Which action would increase the strength of the electric force between the
balloon and the water?
A. Holding a magnet near the stream of water
B. Holding the balloon closer to the stream of water
C. Holding a magnet near the balloon
D. Moving the balloon farther away from the stream of water

Answers

Answer 1

Answer:

B

Explanation:

Holding the balloon closer to the stream of water

Answer 2

Holding the balloon closer to the stream of water is an action which would increase the strength of the electric force between the balloon and the water. Thus, the correct option is B.

What is electric force?

The repulsive or attractive interaction between any two charged bodies is called as the electric force. This is similar to any force, the impact and effects on the given body are described through the Newton's laws of motion. The electric force is one of the various forces which act on objects.

Electric force is the attractive force which is present between the electrons (negatively charged bodies) and the nucleus. Here, a positive charge or a negative charge creates a field in the empty space which is around it, and we call that empty space as an electric field.

Holding the balloon closer to the stream of water is an action which would increase the strength of the electric force between the balloon and the water.

Therefore, the correct option is B.

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Related Questions

What is Newton's First Law of Motion? (2 points)
A For every action there is an equal and opposite reaction.
B An object in motion or at rest will stay that way unless acted on by an unbalanced force.

C The smaller the mass of an object, the greater the acceleration of that object when a force is applied.
D Every force is said to be unbalanced when the force acting against it is less.

Answers

Answer:

B

Explanation:

B is Newton's first law of motion, I got this in class and that was the right answer.

Good Luck!

You want to know how tall a cliff is, so you drop a rock off the edge. You estimate that
the rock has a mass of about 15 kg. You hear a splash 1.00 seconds later. How tall is the
cliff?

Answers

Analysing the Question:

Neglecting air resistance

We are given:

mass (m) = 15 kg

time taken by the rock to reach the water (t) = 1 second

height of the cliff (h) = h m

initial velocity of the rock (u) = 0 m/s  [ the rock was 'dropped' and not 'thrown' downwards]

acceleration of the rock due to gravity (a) = 9.8 m/s²

Finding the height to the cliff:

from the seconds equation of motion

s = ut + 1/2*at²

replacing the variables

h = (0)(1) + 1/2 * (9.8) *(1)²

h = 4.9 m

Therefore, the cliff is 4.9 m high

A skater glides off a frozen pond onto a patch of ground at a speed of 1.8 m/s. Here she is slowed at a constant rate of 3.00 m/s. How fast is the skater moving when she has slid 0.37 m across the ground?​

Answers

Answer:

1.01 m/s

Explanation:

Given

initial speed, u = 1.8m/s

acceleration, a = -3.00m/s² (it's negative because the skater slowed down)

distance, s = 0.37m

Required

Determine the final velocity (v)

This will be solved using the following equation of motion

[tex]v\² = u\² + 2as[/tex]

Substitute values for u, a and s

[tex]v\² = 1.8\² + 2 * -3 * 0.37[/tex]

[tex]v\² = 3.24 - 2.22[/tex]

[tex]v\² = 1.02[/tex]

Take square roots

[tex]v = \sqrt{1.02[/tex]

[tex]v = 1.00995049384[/tex]

[tex]v = 1.01 m/s[/tex] (Approximated)

The final speed of the skater when she has slid 0.37 m is 1.0 m/s.

The given parameters:

Speed of the skater, v = 1.8 m/sacceleration of the skater = 3.0 m/s²Displacement of the skater = -0.37 m

The final speed of the skater when she has slid 0.37 m is calculated as follows;

[tex]v_f^2 = v_i^2 + 2ad\\\\v_f ^2 = 1.8^2 + 2(3)(-0.37)\\\\v_f^2 = 1.02\\\\v_f = \sqrt{1.02} \\\\v_f = 1.0 \ m/s[/tex]

Thus, the final speed of the skater when she has slid 0.37 m is 1.0 m/s.

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A boat with a weight of 547 newtons is floating in a harbor. What is the buoyant force on the boat?
A. 55.7 newtons
B. 5,371.5 newtons
C. 273.5 newtons
D. 547 newtons

Answers

Answer:

a

Explanation:

its correct

One strategy in a snowball at a high angle over level ground. While your opponent is watching this first snowball, you throw a second one at low angle times to arrive before or at the same time as the first one . Assume both snowballs are throw with a speed of 33.2 m/s. The first o e is thrown at an angle of 57 degrees with respect to the horizontal. At what angle should the second snowball be throw to arrive at the same point as the first? Answer in units of degrees

Answers

Answer:

33°

Explanation:

We are given;

Speed at which both snowballs are thrown; v = 33.2 m/s

Angle at which snowballs are thrown with respect to the horizontal; θ = 57°

Now,we want to find out the angle at which the second snowball should be thrown in order to arrive at the same point as the first.

To calculate this angle, we will use complementary angle concept.

Now, because the target is in the same place, there will be two launch angles that will make the snow ball to be placed on the target.

The is calculated from;θ1 = 45° - (57° - 45°) = 33°

What is the common temperature of the final equilibrium state of the system? A 10.0 kg block of ice at 0.0°C is placed into 10 kg water bath at 90.0C in an isolated container. The system eventually comes to equilibrium at a common temperature. Ignore the effect of the container.

Answers

Answer:

The equilibrium temperature is 45°C

Explanation:

Given;

mass of block of ice, m₁ = 10 kg

mass of water, m₂ = 10 kg

initial temperature of the ice, t₁ = 0.0°C

initial temperature of water, t₂ = 90.0°C

let the equilibrium temperature = T

Apply the principles of mixture;

Heat lost by the water = heat gained by the ice

m₂c(90 - T) = m₁c(T - 0)

m₂(90 - T) = m₁(T - 0)

10(90 - T) = 10(T)

90 - T = T

90 = 2T

T = 90 / 2

T = 45°C

Therefore, the equilibrium temperature is 45°C

The number of waves in a given time period is called the ________________ .

Answers

Answer:

frequency.

......... i need more characters so

A bird has a mass of 0.8 kg. Calculate the weight of the bird.​

Answers

Answer:

the mass and wieght means same

A train travels 6.1m/s (E)The train’s destination is 134 km(E), how long until it arrives?

Answers

Answer:

The time taken is  [tex]t =   6.102 \  hours [/tex]

Explanation:

From the question we are told that  

    The velocity of the train is [tex]v_t = 6.1 \ m/s[/tex]

    The distance covered by the train is  [tex]D =  134 \  km  =  134000 \  m[/tex]

   

Generally the time taken is mathematically represented as

     [tex]t =  \frac{  134000}{ 6.1 }[/tex]

=>    [tex]t =  \frac{  134000}{ 6.1 }[/tex]

=>    [tex]t = 21967.21 \ sec [/tex]

Converting to hours

       [tex]t = 21967.21  =  \frac{21967.21 }{3600} =   6.102 \  hours [/tex]

How much kinetic energy does a 1 kg ball
traveling at 5 m/s have?

Answers

Answer:

Okay, first off, the formula for Kinetic Energy is:

KE = 1/2(m)(v)^2

m = mass

v = velcoity (m/s)

Using this formula, we can then calculate the kinetic energy in each scenario:

1) KE = 1/2(100)(5)^2 = 1,250 J

2) KE = 1/2(1000)(5)^2 = 12,500 J

3) KE = 1/2(10)(5)^2 = 125 J

4) KE = 1/2(100)(5)^2 = 1,250 J

ANSWET ASAP

Which term is the rate at which work is done?
energy
power
joules
force

Answers

Answer:

power

Explanation:

Answer:

power

Explanation:

A box is sliding along a frictionless surface and gets to a ramp. Disregarding friction, how fast should the box be going on the ground in order to slide up the ramp to a height of 2.5 meters, where it stops? (Use g = 9.8 m/s2.)

Answers

Answer:

7.0 m/s

Explanation: I just did it

Which of the following statements must be true?

a. The velocity and acceleration are in the direction of the resultant force.
b. The velocity could be in any direction, but the acceleration is in the direction of the resultant force.
c. The velocity and acceleration could be in any direction.
d. None of these choices.

Answers

Answer:

your answer is B. The velocity could be in any direction, but the acceleration is in the direction of the resultant force

A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the page. An external magnetic field of 0.060 T is directed out of the plane of the page. The external field decreases to 0.020 T in 12 milliseconds. What is the magnitude of the total change in the external magnetic flux enclosed by the coil

Answers

Answer:

Δϕ = -2.89 x 10^-4 Wb

Explanation:

Given that A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the page. An external magnetic field of 0.060 T is directed out of the plane of the page. The external field decreases to 0.020 T in 12 milliseconds. What is the magnitude of the total change in the external magnetic flux enclosed by the coil

Let ΔB = Change in magnetic field

A = cross sectional area = 8.5 cm^2

t = time = 12ms

Flux change = Δϕ = (ΔB*A) / t

Δϕ =  [(0.02 - 0.06)x(8.5 x 10^-2)^2 ] / 0.012

Δϕ = (-0.04 x 7.225^-3) / 0.012

Δϕ = -2.89 x 10^-4 Wb

Therefore, the magnitude of the total change in the external magnetic flux enclosed by the coil is -2.89 x 10^-4 Wb

Which of the following best describes the accuracy of a measurement?

A.
the precision of a measuring tool
B.
the amount of uncertainty involved in a measurement
C.
the degree to which the measurement approaches the quantity's true value
D.
the degree to which repeated measurements agree

Answers

Answer:

l think the answer is C. part

The degree to which the measurement approaches the quantity's true value

is what best describes the accuracy of a measurement.

When a quantity is measured, the value is usually unknown by the individual

taking the measurement . This should however be done with a very good

tool/equipment to eliminate inaccuracy.

The accuracy of a quantity is usually measured by how close the difference

is between the measurement taken and the actual value of the quantity.

The different  in value gotten should always be very small.

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A thick conducting spherical shell, with inner radius R1 and outer radius R2, is in electrostatic equilibrium. The area inside and outside the spherical shell is vacuum. The net charge on the spherical shell is Q. Note: for parts (a)-(c), describe the direction of the electric field assuming that Q is positive.
(a) Find the electric field outside the shell, at a distance d > R2 from the center of the shell.
(b) Find the electric field in the shell material, at a distance d from the center of the shell where R1 < d < R2.
(c) Find the electric field at a distance d < R1 from the center of the shell, i.e. in the empty space inside the shell.
(d) How much charge is on the inner surface of the conducting shell

Answers

Answer:

a)     E = k Q / r² ,      k = 1 / 4πε₀,  b)  E=0, C) E=0,   d) Q=0

Explanation:

To find the electric field in this exercise let's use Gauss's law

         Ф = ∫ E. dA = [tex]q_{int}[/tex] /ε₀

Let's use a sphere as a Gaussian surface, in this case the field lines and the radii of the spheres are parallel so the scalar product is reduced to the algebraic product, let's carry out the integral

          E A = q_{int} /ε₀

           

The area of ​​a sphere is

         A = 4π r²

we substitute

          E (4π r²) =q_{int} /ε₀

a) The field is requested outside the spherical shell

        d> R₂

in this case the charge inside the Gaussian surface is

          q_{int} = Q

           E = Q / (4π r²) ε₀

           E = k Q / r²

           k = 1 / 4πε₀

b) In the second case, the field inside the spherical casing is requested.

As the surface is metallic, the charge is located on the surface of it, the electrons repel each other. So the charge inside is zero

           q_{int} = 0

            E = 0

c) The field in the interior region of the shell, all the charge is external, therefore the internal charge is zero

           q_{int} = 0

           E = 0

d) as the shell is conductive, the electrons can move freely, which is why it moves as far away as possible between them, this means that everything is located as close as possible to the outer surface of the conductor. Consequently, THERE ARE NO ELECTRONS ON THE INTERNAL SURFACE

              Q_internal =0

What is the best description of the relationship between emerging scientific ideas and open-mindedness?

HURRY !

Answers

Answer: C ON EDGE 2021

Explanation:

A student wants to determine the density of titanium. The student measures the mass of a solid chunk of titanium, but when he drops it into the water to determine the volume, some water splashes out of the cylinder. Will this cause the calculated density to be higher or lower than it should be? Explain!​

Answers

Answer:

Lower than it should be.

Explanation:

Hello.

In this case, since the density is computed by:

[tex]\rho =\frac{m}{V}[/tex]

And we obtain the volume of the solid by substracting the mass of the water and the solid minus the mass of water:

[tex]V_{solid}=V_{solid\ with \ water}-V_{water}[/tex]

If some water is splashed out of the cylinder, the volume of water will be lower than originally measured, it means that the volume of the solid will be higher than real. In such a way, since the density is in an inversely proportional relationship with volume, as the volume of the solid is wrongly increased, therefore the density of the solid will be lower than in should be.

Regards.

An object increases its velocity from 22 m/s to 36 m/s in 5.0 s. What is the average velocity of the object?

Answers

Answer:

[tex]a=2.8\ m/s^2[/tex]

Explanation:

Given that,

Initial velocity of an object, u = 22 m/s

Final velocity of an object, v = 36 m/s

Time, t = 5 s

It can be assumed to find the average acceleration of the object instead of average velocity.

The change in velocity per unit time is equal to average acceleration of an object. It can be given by :

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{36\ m/s-22\ m/s}{5}\\\\\a=\dfrac{14}{5}\ m/s^2\\\\a=2.8\ m/s^2[/tex]

So, the acceleration of the object is [tex]2.8\ m/s^2[/tex].

If your mass is 72 kg, your textbook's mass is 3.7 kg, and you and your
textbook are separated by a distance of 0.33 m, what is the gravitational
force between you and your textbook? Newton's law of gravitation is
Gm;m2 The gravitational constant G is 6.67 x 10-11 N·m2/kg2.
F gravity
O A. 4.94 x 10-7N
B. 2.45 x 103 N
O C. 5.38 x 10-8N
O D. 1.63 x 10-7N

Answers

Answer:

The answer is 1.63 x 10 -7 N

Explanation:

An appropriate solution is 1.63 x 10 -7 N

Gravitational pressure method: F=G{frac{m_1m_2}{r^2}}

Newton's law of gravitation is Gm;m2

The gravitational regular G is 6.67 x 10-11 N·m2/kg2.

What is gravitational pressure?

Definitions of gravitational pressure. The force of attraction among all masses inside the universe; especially the appeal of the earth's mass for bodies close to its floor.

What law is Newton's regulation of gravitation?

Newton's regulation of typical gravitation states that two bodies in an area pull on every other with a force proportional to their hundreds and the space among them. For large objects orbiting each other—the moon and Earth, for instance—means they without a doubt exert important force on each other.

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A stone is thrown horizontally with an initial speed of 10m/s from the edge of the cliff. A stop watch measures the stone’s trajectory time from the top of the hill to the bottom to be 6.7s. What is the height of the cliff?

Answers

Answer:

h = 219.96 m

Explanation:

Speed of the stone with which it was thrown horizontally, v = 10 m/s

We need to find the height of the cliff if the stone’s trajectory time from the top of the hill to the bottom to be 6.7s.

It means we need to find the distance covered by the stone. As the horizontal speed of the stone is given , it means there is no vertical motion in the stone,u'=0

Using second equation of motion,

[tex]d=u't+\dfrac{1}{2}at^2[/tex]

Put u'=0 and a=g

So,

[tex]d=\dfrac{1}{2}gt^2\\\\\text{Putting all the values to find d}\\\\d=\dfrac{1}{2}\times 9.8\times 6.7^2\\\\d=219.96\ m[/tex]

So, the height of the cliff is 219.96 m.

Different satellites orbit the earth with a vast range of altitudes, from just a couple hundred km, all the way to tens of thousands of km above the surface. The international space station (ISS) is in a low earth orbit, just 400km above the surface (you can see it with the naked eye at sunset and sunrise as a bright, moving dot). At this altitude, the acceleration due to gravity has a value of 8.69.
A. What is the speed of the ISS?
B. What is the orbital period (T) of the ISS in minutes?

Answers

Answer:

Find the answer in the explanation.

Explanation:

Given that the international space station (ISS) is in a low earth orbit, just 400km above the surface. And at this altitude, the acceleration due to gravity has a value of 8.69.

A.) Since the radius of the earth is equal to 6400 km. At it to 400km

The radius of the orbit = 6400 + 400

The radius = 6800 km.

Where orbital speed = sqrt ( g × r )

Orbital speed = sqrt ( 6800 × 8.69 )

Orbital speed = 243.09 km/h

B.) Orbital period acceleration to Kepler third law of planetary motion state that:

The square of the period is directly proportional to the cube of the radius.

T^2 = (4π^2 /GM) r^3

T^2=(4π^2/ 6.67×10^11 × 6.0 × 10^24)r^3

T^2 = (4π^2/ 4.002^36) × 6800^3

T^2 = 9.865×10^-36 × 6800^3

T^2 = 3.10 × 10^-15

T = 5.57 × 10^-8 hours.

Convert the hours to minutes by multiplying it by 60

T = 3.4 × 10^-6 minutes.

Part A:

Since the radius of the earth is equal to 6400 km. At it to 400km

The radius of the orbit = 6400 + 400

The radius = 6800 km.

Where:

Orbital speed = sqrt ( g × r )Orbital speed = sqrt ( 6800 × 8.69 )Orbital speed = 243.09 km/h

The speed of the ISS is 243.09Km/h.

Part B:

The orbital period (T) of the ISS in minutes is :

T^2 = (4π^2 /GM) r^3T^2=(4π^2/ 6.67×10^11 × 6.0 × 10^24)r^3T^2 = (4π^2/ 4.002^36) × 6800^3T^2 = 9.865×10^-36 × 6800^3T^2 = 3.10 × 10^-15T = 5.57 × 10^-8 hours.

Convert the hours to minutes by multiplying it by 60

T = 3.4 × 10^-6 minutes.

The orbital period (T) of the ISS in minutes is 3.4 × 10^-6 minutes.

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A uniform electric field has magnitude E and is directed in the negative x direction. The potential difference between point a (at x = 0.60 m) and point b (at x = 0.90 m) is 240 V.

Required:
a. Which point, a or b, is at the higher potential?
b. Calculate the value of £.
c. A negative point charge q = —0.200 μC is moved from 5 to a. Calculate the work done on the point charge by the electric field.

Answers

Answer:

a)

b is at higher potential

b)

E = 800 V/m

c)

W = -48.0 μ J

Explanation:

Given:

electric field has magnitude E and is directed in the negative x direction

potential difference between point a (at x = 0.60 m) and point b (at x = 0.90 m) = 240 V

a)

Direction of the electric field is in the negative x direction so it is away from the positive charge and the more distance away from direction, the lesser the potential gets. So b is at the higher potential.

b)

When the magnitude of electric field is constant then the potential difference between two points in the field is given by:

Vab = Ed

where

Vab is the potential difference between two points a and b in the field

E is electric field magnitude

d is the distance between two points

Compute distance between two points a and b

d = 0.90 - 0.60

d = 0.30

d = 0.3 m

Since potential difference between two points in the field is given which is 240 V

Compute E:

The above formula becomes:

Vab = Ed

E = Vab/d

  = 240/0.3

E = 800 V/m

c)

The negative charge moved from the higher potential to low so work done on the point charge by electric field is negative.

W = Fd

Electric field = E = F/q

where

E is electric field magnitude

F is electric force on q

q is point charge magnitude

The charge q is either positive or negative and when charge is positive the directions of E and F are same otherwise opposite in case of negative charge.

So to calculate the work done on the point charge by the electric field:

E = F/q

F = Eq

so

W = Fd = Eqd

Now putting the values:

E = 800 V/m

q = - 0.200 μC

d = 0.3 m

W = Eqd

W = 800 * - 0.200 * 0.3

W = 800 * - 0.2 * 10⁻⁶ * 0.3

W = - 48 * 10⁻⁶

W = - 4.8 * 10⁻⁵ J

W = -48.0 μJ

a. The point which is at a higher potential is point a.

b. The magnitude (E) of the uniform electric field is equal to 800 V/m.

c. The work done on the point charge by the electric field is equal to -48 microjoules.

Given the following data:

Point a (at x) = 0.60 meterPoint b (at x) = 0.90 meterPotential difference = 240 Volts.Point charge = -0.200 uC = [tex]0.200 \times 10^{-6}\;C[/tex]

a. Since the uniform electric field with a magnitude (E) is directed in the negative x direction, the point at a higher potential is point a because the electric field is not close to the positive charge and it is further away from point b by 0.90 meter. Also, the potential difference of an electric field is inversely proportional to the square of the distance from a point charge.

b. To calculate the magnitude (E) of the uniform electric field:

At a constant potential difference, the magnitude (E) of this electric field is given by the formula:

[tex]V_{ab} = Ed[/tex]

Where:

V_{ab} is the potential difference between point a and b.E is the magnitude the electric field.d is the distance between point a and b.

First of all, we would determine the distance (d).

[tex]d = d_b - d_a\\\\d = 0.90 -0.60[/tex]

Distance, d = 0.30 meter.

Substituting the given parameters into the formula, we have;

[tex]240 = 0.3E\\\\E=\frac{240}{0.3}[/tex]

E = 800 V/m

c. To calculate the work done on the point charge by the electric field:

Mathematically, the work done on a point charge in an electric field is given by the formula:

[tex]W = Fd = Eqd\\\\W = 800 \times (-0.200 \times 10^{-6}) \times 0.3\\\\W = -240 \times 0.200 \times 10^{-6}\\\\W = -48\times 10^{-6}\;Joules[/tex]

Note: 1 microjoules = [tex]1\times 10^{-6}[/tex] Joules

Work done = -48 microjoules.

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I don’t know what this is

Answers

im pretty sure an earthquake accrues because the other ones don’t make sense... but i think so, so don’t blame me if it’s wrong

What best describes the dropping height of a ball that bounced back up to a height of 45 centimeters?

Less than 45 centimeters, as the ball transforms some of its thermal energy into potential energy

Greater than 45 centimeters, as the ball transforms some of its thermal energy into potential energy

Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy

Greater than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy

Answers

Answer:C:Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy

Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy.

Although the initial energy (potential energy is preserved), the energy of deformation as the ball strikes a surface creates energy dissipation in the form of frictional heat and audible sound energy.

Every time the ball bounces, its height will be less than its previous height.

Explanation:

Answer:

C is incorrect I got it wrong on my test don't go with C

Explanation:

Luis wants to compare the density of three solid objects that are different shapes and sizes. What information does he need to make the comparison?

Answers

Answer:

u need to make sure that comparison is = to shapes and then find the shapes sizes and add them

A driver translation a speed of 115km/hr received a text message on his mobile phone how far is he ,in kilometers,20s later from when he received the text message?

Answers

Answer:

S=d/t.

distance is= speed multiplied by the time

D= 115 multiplied by 20

the answer is 2300km

Driver is approximately 0.63888 kilometers away from where he received the text message after 20 seconds.

To determine the distance the driver travels 20 seconds after receiving the text message, we need to calculate the distance covered during that time interval.

First, we convert the driver's speed from km/h to m/s for consistent units. The conversion factor is 1 km/h = 0.27778 m/s.

Driver's speed = 115 km/h × 0.27778 m/s = 31.944 m/s

Next, we use the formula for distance traveled:

Distance = Speed × Time

Distance = 31.944 m/s × 20 s = 638.88 meters

Therefore, the driver will be 638.88 meters away from the point where he received the text message after 20 seconds.

To convert this distance to kilometers, we divide by 1000:

Distance = 638.88 m ÷ 1000 = 0.63888 km

So, the driver is approximately 0.63888 kilometers away from where he received the text message after 20 seconds.

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Which statement describes oneScientists have changed the model of the atom as they have gathered new evidence. One of the atomic models is shown below.

A purple ball in the center surrounded by overlapping concentric black ovals, each with a small green ball on each of the 6 ovals.

What experimental evidence led to the development of this atomic model from the one before it? feature of Rutherford’s model of the atom?

Answers

Answer:

el continuo movimiento

Explanation:

A car transports its passengers between 3 buildings. It moves from the first building to the second building, 4.76km away, in a direction 37° north of east. It then moves from second building to the third building in a direction 69° west of north. Finally, it returns to the first building, sailing in a direction 28° east of south. Calculate the distance between (a) the second and third buildings and (b) the first and third buildings.​

Answers

Answer:

1) a = 6.14 km

2) b = 4.69 km

Explanation:

Let the first building be A, second building be B and third building be C.

Now, bearing of A = 4.76 km in a direction 37° north of east

Bearing of B = 69° west of north

Bearing of C = 28° east of south

Thus if this 3 points form a triangle, we will have the following angles;

Angle at point A = 28 + (90 - 37) = 81°

Angle at point B = 28 + (90 - 69) = 49°

Angle at point C = 180 - (81 + 49) = 50°

Now, the distance between second and third building is "a" which is represented by BC in the triangle attached while the given distance of 4.76 represents side AB. Thus;

Using sine rule, we can find "a".

a/sin 81 = 4.76/sin 50

a = 6.14 km

B) distance between first and third building is AB in the triangle depicted by "b".

Similar to the first problem, we will use sine rule again.

b/sin49 = 4.76/sin 50

b = 4.69 km

Can someone please help, ty!!
(Will mark brainliest)

Answers

Answer:

graph

noun

a diagram representing a system of connections or interrelations among two or more things by a number of distinctive dots, lines, bars, etc.

Mathematics.

a series of points, discrete or continuous, as in forming a curve or surface, each of which represents a value of a given function.

Also called linear graph . a network of lines connecting points.

a written symbol for an idea, a sound, or a linguistic expression.

verb (used with object)

Mathematics. to draw (a curve) as representing a given function.

to represent by means of a graph.

Explanation:

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