A circular curve of highway is designed for traffic moving at 60 km/h. Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is 150 m, what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at 60 km/h?

Answer:

The charges on the spring are 1.23E-5 C and they have the same sign

Explanation:

Given

Coulomb laws states that the force exerted by a charge q on another charge Q at a distance r is given by

F = kqQ/r²

Where k = 8.99 * 10^9 Nm²/C²

r = 0.43 + 0.013

r = 0.443m

The force on the spring is calculated as;.

F = kx where x is the stretch length of the spring and k is the spring constant

The force acting on the spring = 238 * 0.013

F = 3.094N

By comparison;

F = kqQ/r² becomes

3.094 = F = kqQ/r²

kqQ/r² = 3.094 (Considering that a = Q)

kq²/r² = 3.094

8.99 * 10^9 * q²/0.443 = 3.094 -- make q² the subject of formula

q² = 3.094 * 0.443/8.99*10^9

q² = 1.524629588431E−10

q = √1.524629588431E−10

q = 0.000012347589191545C

q = 1.23E-5 C

The charges on the spring are 1.23E-5 C and they have the same sign

Answer:

[tex]\boxed {3.43 x 10^{3}}[/tex]

Explanation:

We know that speed is defined as distance moved per unit time hence expressed as [tex]v=\frac {d}{t}[/tex] where v is speed in m/s, d is distance in m and t is time in seconds. Making d the subject of the above formula then

[tex]d=vt[/tex]

Substituting 343 m/s for d and 10 s for t then

[tex]d= 343\times10= 3430= 3.43 x 10^{3}[/tex]

Therefore, the distance between speaker and deter is [tex]\boxed {3.43 x 10^{3}}[/tex]

Answer:

The magnitude of the final velocity is 14.14 m/s

Explanation:

The horizontal component of the final velocity and vertical component of the final velocity, forms a perpendicular vector. To determine the magnitude of the final velocity, we sum the two vectors.

To add two perpendicular vector, Pythagoras principle is used.

[tex]V^2 = V_x^2 +V_y^2\\\\V = \sqrt{V_x^2 +V_y^2}\\\\V = \sqrt{(10)^2 +(-10)^2} =14.14 m/s[/tex]

The magnitude of the final velocity is 14.14 m/s

Explanation:

The potential difference of 180 V is applied across resistance = 8.4 ohm

Thus the maximum current flowing I₀ = [tex]\frac{V}{R}[/tex] = [tex]\frac{180}{8.4}[/tex] = 21.4 A

The rms value of the current is = [tex]\frac{I_0}{\sqrt{2} }[/tex]  =  I₀ x 0.7 = 15 A

This value of current is greater than the 12.5 A . Thus the circuit will break .

Answer: Nitrogen gas

Explanation:

Using ideal Gas's law

PV = nRT

where

Pressure of gas, P= 1atm

Volume of gas, V= 5.1L

no of moles of gas, n=

Ideal gas constant, R= 0.0821

Temperature of gas, T= 20°C = 20+273 = 293K

also, n= (mass/molar mass)

mass of the gas m = 5.9g

Molar mass of the gas = ?

So, PV = (mRT/M)

We're looking for molar mass M, then

M = mRT/PV

M = (5.9 * 0.0821 * 293)/(1 * 5.1)

M = 141.93/5.1

M = 27.8g/mol ~ 28g/mol

Since the gas is diatomic, then we say,

Atomic mass of gas = 1/2 * molar mass

Atomic mass = 1/2 * 28

Atomic mass = 14

Therefore, the gas is nitrogen.

Explanation:

When a radioactive substance decays then the fast moving electrons emitted by it is known as beta ray. Basically, a number of beta particles are ejected by a beta ray.

Symbol of a beta particle is [tex]^{0}_{-1}e[/tex]. A beta ray is a natural decay of a radioactive element. As we know that opposite charges get attracted towards each other. So, a beta ray gets attracted towards a positively charged plate.

Therefore, we can conclude that following are the characterizes a beta ray:

Final answer:

Beta rays are negatively charged, attracted to the positively charged plate in an electric field, and are composed of electrons. They are a product of natural radioactive decay and are lighter and much less massive than alpha particles.

Explanation:

Beta rays are characterized by specific properties that distinguish them from other types of radiation produced during natural radioactive decay. According to Ernest Rutherford's research, which involved observing the behavior of radiation in magnetic and electric fields, beta particles are known to be negatively charged and relatively light. This means they are attracted to the positively charged plate in an electric field and are significantly deflected due to their lighter mass compared to alpha particles.

Beta rays are not electromagnetic radiation; this term is reserved for gamma rays, which are uncharged and therefore unaffected by electric fields. Beta rays are indeed a product of natural radioactive decay, specifically during a process known as beta decay, in which a nucleus emits an electron or a positron. Since they are composed of high-energy electrons, the identification that beta rays are composed of electrons is also correct.

The maximum mass of block B, for which block A remains at rest on a 35-degree incline given a static friction coefficient of 0.40 and a 10kg mass of block A, is approximately 1.98 kg.

To solve this question, we will need to use the principles of static friction, inclined planes, and gravitational force. Static friction is what keeps block A from sliding down the incline. It has to overcome the downward force due to gravity on block A which is a component of the weight of block A acting downwards the inclined plane.

To determine the maximum mass of block B, we can equate the static frictional force to the net force acting downward on the inclined plane. The static frictional force is [tex]\mu N[/tex] where μ is the coefficient of static friction and N is the normal force. N = [tex]mAgcos\theta[/tex] where mA and g are the mass and acceleration due to gravity respectively and θ is the angle of the inclined plane. So, static frictional force = [tex]\mu mAgcos\theta[/tex]. The downward force is the sum of components of the weight of block A acting downwards and the weight of block B. So, [tex]mAgcos\theta (\mu)[/tex] = mAg*[tex]sin\theta[/tex] + mBg.

From this equation, you can solve for the mass of Block B: MB = [tex]mA\mu cos\theta[/tex] - mA*[tex]sin\theta[/tex].

Plugging in the given numbers, we get:

MB = 10(0.4*cos(35 degrees) - sin(35 degrees)).

This gives us approximately 1.98 kg as the maximum mass of block B before block A begins to slide.

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To find the largest mass for block MB, we first need to calculate the gravitational force and the static friction on block A, using the given values for mass, gravitational acceleration, coefficient of static friction and incline angle. The tension in the string, created by block MB, has to balance these forces. By setting the net forces equal, we can calculate the value of MB.

The problem revolves around understanding the concept of static friction and forces acting on an inclined plane. Firstly, let's calculate the force due to gravity acting on block A. This is simply F_gravity = mg sin θ where m=10 kg is the mass of the block, g=9.8 m/s² is the gravitational acceleration, and θ is 35°. Next, we need to calculate the frictional force that prevents the block from sliding. Since block A is at rest, static friction is at work here, and we can use the formula F_friction = μN, where μ=0.40 (the coefficient of static friction) and N is the normal force acting on the block, which equals mg cos θ.

To keep block A at rest, the tension T in the string due to the dangling mass MB must balance both gravity and friction. Therefore: T=F_gravity + F_friction. The weight of the dangling mass brings about the tension in the string, and hence, T=MBg. From these, we can calculate the largest mass of MB for which A remains at rest.

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Answer:

0.247 J = 247 mJ

Explanation:

From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁) where m = mass of particle = 5.4 × 10⁻² kg, q = charge of particle = 5.10 × 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁)

          = 1/2 × 5.4 × 10⁻²kg × ((3m/s)² - (2 m/s)²) + 5.10 × 10⁻⁵ C(7850 - 5650)

          = 0.135 J + 0.11220 J

          = 0.2472 J

          ≅ 0.247 J = 247 mJ

Answer:

Layer 3 switch.

Explanation:

A layer 3 switch carrys out both the function of a switch and a router. It acts as a switch that links all the devices that are on the same subnet or virtual LAN at lightning speeds and has IP routing intelligence built into it to carry out the function of a router. It can support routing protocols, check incoming packets, and can also carry out routing decisions based on the source and the destination addresses.

Answer:

Water molecules attracting other water molecules is called cohesive attraction.

Explanation:

They are basically two forces in liquids that determine their wetting characteristics, they are cohesive and adhesive forces.

Cohesion is the attraction between molecules of same liquid example water and water, while adhesion is attraction between molecules of different liquids example alcohol and water.

Therefore, Water molecules attracting other water molecules is called cohesive attraction.

The Potential energy stored in the system is 1 J

Explanation:

Given-

Mass, m = 4 kg

Spring constant, k = 800 N/m

Distance, x = 5cm = 0.05m

Potential energy, U = ?

We know,

Change in potential energy is equal to the work done.

So,

[tex]U = \frac{1}{2} k (x)^2\\\\[/tex]

By plugging in the values we get,

[tex]U = \frac{1}{2} * 800 * (0.05)^2\\ \\U = 400 * 0.0025\\\\U = 1J\\[/tex]

Therefore, Potential energy stored in the system is 1 J

Answer:

3. It is a passive process in which molecules move from a region of higher concentration to a region of lower concentration.

Explanation:

Diffusion can be defined as the movement of solute or gaseous molecules from the region in which they have higher concentration to the regions in which they have lower concentration until they become evenly distributed across the two regions.

Diffusion is a passive process, that is, it requires no energy.

Option 3 is the correct option.

Answer:

It is a passive process in which molecules move from a region of higher concentration to a region of lower concentration

Explanation:

Diffusion is the movement of a substance from an area of high concentration to an area of low concentration. It is an important process for living things.

A single substance tends to move from an area of high concentration to an area of low concentration until the concentration is equal across a space

The total distance travelled by the ball after the fourth impact is 275 feet.

Explanation:

Given-

Height, h = 100 feet

Rebounds half the distance

Distance in feet for the fourth time, x = ?

For the first time, the distance travelled by the ball is, x = 100 feet

For the second time, it will bounce up to 50 feet and fall upto 50 feet( half of 100 feet)

So, the distance travelled after the second impact, x = 100 + 50 + 50 = 200 feet

For the third time, it will bounce up to 25 feet and fall upto 25 feet( half of 50 feet)

So, the distance travelled after the third impact, x = 200 + 25 + 25 = 250 feet

For the fourth time, it will bounce up to 12.5 feet and fall upto 12.5 feet( half of 25 feet)

So, the distance travelled after the fourth impact, x = 250 + 12.5 + 12.5 = 275 feet

Therefore, total distance travelled by the ball after the fourth impact is 275 feet.

Answer:

1.44 s

Explanation:

Since it is a projectile motion, we use the formula for the total time of flight,t

t = 2Usinθ/g where U = initial velocity of ball = 10 m/s, θ = 45 and g = 9.8 m/s²

t = 2Usinθ/g = 2 × 10sin45/9.8 = 1.44 s

So, the time needed for the ball to return to the ground is most nearly: 1.44 s

We have that for the Question it can be said that the time needed for the ball to return to the ground is most nearly

T=1.44

From the question we are told

A ball is thrown into the air with an initial velocity of 10 m/s at an angle of 45 degrees above the horizontal, as represented above. If air resistance is negligible the time needed for the ball to return to the ground is most nearly

Generally the Newton's equation for the vertical displacement  is mathematically given as

[tex]y=ut+1/2at^2\\\\Therefore\\\\T=\frac{2Usin\theta}{g}\\\\T=\frac{2*10*sin45}{9.8}\\\\[/tex]

T=1.44

Therefore

the time needed for the ball to return to the ground is most nearly

T=1.44

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Explanation:

Both distributions describe the number of times an event occurs in a givn number of trials. In the binomial distribution, the probability is the same for each trial. While in the hypergeometric distribution, each trial changes the probability of each subsequent trial, since there is no replacement.

Explanation:

The dart is project with 150 m/s from a point at an angle of 30⁰

The vertical component of velocity = 150 sin 30 = 75 m/s

Thus initial vertical velocity is = 75 m/s

The velocity after 4 s can be calculate by

v = u - g t

here u is the initial velocity and t is the time , g is the acceleration due to gravity .

Thus v = 75 - 10 x 4 = 35 m/s

The velocity after 4.0 seconds will be "35 m/s".

According to the question,

Speed,

Angle,

Vertical component of velocity,

After 4 seconds, the velocity will be:

→ [tex]v = u-gt[/tex]

By substituting the values, we get

     [tex]= 75-10\times 4[/tex]

     [tex]= 75-40[/tex]

     [tex]= 35 \ m/s[/tex]

Thus the answer above is right.    

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Answer:

0.6m/s

Explanation:

Totally inelastic collision

m₁v₁ + m₂v₂ = ( m₁ + m₂)v(final)

Where v (v₁ and v₂) is the initial velocity of the objects

v(final) is the  final velocity of the objects stuck together

Toy car a , m₁ = 3kg, v₁ = 1m/s

Toy car b , m₂ = 2kg, v₂ = 0m/s

m₁v₁ + m₂v₂ = ( m₁ + m₂)v(final)

3(1) + 2(0) = (3 + 2) v(final)

3 = 5 v(final)

v(final) = 0.6m/s

Explanation:

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Answer:

Valence shell electron pair repulsion theory

Explanation:

It is also called the Gillespie-Nyholm theory after its two main discoverers, Ronald Gillespie and Ronald Nyholm.

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Answer:

160N/m

Explanation:

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

k is the spring constant

e is the extension

From the formula k = F/e

Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.

The spring constant k = ma/e where

m is the mass of the block = 0.4kg

a is the acceleration = 8.0m/s²

e is the extension of the spring = 2.0cm = 0.02m

K = 0.4×8/0.02

K = 3.2/0.02

K = 160N/m

The spring constant of the spring is therefore 160N/m

Final answer:

The spring constant of the spring is calculated using Hooke's Law and Newton's second law of motion. By multiplying the mass of the block by its acceleration, we found the force, and then divided the force by the displacement to get the spring constant, which is 160 N/m.

Explanation:

To determine the spring constant of the spring, we need to apply Hooke's Law, which states that the force exerted by an ideal spring is directly proportional to its displacement from the equilibrium position (F = -kx), where 'F' is the force, 'k' is the spring constant, and 'x' is the displacement. Since the block is on a frictionless surface and we know the acceleration (a = 8.0 m/s2) and the mass (m = 0.40 kg), we can first find the force using Newton's second law (F = ma), and then use that force to calculate the spring constant 'k'.

The force exerted by the spring can be calculated as:

F = m * a
= 0.40 kg * 8.0 m/s²
= 3.2 N

The displacement (x) from the equilibrium position is given as 2.0 cm, which is 0.020 m in SI units. Using Hooke's Law, the spring constant can be calculated:

k = F / x
= 3.2 N / 0.020 m
= 160 N/m

Answer: non-shifting lines indicate non moving star.

Explanation: when a star is moving toward the detector, the wavelength will decrease - there will be a blue shift.

When it's moving away from the earth or detector, the wavelength will increase - there will be a red shift.

Identifiable patterns of absorption lines that appear shorter or longer wavelengths than normal indicate that the star is moving

In visualisation, the bottom spectrum shows the normal position of absorption line for a star that is not moving toward or away from the earth.

Non-shifting lines in the binary star system's spectrum are attributed to absorption by interstellar clouds, which unlike the stars, do not move relative to us. The narrowness of these lines indicates the low pressure of the absorbing gas.

When observing a binary star system and noting the Doppler Effect in the spectral lines, if certain lines do not shift, it indicates they originate from something that is not moving with respect to us. Most of the absorption lines shift due to the motion of the binary stars, which causes Doppler shifts as the stars move toward or away from us. This movement results in the spectral lines being blue-shifted when the star is approaching us, and red-shifted when it's receding. However, the lines that remain constant are likely due to the absorption by interstellar clouds located between Earth and the stars. The non-shifting lines are also much narrower, suggesting that the absorbing gas is at a very low pressure. This non-movement of specific lines helped in discovering the presence of interstellar materials, as their spectral lines do not participate in the Doppler shifts associated with the stars' orbits.

Answer:

The boat moves 0.244 m towards the end where the 59 kg person was at the start of the calculations.

Explanation:

The boat only moves because the centre of mass changes a bit if the two people on opposite ends of the boat exchange seats.

The boat moves a distance of the change in centre of mass

Noting that the weight of the boat acts at the centre of the boat at 1.45m from both ends.

For convention, we call the original position of the 59 kg person as x=0

This means,

59 kg person is at x = 0 m

88 kg of the boat acts at x = 1.45 m from the end of the 59 kg person.

78 kg person is at x = 2.90 m

Centre of mass = X = (Σ mᵢxᵢ)/(Σ mᵢ)

For the initial setup,

X = [(59×0) + (88×1.45) + (78×2.90)]/(59+88+78)

X = (353.8/225)

X = 1.572 m

(Don't forget that this is 1.572 m from the end we designated x=0 m)

When the people exchange positions,

59 kg person is now at the other end of the boat with x = 2.90 m

88 kg of the boat still acts at the centre of the boat at x = 1.45 m

And 78 kg person is now at the end of the boat with x = 0 m

Centre of mass = X = (Σ mᵢxᵢ)/(Σ mᵢ)

X = [(59×2.90) + (88×1.45) + (78×0)]/(59+88+78)

X = (298.7/225)

X = 1.328 m

(This is 1.328 m from the end we designated x=0 m from the start)

So, the centre of mass moves a distance of (1.572 - 1.328) towards the end of the boat we designated x=0 m from the start.

Hence, the boat moves 0.244 m towards the end where the 59 kg person was at the start of the calculations.

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Answer:

The wire should be tightened because the present tension is 532.54 N where the required tension is 800 N and the higher the tension the more tightening is required.

Explanation:

To solve the question

v = [tex]\sqrt{\frac{F_t}{\mu} } = \sqrt{\frac{L*F_t}{m} }[/tex] where

v = velocity of the pulse in the string = 46.154 m/s

[tex]F_t[/tex] = Required tension force = 800 N

m = Mass of the wire = 15 kg

L = length of the wire to be tension-ed = 60 m

Since the pulse travels twice the distance of 60 m in 2.6 s the velocity is given by

v = 2×60/2.6 = 46.154 m/s

Therefore making [tex]F_t[/tex] the subject of the formula and substituting the values, we have

[tex]F_t[/tex] = [tex]\frac{v^2m}{L}[/tex] =[tex]\frac{46.154^{2*15} }{60}[/tex] = 532.54 N

This means that, as it is, the present tension in the wire is 532.54 N which is less than  the required 800 N, therefore the wire should be tightened

The wire should be tightened because the current tension on the wire is less than the required tension.

The given parameters;

The speed of the wave as the pulse traveled from one pole to the other two times, is calculated as follows;

[tex]v = \frac{2d}{t} \\\\v = \frac{2 \times 60}{2.6} \\\\v = 46.154 \ m/s[/tex]

The tension created on the wire during the pulse motion is calculated as follows;

[tex]v = \sqrt{\frac{T}{m/L} } \\\\v ^2 = \frac{TL}{m} \\\\T = \frac{v^2 m}{L} \\\\T = \frac{(46.154)^2 \times 15}{60} \\\\T = 532.55 \ N[/tex]

The current tension on the wire (532.55 N) is less than the required tension of 800 N. Thus, the wire should be tightened.

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Answer:

(a). The ball's centripetal acceleration is [tex]16.17\times10^{2}\ m/s^2[/tex]

(b). The magnitude of the net force is 232.9 N.

Explanation:

Given that,

Mass of baseball = 144 g

Speed = 81 mph = 36.2 m/s

Distance = 81 cm

(a). We need top calculate the ball's centripetal acceleration just before it is released

Using formula of centripetal acceleration

[tex]a=\dfrac{v^2}{r}[/tex]

Where, v = speed

r  = radius

Put the value into the formula

[tex]a=\dfrac{(36.2)^2}{81\times10^{-2}}[/tex]

[tex]a=1617.82\ m/s^2[/tex]

[tex]a=16.17\times10^{2}\ m/s^2[/tex]

(b). We need to calculate the magnitude of the net force that is acting on the ball just before it is released

Using formula of force

[tex]F=\dfrac{mv^2}{r}[/tex]

Put the value into the formula

[tex]F=\dfrac{144\times10^{-3}\times(36.2)^2}{81\times10^{-2}}[/tex]

[tex]F=232.9\ N[/tex]

Hence, (a). The ball's centripetal acceleration is [tex]16.17\times10^{2}\ m/s^2[/tex]

(b). The magnitude of the net force is 232.9 N.

Answer:

Main sequence

Giant

Planetary nebula

White dwarf

Black dwarf

Explanation:

Main Sequence:

Stars are called main sequence stars when their core temperature reaches up to 10 million kelvin and their start the nuclear fusion reactions of hydrogen into helium in the core of the star. For example sun is known as to be in the stage of main sequence as the nuclear fusion reactions are happening in its core.

Giant:

Next step is the Giant phase. When the stars run out of their fuel that is hydrogen for the nuclear fusion reactions then they convert into Giant stars. Giant stars have the larger radius and luminosity then the main sequence stars.

Planetary nebula:

Planetary nebula consists of glowing gases and plasma, it ejects from the red giant stars that run out of their fuel.

White dwarf:

When the stars run out of their fuel then they shed the outer layer planetary nebula, the remaining core part that left behind is called as white dwarf. It's the most dense part as the most of the mass is concentrated in this part.

Black dwarf:

When the white dwarf cool down completely that it no longer emitt heat and light then it is called as black dwarf.

These were the possible stages that includes in the evolution of stars

Answer: 65mph, 75mph

Explanation:

Let us assume x to be the speed of the slower train, in mph (miles per hour).

Then the speed of the other train is (x+10) mph, according to the question.

We then would have an equation like this

1.6x + 1.6(x+10) = 224.

This is because, the first addend in the left side is the distance covered by the slower train.

The second addend in the left side is the distance covered by the faster train.

The sum is 224 miles, because they together covered all the distance to the moment when they meet each other.

1.6x + 1.6x + 16 = 224

3.2x + 16 = 224

3.2x = 224 - 16

3 2x = 208

x = 208/3.2

x = 65

Thus the speed of the slower train is 65mph, and that of the other train is 65 + 10 = 75mph

Final answer:

By setting up and solving a linear equation, the speed of the slower train is determined to be 65 mph, while the faster train's speed is 75 mph as one travels 10 mph faster than the other.

Explanation:

To solve the problem of the two trains travelling on parallel tracks towards each other, we must first define the variables for their speeds. Let's say one train travels at x mph and the other at (x + 10) mph. Since they are moving towards each other, you can add their speeds together to find how fast the distance between them is closing. The total closing speed is then x + (x + 10) mph, which simplifies to 2x + 10 mph.

The total distance to be covered by the two trains until they pass each other is 224 miles. We know they pass each other after 1.6 hours, so using the formula Distance = Speed x Time, we can set up the equation: 224 miles = (2x + 10 mph) x 1.6 hours. Solving for x gives us the speed of the slower train.

Performing the algebraic steps:

224 = (2x + 10) x 1.6

224 = 3.2x + 16

208 = 3.2x

x = 65 mph.

Therefore, the speed of the slower train is 65 mph and the speed of the faster train is 75 mph.

Answer:

Explanation:

Given

balloon is rising with a speed of [tex]u_y=6\ m/s[/tex]

Person throws a ball out of basket with a horizontal velocity of [tex]u_x=10\ m/s[/tex]

Considering upward direction to be positive

When ball is thrown it has two velocity i.e. in upward direction and in horizontal direction so net velocity is

[tex]v_{net}=\sqrt{(u_x)^2+(u_y)^2}[/tex]

[tex]v_{net}=\sqrt{(6)^2+(10)^2}[/tex]

[tex]v_{net}=\sqrt{36+100}[/tex]

[tex]v_{net}=\sqrt{136}[/tex]

[tex]v_{net}=11.66\ m/s[/tex]

Direction of velocity

[tex]\tan \theta =\dfrac{u_y}{u_x}[/tex]

[tex]\tan \theta =\dfrac{6}{10}[/tex]

[tex]\theta =30.96^{\circ}[/tex]

where [tex]\theta [/tex] is angle made by net velocity with horizontal .

THE ANSWER IS : THERE ARE ONLY ABOUT 100 DIFFERENT KINDS OF ATOMS THAT COMBINE TO FORM ALL SUBSTANCES

Explanation:

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude [tex]g[/tex].

If we have the initial velocity [tex]v_o[/tex] and its angle [tex]\theta[/tex], we can obtain the vertical component of the velocity [tex]v_{oy}[/tex] using trigonometry:

[tex]v_{oy}=v_osin\theta[/tex]

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

[tex]v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }[/tex]

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

[tex]g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}[/tex]

Finally, from the equation of horizontal motion with constant speed, we have that:

[tex]x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}[/tex]

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

[tex]v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m[/tex]

In words, the projectile travels 7.49m horizontally before it lands.

Answer:

Option B, it is strongly attracted

Explanation:

A Test Charge Determines Charge on Insulating and Conducting Balls, and the points made regarding conductors, it can be ascertained that in conductors, the electrons are free to move about. This means that when a charge is brought near to a conductor, the opposite charges all navigate to the sharpest point closest the charge and a strong attraction is created.

This shows that the rod will be strongly attracted. The density of the charges on the rod is mostly concentrated at the sharpest point.

End A of the rod will be strongly attracted to the negatively charged metal ball because of the process of charge induction, where opposite charges attract.

When the negatively charged metal ball is brought near to end A of the rod, end A of the rod will be strongly attracted to the negatively charged ball. This is because of the principle of charge induction. When a charged body is brought near to another body, it will cause the charges in that body to redistribute. Opposite charges attract, so the near side of the rod (end A) will have a positive charge induced on it, and this positive charge will be attracted to the negative charge on the ball. So, the correct answer is option b. It is strongly attracted.

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