zero mark on the vernier scale is to the left of the zero mark on the main scale. This means that the caliper reads a value less than the actual value, leading to a negative zero error.

Answer:

Friction is the resistance of motion when two objects rub against each other.

Explanation:

Hope this helps.

Every action have equal and opposite reaction. for example when we fire bullet from a gun, the gun will recoil back and bullet moves forward. In case of rocket, rocket is fired, thrust is reaction of force applied by the gas on the floor.

A push or a pull that an object experiences as a result of interacting with another item is known as a force. Interactions result in forces! As was covered in Lesson 2, certain forces are the result of contact interactions (reaction, frictional, tensional, and applied forces are examples of contact forces), whilst other forces (gravitational, electrical, and magnetic forces) are the consequence of action-at-a-distance interactions. Newton postulated that whenever objects A and B interact, they exert forces on one another. You put a downward force on the chair when you sit on it, and the chair responds by exerting an upward force on your body. This contact creates two forces: one force on the chair and one force on your body.

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Answer:

0.514 m/s

Explanation:

The knot is a unit of nautical speed used in maritime navigation.

1 knot is equal to one nautical mile per hour,

1 nautical mile per hour = 1.852 km/h

The basic SI system of units of speed is in 'm/s'

1.852 km/h = (1.852 x 1000)/3600 = 0.514 m/s

Answer:

a)  ρ = 6.25 10⁵ μg / m³, b) ρ  = 1 10⁷ μg / m³

Explanation:

Let's analyze the exercise a little before starting, we must know the amount of pollutant in the box, that the one that enters less the one that degrades and with this value find the density or concentration.

Let's start by finding the volume of air that goes into the box

               V = Lh x

Let's find the distance of air that enters per unit of time, as it goes at constant speed

               x = v₀ t

we substitute

               V₀ = Lh v₀ t

At this same time, a quantity of pollutant is distributed

              Q₀ = r t  

the contaminant that is entering reaches the entire box, therefore the total amount of contaminant is

               Q = Qo t

we substitute

               Q = r t²

the net amount of pollutant that remains is that less enters the one that degraded in the same time, as they ask for the steady state

              \(Q_{net}\)= Q - k t

the pollutant concentration is

              ρ = Q_net / V

              V = L L h

              ρ =\(\frac{r \ t^2 - k \ t}{ L^2 h}\)

              ρ = \((r \frac{ L^2}{v_o^2} - k \frac{L}{v_o} ) \frac{1}{L^2 h}\)

               ρ = \(\frac{r}{ v_o h} -\frac{k}{v_o L h}\)

let's reduce the magnitudes to the SI system

           r = 10 kg / s

           L = 100 km = 100 10³ m

           h = 1 km = 1 10³ m

           k = dq / dt = 0.20 1/h ( 1h/3600 s) = 5.5555 10⁻⁵  1/s

           v₀ = 4 m / s

let's calculate

The volume of the box

             V = (100 100 1) 109

             V = 1 10¹³ m³

            ρ = \(\frac{10}{ 4^2 \ 1\ 10^3 } - \frac{5.5556 \ 10^{-5}}{ 4 \ 100 \ 10^3 1 \ 10^3}\)

            ρ = \(6.25 10^{-4} - 1.389 ^{-13}\)

            ρ = 6.25 10⁻⁴ kg / m³

let's reduce to μg / m³

               ρ = 6.25 10⁻⁻⁴ kg / m³ (10⁹ μg / 1kg)

               ρ = 6.25 10⁵ μg / m³

b) in case the air speed decreases to v₀ = 1 m / s

             ρ= \frac{10}{ 1^2 \  1\  10^3 } - \frac{5.5556 \ 10^{-5}}{ 1 \ 100 \ 10^3  1 \ 10^3}

             ρ = 1 10⁻² - 5.5556 10⁻¹³

             ρ =  1 10⁻² kg / m³

             ρ  = 1 10⁷ μg / m³

The skater's final angular velocity is approximately 9.86 rad/s.

The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Initially, the skater has an angular momentum of:

L_initial = I_initial * ω_initial

Substituting the given values:

L_initial = 2.12 kg m² * 3.25 rad/s

The skater's final angular momentum remains the same, as angular momentum is conserved:

L_final = L_initial

The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:

L_final = I_final * ω_final

0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s

Solving for ω_final:

ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²

Hence, the skater's final angular velocity is approximately 9.86 rad/s.

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When the teacher asked the student to make words out of the jumbled word gzeysktqix, the student was being tested on his ability to unscramble words. Unscrambling words is the process of taking a word or series of letters that are out of order and rearranging them to form a word that makes sense.

When trying to unscramble a word, it is important to look for any patterns that can help identify smaller words within the jumbled letters. This can help make the process easier and quicker. For example, in the jumbled word gzeysktqix, one might notice that the letters "sktqix" appear together.

This could indicate that these letters could potentially form a word. By looking at the remaining letters, one could notice that the letters "g", "z", "e", and "y" could also form smaller words. After some rearranging, the letters can be unscrambled to form the words "sky", "zig", "sex", and "yet". These are just a few examples, as there are likely many other words that can be formed from this jumbled word.

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Answer: Conversation of Matter

Explanation: According to Quizlet, Converstion of Matter is the answer.