How does your education contribute to comunity development

Answer:

2

Explanation:

Answer:

The answer is B

Explanation:

Answer:

a) 3.66 s

b) 124.4 m

c) 3.12s

Explanation:

Given that

Speed of the Red Car, v₁ = 34 m/s

Speed of the Blue Car, v₂ = 28 m/s

Distance between the two cars, d = 22 m

The difference between the speed of the cars is: 34 - 28 = 6 m/s

From this, we can deduce that the red car will be catching up to the blue car at a speed of 6 m/s.

1)

If we divide the distance by the difference in speed. This becomes

d/v = 22/6 = 3.66 s. Which means, It takes 3.66 seconds for the red car to meet up with the blue car.

2

From the previous part, we were able to confirm that it took 3.66 seconds for the red car to meet up the blue car. Also, the speed with which it were traveling was travelling at, was constant, so we only need to multiply it by the time from (1) to get the distance.

v = d * t = 34 * 3.66 = 124.4

Therefore the red car travels at 124.4 m before catching up to the blue car.

3

If the red car starts to accelerate the moment we see it, the time it will take, to get to the blue car will be less than what we had gotten. We can find this using one of the equations of motion.

S = ut + ½gt², where

S = 22

u = 6

t = ?

g = 2/3

22 = 6t + 1/3t²

By using the quadratic formula, we find out the two answers listed below

t1 = 3.12 s

t2 = - 21.12 s

We all know that negative time is not possible, so the answer is t1. At 3.12 seconds

Answer:

Lower

Lower

gsintheta (gsinθ)

Explanation:

The sum of forces resolved parallel to the inclined plane is given by;

F - mgsinθ = 0

ma - mgsinθ = 0

ma = mgsinθ

a = gsinθ

Acceleration is proportional to angle of inclination, thus the lower the angle of the slope, lower the acceleration along the ramp.

therefore, the speed at the bottom of a slope will be lower, (velocity is directly proportional to acceleration) and, consequently, the control will be better.

The acceleration along the ramp, is gsintheta (gsinθ)

Answer:

C

Explanation:

Measuring your weight with a scale that measures to the nearest

tenth of a kg but adds 2 kg to each measurement

Answer:

A.) 22.12 m/s

B.) 3.27 m

C.) 0.57

Explanation:

Given that a mass of 0.4 kg is dropped from rest in a medium offering a resistance of 0.2|v|, where is measured in meters per second. Acceleration due to gravity is 9.8 m/s^2.

a. If the mass is dropped from a height of 25 m, its initial velocity U will be equal to zero. when it hits the ground, the final velocity V will be:

V^2 = U^2 + 2gH

Substitute g and the height into the formula

V^2 = 2 × 9.8 × 25

V^2 = 490

V = 22.12 m/s

b. If the mass is to attain a velocity of no more than 8 m/s, the maximum height from which it can be dropped will be calculated by using the same formula.

V^2 = U^2 + 2gH

Where V = 8 m/s and U = 0

8^2 = 0 + 2 × 9.8 × H

64 = 19.6H

H = 64/19.6

H = 3.27 m

c. Suppose that the resistive force is k|v|, where v is measured in m/s, and k is constant. If the mass is dropped from a height of 25 m and must hit the ground with a velocity of no more than 8 m/s, determine the coefficient of resistance that is required.

The energy at 25m heigh will be:

E = mgh

E = 0.4 × 9.8 × 25

E = 98 J

1/2mv^2 = 98

1/2 × 0.4 × ( 8 - v )^2 = 98

0.2( 8 - v )^2 = 98

( 8-v)^2 = 98/0.2

( 8 - v )^2 = 490

8 - v = 22.12

v = 22.14 - 8

v = 14.14

Coefficient of resistance required will be 8/ 14.14 = 0.57

Answer:

27%

Explanation:

15.999 divided by 58.32 = .27433128

Move the decimal place over 2 places.

27%

Make use of percentages. Given that the molecular mass of magnesium hydroxide (MOH.) is 58.32 amu and the atomic mass of an oxygen atom is 15.999 amu, oxygen accounts for 27% of this compound.

Chemical substances known as molecular compounds assume the shape of distinct molecules. Examples include common compounds like carbon dioxide (CO2) and water (H2O) (Figure 3.1.1 ).

These substances differ significantly from ionic substances like sodium chloride (NaCl). When metal atoms give up one or more of their electrons to non-metal atoms, ionic compounds are created. The cations and anions that result

Therefore,

58.32 / 15.999 (27.3%) =.27433128

Add two places to the decimal place.

27%

The molar mass, which is stated in g/mol, is defined as the mass of a specific material divided by the amount of a substance. In light of this, the molar mass is average. Amu, short for atomic mass units, is the unit used to quantify molecular mass.

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Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

[tex]E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}[/tex]

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

[tex]r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm[/tex]

The magnitude of the electric field is now given as;

[tex]E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m[/tex]

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Answer:

more/ faster

Explanation:

I got it right on my test

Particles in a warmer block will have more kinetic energy and move faster than the particles in a colder block option (D) more/ faster is correct.

The energy which an item has as a function of motion is determined by the initial energy. It is described as the effort required to propel a material body from rest to the given velocity.

As we know,

As the temperature rises, the particles' kinetic energy rises and they begin to move more quickly. Particles have higher kinetic energy in gases than in solids.

Thus, particles in a warmer block will have more kinetic energy and move faster than the particles in a colder block option (D) more/ faster is correct.

The question is incomplete.

The complete question is:

Particles in a warmer block will have ________ kinetic energy and move ______than the particles in the colder block.

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The initial angular acceleration is 5.78 rad/s^2.

Since

A gymnast on the uneven parallel bars is at rest, tipped at a 45∘ angle from the vertical. The distance from her hands to her feet is 1.8 m.

So,

[tex]Torque \tau = r \times F\\\\ \tau = l\alpha\\\\so, r \times F = l\alpha\\\\mg\ cos\ 45 \times r = (1/3)mR^2 \times \alpha\\\\9.8\times cos\ 45 \times 0.9 = (1/3)\times (1.8)^2 \times \alpha\\\\\alpha = 5.78\ rad/s^2[/tex]

hence, The initial angular acceleration is 5.78 rad/s^2.

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Answer:

Average speed is  6m/s

Average velocity is 5.859m/s

Explanation:

Average speed, s = d/t

d is distant

t is time

speed north is 800m and 110s

speed due south is 400m and 90s  

Average speed = (800+400) / (110+90)

                          = 1200/200

                          =6m/s

Average velocity V = (v + u)/2

v = final velocity

u = initial velocity

V = (7.273 + 4.444)/2 = 11.717/2

V = 5.859m/s

Answer:

dail 911

tell problem and location

plz mark brainliest

Answer:

θ = 720º

Explanation:

Let's use the rotational kinematics relations for this exercise, remember that all angles must be given in radians.

Let's reduce the magnitudes to the SI system

         θ = 2 rev (2π rad / 1 rev) = 4π rad

we assume that the pulley has a radius r

linear and angular coordinates are related

         y = θ r

let's calculate

         y = 4π r

to convert the distance in angles to radians

        θ = y / r

        θ = 4π

let's change this angle to degrees

        π rad = 180º

         θ = 4π rad (180º /π rad)

         θ = 720º

Answer:

Same

Explanation:

Mass is the quantity of matter in a certain object.

WHEREVER you take a 2kg object, the mass will remain 2kg. All that changes is the Weight ..Weight the force which the centre of a Planet uses to pull everything towards itself.

On earth, it is 9.81 whereas on the Moon it is 1.6

Here, we are required to determine the mass of a 2kg object on the Earth and on the moon.

The mass of an object is simply defined as the quantity of matter present in the object and this quantity remains constant.

The mass of an object is simply defined as the quantity of matter present in the object and this quantity remains constant.In essence, if the mass of an object is 2kg, this mass remains constant on planet earth and the moon.

PS: The weight of an object which is a measure of the acceleration due to gravity which differs with location and position is the constant which may be different about the object on the earth and the moon respectively.

The acceleration due to gravity on earth and the moon is 9.8m/s² and 1.6m/s² respectively.

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Answer:

Yellow

Explanation:

It is for sure not red

Blackbody radiation graph for Star A, which has a surface temperature of 5,700 degrees Celsius. At the point where the curve peaks, the wavelength is approximately 550 nanometers, in the middle of the visible light spectrum. The color will Star A be Yellow.

A waveform signal that is carried in space or down a wire has a wavelength, which is the separation between two identical places (adjacent crests) in the consecutive cycles. This length is typically defined in wireless systems in meters (m), centimeters (cm), or millimeters (mm).

A blackbody's spectrum is continuous (it emits some light at all wavelengths), and it peaks at a particular wavelength. For hotter objects, the peak of the blackbody curve shifts to shorter wavelengths in a spectrum. If you conceive of a blackbody in terms of visible light, the hotter it is, the bluer its peak emission wavelength will be. For instance, the temperature of the sun is about 5800 Kelvin. With this temperature, a blackbody's peak is roughly 500 nanometers away, or the wavelength of yellow.

The color will Star A be Yellow.

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Answer:

Third Option: It is decreasing by 10m/s^2

Explanation:

When a ball is thrown in the air, it's fastest velocity occurs just as it leaves its source (e.g. the hand that throws it), and just before it hits the starting point. This is because this is when it has the most kinetic energy which translated into speed. As it rises, it starts to lose kinetic energy as it transforms into gravitational potential energy as the ball is fighting against the force of gravity as it moves further away from the Earth's surface. This means it begins to slow down until its speed is 0, its highest point, before falling and gaining kinetic energy again. Because of this, we can eliminate options 1, 2 and 5 as we know the ball's velocity decreases as it rises.

As the ball rises it's velocity decreases at a constant rate, as the amount of kinetic energy it has is proportional to its height. Therefore, the best option is option 3, as it states it is decreasing and gives a constant deceleration rate.

Hope this helped!

Answer:

The acceleration of the car is -4.166 m/s²

The time between applying the brakes and hitting the obstacle is 6.4 s

Explanation:

Given;

initial velocity of the car when the brake was applied, u = 31.3 m/s

final velocity of the car after hitting an obstacle, v = 4.65 m/s

distance traveled before stopping, s = 115 m

The acceleration of the car is given by;

v² = u² + 2as

2as = v² - u²

a = (v² - u²) / 2s

a = (4.65² - 31.3²) / (2 x 115)

a = -4.166 m/s²

The time between applying the brakes and hitting the obstacle;

v = u + at

at = v - u

t = (v - u) / a

t = (4.65 - 31.3) / -4.166

t = 6.4 s

Answer:

How big your foot is. (Big Brain)

Explanation:

Answer:

If Earth were to stop spinning on its axis, gradually the oceans would migrate towards the poles from the equator.  The atmosphere would still be in motion with the Earth's original 1100 mile per hour rotation speed at the equator. And this means, rocks, topsoil, trees, buildings,  and so on, would be swept away into the atmosphere.

Hope that helped <3

Answer:

It is gaining potential energy

Answer: it could also be gravitational potential energy

This is because as it rises in the air it will gain more energy to fall back down

Answer:

a

[tex]P =0.1094 \ m[/tex] past the goal post

Yes the Flyers won the match

Explanation:

Generally the distance covered by the pluck during the last 1.25 seconds is mathematically represented as

     [tex]D = ut + \frac{1}{2} * a * t^2[/tex]

=>  [tex]D =30  * 1.25 + \frac{1}{2} * (-0.5) * (1.25)^2[/tex]

=>   [tex]D =37.1094 \ m  [/tex]

Generally the position of the puck when the game clock reaches zero is mathematically represented as

     [tex]P = 37.1094 -37[/tex]

       [tex]P =0.1094 \ m[/tex] past the goal post

Given that the Bruins where one point down and Flyers scored another goal  it means that the flyer are now two point up hence they won the match

Answer:

13.72 m/s

Explanation:

Here, we are to determine the final velocity of the coconut. Applying first equation of free fall;

v = u + gt

where: v is the final velocity of the object, u is the initial velocity, g is the force of gravity and t is the time.

Since the coconut falls in the direction of gravity, then g = 9.8 m/[tex]s^{2}[/tex].

t = 1.4 s, u = 0 m/s

So that;

v = 0 + 9.8(1.4)

  = 13.72

The coconut hits Chief Boolie's toe with a velocity of 13.72 m/s

Answer:

a)     Eₓ = 0  ,   b)    c)    Ex_total = 3.96 10⁻⁶ N

Explanation:

For this problem we will look for the expression for the electric field of a ring of charge, at a point on its central axis.

From the symmetry of the ring, the resulting field is in the direction of the axis, we suppose that it corresponds to the x and y axis and the perpendicular direction but since the two sides of the annulus are offset.

       dE = k dq / r²

The component of this field in the direction to x is

       d Eₓ = dE cos θ

the distance of ring to the point

        r² = x² + a²

where a is the radius of each ring

the cosine is the adjacent leg between the hypotenuse

       Cos θ = x /r

we substitute

   d Ex = k / (x² + a²)  dq      x /√ (x² + a²)

we integrate

      Ex = ∫  k / (x² + a²)^{3/2}  x dq

      Ex = k x/ (x² + a²)^{1.5}   Q

the total field is the sum of the fields created by each ring, since they are on the same line (x-axis), you can perform an algebraic sum

     Eₓ_total = Eₓ₁ + Eₓ₂

a) at the origin x = 0

           in field is zero, since the numerator e makes zero

              Eₓ = 0

b) for point x = 40.0cm = 0.400 m

we substitute the values ​​in our equation

       Ex_total = k x [Q (x² + 0.1²)^{1.5} + 2Q / (x² + 0.2²2) ^3/2]

        Ex_total = k Q x [1 / (x² + 0.01)^1.5 + 2 / (x² + 0.04)^1.52]

let's calculate

       Ex_total = 9 10⁹ 30 10⁻⁹ 0.40 [1 / (0.4 2 + 0.01) 3/2 + 2 / (0.4 2 + 0.04) 3/2] 10⁻⁹

       Ex_total = 108 [14.2668 + 22.36] 10⁻⁹

       Ex_total = 3.96 10⁻⁶ N

Answer:

no

Explanation:

ok

Answer:

Pull towards itself.

Explanation:

That's what I know.

Tension is a force that tends to pull or stretch an object. A flexible material, like a rope, cable, or string, is used to transmit the force.

A force known as tension has a tendency to pull or stretch an object. When a flexible medium, like a rope, cable, or string, is under strain and supporting an imposed load or weight, it transmits force via the flexible medium.

A rope or cable that is under tension receives equal and opposing forces pulling it in different directions at either end. For instance, the tension in a rope will be the same along its whole length if both ends are pulled with equal effort.

When you use a pulley system to hoist a weight, suspend an object from a rope, or support the weight of traffic on a bridge, tension can be present.

Therefore, Tension is a force that tends to pull or stretch an object.

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It depends on the circuit. Sometimes it becomes a bit weaker, sometimes it stays the same.

Answer:

In a series circuit, the current that flows through each of the components is the same, and the voltage across the circuit is the sum of the individual voltage drops across each component.

Answer:

a) F= 0,19  [N]   according to problem statement

b) F = 0,19*10⁹ [N]  using the right value of K

Explanation:

The force between two electric charges is according to Coulomb´s law is:

F = K * q₁*q₂ / d²    where  q₁  and q₂ are the charges on body one and body 2 respectively, d is the distance between the two bodies and K is a constant  K = 8,988100*10⁹ N.m²/C². The problem establishes to use        K = 8,988100 N.m²/C².

NOTE: To value of is :  K = 8,988100*10⁹ N.m²/C². I am going to solve the problem using K = 8,988100 N.m²/C² if that information was an error, all we need to get the right answer is multiply the result by 10⁹

Then:

F = 8,988100 * 1,2* 0,36 / (4,5)²     [ N*m²/C² ] * [ C*C*/m²]

F = 3,882859/ 20,25  [N]

F= 0,19  [N]

The force is of repulsion since the two charges are positive and in the direction of the straight line which passes through the centers of the bodies

Answer:

B

Explanation:

Answer:

B

Explanation:

Just took the quiz

Answer:

The values are [tex]I_L  =  500.37 \ mL / minutes [/tex]

and [tex]I_r  = 299.63 \ mL / minutes [/tex]  

Explanation:

From the question we are told that  

   The pressure at the bifurcation of the femoral artery is  [tex]P = 100\ mm\ Hg[/tex]

   The pressure at the left or right femoral artery is  [tex]P_r = P_l  =  10 \  mm \  Hg[/tex]

   The resistance in the  left femoral artery is  [tex]R_L[/tex]

   The resistance in the  right  femoral artery is  [tex]R_r =  1.67 R_L[/tex]

   The total  flow rate is  [tex]I = 800 mL/minute[/tex]

The diagram illustrating this question is shown on the first uploaded image

Generally this flow of blood through the artery can be compared to the a circuit as shown on the first uploaded image

Generally the rate at which blood flows through the  left femoral artery is mathematically represented as

      [tex]I_L  =  I * \frac{R_r }{R_L + R_r}[/tex]

=>      [tex]I_L  =  800 * \frac{1.67 R_L }{R_L + 1.67 R_L}[/tex]

=>    [tex]I_L  =  800 * \frac{1.67 R_L }{2.67 R_L}[/tex]

=>    [tex]I_L  =  500.37 \ mL / minutes [/tex]

Generally the rate at which blood flows through the  right  femoral artery is mathematically represented as

      [tex]I_r  =  I * \frac{R_L }{R_L + R_r}[/tex]

=>      [tex]I_r  =  800 * \frac{R_L }{R_L + 1.67 R_L}[/tex]

=>    [tex]I_r  =  800 * \frac{ R_L }{2.67 R_L}[/tex]

=>    [tex]I_r  = 299.63 \ mL / minutes [/tex]    

Answer: false

Explanation: They are measured by cubic ______. ______ is feet, meters, inches, any way to measure